Signals and Systems - Electrical Engineering

212 C H A P T E R 3: The Laplace Transform

where

F(s)=

1 −e−s

s+ 1

The inverse ofF(s)is

f(t)=e−tu(t)−e−(t−^1 )u(t− 1 )

and the inverse ofX(s)is thus given by

x(t)=f(t)+f(t− 2 )+f(t− 4 )+··· n

3.4.3 Inverse of Two-Sided Laplace Transforms

When finding the inverse of a two-sided Laplace transform we need to pay close attention to the

region of convergence and to the location of the poles with respect to thejaxis. Three regions of

convergence are possible:

n A plane to the right of all the poles, which corresponds to a causal signal.

n A plane to the left of all poles, which corresponds to an anti-causal signal.

n A region that is in between poles on the right and poles on the left (no poles included in it),

which corresponds to a two-sided signal.

If thejaxis is included in the region of convergence, bounded-input bounded-output (BIBO) sta-

bility of the system, or absolute integrability of the impulse response of the system, is guaranteed.

Furthermore, the system with that region of convergence would have a frequency response, and the

signal a Fourier transform. The inverses of the causal and the anti-causal components are obtained

using the one-sided Laplace transform.

nExample 3.20

Find the inverse Laplace transform of

X(s)=

1

(s+ 2 )(s− 2 )

ROC:− 2 <Re(s) < 2

Solution

The ROC− 2 <Re(s) <2 is equivalent to {(σ,):− 2 < σ <2,−∞<  <∞}. The partial

fraction expansion is

X(s)=

1

(s+ 2 )(s− 2 )

=

−0.25

s+ 2

+

0.25

s− 2

− 2 <Re(s) < 2

where the first term with the pole ats=−2 corresponds to a causal signal with a region of con-

vergenceRe(s) >−2, and the second term corresponds to an anti-causal signal with a region of