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Lecture 21: Tue Oct 27, 2020
Reminder:
• Homework 9 due Thursday
Lecture
• Laplace transforms
1
3084 Reading
2
Analogy
x( t )
x[n ]DTFT Z Transform2026
3084FT Laplace
finite-energy signals general signals
3
2 Laplace Transforms
Bilateral: X( s ) =Z
–∞
∞x( t )e–stdt
• where s = + j
• Useful property: reduces to Fourier integral when s = j
Unilateral: X( s ) =Z
0–
∞x( t )e–stdt
Our choice, because:
• signals are often causal anyway
• less concern about “ROC”
• it helps understand diff-eq systems with init conditions
only reduces to Fourier when signal is causal
identical for“causal” signals
4
Function vs Surface
Im{s}
Re{s}
|X( s )|
|X( j)|
0
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The (1-Sided) Laplace Transform
X( s ) =Z
0–
∞
x( t )e–stdt
Plug in real and imaginary parts of s = + j
⇒ X( s ) =Z
0–
∞
x( t )e–( + j)tdt
=Z
0–
∞
x( t )e–te–jtdt
= F{x( t )e–t}. (when causal.)
Even when x( t ) does not have a FT, x( t )e–t will for large enough !
6
Back to Pop Quiz!
What is the transform of x( t ) = e3tu( t )?
0 t
Fourier
7
Back to Pop Quiz!
What is the transform of x( t ) = e3tu( t )?
0 t
FourierLaplace
8
Back to Pop Quiz!
What is the transform of x( t ) = e3tu( t )?
⇒ X( s ) =Z
0–
∞
x( t )e–stdt
= F{x( t )e–t} = F{u( t )e(3 – )t}
= (using FT Table 1)
=
only when = Re{s} > 3
0 t
FourierLaplace
1
j 3–+---------------------------
1s 3–-----------
9
About Laplace: X( s ) =Z0–
∞
x( t )e–stdt
It’s better than Fourier when:
signals might not have Fourier transforms
analyzing systems that may or may not be unstable
solving diff. eq’s
Setting s = j reduces LT to FT (when signal is causal). (This explains why we bothered to carry the “j” around in X(j)!)
Useful interpretation relating Laplace to Fourier:
X( s ) = F{x( t )e–t} , where = Re{ s}
FT integral will converge for large enough even when x( t ) has no FT
10
Unit Stepx( t ) = u( t )
↔ X( s ) = Z
0
∞
x( t )e–stdt
= F{u( t )e–t}
=
=
only when = Re{s} > 0
0 t
1
j +------------------
1s---
this is the “region of convergence” (ROC)
11
Example
x( t ) = e–atu( t ) ↔ X( s ) = x( t )e–stdt =
only when = Re{s} > –a
0 t
0
1
s a+-------------
ROC
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Intuition
Im{s}
Re{s}
Each slice along line Re{s} = is |F{x( t )e–t}|
|X( s )|
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Integration Property
INTEGRATORx( t ) y( t ) = x( )d
0
t
↔ Y( s ) = ?
use convolution property:what is impulse response of integrator?
14
Integration Property
INTEGRATORx( t ) y( t ) = x( )d
0
t
↔ Y( s ) = X( s )1s---
15
Unit Ramp
x( t ) = tu( t ) ↔ X( s ) =
Can solve in at least 3 ways:
convolution property
Integration by parts
integration property
1s2-----
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Triangle
↔ X( s ) =
Can solve in at least 2 ways:
convolution property ⇒ X( s ) =
express x( t ) = r( t ) – 2r(t – 1) + r(t – 2), where r( t ) = tu( t ) is unit ramp, and use delay property
⇒ X( s ) =
1
20 t
x( t )
1 e s––s
---------------- 2
1 e s––s
---------------- 2
1s2----- 2e s–
s2-----------– e 2s–
s2----------+
17
Pop Quiz
If X( s ) = , what is x( t )?
Hint: Consider this picture:
1
s5-----
INTEGRATORu( t )
INTEGRATOR INTEGRATOR INTEGRATOR
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Table 1: Laplace Pairs
( t )
u( t )
e–atu( t )
↔ 1
1s---↔
↔ 1s a+-------------
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Table 1: Pairs
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Table 2: Properties
–tx( t )dds------X( s )s-domain differentiation
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