Question

A discrete-time system is described by the following rule
y(n)=(-1)^n x(n) + 2x(n -1)
where x is the input signal, and y the output signal.

Classify the system as:
i. causal/non-causal
ii. linear/nonlinear
iii. time-invariant/time-varying

Answer 1

The function is ,

$y(n) = (-1)^n x(n)+2x(n-1)$

(i) Causal/Non Causal:

The impulse response of the system can be written as,

$h(n) = (-1)^n \delta(n) + 2\delta(n-1 )$

The impulse response is zero zero for all $n<0$ which implies the system is Causal.

ii) Linear/Non Linear:

For input $x_1(n)$ let output be $y_1(n)$

For input $x_2(n)$ let output be $y_2(n)$

We have, $y_1(n) = (-1)^n x_1(n) + 2x_1(n-1)$

$y_2(n) = (-1)^n x_2(n) + 2x_2(n-1)$

If the input was $ax_1(1)+bx_2(n)$,out put is

$y_{1,2} = (-1)^n[ax_1(n)+bx_2(n)]+2[ax_1(n-1)+bx_2(n-1)]$

$= (-1)^nax_1(n)+2ax_1(n-1)+(-1)^nbx_2(n)+2bx_2(n-1)$

$= ay_1(n) + by_2(n)$

Thus we observe the system satisfy the superposition principle,hence it is Linear.

iii) Time invariant/Time varying:

For a delayed input $x(n-n_0)$ the output is

$y(n,n_0) = (-1)^nx(n-n_0) + 2x(n-n_0-1)$ ......................(1)

For a delayed time $n-n_0$, The output equation is,

$y(n-n_0) = (-1)^{n-n_0} x(n-n_0) + 2x(n-n_0-1)$ .............(2)

We observe that equation (1) and (2) are not equal. Which implies it is Time varying.