Signals and Systems - Electrical Engineering

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

Solution

The partial fraction expansion is

X(s)=

A

s+ 1 −j

√

3

+

A∗

s+ 1 +j

√

3

+

B

s

We then have

B=sX(s)|s= 0 = 1

A=X(s)(s+ 1 −j

√

3 )|s=− 1 +j√ 3 =0.5

(

− 1 +

j

√

3

)

=

1

√

3

∠ 150 ◦

so that

x(t)=

2

√

3

e−tcos(

√

3 t+ 150 ◦)u(t)+u(t)

=−[cos(

√

3 t)+0.577 sin(

√

3 t)]e−tu(t)+u(t) n

Remarks

n Following the above development, when the poles are complex conjugate and double the procedure for the

double poles is repeated. Thus, the partial expansion is given as

X(s)=

N(s)

(s+α−j 0 )^2 (s+α+j 0 )^2

=

a+b(s+α−j 0 )

(s+α−j 0 )^2

+

a∗+b∗(s+α+j 0 )

(s+α+j 0 )^2

(3.28)

so that finding a and b we obtain the inverse.

n The partial fraction expansion for second- and higher-order poles should be done with MATLAB.

nExample 3.18

In this example we use MATLAB to find the inverse Laplace transform of more complicated func-

tions than the ones considered before. In particular, we want to illustrate some of the additional

information that our functionpfeLaplacegives. Consider the Laplace transform

X(s)=

3 s^2 + 2 s− 5

s^3 + 6 s^2 + 11 s+ 6

Find poles and zeros ofX(s), and obtain the coefficients of its partial fraction expansion (also

called the residues). Useilaplaceto find its inverse and plot it usingezplot.

Solution

The following is the functionpfeLaplace.