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6.003: Signals and Systems
Laplace and Z Transforms
October 1, 2009
Mid-term Examination #1
Wednesday, October 7, 7:30-9:30pm, Walker Memorial.
No recitations on the day of the exam.
Coverage: DT Signals and Systems
Lectures 1–5
Homeworks 1–4
Homework 4 will include practice problems for mid-term 1.
However, it will not collected or graded. Solutions will be posted.
Closed book: 1 page of notes (812 × 11 inches; front and back).
Designed as 1-hour exam; two hours to complete.
Review sessions during open office hours.
Conflict? Contact [email protected] before Friday, October 2, 5pm.
Last Time
Many continuous-time systems can be represented with differential
equations.
Example: leaky tank
r0(t)
r1(t)h1(t)
Differential equation representation:
τ r1(t) = r0(t)− r1(t)
Last time we considered two methods to solve differential equations:
• solving homogeneous and particular equations
• singularity matching
Solving Differential Equations with Laplace Transform
The Laplace transform provides a particularly powerful method of
solving differential equations — it transforms a differential equation
into an algebraic equation.
Method (where L represents the Laplace transform):
differential algebraic
algebraic differential equation
−→
↓ solve
−→
differentialequation
algebraic
equation
algebraicanswer
solution todifferential equation
L−→
↓ solve
L−1−→
Laplace Transform: Definition
Laplace transform maps a function of time t to a function of s.
X(s) =∫x(t)e−stdt
There are two important variants:
Unilateral (18.03)
X(s) =∫ ∞
0x(t)e−stdt
Bilateral (6.003)
X(s) =∫ ∞−∞x(t)e−stdt
Both share important properties — will discuss differences later.
Laplace Transforms
Example: Find the Laplace transform of x1(t):
0t
x1(t)
x1(t) ={Ae−σt if t ≥ 00 otherwise
X1(s) =∫ ∞−∞x1(t)e−stdt =
∫ ∞0Ae−σte−stdt = Ae
−(s+σ)t
−(s+ σ)
∣∣∣∣∣∞
0= A
s+ σ
provided Re{s+ σ} > 0 which implies that Re{s} > −σ.
−σ
s-plane
ROCA
s+ σ; Re{s} > −σ
Check Yourself
0t
x2(t)
x2(t) ={e−t − e−2t if t ≥ 00 otherwise
Which of the following is the Laplace transform of x2(t)?
1. X2(s) = 1(s+1)(s+2) ; Re{s} > −1
2. X2(s) = 1(s+1)(s+2) ; Re{s} > −2
3. X2(s) = s(s+1)(s+2) ; Re{s} > −1
4. X2(s) = s(s+1)(s+2) ; Re{s} > −2
5. none of the above
Check Yourself
X2(s) =∫ ∞
0(e−t − e−2t)e−stdt
=∫ ∞
0e−te−stdt−
∫ ∞0e−2te−stdt
= 1s+ 1
− 1s+ 2
= (s+ 2)− (s+ 1)(s+ 1)(s+ 2)
= 1(s+ 1)(s+ 2)
These equations converge if Re{s + 1} > 0 and Re{s + 2} > 0, thus
Re{s} > −1.
−1−2
s-plane
ROC1(s+ 1)(s+ 2)
; Re{s} > −1
Check Yourself
0t
x2(t)
x2(t) ={e−t − e−2t if t ≥ 00 otherwise
Which of the following is the Laplace transform of x2(t)?
1. X2(s) = 1(s+1)(s+2) ; Re{s} > −1
2. X2(s) = 1(s+1)(s+2) ; Re{s} > −2
3. X2(s) = s(s+1)(s+2) ; Re{s} > −1
4. X2(s) = s(s+1)(s+2) ; Re{s} > −2
5. none of the above
Regions of Convergence
Left-sided signals have left-sided Laplace transforms (bilateral only).
Example:
t
x3(t)
−1x3(t) =
{−e−t if t ≤ 00 otherwise
X3(s) =∫ ∞−∞x3(t)e−stdt =
∫ 0
−∞−e−te−stdt = −e
−(s+1)t
−(s+ 1)
∣∣∣∣∣0
−∞= 1s+ 1
provided Re{s+ 1} < 0 which implies that Re{s} < −1.
−1
s-plane
RO
C1s+ 1
; Re{s} < −1
Left- and Right-Sided Signals
We can concisely express left- and right-sided signals by multiplica-
tion with step functions.
0t
x1(t)x1(t) =
{e−t if t ≥ 00 otherwise
≡ e−tu(t)
t
x3(t)
−1
x3(t) ={−e−t if t ≤ 00 otherwise
≡ −e−tu(−t)
Left- and Right-Sided ROCs
Laplace transforms of left- and right-sided exponentials have the
same form (except −); with left- and right-sided ROCs, respectively.
0t
e−tu(t)time function Laplace transform
−1
s-plane
ROC1s+ 1
t
−e−tu(−t)
−1 −1
s-plane
RO
C
1s+ 1
Check Yourself
Find the Laplace transform of x4(t).
0t
x4(t)
x4(t) = e−|t|
1. X4(s) = 21−s2 ; −∞ < Re{s} <∞
2. X4(s) = 21−s2 ; −1 < Re{s} < 1
3. X4(s) = 21+s2 ; −∞ < Re{s} <∞
4. X4(s) = 21+s2 ; −1 < Re{s} < 1
5. none of the above
Check Yourself
X4(s) =∫ ∞−∞e−|t|e−stdt
=∫ 0
−∞e(1−s)tdt+
∫ ∞0e−(1+s)tdt
= e(1−s)t
(1− s)
∣∣∣∣∣0
−∞+ e−(1+s)t
−(1 + s)
∣∣∣∣∣∞
0
= 11− s︸ ︷︷ ︸
Re{s}<1
+ 11 + s︸ ︷︷ ︸
Re{s}>−1
= 1 + s+ 1− s(1− s)(1 + s)
= 21− s2
; −1 < Re{s} < 1
The ROC is the intersection of Re{s} < 1 and Re{s} > −1.
Check Yourself
The Laplace transform of a signal that is both-sided a vertical strip.
0t
x4(t)
x4(t) = e−|t|
−1 1
s-plane
ROCX4(s) = 21− s2
−1 < Re{s} < 1
Check Yourself
Find the Laplace transform of x4(t). 2
0t
x4(t)
x4(t) = e−|t|
1. X4(s) = 21−s2 ; −∞ < Re{s} <∞
2. X4(s) = 21−s2 ; −1 < Re{s} < 1
3. X4(s) = 21+s2 ; −∞ < Re{s} <∞
4. X4(s) = 21+s2 ; −1 < Re{s} < 1
5. none of the above
Solving Differential Equations with Laplace Transforms
Solve the following differential equation:
y(t) + y(t) = δ(t)
Take the Laplace transform of this equation.
L{y(t) + y(t)} = L{δ(t)}
The Laplace transform of a sum is the sum of the Laplace transforms
(prove this as an exercise).
L{y(t)}+ L{y(t)} = L{δ(t)}
What’s the Laplace transform of a derivative?
Laplace transform of a derivative
Assume that X(s) is the Laplace transform of x(t):
X(s) =∫ ∞−∞x(t)e−stdt
Find the Laplace transform of y(t) = x(t).
Y (s) =∫ ∞−∞y(t)e−stdt
=∫ ∞−∞x(t)e−stdt
= x(t)e−st∣∣∞−∞ −
∫ ∞−∞x(t)(−se−st)dt
The first term must be zero since X(s) converged. Thus
Y (s) = s∫ ∞−∞x(t)e−stdt = sX(s)
Solving Differential Equations with Laplace Transforms
Back to the previous problem:
L{y(t)}+ L{y(t)} = L{δ(t)}
Let Y (s) represent the Laplace transform of y(t).
Then sY (s) is the Laplace transform of y(t).
sY (s) + Y (s) = L{δ(t)}
What’s the Laplace transform of the impulse function?
Laplace transform of the impulse function
Let x(t) = δ(t).
X(s) =∫ ∞−∞δ(t)e−stdt
=∫ ∞−∞δ(t) e−st
∣∣t=0 dt
=∫ ∞−∞δ(t) 1 dt
= 1
Sifting property: δ(t) sifts out the value of the integrand at t = 0.
Solving Differential Equations with Laplace Transforms
Back to the previous problem:
sY (s) + Y (s) = L{δ(t)} = 1
This is a simple algebraic expression. Solve for Y (s):
Y (s) = 1s+ 1
We’ve seen this Laplace transform previously.
y(t) = e−tu(t) (why not y(t) = −e−tu(−t) ?)
Notice that we solved the differential equation y(t) + y(t) = δ(t) with-
out computing homogeneous and particular solutions and without
singularity matching.
Solving Differential Equations with Laplace Transforms
Summary of method.
Start with differential equation:
y(t) + y(t) = δ(t)
Take the Laplace transform of this equation:
sY (s) + Y (s) = 1
Solve for Y (s):
Y (s) = 1s+ 1
Take inverse Laplace transform (by recognizing form of transform):
y(t) = e−tu(t)
Solving Differential Equations with Laplace Transforms
Recognizing the form ...
Is there a more systematic way to take an inverse Laplace transform?
Yes ... and no.
Formally,
x(t) = 12πj
∫ σ+j∞σ−j∞
X(s)estds
but this integral is not generally easy to compute.
This equation can be useful to prove theorems.
We will find better ways (e.g., partial fractions) to compute inverse
transforms for common systems.
Solving Differential Equations with Laplace Transforms
Example 2:
y(t) + 3y(t) + 2y(t) = δ(t)
Laplace transform:
s2Y (s) + 3sY (s) + 2Y (s) = 1
Solve:
Y (s) = 1(s+ 1)(s+ 2)
= 1s+ 1
− 1s+ 2
Inverse Laplace transform:
y(t) =(e−t − e−2t)u(t)
These forward and inverse Laplace transforms are easy if
• differential equation is linear with constant coefficients, and
• the input signal is an impulse function.
Properties of Laplace Transforms
The use of Laplace Transforms to solve differential equations de-
pends on several important properties.
Property x(t) X(s) ROC
Linearity ax1(t) + bx2(t) aX1(s) + bX2(s) ⊃ (R1 ∩R2)
Delay by T x(t− T ) X(s)e−sT R
Multiply by t tx(t) −dX(s)ds
R
Multiply by e−αt x(t)e−αt X(s+ α) shift R by −α
Differentiate in tdx(t)dt
sX(s)
Integrate in t
∫ t−∞x(τ) dτ X(s)
s⊃ (R ∩ {Re{s} > 0})
Convolve in t
∫ ∞−∞x1(τ)x2(t− τ) dτ X1(s)X2(s) ⊃ (R1 ∩R2)
Z Transform
The Z transform is the DT analog of the Laplace transform.
The Z transform converts a difference (recurrence) equation into a
simple algebraic equation.
Method (where Z represents the Z transform):
differential algebraic
algebraic differential equation
−→
↓ solve
−→
differenceequation
algebraic
equation
algebraicanswer
solution todifference equation
Z−→
↓ solve
Z−1−→
Z Transform: Definition
Z transform maps a function of discrete time n to a function of z.
X(z) =∑x(n)z−n
There are two important variants:
Unilateral
X(z) =∞∑n=0x[n]z−n
Bilateral
X(z) =∞∑n=−∞
x[n]z−n
Differences are analogous to those for the Laplace transform.
Z Transforms
Example: Find the Z transform of x1[n]:
−4−3−2−1 0 1 2 3 4n
x1[n]x1[n] =
{(78
)nif n ≥ 0
0 otherwise
≡(
78
)nu[n]
X1(z) =∞∑n=−∞
(78
)nz−nu[n] =
∞∑n=0
(78
)nz−n = 1
1− 78z−1 = z
z − 78
provided |z| > 78 .
−4−3−2−1 0 1 2 3 4n
x1[n] =(
78
)nu[n]
78
z-plane ROC
z
z − 78
Check Yourself
Let X2(z) represent the Z transform of x2[n]:
−4 −3 −2 −1 0 1 2n
x2[n] = −(
78
)nu[−1− n]
What is the relation between X2(z) and X1(z)?
1. X2(z) = −X1(z)2. X2(z) = X1(z)
3. X2(z) = 1X1(z)
4. none of the above
Check Yourself
Let X2(z) represent the Z transform of x2[n].What is the relation between X2(z) and X1(z)?
−4 −3 −2 −1 0 1 2n
x2[n] = −(
78
)nu[−1− n]
X2(z) =∞∑n=−∞
−(
78
)nu[−1− n]z−n = −
−1∑n=−∞
(78
)nz−n
Let l = −1− n:
X2(z) = −∞∑l=0
(78
)−1−lz1+l = −8z
7
∞∑l=0
(8z7
)l= −8z
71
1− 8z7
= z
z − 78
provided∣∣8z
7∣∣ < 1 or |z| < 7
8 .
Z Transforms: Check Yourself
−4 −3 −2 −1 0 1 2n
x2[n] = −(
78
)nu[−1− n]
−(8
7)1−
(87)2−
(87)3
−(8
7)4 7
8
z-plane ROC
z
z − 78
−4−3−2−1 0 1 2 3 4n
x1[n] =(
78
)nu[n]
78
z-plane ROC
z
z − 78
Check Yourself
Let X2(z) represent the Z transform of x2[n]:
−4 −3 −2 −1 0 1 2n
x2[n] = −(
78
)nu[−1− n]
What is the relation between X2(z) and X1(z)? 2
1. X2(z) = −X1(z)2. X2(z) = X1(z)
3. X2(z) = 1X1(z)
4. none of the above
Solving Difference Equations with Z Transforms
Solve the following difference equation:
y[n]− 12y[n− 1] = δ[n]
Take the Z transform of this equation.
Z{y[n]− 1
2y[n− 1]
}= Z {δ[n]}
The Z transform of a sum is the sum of the Z transforms (prove
this as an exercise).
Z {y[n]} − Z{
12y[n− 1]
}= Z {δ[n]}
What’s the Z transform of a delay?
Z transform of a delay
Assume that X(z) is the Z transform of x[n]:
X(z) =∞∑n=−∞
x[n]z−n
Find the Z transform of y[n] = x[n− 1].
Y (z) =∞∑n=−∞
y[n]z−n =∞∑n=−∞
x[n− 1]z−n
Let l = n− 1:
Y (z) =∞∑l=−∞
x[l]z−1−l = z−1∞∑l=−∞
x[l]z−l = z−1X(z)
Solving Difference Equations with Z Transforms
Back to the previous problem:
Z {y[n]} − Z{
12y[n− 1]
}= Z {δ[n]}
Let Y (z) represent the Z transform of y[n]. Then z−1Y (z) is the Z
transform of y[n− 1].
Y (z)− 12z−1Y (z) = Z {δ[n]}
What’s the Z transform of the unit sample function?
Z transform of the unit sample function
Let x[n] = δ[n]. Substituting in the definition of the Z transform:
X(z) =∞∑n=−∞
δ[n]z−n = z−0 = 1
Solving Difference Equations with Z Transforms
Back to the previous problem:
Y (z)− 12z−1Y (z) = Z {δ[n]} = 1
This is a simple algebraic expression. Solve for Y (z):
Y (z) = 11− 1
2z−1
By analogy with the Z transform of x1[n]:
y[n] =(
12
)nu[n]
This method for solving difference equations is analogous to using
the Laplace transform to solve a differential equation.
Solving Difference Equations with Z Transforms
Summary of method.
Start with difference equation:
y[n]− 12y[n− 1] = δ[n]
Take the Z transform of this equation:
Y (z)− 12z−1Y (z) = 1
Solve for Y (z):
Y (z) = 11− 1
2z−1
Take the inverse Z transform (by recognizing the form of the trans-
form):
y[n] =(
12
)nu[n]
Inverse Z transform
The inverse Z transform has defined by an integral that is not par-
ticularly easy to solve.
Formally,
x[n] = 12πj
∫CX(z)zn−1ds
were C represents a closed contour that circles the origin by running
in a counterclockwise direction through the region of convergence.
This integral is not generally easy to compute.
This equation can be useful to prove theorems.
We will find better ways (e.g., partial fractions) to compute inverse
transforms for the kinds of systems that we most frequently en-
counter.
Properties of Z Transforms
The use of Z Transforms to solve differential equations depends on
several important properties.
Property x[n] X(z) ROC
Linearity ax1[n] + bx2[n] aX1(z) + bX2(z) ⊃ (R1 ∩R2)
Delay x[n− 1] z−1X(z) R
Multiply by n nx[n] −z dX(z)dz
R
Convolve in n∞∑
m=−∞x1[m]x2[n−m] X1(z)X2(z) ⊃ (R1 ∩R2)
Concept Map: Discrete-Time Systems
Topics
+
Delay
Delay
X Y
Block Diagram
H = 11−R−R2
System Functional
h : 1, 1, 2, 3, 5, 8, 13, . . .
Unit-Sample Response
y[n] = x[n] + y[n−1] + y[n−2]
Difference Equation
H(z) = z2
z2 − z − 1
System Function
Concept Map: Discrete-Time Systems
Connections
+
Delay
Delay
X Y
Block Diagram
H = 11−R−R2
System Functional
h : 1, 1, 2, 3, 5, 8, 13, . . .
Unit-Sample Response
y[n] = x[n] + y[n−1] + y[n−2]
Difference Equation
H(z) = z2
z2 − z − 1
System Function
Concept Map: Discrete-Time Systems
Example of connections
+
Delay
Delay
X Y
Block Diagram
H = 11−R−R2
System Functional
h : 1, 1, 2, 3, 5, 8, 13, . . .
Unit-Sample Response
y[n] = x[n] + y[n−1] + y[n−2]
Difference Equation
H(z) = z2
z2 − z − 1
System Function
Delay ↔ R
Check Yourself
Where is the Z transform ?
+
Delay
Delay
X Y
Block Diagram
H = 11−R−R2
System Functional
h : 1, 1, 2, 3, 5, 8, 13, . . .
Unit-Sample Response
y[n] = x[n] + y[n−1] + y[n−2]
Difference Equation
H(z) = z2
z2 − z − 1
System Function
1
2
3
4
5 6
7 8
Check Yourself
Where is the Z transform ?
+
Delay
Delay
X Y
Block Diagram
H = 11−R−R2
System Functional
h : 1, 1, 2, 3, 5, 8, 13, . . .
Unit-Sample Response
y[n] = x[n] + y[n−1] + y[n−2]
Difference Equation
H(z) = z2
z2 − z − 1
System Function
Z transform
Concept Map: Discrete-Time Systems
Examples of connections
+
Delay
Delay
X Y
Block Diagram
H = 11−R−R2
System Functional
h : 1, 1, 2, 3, 5, 8, 13, . . .
Unit-Sample Response
y[n] = x[n] + y[n−1] + y[n−2]
Difference Equation
H(z) = z2
z2 − z − 1
System Function
Delay ↔ R
Delay ↔ z−1
Delay ↔ [n− 1]
R ↔ z−1
Step-by-Step Series,
Partial Fractions
Step-by-Step Z transform
Summary
Introduced Laplace transform
Applied Laplace transform to solving differential equations
Introduced Z transform
Applied Z transform to solving difference equations
Found lots of connections between representations of DT systems