Laplace transform - Continuous and Discrete Time Signals and Systems - page 292

which reduces to

5 s 2 + 40 s + 59
( s + 1) ( s 2 + 7 s + 12)

= 5 s 2 + 40 s + 59

( s + 1)( s + 3)( s + 4) .
Taking the partial fraction expansion, we obtain
5 s 2 + 40 s + 59
( s + 1)( s + 3)( s + 4)

W ( s ) =

k 1
( s + 1)

k 2
( s + 3)

k 3
( s + 4) ,

≡ + +

where the partial fraction coefficients are given by

5 s 2 + 40 s + 59
( s + 1)( s + 3)( s + 4)

= 5 − 40 + 59
(2)(3)

k 1 = ( s + 1) = 4 ,

s =− 1

5 s 2 + 40 s + 59
( s + 1)( s + 3)( s + 4)

= 45 − 120 + 59
( − 2)(1)

k 2 = ( s + 3) = 8 ,

s =− 3

and

5 s 2 + 40 s + 59
( s + 1)( s + 3)( s + 4)

= 80 − 160 + 59
( − 3)( − 1)

k 3 = ( s + 4) =− 7 .

s =− 4

Substituting the values of the partial fraction coefficients k 1 , k 2 , and k 3 ,we
obtain

4
( s + 1)

8
( s + 3)

− 7

( s + 4) .
Calculating the inverse Laplace transform of both sides, we obtain the output
signal as follows:

W ( s ) ≡ +

− t − 3 t − 4 t ] u ( t ) .

w ( t ) ≡ [4e + 8e − 7e

Zero-input response To calculate the zero-input output, the input signal is
assumed to be zero. Equation (6.32) reduces to
d 2 w zi
d t 2

+ 7 d w zi
d t

+ 12 w zi ( t ) = 0 , (6.33)

− −

with initial conditions w (0 ) = 5 and w (0 ) = 0. Calculating the Laplace

transform of Eq. 6.33 yields

[ s 2 W zi ( s ) − 5 s ] + 7[ sW zi ( s ) − 5] + 12 W zi ( s ) = 0

or

W zi ( s ) = 5 s + 35

s 2 + 7 s + 12 .
Using the partial fraction expansion, the above equation is expressed as follows:

5 s + 35
s 2 + 7 s + 12

20
s + 3

− 15

s + 4 .
Taking the inverse Laplace transform, the zero-input response is given by

W zi ( s ) = ≡

− 3 t − 4 t ] u ( t ) .

w zi ( t ) ≡ [20e − 15e