Laplace transform - Continuous and Discrete Time Signals and Systems - page 277

g ( t )

Proof
Calculating the Laplace transform of { a 1 x 1 ( t ) + a 1 x 2 ( t ) } using Eq. (6.9) yields

4

t

−4 −3 −2 −1 0 2

∞

a 1 x 1 ( t ) + a 2 x 2 ( t ) e − st d t L a 1 x 1 ( t ) + a 2 x 2 ( t ) =

Fig. 6.5. Causal function g ( t )
considered in Example 6.8.

0 −

∞ ∞

a 1 x 1 ( t )e − st d t + a 2 x 2 ( t )e − st d t

=

0 − 0 −

∞ ∞

x 1 ( t )e − st d t + a 2 x 2 ( t )e − st d t

= a 1

0 − 0 −

= a 1 X 1 ( s ) + a 2 X 2 ( s ) ,

which proves Eq. (6.17).

By definition of the ROC, the Laplace transform X 1 ( s ) is finite within the
specified region R 1 . Similarly, X 2 ( s ) is finite within its ROC R 2 . Therefore, the
linear combination a 1 X 1 ( s ) + a 2 X 2 ( s ) must at least be finite in region R that
represents the intersection of the two regions i.e. R = R 1

∩ R 2 . If there is no
common region between R 1 and R 2 , then the Laplace transform of { a 1 x 1 ( t ) +
a 1 x 2 ( t ) } does not exist. Due to the cancellation of certain terms in a 1 X 1 ( s ) +
a 2 X 2 ( s ), it is also possible that the overall ROC of the linear combination is
larger than R 1 ∩ R 2 . To illustrate the application of the linearity property, we

consider the following example.

Example 6.8
Calculate the Laplace transform of the causal function g ( t ) shown in Fig. 6.5.

Solution
The causal function g ( t ) is expressed as the linear combination

g ( t ) = 4 x 3 ( t ) + 2 x 7 ( t ) ,

where the CT functions x 3 ( t ) and x 7 ( t ) are defined in Example 6.4. Based on
the results of Example 6.4, the Laplace transforms for x 3 ( t ) and x 7 ( t ) are given
by

2

for s = 0

X 3 ( s ) = with ROC: entire s-plane

1
s [e

− 2 s − e − 4 s ]

for s = 0

and

3

for s = 0

X 7 ( s ) = 2

s 2 [1 − e

with ROC: entire s-plane

− s − s e − 2 s ]

for s = 0 .