feedback, poles, and fundamental modes (pdf)
TRANSCRIPT
6003 Signals and Systems
Feedback Poles and Fundamental Modes
September 15 2011 1
Homework
Doing the homework is essential to understanding the content
Weekly Homework Assigments
bull tutor (exam-type) problems
answers are automatically checked to provide quick feedback
bull engineering design (real-world) problems
graded by a human
Learning doesnrsquot end when you have submitted your work
bull solutions will be posted on Wednesdays at 5pm
bull read solutions to find errors and to see alternative approaches
bull mark the errors in your previously submitted work
bull submit the markup by Friday at 5pm
bull identify ALL errors and get back half of the points you lost
2
Last Time Multiple Representations of DT Systems
Verbal descriptions preserve the rationale
ldquoTo reduce the number of bits needed to store a sequence of
large numbers that are nearly equal record the first number
and then record successive differencesrdquo
Difference equations mathematically compact
y[n] = x[n] minus x[n minus 1]
Block diagrams illustrate signal flow paths
minus1 Delay
+x[n] y[n]
Operator representations analyze systems as polynomials
Y = (1 minus R) X
3
Last Time Feedback Cyclic Signal Paths and Modes
Systems with signals that depend on previous values of the same
signal are said to have feedback
Example The accumulator system has feedback
Delay
+X Y
minus1 Delay
+X Y
By contrast the difference machine does not have feedback
4
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
5
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
6
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
7
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
The response will persist even though the input is transient
8
minus1 0 1 2 3 4n
y[n]
Geometric Growth Poles
These unit-sample responses can be characterized by a single number
mdash the pole mdash which is the base of the geometric sequence
Delay
+
p0
X Y
y[n] =
pn0 if n gt= 0
0 otherwise
minus1 0 1 2 3 4n
y[n]
minus1 0 1 2 3 4n
y[n]
p0 = 05 p0 = 1 p0 = 12 9
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Homework
Doing the homework is essential to understanding the content
Weekly Homework Assigments
bull tutor (exam-type) problems
answers are automatically checked to provide quick feedback
bull engineering design (real-world) problems
graded by a human
Learning doesnrsquot end when you have submitted your work
bull solutions will be posted on Wednesdays at 5pm
bull read solutions to find errors and to see alternative approaches
bull mark the errors in your previously submitted work
bull submit the markup by Friday at 5pm
bull identify ALL errors and get back half of the points you lost
2
Last Time Multiple Representations of DT Systems
Verbal descriptions preserve the rationale
ldquoTo reduce the number of bits needed to store a sequence of
large numbers that are nearly equal record the first number
and then record successive differencesrdquo
Difference equations mathematically compact
y[n] = x[n] minus x[n minus 1]
Block diagrams illustrate signal flow paths
minus1 Delay
+x[n] y[n]
Operator representations analyze systems as polynomials
Y = (1 minus R) X
3
Last Time Feedback Cyclic Signal Paths and Modes
Systems with signals that depend on previous values of the same
signal are said to have feedback
Example The accumulator system has feedback
Delay
+X Y
minus1 Delay
+X Y
By contrast the difference machine does not have feedback
4
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
5
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
6
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
7
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
The response will persist even though the input is transient
8
minus1 0 1 2 3 4n
y[n]
Geometric Growth Poles
These unit-sample responses can be characterized by a single number
mdash the pole mdash which is the base of the geometric sequence
Delay
+
p0
X Y
y[n] =
pn0 if n gt= 0
0 otherwise
minus1 0 1 2 3 4n
y[n]
minus1 0 1 2 3 4n
y[n]
p0 = 05 p0 = 1 p0 = 12 9
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Last Time Multiple Representations of DT Systems
Verbal descriptions preserve the rationale
ldquoTo reduce the number of bits needed to store a sequence of
large numbers that are nearly equal record the first number
and then record successive differencesrdquo
Difference equations mathematically compact
y[n] = x[n] minus x[n minus 1]
Block diagrams illustrate signal flow paths
minus1 Delay
+x[n] y[n]
Operator representations analyze systems as polynomials
Y = (1 minus R) X
3
Last Time Feedback Cyclic Signal Paths and Modes
Systems with signals that depend on previous values of the same
signal are said to have feedback
Example The accumulator system has feedback
Delay
+X Y
minus1 Delay
+X Y
By contrast the difference machine does not have feedback
4
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
5
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
6
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
7
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
The response will persist even though the input is transient
8
minus1 0 1 2 3 4n
y[n]
Geometric Growth Poles
These unit-sample responses can be characterized by a single number
mdash the pole mdash which is the base of the geometric sequence
Delay
+
p0
X Y
y[n] =
pn0 if n gt= 0
0 otherwise
minus1 0 1 2 3 4n
y[n]
minus1 0 1 2 3 4n
y[n]
p0 = 05 p0 = 1 p0 = 12 9
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Last Time Feedback Cyclic Signal Paths and Modes
Systems with signals that depend on previous values of the same
signal are said to have feedback
Example The accumulator system has feedback
Delay
+X Y
minus1 Delay
+X Y
By contrast the difference machine does not have feedback
4
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
5
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
6
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
7
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
The response will persist even though the input is transient
8
minus1 0 1 2 3 4n
y[n]
Geometric Growth Poles
These unit-sample responses can be characterized by a single number
mdash the pole mdash which is the base of the geometric sequence
Delay
+
p0
X Y
y[n] =
pn0 if n gt= 0
0 otherwise
minus1 0 1 2 3 4n
y[n]
minus1 0 1 2 3 4n
y[n]
p0 = 05 p0 = 1 p0 = 12 9
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
5
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
6
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
7
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
The response will persist even though the input is transient
8
minus1 0 1 2 3 4n
y[n]
Geometric Growth Poles
These unit-sample responses can be characterized by a single number
mdash the pole mdash which is the base of the geometric sequence
Delay
+
p0
X Y
y[n] =
pn0 if n gt= 0
0 otherwise
minus1 0 1 2 3 4n
y[n]
minus1 0 1 2 3 4n
y[n]
p0 = 05 p0 = 1 p0 = 12 9
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
6
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
7
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
The response will persist even though the input is transient
8
minus1 0 1 2 3 4n
y[n]
Geometric Growth Poles
These unit-sample responses can be characterized by a single number
mdash the pole mdash which is the base of the geometric sequence
Delay
+
p0
X Y
y[n] =
pn0 if n gt= 0
0 otherwise
minus1 0 1 2 3 4n
y[n]
minus1 0 1 2 3 4n
y[n]
p0 = 05 p0 = 1 p0 = 12 9
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
7
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
The response will persist even though the input is transient
8
minus1 0 1 2 3 4n
y[n]
Geometric Growth Poles
These unit-sample responses can be characterized by a single number
mdash the pole mdash which is the base of the geometric sequence
Delay
+
p0
X Y
y[n] =
pn0 if n gt= 0
0 otherwise
minus1 0 1 2 3 4n
y[n]
minus1 0 1 2 3 4n
y[n]
p0 = 05 p0 = 1 p0 = 12 9
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Last Time Feedback Cyclic Signal Paths and Modes
The effect of feedback can be visualized by tracing each cycle
through the cyclic signal paths
Delay
+
p0
X Y
minus1 0 1 2 3 4n
x[n] = δ[n]
minus1 0 1 2 3 4n
y[n]
Each cycle creates another sample in the output
The response will persist even though the input is transient
8
minus1 0 1 2 3 4n
y[n]
Geometric Growth Poles
These unit-sample responses can be characterized by a single number
mdash the pole mdash which is the base of the geometric sequence
Delay
+
p0
X Y
y[n] =
pn0 if n gt= 0
0 otherwise
minus1 0 1 2 3 4n
y[n]
minus1 0 1 2 3 4n
y[n]
p0 = 05 p0 = 1 p0 = 12 9
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
minus1 0 1 2 3 4n
y[n]
Geometric Growth Poles
These unit-sample responses can be characterized by a single number
mdash the pole mdash which is the base of the geometric sequence
Delay
+
p0
X Y
y[n] =
pn0 if n gt= 0
0 otherwise
minus1 0 1 2 3 4n
y[n]
minus1 0 1 2 3 4n
y[n]
p0 = 05 p0 = 1 p0 = 12 9
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole
n n
n n
n
10
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
How many of the following unit-sample responses can be
represented by a single pole 3
n n
n n
n
11
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Geometric Growth
The value of p0 determines the rate of growth
y[n] y[n] y[n] y[n]
minus1 0 1z
p0 lt minus1 magnitude diverges alternating sign
minus1 lt p0 lt 0 magnitude converges alternating sign
0 lt p0 lt 1 magnitude converges monotonically
p0 gt 1 magnitude diverges monotonically 12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are
more complicated
R
R
16
minus063
+X Y
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Not geometric This response grows then decays 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler
systems (divide and conquer)
R
R
16
minus063
+X Y
Y = X + 16RY minus 063R2Y
(1 minus 16R + 063R2) Y = X
(1 minus 07R)(1 minus 09R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems
(1 minus 07R)(1 minus 09R) Y = X
+
07 R
+
09 R
X YY2
(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2
+
09 R
+
07 R
X YY1
(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1
The order doesnrsquot matter (if systems are initially at rest) 15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by
multiplying the polynomial representations of the subsystems
Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _
_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)
Multiply then collect terms of equal order
Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X
+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot
16
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Multiplying Polynomial
Graphical representation of polynomial multiplication
Y
Y X
= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1
a R
a2 R2
a3 R3
+
1
b R
b2 R2
b3 R3
+
X Y
Collect terms of equal order
= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
17
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
minus1 0 1 2 3 4 5 6 7 8n
y[n]
Multiplying Polynomials
Tabular representation of polynomial multiplication
(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)
1 bR b2R2 b3R3 middot middot middot
1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot
a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot
middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot
Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X
18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts
R
R
16
minus063
+X Y
Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R
19
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Second-Order Systems Equivalent Forms
The sum of simpler parts suggests a parallel implementation
Y X
= 45
1 minus 09R minus
35 1 minus 07R
+
09 R
45 +
minus35
R07
+
X YY1
Y2
If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0
Thus y[n] = 45(09)n minus 35(07)n for n ge 0
20
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Partial Fractions
Graphical representation of the sum of geometric sequences
minus1 0 1 2 3 4 5 6 7 8n
y1[n] = 09n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y2[n] = 07n for n ge 0
minus1 0 1 2 3 4 5 6 7 8n
y[n] = 45(09)n minus 35(07)nfor n ge 0
21
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Partial Fractions
Partial fractions provides a remarkable equivalence
R
R
16
minus063
+X Y
+
09 R
45 +
minus35
R07
+
X YY1
Y2
rarr follows from thinking about system as polynomial (factoring) 22
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Poles
The key to simplifying a higher-order system is identifying its poles
Poles are the roots of the denominator of the system functional
when R rarr 1 z
Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)
p0=07 p1=09
1 Substitute R rarr and find roots of denominator
z 2 2Y 1 z z= = =
X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2
z0=07 z1=09
The poles are at 07 and 09
23
︸ ︷︷ ︸ ︸ ︷︷ ︸
︸ ︷︷ ︸ ︸ ︷︷ ︸
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
24
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z == = 14
1 18
1 2 + 14
18
121 + z +minus z minus z minus2 zz z
1 The unit sample response converges to zero 12 and z = 1
42 There are poles at z =123 There is a pole at z =
4 There are two poles
5 None of the above
25
14
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]
1 1 11 + 4R minus 8R2 Y = R minus 2R2 X
12R
2R minusH(R) =
Y = X 1
418R21 + R minus
1 12
1 12
12
minus z minus z minus2z z= = = 14
1 18
1 2 + 14
18
12
141 + z +minus z minus z minus2 zz z
radic 1 The unit sample response converges to zero
12 and z = 1
4 X2 There are poles at z =12 Xradic3 There is a pole at z =
4 There are two poles
5 None of the above X
26
( ) ( )
( )( )
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
Consider the system described by
y[n] = minus 1 4y[n minus 1] +
1 8y[n minus 2] + x[n minus 1] minus
1 2x[n minus 2]
How many of the following are true 2
1 The unit sample response converges to zero
2 There are poles at z = 12 and z = 1
4
3 There is a pole at z = 12
4 There are two poles
5 None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
What are the pole(s) of the Fibonacci system
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
What are the pole(s) of the Fibonacci system
Difference equation for Fibonacci system
y[n] = x[n] + y[n minus 1] + y[n minus 2]
System functional Y 1
H = = X 1 minus R minus R2
Denominator is second order rarr 2 poles
34
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z
H = = = X 1 minus R minus R2 rarr
1 minus 1 minus 1 z2 minus z minus 1 z 2z
Poles are at radic1 plusmn 5 1
z = = φ or minus2 φ
where φ represents the ldquogolden ratiordquo radic1 + 5
φ = asymp 16182 The two poles are at
1 z0 = φ asymp 1618 and z1 = minus asymp minus0618
φ
35
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
What are the pole(s) of the Fibonacci system 4
1 1
2 1 and minus1
3 minus1 and minus2
4 1618 and minus0618
5 none of the above
36
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Example Fibonaccirsquos Bunnies
Each pole corresponds to a fundamental mode
1 φ asymp 1618 and minus asymp minus0618
φ
One mode diverges one mode oscillates
minus1 0 1 2 3 4n
φn
minus1 0 1 2 3 4n
(minus 1φ
)n
37
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Example Fibonaccirsquos Bunnies
The unit-sample response of the Fibonacci system can be written
as a weighted sum of fundamental modes
φ 1 Y 1 radic
5 φ radic
5H = = = +
X 1 minus R minus R2 1 minus φR 1 + 1 Rφ
φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0
5 φ 5
But we already know that h[n] is the Fibonacci sequence f
f 1 1 2 3 5 8 13 21 34 55 89
Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]
38
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Complex Poles
What if a pole has a non-zero imaginary part
Example Y 1 = X 1 minus R + R2
1 z2 = =
1 minus 1 z + 1
z2 z2 minus z + 1
1 radic
3 plusmnjπ3Poles are z = 2 plusmn 2 j = e
What are the implications of complex poles
39
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Complex Poles
Partial fractions work even when the poles are complex
jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e
There are two fundamental modes (both geometric sequences)
e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)
n n
40
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Complex Poles
Complex modes are easier to visualize in the complex plane
e jnπ3 = cos(nπ3) + j sin(nπ3)
Re
Im
e j0π3
e j1π3e j2π3
e j3π3
e j4π3 e j5π3
Re
Im
e j0π3
e j5π3e j4π3
e j3π3
e j2π3 e j1π3
n
eminusjnπ3 = cos(nπ3) minus j sin(nπ3)
n
41
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Complex Poles
The output of a ldquorealrdquo system has real values
y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1
H = = X 1 minus R + R2
1 1 = jπ3R
times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=
j radic
3 1 minus e jπ3Rminus
1 minus eminusjπ3R 1 h[n] = radic
j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π
3 3
1
minus1
n
h[n]
42
( )
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
Unit-sample response of a system with poles at z = re plusmnjΩ
n
Which of the following isare true 2
1 r lt 05 and Ω asymp 05
2 05 lt r lt 1 and Ω asymp 05
3 r lt 05 and Ω asymp 008
4 05 lt r lt 1 and Ω asymp 008
5 none of the above
44
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
R R R+X Y
How many of the following statements are true
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
45
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Check Yourself
R R R+X Y
How many of the following statements are true 4
1 This system has 3 fundamental modes
2 All of the fundamental modes can be written as geometrics
3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1
4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1
5 One of the fundamental modes of this system is the unit
step
46
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
Summary
Systems composed of adders gains and delays can be characterized
by their poles
The poles of a system determine its fundamental modes
The unit-sample response of a system can be expressed as a weighted
sum of fundamental modes
These properties follow from a polynomial interpretation of the sysshy
tem functional
47
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms
MIT OpenCourseWarehttpocwmitedu
6003 Signals and SystemsFall 2011
For information about citing these materials or our Terms of Use visit httpocwmiteduterms