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Lecture 21: Tue Oct 27, 2020

Reminder:

• Homework 9 due Thursday

Lecture

• Laplace transforms

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3084 Reading

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Analogy

x( t )

x[n ]DTFT Z Transform2026

3084FT Laplace

finite-energy signals general signals

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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

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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 !

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Back to Pop Quiz!

What is the transform of x( t ) = e3tu( t )?

0 t

Fourier

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Back to Pop Quiz!

What is the transform of x( t ) = e3tu( t )?

0 t

FourierLaplace

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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

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j 3–+---------------------------

1s 3–-----------

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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

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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)

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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?

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Integration Property

INTEGRATORx( t ) y( t ) = x( )d

0

t

↔ Y( s ) = X( s )1s---

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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----------+

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Pop Quiz

If X( s ) = , what is x( t )?

Hint: Consider this picture:

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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