3.5 Analysis of LTI Systems 221
and substitutingX(s)=1, then
H(s)=Y(s)=
1
sT
[1−e−sT]
The impulse response is then
h(t)=
1
T
[u(t)−u(t−T)]. n
3.5.2 Computation of the Convolution Integral
From the point of view of signal processing, the convolution property is the most important
application of the Laplace transform to systems. The computation of the convolution integral is
difficult even for simple signals. In Chapter 2 we showed how to obtain the convolution integral
analytically as well as graphically. As we will see in this section, it is not only that the convolution
property of the Laplace transform gives an efficient solution to the computation of the convolution
integral, but that it introduces an important representation of LTI systems, namely thetransfer func-
tion of the system. A system, like signals, is thus represented by the poles and zeros of the transfer
function. But it is not only the pole-zero characterization of the system that can be obtained from the
transfer function. The system’s impulse response is uniquely obtained from the poles and zeros of
the transfer function and the corresponding region of convergence. The way the system responds to
different frequencies will be also given by the transfer function. Stability and causality of the system
can be equally related to the transfer function. Design of filters depends on the transfer function.
The Laplace transform of the convolutiony(t)=[x∗h](t)is given by the product
Y(s)=X(s)H(s) (3.38)
whereX(s)=L[x(t)]andH(s)=L[h(t)]. The transfer functionof the systemH(s)is defined as
H(s)=L[h(t)]=
Y(s)
X(s)
(3.39)
H(s)transfers the Laplace transformX(s)of the input into the Laplace transform of the outputY(s). OnceY(s)
is found,y(t)is computed by means of the inverse Laplace transform.
nExample 3.25
Use the Laplace transform to find the convolutiony(t)=[x∗h](t)when
(1) the input isx(t)=u(t)and the impulse response is a pulseh(t)=u(t)−u(t− 1 ), and
(2) the input and the impulse response of the system arex(t)=h(t)=u(t)−u(t− 1 ).
Solution
n The Laplace transforms areX(s)=L[u(t)]= 1 /sandH(s)=L[h(t)]=( 1 −e−s)/s, so that
Y(s)=H(s)X(s)=
1 −e−s
s^2
