Signals and Systems - Electrical Engineering

3.4 Inverse Laplace Transform 207

FIGURE 3.15

Inverse Laplace

transform of

X(s)= 4 /(s(s+ 2 )^2 ):

(a) poles and zeros and

(b)x(t).

− 2 0 2

− 1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

σ

jΩ

0 5 10

0

0.2

0.4

0.6

0.8

1

t

x(t)

(a) (b)

To findCwe compute the partial fraction expansion for a value ofsfor which no division by zero

is possible. For instance, if we lets=1 we can find the value ofC, after which we can find the

inverse.

The initial valuex( 0 )=0 coincides with

lim

s→∞

[

sX(s)=

4 s

s(s+ 2 )^2

]

=lim

s→∞

4 /s^2

( 1 + 2 /s)^2

= 0

To find the inverse Laplace transform with MATLAB we use a similar script to the one used before;

only the numerator and denominator description needs to be changed. The plots are shown in

Figure 3.15. n

nExample 3.17

Find the inverse Laplace transform of the function

X(s)=

4

s((s+ 1 )^2 + 3 )

which has a simple real poles=0, and complex conjugate poless=− 1 ±j

√

3.