2.3 LTI Continuous-Time Systems 145
n If the input signalx(t)is causal (i.e.,x(t)=0 fort<0), we can simplify further the above
equation. Indeed
y(t)=
∫t
0
x(τ)h(t−τ)dτ
where the lower limit of the integral is set by the causality of the input signal, and the upper
limit is set by the causality of the system. This equation clearly indicates that the system is causal,
as the outputy(t)depends on present and past values of the input (considering the integral an
infinite sum, the integrand depends continuously onx(τ), fromτ=0 toτ=t, which are past
and present input values). Also ifx(t)=0 the output is also zero.
2.3.7 Graphical Computation of Convolution Integral
Graphically, the computation of the convolution integral, Equation (2.18), consists in multiplying
x(τ)(as a function ofτ) by a reflected (again as function ofτ) and shifted to the righttsec impulse
responseh(t−τ). Once this product is obtained we integrate it from 0 tot(the time at which we
are computing the convolution). The computational cost of this operation is rather high considering
that these operations need to be done for each value oftfor which we are interested in finding the
outputy(t). A more efficient way will be by using the Laplace transform as we will see in the next
chapter.
nExample 2.12
Graphically find the unit-stepy(t)response of an averager, withT=1 sec, which has an impulse
response
h(t)=u(t)−u(t− 1 )
Solution
Plotting the input signalx(τ)=u(τ)and the reflected and delayed impulse responseh(t−τ), both
as functions ofτ, for some value oft(notice that whent=0,h(−τ)is the reflected version of
the impulse response, and fort>0,h(t−τ)ish(−τ)shifted bytto the right) are as shown in
Figure 2.11. Notice the position ofh(t−τ)with respect tox(τ)as it moves from left to right ast
goes from−∞to∞.
We then have the following results for different values oft:
n Ift<0, thenh(t−τ)andx(τ)do not overlap and so the convolution integral is zero, or
y(t)=0 fort<0. That is, the system fort<0 has not yet been affected by the input.
n Fort≥0 andt− 1 <0, or equivalently 0≤t<1,h(t−τ)andx(τ)increasingly overlap, and
as such the integral increases linearly from 0 att=0 to 1 whent=1. So thaty(t)=tfor
0 ≤t<1. That is, for this period of time the system starts reacting slowly to the input.
