i) memory less,

ii) casual,

iii) time –invariant

$ Y[n] = nx [n].$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 05

Year : DEC 2015

(i) From the given system equation, the output depends upon present input only. Hence the system is static. i.e. memory less.

(ii) The system equation is y[n] = n x[n]. The output depends upon present input only. Hence the system is causal.

(iii) The given discrete time system equation is

$y[n] = T\{x[n]\} = n x[n] ---------- 1$

When input x[n] is delayed by ‘k’ samples, the response is

$y[n, k] = T\{x[n - k]\} \\ ∴y[n, k] = n x[n - k] ------------ 2$

Here, it is seen that only input x[n] is delayed. The multiplier ‘n’ is not part of the input. Hence it cannot be written as (n - k).

Now let us shift or delay the output y[n] given by equation 1 by ‘k’ samples i.e.

$∴y[n - k] = [n – k] x[n – k] ------------- 3$

Here both n and x[n] in the equation y[n] = n x[n] will be shifted by ‘k’ samples since they are part of the output sequence.

Therefore from equations 2 and 3

$y[n, k] ≠ y[n - k]$

Hence the system is time variant.