Question #213518

Consider the LTI system

ð‘¦ ð‘› = ð‘¥ ð‘› + 1 + 2ð‘¥ ð‘› + ð‘¥ ð‘› âˆ’ 1 âˆ’ ð‘¥ ð‘› âˆ’ 2

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

Expert's answer

"\ud835\udc66_ \ud835\udc5b = \ud835\udc65 _\ud835\udc5b + 1 + 2\ud835\udc65 _\ud835\udc5b + \ud835\udc65_ \ud835\udc5b \u2212 1 \u2212 \ud835\udc65 _\ud835\udc5b \u2212 2 \\implies y(n) = (-1)^n x(n)+2x(n-1)\n"

The function is ,

"y(n) = (-1)^n x(n)+2x(n-1)\n"

(i) Causal/Non-Causal:

The impulse response of the system can be written as,

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

The impulse response is zero for allÂ "n<0",Â implying the Causal system**.**

ii) Linear/Non-Linear:

For inputÂ "x_1(n) "Â let output beÂ "y_1(n)"

For inputÂ "x_2(n) "Â let output beÂ "y_2(n)"

We have,Â "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 wasÂ "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 isÂ **Linear.**

iii) Time invariant/Time-varying:

For a delayed inputÂ "x(n-n_0)"Â the output is

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

For a delayed timeÂ "n-n_0 ", The output equation is,

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

We observe that equations (1) and (2) are not equal. Which implies it isÂ **Time-varying.**

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