Signals and Systems - Electrical Engineering

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

which satisfies the initial conditions and attempts to follow the input signal. This is the unit-step

response.

In the Laplace domain, the above can be easily computed as follows. From the transfer function,

we have that

Y(s)=H(s)X(s)

=

1

(s+ 1 )^2

1

s

where we replaced the transfer function and the Laplace transform ofx(t)=u(t). The partial

fraction expansion ofY(s)is then

Y(s)=

A

s

+

B

s+ 1

+

C

(s+ 1 )^2

and after obtaining thatA=1,C=−1, andB=−1, we get

y(t)=s(t)=u(t)−e−tu(t)−te−tu(t)

which coincides with the solution of the convolution integral. It has been obtained, however, in a

much easier way. n

nExample 3.27

Consider the positive feedback system created by a microphone close to a set of speakers that are

putting out an amplified acoustic signal (see Figure 3.18), which we considered in Example 2.18

in Chapter 2. Find the impulse response of the system using the Laplace transform, and use it to

express the output in terms of a convolution. Determine the transfer function and show that the

system is not BIBO stable. For simplicity, letβ=1,τ=1, andx(t)=u(t). Connect the location of

the poles of the transfer function with the unstable behavior of the system.

Solution

As we indicated in Example 2.18 in Chapter 2, the impulse response of a feedback system cannot

be explicitly obtained in the time domain, but it can be done using the Laplace transform. The

input–output equation for the positive feedback is

y(t)=x(t)+βy(t−τ)

FIGURE 3.18

Positive feedback created by closeness

of a microphone to a set of speakers.

+

×

x(t) y(t)

Delay τ

βy(t −τ)

β