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10. Laplace transforms 1
Agenda for transforms (1 of 2)1. System response2. Transforms3. Partial fractions4. Laplace transforms5. Transfer functions6. Laplace applications7. Frequency response
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10. Laplace transforms 2
1. System responseIntroductionExampleDiscrete convolutionContinuous convolution
1. System response
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10. Laplace transforms 3
Introduction
The response of the system to inputs and disturbances is important in design
Differential equations provide insight into this response
Transform methods provide a simpler way of solving differential equations
We will assume linear differential equations with constant coefficients
1. System response
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10. Laplace transforms 4
Example (1 of 4)
system, hinput, r output, y
h(t)
0 1 2 3 4 5 6time0
1
Response of system to a unit impulse (t) = 1 at t=0
.99 .95.75
.4 .3 .25
1. System response
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10. Laplace transforms 5
Example (2 of 4)
What is the response of the system at t = 6 to the following inputs• At t=1, input(1) = 2 (1) • At t=3, input(3) = 3 (3) • At t=5, input(5) = 1 (5)
input
0 1 2 3 4 5 6time
3210
y = ?
1. System response
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10. Laplace transforms 6
Example (3 of 4)
y(6) = 2 * 0.3 = 0.6
0 1 2 3 4 5 6
time
0
1.99 .95
.75.4 .3
.25
0 1 2 3 4 5 60
1.99 .95
.75.4 .3
.25
0 1 2 3 4 5 60
1.99 .95
.75.4 .3 .25
time
time
y(6) = 3 * 0.75 = 2.25
y(6) = 1 * 0.99 = 0.99
(1) = 2
(3) = 3
(5) = 1
y(6) = 0.6 + 2.25 + 0.99 = 3.84
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10. Laplace transforms 7
Example (4 of 4)
Response of system• y(6) = h(6 -1) * 2 (1) + h(6 -3) * 3 (3) +
h(6 -5) * 1 (5)• This relationship is an example of
convolution
1. System response
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10. Laplace transforms 8
Discrete convolution
y(n) = h(n - k) r(k)k = -
1. System response
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10. Laplace transforms 9
Continuous convolution
y(t) = h(t-) r() d
0
t
1. System response
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10. Laplace transforms 10
2. Transforms
DefinitionExamples
2. Transforms
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10. Laplace transforms 11
Definition
Transforms -- a mathematical conversion from one way of thinking to another to make a problem easier to solve
transformsolution
in transformway of
thinking
inversetransform
solution in original
way of thinking
problem in original
way of thinking
2. Transforms
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10. Laplace transforms 12
Example 1
English to algebra solution
in algebra
algebra toEnglish
solution in English
problem in English
2. Transforms
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10. Laplace transforms 13
Example 2
English tomatrices solution
in matrices
matrices toEnglish
solution in English
problem in English
2. Transforms
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10. Laplace transforms 14
Example 3
Laplace transform
solutionin
s domain
inverse Laplace
transform
solution in timedomain
problem in time domain
• Other transforms• Fourier• z-transform• wavelets
2. Transforms
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10. Laplace transforms 15
3. Partial fractions
DefinitionDifferent terms of 1st degreeRepeated terms of 1st degreeDifferent quadratic termsRepeated quadratic terms
3. Partial fractions
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10. Laplace transforms 16
Definition
Definition -- Partial fractions are several fractions whose sum equals a given fraction
Example -- • (11x - 1)/(x2 - 1) = 6/(x+1) + 5/(x-1)• = [6(x-1) +5(x+1)]/[(x+1)(x-1))]• =(11x - 1)/(x2 - 1)
Purpose -- Working with transforms requires breaking complex fractions into simpler fractions to allow use of tables of transforms
3. Partial fractions
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10. Laplace transforms 17
Different terms of 1st degree
To separate a fraction into partial fractions when its denominator can be divided into different terms of first degree, assume an unknown numerator for each fraction
Example -- • (11x-1)/(x2 - 1) = A/(x+1) + B/(x-1)• = [A(x-1) +B(x+1)]/[(x+1)(x-1))]• A+B=11• -A+B=-1• A=6, B=5
3. Partial fractions
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10. Laplace transforms 18
Repeated terms of 1st degree (1 of 2)
When the factors of the denominator are of the first degree but some are repeated, assume unknown numerators for each factor• If a term is present twice, make the
fractions the corresponding term and its second power
• If a term is present three times, make the fractions the term and its second and third powers
3. Partial fractions
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10. Laplace transforms 19
Repeated terms of 1st degree (2 of 2)
Example -- • (x2+3x+4)/(x+1)3= A/(x+1) + B/(x+1)2 +
C/(x+1)3 • x2+3x+4 = A(x+1)2 + B(x+1) + C• = Ax2 + (2A+B)x + (A+B+C)• A=1• 2A+B = 3• A+B+C = 4• A=1, B=1, C=2
3. Partial fractions
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10. Laplace transforms 20
Different quadratic termsWhen there is a quadratic term, assume a
numerator of the form Ax + BExample --
• 1/[(x+1) (x2 + x + 2)] = A/(x+1) + (Bx +C)/ (x2 + x + 2)
• 1 = A (x2 + x + 2) + Bx(x+1) + C(x+1) • 1 = (A+B) x2 + (A+B+C)x +(2A+C)• A+B=0• A+B+C=0• 2A+C=1• A=0.5, B=-0.5, C=0
3. Partial fractions
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10. Laplace transforms 21
Repeated quadratic termsExample --
• 1/[(x+1) (x2 + x + 2)2] = A/(x+1) + (Bx +C)/ (x2 + x + 2) + (Dx +E)/ (x2 + x + 2)2
• 1 = A(x2 + x + 2)2 + Bx(x+1) (x2 + x + 2) + C(x+1) (x2 + x + 2) + Dx(x+1) + E(x+1)
• A+B=0• 2A+2B+C=0• 5A+3B+2C+D=0• 4A+2B+3C+D+E=0• 4A+2C+E=1• A=0.25, B=-0.25, C=0, D=-0.5, E=0
3. Partial fractions
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10. Laplace transforms 22
4. Laplace transform
Laplace transformationDefinitionTransforms
4. Laplace transforms
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10. Laplace transforms 23
Laplace transformation
linear differential equation
timedomainsolution
Laplacetransformed
equation
Laplacesolution
time domain
Laplace domain orcomplex frequency domain
integration
algebra
Laplace transform
inverse Laplace transform
4. Laplace transforms
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10. Laplace transforms 24
Definition
The Laplace transform of the function f(t) is
F(s) =
0
e-st f(t) dt
4. Laplace transforms
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10. Laplace transforms 25
Transforms (1 of 11)
Impulse -- (to)
F(s) =
0
e-st (to) dt
= e-sto
f(t)
t
(to)
4. Laplace transforms
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10. Laplace transforms 26
Transforms (2 of 11)
Step -- u (to)
F(s) =
0
e-st u (to) dt
= e-sto/sf(t)
t
u (to)1
4. Laplace transforms
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10. Laplace transforms 27
Transforms (3 of 11)
tn
F(s) =
0
e-st tn dt
= n!/sn+1
4. Laplace transforms
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10. Laplace transforms 28
Transforms (4 of 11)
e-at
F(s) =
0
e-st e-at dt
= 1/(s+a)
4. Laplace transforms
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10. Laplace transforms 29
Transforms (5 of 11)
e-atcos t
F(s) =
0
e-st e-atcos t dt
= (s+a)/[(s+a)2 + 2]
4. Laplace transforms
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10. Laplace transforms 30
Transforms (6 of 11)
e-atsin t
F(s) =
0
e-st e-atsin t dt
= /[(s+a)2 + 2]
4. Laplace transforms
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10. Laplace transforms 31
Transforms (7 of 11)
f1(t) f2(t)
a f(t)
eat f(t)
f(t +T)
f(t/a)
F1(s) ± F2(s)
a F(s)
F(s-a)
eTs F(s)
a F(as)
Linearity
Constant multiplication
Complex shift
Real shift
Scaling
4. Laplace transforms
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10. Laplace transforms 32
Transforms (8 of 11)
n-th derivative
Dn f(t) sn F(s) - Dn-1 f(0) - s Dn-2 f(0) - … - sn-1 f(0)
4. Laplace transforms
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10. Laplace transforms 33
Transforms (9 of 11)
first derivative
f(t) dt 1/s F(s)
0
t
4. Laplace transforms
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10. Laplace transforms 34
Transforms (10 of 11)
convolution integral
f1() f2(t-) t) d F1(s)
0
t
F2(s)
4. Laplace transforms
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10. Laplace transforms 35
Transforms (11 of 11)
Most mathematical handbooks have tables of Laplace transforms
4. Laplace transforms
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10. Laplace transforms 36
Solution process (1 of 8)
Any nonhomogeneous linear differential equation with constant coefficients can be solved with the following procedure, which reduces the solution to algebra
4. Laplace transforms
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10. Laplace transforms 37
Solution process (2 of 8)Step 1: Put differential equation into
standard form• D2 y + 2D y + 2y = cos t• y(0) = 1• D y(0) = 0
4. Laplace transforms
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10. Laplace transforms 38
Solution process (3 of 8)
Step 2: Take the Laplace transform of both sides• L{D2 y} + L{2D y} + L{2y} = L{cos t}
4. Laplace transforms
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10. Laplace transforms 39
Solution process (4 of 8)Step 3: Use table of transforms to express
equation in s-domain• L{D2 y} + L{2D y} + L{2y} = L{cos t}• L{D2 y} = s2 Y(s) - sy(0) - D y(0)• L{2D y} = 2[ s Y(s) - y(0)]• L{2y} = 2 Y(s)• L{cos t} = s/(s2 + 1) • s2 Y(s) - s + 2s Y(s) - 2 + 2 Y(s) = s /(s2 + 1)
4. Laplace transforms
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10. Laplace transforms 40
Solution process (5 of 8)
Step 4: Solve for Y(s)• s2 Y(s) - s + 2s Y(s) - 2 + 2 Y(s) = s/(s2 + 1)• (s2 + 2s + 2) Y(s) = s/(s2 + 1) + s + 2 • Y(s) = [s/(s2 + 1) + s + 2]/ (s2 + 2s + 2) • = (s3 + 2 s2 + 2s + 2)/[(s2 + 1) (s2 + 2s + 2)]
4. Laplace transforms
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10. Laplace transforms 41
Solution process (6 of 8)Step 5: Expand equation into format covered by
table• Y(s) = (s3 + 2 s2 + 2s + 2)/[(s2 + 1) (s2 + 2s + 2)]• = (As + B)/ (s2 + 1) + (Cs + E)/ (s2 + 2s + 2)• (A+C)s3 + (2A + B + E) s2 + (2A + 2B + C)s +
(2B +E)• 1 = A + C• 2 = 2A + B + E• 2 = 2A + 2B + C• 2 = 2B + E• A = 0.2, B = 0.4, C = 0.8, E = 1.2
4. Laplace transforms
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10. Laplace transforms 42
Solution process (7 of 8)
• (0.2s + 0.4)/ (s2 + 1) • = 0.2 s/ (s2 + 1) + 0.4 / (s2 + 1)• (0.8s + 1.2)/ (s2 + 2s + 2)• = 0.8 (s+1)/[(s+1)2 + 1] + 0.4/ [(s+1)2 + 1]
4. Laplace transforms
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10. Laplace transforms 43
Solution process (8 of 8)
Step 6: Use table to convert s-domain to time domain• 0.2 s/ (s2 + 1) becomes 0.2 cos t• 0.4 / (s2 + 1) becomes 0.4 sin t• 0.8 (s+1)/[(s+1)2 + 1] becomes 0.8 e-t cos
t• 0.4/ [(s+1)2 + 1] becomes 0.4 e-t sin t• y(t) = 0.2 cos t + 0.4 sin t + 0.8 e-t cos t +
0.4 e-t sin t
4. Laplace transforms
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10. Laplace transforms 44
5. Transfer functions
IntroductionExampleBlock diagram and transfer functionTypical block diagramBlock diagram reduction rules
5. Transfer functions
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10. Laplace transforms 45
Introduction
Definition -- a transfer function is an expression that relates the output to the input in the s-domain
differentialequation
r(t) y(t)
transferfunction
r(s) y(s)
5. Transfer functions
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10. Laplace transforms 46
Example
v(t)
R
C
L
v(t) = R I(t) + 1/C I(t) dt + L di(t)/dt
V(s) = [R I(s) + 1/(C s) I(s) + s L I(s)]
Note: Ignore initial conditions5. Transfer functions
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10. Laplace transforms 47
Block diagram and transfer function
V(s) • = (R + 1/(C s) + s L ) I(s)• = (C L s2 + C R s + 1 )/(C s) I(s)
I(s)/V(s) = C s / (C L s2 + C R s + 1 )
C s / (C L s2 + C R s + 1 )V(s) I(s)
5. Transfer functions
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10. Laplace transforms 48
Typical block diagram
controlGc(s)
plantGp(s)
feedbackH(s)
pre-filterG1(s)
post-filterG2(s)
reference input, R(s)
error, E(s)
plant inputs, U(s)
output, Y(s)
feedback, H(s)Y(s)
5. Transfer functions
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10. Laplace transforms 49
Block diagram reduction rules
G1 G2 G1 G2
U Y U Y
G1
G2
U Y+
+ G1 + G2
U Y
G1
G2
U Y+
- G1 /(1+G1 G2)U Y
Series
Parallel
Feedback
5. Transfer functions
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10. Laplace transforms 50
6. Laplace applicationsInitial valueFinal valuePolesZerosStability
6. Laplace applications
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10. Laplace transforms 51
Initial value
In the initial value of f(t) as t approaches 0 is given by
f(0 ) = Lim s F(s)s
f(t) = e -t
F(s) = 1/(s+1)
f(0 ) = Lim s /(s+1) = 1s
Example
6. Laplace applications
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10. Laplace transforms 52
Final value
In the final value of f(t) as t approaches is given by
f() = Lim s F(s)s 0
f(t) = e -t
F(s) = 1/(s+1)
f( ) = Lim s /(s+1) = 0s 0
Example
6. Laplace applications
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10. Laplace transforms 53
Poles
The poles of a Laplace function are the values of s that make the Laplace function evaluate to infinity. They are therefore the roots of the denominator polynomial
10 (s + 2)/[(s + 1)(s + 3)] has a pole at s = -1 and a pole at s = -3
Complex poles always appear in complex-conjugate pairs
The transient response of system is determined by the location of poles
6. Laplace applications
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10. Laplace transforms 54
Zeros
The zeros of a Laplace function are the values of s that make the Laplace function evaluate to zero. They are therefore the zeros of the numerator polynomial
10 (s + 2)/[(s + 1)(s + 3)] has a zero at s = -2Complex zeros always appear in complex-
conjugate pairs
6. Laplace applications
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10. Laplace transforms 55
StabilityA system is stable if bounded inputs produce bounded
outputsThe complex s-plane is divided into two regions: the
stable region, which is the left half of the plane, and the unstable region, which is the right half of the s-plane
s-plane
stable unstable
x
x
xx x
x
x
j
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10. Laplace transforms 56
7. Frequency response
IntroductionDefinitionProcessGraphical methodsConstant KSimple pole at origin, 1/ (j)n
Simple pole, 1/(1+j )Simple pole, 1/(1+j )Error in asymptotic approximationQuadratic pole
7. Frequency response
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10. Laplace transforms 57
Introduction
Many problems can be thought of in the time domain, and solutions can be developed accordingly.
Other problems are more easily thought of in the frequency domain.
A technique for thinking in the frequency domain is to express the system in terms of a frequency response
7. Frequency response
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10. Laplace transforms 58
Definition
The response of the system to a sinusoidal signal. The output of the system at each frequency is the result of driving the system with a sinusoid of unit amplitude at that frequency.
The frequency response has both amplitude and phase
7. Frequency response
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10. Laplace transforms 59
ProcessThe frequency response is computed by
replacing s with j in the transfer function
f(t) = e -t
F(s) = 1/(s+1)
Example
F(j ) = 1/(j +1)
Magnitude = 1/SQRT(1 + 2)
Magnitude in dB = 20 log10 (magnitude)
Phase = argument = ATAN2(- , 1)
magnitude in dB
7. Frequency response
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10. Laplace transforms 60
Graphical methods
Frequency response is a graphical methodPolar plot -- difficult to constructCorner plot -- easy to construct
7. Frequency response
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10. Laplace transforms 61
Constant K
+180o
+90o
0o
-270o
-180o
-90o
60 dB40 dB20 dB 0 dB
-20 dB-40 dB-60 dB
magnitude
phase
0.1 1 10 100, radians/sec
20 log10 K
arg K
7. Frequency response
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10. Laplace transforms 62
Simple pole, 1/ (j)n
+180o
+90o
0o
-270o
-180o
-90o
60 dB40 dB20 dB 0 dB
-20 dB-40 dB-60 dB
magnitude
phase
0.1 1 10 100, radians/sec
1/
1/ 21/ 3
1/ 1/ 2
1/ 3
G(s) = n2/(s2 + 2 ns + n
2)
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10. Laplace transforms 63
Simple pole, 1/(1+j)
+180o
+90o
0o
-270o
-180o
-90o
60 dB40 dB20 dB 0 dB
-20 dB-40 dB-60 dB
magnitude
phase
0.1 1 10 100T
7. Frequency response
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10. Laplace transforms 64
Error in asymptotic approximation
T
0.01
0.1
0.5
0.76
1.0
1.31
1.73
2.0
5.0
10.0
dB
0
0.043
1
2
3
4.3
6.0
7.0
14.2
20.3
arg (deg)
0.5
5.7
26.6
37.4
45.0
52.7
60.0
63.4
78.7
84.37. Frequency response
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10. Laplace transforms 65
Quadratic pole
+180o
+90o
0o
-270o
-180o
-90o
60 dB40 dB20 dB 0 dB
-20 dB-40 dB-60 dB
magnitude
phase
0.1 1 10 100T
7. Frequency response