Signals and Systems - Electrical Engineering

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

When we have double real poles we need to express the numeratorN(s)as a first-order polynomial,

just as in the case of a pair of complex conjugate poles. The values ofaandbcan be computed in

different ways, as we illustrate in the following examples.

nExample 3.16

Typically, the Laplace transforms appear as combinations of the different terms we have consid-

ered, for instance a combination of first- and second-order poles gives

X(s)=

4

s(s+ 2 )^2

which has a pole ats=0 and a double pole ats=−2. Find the causal signalx(t). Use MATLAB to

plot the poles and zeros ofX(s)and to find the inverse Laplace transformx(t).

Solution

The partial fraction expansion is

X(s)=

A

s

+

a+b(s+ 2 )

(s+ 2 )^2

The value ofA=X(s)s|s= 0 =1, and so

X(s)−

1

s

=

4 −(s+ 2 )^2

s(s+ 2 )^2

=

−(s+ 4 )

(s+ 2 )^2

=

a+b(s+ 2 )

(s+ 2 )^2

Comparing the numerators ofX(s)− 1 /sand the one in the partial fraction expansion givesb=− 1

anda+ 2 b=−4 ora=−2. We then have

X(s)=

1

s

+

− 2 −(s+ 2 )

(s+ 2 )^2

so that

x(t)=[1− 2 te−^2 t−e−^2 t]u(t)

Another way to do this type of problem is to expressX(s)as

X(s)=

A

s

+

B

(s+ 2 )^2

+

C

s+ 2

We find theAas before, and then findBby multiplying both sides by(s+ 2 )^2 and lettings=−2,

which gives

X(s)(s+ 2 )^2 |s=− 2 =

[

A(s+ 2 )^2

s

+B+C(s+ 2 )

]

s=− 2

so that

B=X(s)(s+ 2 )^2 |s=− 2