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According to my solution,

$$y(n) = x(n^2) \tag{1}$$

for $y(n,k)$ i.e output for delayed input $x(n-k)$

$$x_2(n) = x(n-k)$$

so $$y_2(n) = x_2(n) = x\big((n-k)^2\big)$$

and for delayed output signal $y_1(n)$, replace $n$ by $n-k$ in equation (1), so we get,

$$y_1(n) = x\big((n-k)^2\big)$$ and therefore system is time invariant

But in the answers to the book in which this question it says the system is time variant. Can anyone point out the mistake in my steps, and give correct way of solving this sum?

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The system is obviously time-varying. However you make the following usual mistake in the first case of showing the response to the delayed input $x_2[n] = x[n-k]$ as follows.

Writing the output $y[n]$ in the formal manner: $$ y[n] = \mathcal{T}\{x[n]\} = x[n^2] $$ and denoting the shifted input as $$x_2[n] = x[n-k]$$ you would have the corresponding output $y_k[n]$ as:

$$y_k[n] = \mathcal{T}\{ x_2[n]\} = x_2[n^2] = x[n^2 -k] $$ which is defintely not equal to $y[n-k] = x[(n-k)^2]$ hence the system is time-varying.

You should have defined the output as $$ x_2[n] = x[n-k] \implies x_2[n^2] = x[n^2 -k] $$ but you didn't. See that the equality sign between $x[n]$ and $x_2[n]$ enables you to operate on $n$ only, not the whole argument of both sides.

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  • $\begingroup$ Okay I understand my mistake, but during this time I was searching for the solution manual for the book(proakis) online and in it even though it also comes to the same conclusion(time variant), their proof states for input x(n-k) ---> $y_2(n) = x((n-k)^2) = x(n^2 + k^2 - 2nk)$ != y(n-k). They reversed what they had to do with y(n-k) and $y_2(n)$ didn't they? Even if the conclusion is time variant. $\endgroup$ – nino Nov 13 '17 at 14:12
  • $\begingroup$ @nino if that's what they did, so yes they reversed it. But I hope the corrected version here helped you... $\endgroup$ – Fat32 Nov 13 '17 at 20:58
  • $\begingroup$ Yes this answer does help. They sketched the signals $y(n-k)$ and $y_2(n)$ for k=2 and $x(n)$ = {$0,1^\star,1,1,1,0$..} (asterisk denoting origin) and did the following : $y(n) = x(n^2) $= {...,$0,1,1^\star,1,0,$...}, $y(n-2)$ = {...,$0,0^\star,1,1,1,0,$..}, $x(n-2)$ = {....,$0^\star,0,1,1,1,1,0,$...}, $y_2(n) = T[x(n-2)]$ = {...,$0,1,0,0^\star,0,1,0,$...} so that probably added to my confusion $\endgroup$ – nino Nov 14 '17 at 1:20
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@Fat32 followed a generic approach. It can be completed with something more ad-hoc, depending on the specific problem at hand. Here, the $n^2$ indices entail that $y$ will be limited to values of $x$ with positive (and square) indices.

Now, we suspect a counter-example will do the job, by chosing some with given values at squared indices, and translating them. Sometimes, very simple examples are working. take $x[n]$ to be the Kronecker symbol $\delta_0$, equal to 1 at $n =0$ and zero elsewhere. Its output to the system is itself. Translate it to the left: $x_{-1}[n] = \delta_{-1}$. Since the latter is zero for positive indices, the output is identically zero, hence not a shift of the previous output $\delta_0$.

While not generic, such examples can provide insights to the properties of the system. In difficulty, do not hesitate to try with simple signals (a ramp, a comb, a triangle) to see what happens, before attacking the problem globally.

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