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?
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Sign up to join this communityI 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?
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]\$.
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.