Laplace transform - Continuous and Discrete Time Signals and Systems - page 297

Example 6.19
Consider the LTIC systems in Example 6.18(i) and (ii). Calculate the impulse
response if the specified LTIC systems are causal. Repeat for non-causal
systems.

Solution
(i) Using the partial fraction expansion, H 1 ( s ) can be expressed as follows:

( s + 4)( s + 5)
s 2 ( s + 2)( s − 2)

≡ k 1
s

+ k 2
s 2

k 3
( s + 2)

k 4
( s − 2) ,

H 1 ( s ) = + +

where

d
d s

( s + 4)( s + 5)
( s + 2)( s − 2)

2 s + 9
s 2 − 4

− ( s + 4)( s + 5)
( s 2 − 4) 2

=− 9

k 1 = ≡ 2 s

4 ,

s = 0 s = 0

( s + 4)( s + 5)
s 2 ( s + 2)( s − 2)

≡ (4)(5)
2( − 2)

s 2

k 2 = =− 5 ,

s = 0

( s + 4)( s + 5)
s 2 ( s + 2)( s − 2)

≡ (2)(3)
4( − 4)

=− 3 k 3 = ( s + 2)

8 ,

s =− 2

and

( s + 4)( s + 5)
s 2 ( s + 2)( s − 2)

≡ (6)(7)
4(4)

= 21
8

k 4 = ( s − 2) .

s = 2

Therefore,

H 1 ( s ) ≡− 9
4 s

− 5
s 2

3
8( s + 2)

21
8( s − 2) .

− +

If H 1 ( s ) represents a causal LTIC system, then its ROC, based on Property 7,
is given by R c :Re { s > 2. Based on the linearity property, the overall ROC R c
is only possible if the ROCs for the individual terms in H 1 ( s ) are given by

=− 9
4 s

5
s 2

3
8( s + 2 )
ROC:Re s >− 2

21
8 ( s − 2) .
ROC:Re s > 2

H 1 ( s ) − − +

ROC:Re s > 0 ROC:Re s > 0

By calculating the inverse Laplace transform, the impulse response for a causal
LTIC system is obtained as follows:

− 9
4

− 5 t − 3 + 21

8

− 2 t

e 2 t

h 1 ( t ) =

8 e

u ( t ) .

If H 1 ( s ) represents a non-causal system, then its ROC can have three different
values: Re { s < − 2; − 2 < Re s < 0; or 0 < Re s < 2 in the s-plane. Select-
ing Re { s < − 2 as the ROC will lead to a left-sided signal. The remaining two
choices will lead to a double-sided signal. Assuming that we select the overall
ROCtobe R nc :Re { s < − 2 , the ROCs for the individual terms in H 1 ( s ) are