Signals and Systems - Electrical Engineering

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

where the{pk}are simple real poles ofX(s), its partial fraction expansion and its inverse are given by

X(s)=

∑

k

Ak

s−pk

⇔ x(t)=

∑

k

Akepktu(t) (3.22)

where the expansion coefficients are computed as

Ak=X(s)(s−pk)

∣

∣s=pk

According to Laplace transform tables the time function corresponding toAk/(s−pk)isAkepktu(t),

thus the form of the inversex(t). To find the coefficients of the expansion, sayAj, we multiply both

sides of the Equation (3.22) by the corresponding denominator(s−pj)so that

X(s)(s−pj)=Aj+

∑

k6=j

Ak(s−pj)

s−pk

If we lets=pj, ors−pj=0, in the above expression, all the terms in the sum will be zero and we

find that

Aj=X(s)(s−pj)

∣

∣

∣s=pj

nExample 3.14

Consider the proper rational function

X(s)=

3 s+ 5

s^2 + 3 s+ 2

=

3 s+ 5

(s+ 1 )(s+ 2 )

Find its causal inverse.

Solution

The partial fraction expansion is

X(s)=

A 1

s+ 1

+

A 2

s+ 2

Given that the two poles are real, the expected signalx(t)will be a superposition of two decaying

exponentials, with damping factors−1 and−2, or

x(t)=[A 1 e−t+A 2 e−t]u(t)

where as indicated above,

A 1 =X(s)(s+ 1 )|s=− 1 =

3 s+ 5

s+ 2

|s=− 1 = 2