Signals and Systems - Electrical Engineering

8.3 Discrete-Time Systems 493

from which a general expression for the impulse response is obtained ash[n]=0.5nu[n].

The response tox[n]=u[n]−u[n−3] using the convolution sum is then given by

y[n]=

∑∞

k=−∞

x[k]h[n−k]=

∑∞

k=−∞

(u[k]−u[k−3])0.5n−ku[n−k]

Since as functions ofk,u[k]u[n−k]=1 for 0≤k≤n, zero otherwise, andu[k−3]u[n−k]= 1
for 3≤k≤n, zero otherwise (in the two cases, draw the two signals as functions ofkand verify
this is true),y[n] can be expressed as

y[n]=0.5n

[n

∑

k= 0

0.5−k−

∑n

k= 3

0.5−k

]

u[n]

=











0 n< 0

0.5n

∑n

k= 0 0.5

−k n=0, 1, 2

0.5n

∑ 2

k= 0 0.5

−k n≥3.

Another way to solve this problem is to notice that the input can be rewritten as

x[n]=δ[n]+δ[n−1]+δ[n−2]

and since the system is LTI, the output can be written as

y[n]=h[n]+h[n−1]+h[n−2]=0.5nu[n]+0.5n−^1 u[n−1]+0.5n−^2 u[n−2]

which gives

y[0]=0.5^0 = 1

y[1]=0.5^1 +0.5^0 =

3

2

y[2]=0.5^2 +0.5^1 +0.5^0 =

7

4

y[3]=0..5^3 +0.5^2 +0.5=

7

8

···

which coincides with the above more general solution. It should be noticed that even in a simple
example like this the computation required by the convolution sum is quite high. We will see
that the Z-transform simplifies these types of problems, just like the Laplace transform does in the
computation of the convolution integral.

The following MATLAB script is used to verify the above results. The MATLAB functionfilteris
used to compute the impulse response and the response of the filter to the pulse. The output
obtained then withfiltercoincided with the output computed usingconv, as it should. Figure 8.11
displays the results.