Laplace transform - Continuous and Discrete Time Signals and Systems - page 280

f ( t )

Example 6.10
Calculate the Laplace transform of the causal function f ( t ) shown in Fig. 6.7.

6

t

−1 0 1 2 3 5

Solution
In terms of the waveform h ( t ) shown in Fig. 6.6, f ( t ) is expressed as follows:

Fig. 6.7. Waveform f ( t ) used in
Example 6.10.

f ( t ) = 2 h ( t − 3) .

In Example 6.9, the Laplace transform of h ( t )isgivenby

9

for s = 0

H ( s ) = 1 . 5

s 2

− 2 s − 2 s e − 4 s ]

[1 − e for s = 0 ,

with the entire s-plane as the ROC. Using the time-shifting property, the Laplace
transform of f ( t )is

←→

− 3 s H ( s )

f ( t ) = 2 h ( t − 3) 2e with ROC: R ,

which results in the following Laplace transform for f ( t ):

− 3 s ] s = 0

[18e

= 18 for s = 0

H ( s ) = 3

s 2 [e

− 3 s − e − 5 s − 2 s e − 7 s ]

for s = 0 ,

with the entire s-plane as the ROC.

6.4.4 Shifting in the s-domain

←→

If x ( t ) X ( s ) with ROC: R , then the Laplace transform of

←→

e s 0 t x ( t )

+ Re s 0 (6.20)

X ( s − s 0 ) with ROC: R

for both unilateral and bilateral Laplace transforms. Shifting a signal in the
complex s-domain by s 0 causes the ROC to shift by Re { s 0 } . Although the
amount of shift s 0 can be complex, the shift in the ROC is always a real number.
In other words, the ROC is always shifted along the horizontal axis, irrespective
of the value of the imaginary component in s 0 .
The shifting property can be proved directly from Eq. (6.9) by considering
the CTFT of the signal exp( s 0 t ) x ( t ). The proof is left as an exercise for the
reader (see Problem 6.6).

Example 6.11
Using the Laplace transform pair

←→ 1
s

u ( t ) with ROC: Re { s }> 0 ,

calculate the Laplace transform of (i) x 1 ( t ) = cos( ω 0 t ) u ( t ) and (ii) x 2 ( t ) =
sin( ω 0 t ) u ( t ).