Laplace transform - Continuous and Discrete Time Signals and Systems - page 264

Im{ s }

x ( t ) = e − at u ( t )

t Re{ s }

0 − a 0

(a) (b)

Example 6.1 shows that the bilateral Laplace transform of the decaying expo-
nential function x ( t ) = e

Fig. 6.1. (a) Exponential
decaying function
x ( t ) = e − at u ( t ); (b) its
associated ROC, Re{ s > − a ,
over which the bilateral Laplace
transform exists.

− at u ( t ) will converge to a finite value X ( s ) = 1 / ( s + a )
within the ROC (Re { s > − a ). In other words, the bilateral Laplace transform
of x ( t ) = e

− at u ( t ) exists for all values of a within the specified ROC. No restric-
tion is imposed on the value of a for the existence of the Laplace transform.
On the other hand, the CTFT of the decaying exponential function exists only
for a > 0. For a < 0, the exponential function x ( t ) = e

− at u ( t ) is not absolutely
integrable, and hence its CTFT does not exist. This is an important distinction
between the CTFT and the bilateral Laplace transform. The CTFT exists for a
limited number of absolutely integrable functions. By associating an ROC with
the bilateral Laplace transform, we can evaluate the Laplace transform for a
much larger set of functions.

Example 6.2
Calculate the bilateral Laplace transform of the non-causal exponential function
g ( t ) =− e

− at u ( − t ).

Solution
Substituting g ( t ) =− e

− at u ( − t ) in Eq. (6.5), we obtain

0
−∞

∞ 0

1
( s + a ) e − ( s + a ) t

− e − at u ( − t )e − st d t =− e − ( s + a ) t d t =

G ( s ) = .

−∞ −∞

At the upper limit, t → 0, e − ( s + a ) t = 1. At the lower limit, t →−∞ ,e − ( s + a ) t
is finite only if Re { s + a < 0, where it equals zero. The bilateral Laplace
transform is therefore given by

1
( s + a )

for Re s < − a

G ( s ) =

undefined for Re s ≥− a .

Figure 6.2 illustrates the ROC, Re s < − a , for the bilateral Laplace transform
of g ( t ) =− e

− at u ( − t ).