Question

10) A continuous-time system is described by the following relation, y(t) = (cos 3t) x (t) where y(t) is the output of the system and x(t) is the input to the system. D

etermine by sufficient explanation whether the system is: a) Stable, b) Memoryless, c) Causal, d) Time invariant, e) Linear.

Answer 1

a) yes this system is stable because for any finite input value it is giving finite oitput value.

and value of cos(3t) will always remain between - 1 and 1

b) yes this system is a memoryless system because output depends on the present input and not on this past or input values of the signal.

c) yes the system is causal because to be a causal signal system output should depend on present and past input values or present or past input values so here this system is depending on present input values of the signal.

d) Time invariant

Let’s assume x1(t)=x(t-to), so, for this input we have the output as;
y1(t)= [cos(3t)] x1(t)i

if we substitute x1(t)=x(t-to), we get:

y1(t)= [cos(3t)] x(t-to)

On the other hand, let’s compute y(t-to) as;
y(t-to)= [cos(3(t-to))] x(t-to)
Since y1(t) is not equal to y(t-to), so the system is not time invariant, it is a time
variant system

e) Linearity

For input x1(t), we have y1(t) = [cos(3t)] x1(t), and
For input x2(t), we have y2(t) = [cos(3t)]x2(t), and,
For input x3(t)=a x1(t)+b x2(t), we have
y3(t) = [cos(3t)] x3(t)
=[cos(3t)] [ax1(t)) + bx2(t)]
= a [cos(3t)] x1(t)+ b [cos(3t)] x2(t)
= a y1(t)+b y2(t);

system is linear