Answer to Question #213518 in Electrical Engineering for KHALID

Question #213518

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


1
Expert's answer
2021-07-05T04:53:46-0400

"\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)"

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\n)"


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.

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