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The question is this:

$$ y[n] = \cos\left(\frac{5\pi}{8}n + \frac{\pi}{4}\right)$$

This is what my teacher said when I asked him for help-:

In any system, inputs are not given, then we have to assume input is impulse signal. Then the response is impulse response

I really couldn't comprehend much of what he said, but I still tried my best, and this is what I got:

Working by OP

Am I correct? (I Don't Think so as I believe this should be a non linear system, I have a gut feeling).

Can you help me here?

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    $\begingroup$ Please include the question and your answer, instead of linking to images. $\endgroup$
    – MBaz
    Jul 4, 2021 at 15:27
  • $\begingroup$ I don't know how to type mathematics. $\endgroup$
    – achhainsan
    Jul 4, 2021 at 15:51
  • $\begingroup$ dsp.stackexchange.com/editing-help $\endgroup$
    – MBaz
    Jul 4, 2021 at 16:09
  • $\begingroup$ Maybe test one of the properties of linearity. If you multiply the input by a factor the output should increase by that factor: $f(c*x[n]) = c * f(x[n])$. $\endgroup$
    – IanJ
    Jul 4, 2021 at 19:18

1 Answer 1

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$y[n] = \cos(\frac{5\pi}{8}n + \frac{\pi}{4})$ is not a system, but a signal.

Anyway, if you insist that you have a system with a fixed output $y[n] = \cos(\frac{5\pi}{8}n + \frac{\pi}{4})$, then this will be a non-linear system.

Proof is easy: assume an arbitrary input $x_1[n]$ to your system. The output will be $y_1[n] = \cos(\frac{5\pi}{8}n + \frac{\pi}{4})$. Then, assume another input $x_2[n] = a ~ x_1[n]$ , the output will be $y_2[n] = \cos(\frac{5\pi}{8}n + \frac{\pi}{4})$. Then since $y_2[n] \neq a ~y_1[n]$, the system cannot be linear; thus it must be nonlinear.

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