Page 97 - Schaum's Outline of Theory and Problems of Signals and Systems

LINEAR TIME-INVARIANT SYSTEMS [CHAP. 2

For t < 0, we have x(t) = 0, and Eq. (2.101 becomes Eq. (2.103). Hence,
y(r) = Bepa' 1 <O
From the auxiliary condition y(0) = y,, we obtain

~(r) =yoe-"' r<O ( 2.108)
(b) Combining Eqs. (2.107) and (2.108), y(r) can be expressed in terms of y,,(t) (zero-input
response) and y,,(t) (zero-state response) as

where

2.21. Consider the system in Prob. 2.20.
(a) Show that the system is not linear if y(0) = y,, # 0.
(b) Show that the system is linear if y(0) = 0.

(a) Recall that a linear system has the property that zero input produces zero output (Sec.
1.5E). However, if we let K = 0 in Eq. (2.102), we have x(r) = 0, but from Eq. (2.109) we
see that
y(t) = y,,e-" + 0 Yo + 0
Thus, this system is nonlinear if y(0) = yo # 0.
(b) If y(0) = 0, the system is linear. This is shown as follows. Let xJt) and x,(t) be two input
signals and let y ,(t) and y,(r) be the corresponding outputs. That is,

with the auxiliary conditions

= ~2(0) = 0
Consider
x(r) =a,x,(t) +a,x*(t)
where a, and a, are any complex numbers. Multiplying Eq. (2.111) by a, and Eq. (2.112)
by a? and adding, we see that

)'(r) =aI~l(t) +azy,(t)
satisfies the differential equation

dy(t) +ay(t) =x(I)
dr
and also, from Eq. (2.113)
~(0)
= a,y,(O) + ~,Y,(O) = 0
Therefore, y(t) is the output corresponding to x(t), and thus the system is linear.