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?

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

Suppose you had another input

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

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]\$.

  • \$\begingroup\$ 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? \$\endgroup\$ – Lerbi Dec 17 '16 at 2:51
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    \$\begingroup\$ 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. \$\endgroup\$ – The Photon Dec 17 '16 at 3:09
  • \$\begingroup\$ Interesting. If that pops up on the exam Ill try the Delta function method! Thanks a lot \$\endgroup\$ – Lerbi Dec 17 '16 at 3:13

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


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


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