Page 114 - Schaum's Outline of Theory and Problems of Signals and Systems

CHAP. 21 LINEAR TIME-INVARIANT SYSTEMS 1 03

For n < 0, we have x[n] = 0, and Eq. (2.149) becomes Eq. (2.151). Hence.
y[n] =Ban (2.159)
From the auxiliary condition y[ - 11 = y - ,, we have

y[- 11 =y-, = BU-'
from which we obtain B = y - ,a. Thus,

Combining Eqs. (2.158) and (2.160), y[n] can be expressed as

Note that as in the continuous-time case (Probs. 2.21 and 2.221, the system described by
Eq. (2.149) is not linear if y[- 11 # 0. The system is causal and time-invariant if it is initially at
rest, that is, y[- 11 = 0. Note also that Eq. (2.149) can be solved recursively (see Prob. 2.43).

2.43. Consider the discrete-time system in Prob. 2.42. Find the output y[n] when x[n] =
KS[n] and y[-l] =y-, =a.

We can solve Eq. (2.149) for successive values of y[n] for n r 0 as follows: rearrange Eq.
(2.149) as
y[n] =ay[n - 11 +x[n]
Then

y[n] =ay[n - 11 +x[n] =an(acu + K) =an+'cu + anK ( 2.16.3)
Similarly, we can also determine y[n] for n < 0 by rearranging Eq. (2.149) as

Then y[-11 =a

1
y[-n] = -{y[-n + 11 -x[-n + 11) =a-"+'a
a
Combining Eqs. (2.163) and (2.169, we obtain
y[n] = an+'a + Kanu[n]