# How to check a discrete time system if the following is linear and time invariant?

I have the following system:

$$y(n) = x(n^2+n)$$ where x(n) = 1 if 0<=n<=3; 0 otherwise.

I tried doing the usual check if yd(n) = y(n+d), but that's not giving me the right answer.

Anyone know how to attempt this?

• - until n is related to t it is time invariant but since y(n) obviously has a n^2 term, it is non-linear – Tony Stewart Sunnyskyguy EE75 Dec 17 '16 at 2:12
• @TonyStewart.EEsince'75, presumably we're talking about a discrete-time system. – The Photon Dec 17 '16 at 2:12
• Sorry, this is a discrete system! – Lerbi Dec 17 '16 at 2:13
• LTI system means it independent of absolute time, as long as initial conditions are given, – Tony Stewart Sunnyskyguy EE75 Dec 17 '16 at 2:23
• I don't understand your notation when you said you checked if "yd(n) = y(n+d)"... what is meant by "yd(n)"? – The Photon Dec 17 '16 at 2:53

$$x[n] = \left\lbrace\begin{matrix}1 & 1\le n \le 4 \\0 & \rm{otherwise} \end{matrix}\right.$$

Does the output look the same as it did for your example input, but only shifted in time?

This is testing the property of an LTI system that when $y[n]$ is the output for input $x[n]$, then the output should be $y[n-n_0]$ for the input $x[n-n_0]$.

• I understand this, and I should clarify that I did get the right answer going about it like this. What you would do in this case is have your new y'(n) = x(n^2+n) function, and plug in values in for n and compare the input n^2+n to values between 1 and 4. However, this method doesn't seem... "mathematical" enough, I guess? Is this the only way? – Lerbi Dec 17 '16 at 2:51
• It would be quicker to test with delta functions, say $\delta[n-1]$ and $\delta[n-4]$ or something. Basically, seeing that $n^2$ should set off huge red flags right away. – The Photon Dec 17 '16 at 3:09
• Interesting. If that pops up on the exam Ill try the Delta function method! Thanks a lot – Lerbi Dec 17 '16 at 3:13

In order to check if something is linear, the following condition needs to be checked.

$y(\alpha_1n_1+\alpha_2n_2)=\alpha_1y(n_1)+\alpha_2y(n_2)$

It is clear that this is not the case for arbitrary inputs $x(n)$. E.g. $x(n)$ does not satisfy this equation.