# How to find a system without an input $x[n]$ is linear or non linear

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: 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?

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

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