2

3

How Do We Perform Inverse Laplace Transform?

• In 6.003, we will only deal with Laplace transform that are:

1) Rational, i.e. X(s) = N(s)/D(s) ;

And/or:

2) exponential, i.e. X(s) = e-sT .

•For case 2), use shift property

•For case 1), use PFE.

•For case 1) & 2), i.e.

x(t!T)" # $ e!sT X(s) (Similar to the FT property

x(t!T)" # $ e!j

%

TX(j

%

)

X(s)=e!sT N(s) / D(s)

( )

X1(s)"x1(t)

1 2 4 4 3 4 4

Then b

x(t)=x1(t)t!t"T=x1(t"T)

4

Inverse Laplace Transforms Via Partial Fraction

Expansion and Properties

Example:

Three possible ROC’s — corresponding to three different signals

Recall

X(s)=s+3

s+1

( )

s!2

( )

=A

s+1+B

s!2

A=!2

3,B=5

3

1

s+a, Re(s)<!a" # $ !e!at u(!t) left - sided

and 1

s+a, Re(s)>!a" # $ e!at u(t) right - sided

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3

5

Example (cont.)

ROC I: — Left-sided signal.

x(t) =

ROC II: — Two-sided signal, has Fourier Transform.

x(t) =

ROC III: — Right-sided signal.

x(t) =

!Ae!tu(!t)!Be 2tu(!t)

!2

3

e!tu(t)+5

3

e2tu(!t)

"

#

$ %

&

' (0 as t(±)

Ae!tu(t)+Be 2tu(t)

=2

3

e!t!5

3

e2t"

#

$ %

&

' u(!t) Diverges as t( ! )

=!2

3

e!t+5

3

e2t"

#

$ %

&

' u(t) Diverges as t(+)

6

Properties of Laplace Transforms

•Many parallel properties of the CTFT, but for Laplace

transforms we need to determine implications for the ROC

•For example:

Linearity

ROC at least the intersection of ROC’s of X1(s) and X2(s)

ROC can be bigger (due to pole-zero cancellation)

E.g. x1(t) = x2(t) and a = -b

Then a x1(t) + b x2(t) = 0 → X(s) = 0

⇒ ROC entire s-plane

ax1(t)+bx2(t)! " # aX1(s)+bX2(s)

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4

7

Time Shift

x(t!T)" # $ e!sT X(s) , same ROC as X(s)

e!sT

s+2, Re(s)>!2" # $ e!2tu(t)t#t!T

!T="3

e3s

s+2, Re(s)>"2# $ % e"2(t+3)u(t+3)

E.g.,

e3s

s+2, Re(s)>!2" # $ ?

Causal or not?

8

Time-Domain Differentiation

x(t)=1

2

!

jX(s)est ds

"

#j$

"

+j$

%,dx(t)

dt =1

2

!

jsX(s)estds

"

#j$

"

+j$

%

ROC could be bigger than the ROC of X(s), if there is pole-zero

cancellation. E.g.

x(t)=u(t)! " # 1

s, Re(s)>0

s-Domain Differentiation

!tx(t)" # $ dX(s)

ds , with same ROC as X(s)

(Derivation is

similar to d

dt

!s)

!

dx(t)

dt " # $ sX(s), with ROC containing the ROC of X(s)

dx(t)

dt =

!

(t)" # $ 1=s%1

sROC =entire s&plane

E.g.te!atu(t)" # $ !d

ds

1

s+a

%

&

' (

)

* =1

s+a

( )

2, Re(s)>!a

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5

9

Convolution Property

x(t)h(t)y(t)=h(t)!x(t)

For x(t)! " # X(s), y(t)! " # Y(s), h(t)! " # H(s)

•This is the main reason why Laplace transform is useful

in solving problems involving LTI systems, just like the

Fourier transform. However, because Laplace transform

and its inverse transform are not symmetric, the reverse

relation

x(t)•h(t) X(s)∗H(s)

is not true, different from Fourier transform. Thus,

L-transform is not useful in modulation and sampling.

×

Then Y(s)=H(s)!X(s)

10

Convolution Property (cont.)

ROC of Y(s) = H(s)X(s), at least the overlap of the ROC’s

of H(s) & X(s)

a) ROC could be empty if there is no overlap between the

two ROCs

E.g. x(t) = etu(t) Re(s) > 1 and h(t) = -e-tu(-t) Re(s) < -1

b) ROC could be larger than the overlap of the two.

E.g. x(t)∗h(t) =

δ

(t) ⇒ ROC entire s-plane.

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pakhi

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University:
Birla Institute of Technology and Science

Subject:
Signals and Systems

Upload date:
18/07/2012