# Is this system linear or non linear? $y[n]=cos(\frac{5\pi}{8}*n+\frac{\pi}{4})$ [closed]

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

If I let $$(\frac{5\pi}{8}*n+\frac{\pi}{4})=x[n]$$, I get the system as non linear. But one of my teacher says it is linear, whereas another teacher says it is non linear.

• systems are classified as "linear" or "nonlinear". not signals. signals are what they are. Sep 13, 2021 at 5:17
• Oh so is it linear system or non linear system? Thanks for the input tho. Sep 13, 2021 at 5:34
• there is no system. it's a signal. Sep 13, 2021 at 5:48
• So the system's output is always the same, regardless of the input? There's no $x[n]$ in your equation ... Sep 13, 2021 at 6:22
• @MattL. What about a system that maps everything to 0 ... ;-). Sep 13, 2021 at 9:03

If the input is $$x[n] = (\frac{5\pi}{8}*n+\frac{\pi}{4}).$$ Then the system is $$y[n]=cos(x[n])$$, which is a non-linear system.
If the input is $$x[n]=cos(\frac{5\pi}{8}*n+\frac{\pi}{4}).$$ Then the system is $$y[n] = x[n]$$, which is a linear system.
Suppose $$y_1[n] = f(x_1[n]), y_2[n] = f(x_2[n]),$$ then for a linear system, $$y_1[n]+y_2[n]=f(x_1[n])+f(x_2[n])=f(x_1[n]+x_2[n])$$ which means, if $$x[n] = x_1[n]+x_2[n]$$, then $$y[n]=f(x[n])=y_1[n]+y_2[n]$$.