Signals and Systems - Electrical Engineering

204 C H A P T E R 3: The Laplace Transform

nExample 3.15

Consider the Laplace function

X(s)=

2 s+ 3

s^2 + 2 s+ 4

=

2 s+ 3

(s+ 1 )^2 + 3

Find the corresponding causal signalx(t), then use MATLAB to validate your answer.

Solution

The poles are at− 1 ±j

√

3, so that we expect thatx(t)is a decaying exponential with a damping

factor of−1 (the real part of the poles) multiplied by a causal cosine of frequency

√

3. The partial
   fraction expansion is of the form

X(s)=

2 s+ 3

s^2 + 2 s+ 4

=

a+b(s+ 1 )

(s+ 1 )^2 + 3

so that 3+ 2 s=(a+b)+bs, orb=2 anda+b=3 ora=1. Thus,

X(s)=

1

√

3

√

3

(s+ 1 )^2 + 3

+ 2

s+ 1

(s+ 1 )^2 + 3

which corresponds to

x(t)=

[

1

√

3

sin(

√

3 t)+2 cos(

√

3 t)

]

e−tu(t)

The valuex( 0 )=2 and according to the initial value theorem the following limit should equal it:

lim

s→∞

[

sX(s)=

2 s^2 + 3 s

s^2 + 2 s+ 4

]

=lim

s→∞

2 + 3 /s

1 + 2 /s+ 4 /s^2

= 2

which is the case, indicating the result is probably correct (satisfying the initial value theorem is

not enough to indicate the result is correct, but if it does not the result is wrong).

We use the MATLAB functionilaplaceto compute symbolically the inverse Laplace transform and

plot the response usingezplot, as shown in the following script.

%%%%%%%%%%%%%%%%%

% Example 3.15

%%%%%%%%%%%%%%%%%

clear all; clf

syms s t w

num = [0 2 3]; den = [1 2 4]; % coefficients of numerator and denominator

subplot(121)

splane(num,den) % plotting poles and zeros

disp(’>>>>>Inverse Laplace<<<<<’)

x = ilaplace((2 * s + 3)/(s ˆ 2 + 2 * s + 4)); % inverse Laplace transform

subplot(122)