Page 61 - Schaum's Outline of Theory and Problems of Signals and Systems

SIGNALS AND SYSTEMS [CHAP. 1

Consider a discrete-time system represented by an operator T such that
y[n] = T(x[n]) = x*[n] (1.116)
where x*[n] is the complex conjugate of x[n]. Then

T{xl[nI +x2[nI} = {x~[nI +~2[nI}* =xT[nI +x?[nI =~l[nI +~2[nI
Next, if a is any arbitrary complex-valued constant, then
T{ax[n]} = {ax[n]) * = cu*x*[n] = a*y[n] # ay[n]
Thus, the system is additive but not homogeneous.

Show that the causality for a continuous-time linear system is equivalent to the
following statement: For any time to and any input x(t) with x(t) = 0 for t r r,,,
to.
the output y(t) is zero for t I
Find a nonlinear system that is causal but does not satisfy this condition.
Find a nonlinear system that satisfies this condition but is not causal.

Since the system is linear, if x(t) = 0 for all I, then y(t) = 0 for all t. Thus, if the system is
causal, then x(r) = 0 for t _< to implies that y(t) = 0 for 1 I I,. This is the necessary
condition. That this condition is also sufficient is shown as follows: let x,(t) and x,(t) be
two inputs of the system and let y,(t) and y,(t) be the corresponding outputs. If
x,(t) =x,(t) for t I to, or x(r) =x,(t) -x2(t) = 0 for I st,, then y,(t) =y2(t) for I st,,
or y(t)=y,(~)-y2(t)=0 for ts t,.
Consider the system with the input-output relation
y(t) =x(t) + l
This system is nonlinear (Prob. 1.40) and causal since the value of y(t) depends on only
the present value of x(t). But with x(t) = 0 for I I I,, y(t) = I for t s t,.
Consider the system with the input-output relation
y(t) =x(t)x(t + 1)
It is obvious that this system is nonlinear (see Prob. 1.35) and noncausal since the value of
y(t) at time I depends on the value of x(t + I) of the input at time I + 1. Yet x(t) = 0 for
t,
t I implies that y(t) = 0 for I I I,.

1.44. Let T represent a continuous-time LTI system. Then show that
T{es'} = ks'

where s is a complex variable and h is a complex constant.
Let y(t) be the output of the system with input x(t) = e". Then
T{eSt) = y(t)
Since the system is time-invariant, we have

T(es('+'(l)) = y (I + to)
for arbitrary real to. Since the system is linear, we have
T{~~('+'II)) T{eS' es'~} = e"~T{eS') = eS'~y(t)
=
Hence, y(r +I,) =eH0y(t)