Answer to Question #213518 in Electrical Engineering for KHALID

Question

Consider the LTI system

𝑦 𝑛 = 𝑥 𝑛 + 1 + 2𝑥 𝑛 + 𝑥 𝑛 − 1 − 𝑥 𝑛 − 2

What would be the response for the input as shown in figure

Answer 1

$𝑦_ 𝑛 = 𝑥 _𝑛 + 1 + 2𝑥 _𝑛 + 𝑥_ 𝑛 − 1 − 𝑥 _𝑛 − 2 \implies y(n) = (-1)^n x(n)+2x(n-1)$

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 for all $n<0$, implying the Causal system.

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 satisfies 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 equations (1) and (2) are not equal. Which implies it is Time-varying.