I followed these steps, but the answer still says that this system is time-invariant

let: $x_2[n] = x[n-k]$

$$\begin{align} y_2[n] &= x_2[-n] \\ &= x[-(n-k)] \\ &= x[k-n] \\ \end{align}$$


$$\begin{align} y[n-k] &= x[(-(n-k)] \\ &= x[k-n] \end{align}$$

where am I wrong in the analysis?

  • 3
    $\begingroup$ it's not time invariant because it matters where $n=0$ is. $\endgroup$ Commented Apr 4, 2018 at 5:31

3 Answers 3


A time-invariant system is one that, when you shift the input signal, the output is shifted by the same amount.

A system that reverses the signal cannot be time-invariant because when you shift the input, the output is shifted the other way. $k$ and $-k$ are not the same amount.

$$ y[n-k] = x[k-n] = x[-n \mathbf{+} k] $$


I would suggest to first try with simple examples, to get more insight. And a counter example might suffice to disprove the claim.

Here, take the discrete delta $\delta_k(n)$, zero everywhere, except at $n=k$, where $\delta(k)=1$. This discrete input is the paragon for time-invariant linear systems.

So: with your system $\mathcal{S}$, you get $$\mathcal{S} \delta_0(n) = \delta_0(-n)= \delta_0(n)$$ a mirror-symmetry invariant. But $$\mathcal{S} \delta_1(n) = \delta_{-1}(n)$$

So, time-invariant here is doomed.


You have asked what is wrong in your analysis. First let me explain the intuition: You have a sequence X(n). At n=1 the sequence takes the value X(1). At n=2 the sequence takes X(2) and so on. You now define Y(n) as X(-n). This means at n=1, Y(n) takes the value of X(-1), at n=2, Y(n) takes the value X(-2) and so on. Now, you delay the input sequence by k. This means at n=1 the value of the sequence is X(1-k) at n=2 the value is X(2-k) and so on. Now for the same system when this is the input sequence, at n=1, Y(1) = the value of the input sequence at n=-1 which is Y(-1-k). Now can you see the big picture?

There are many ways to understand this: When you take X(n-k) and then flip it as X(-(n-k)) then you are reflecting the input sequence about X=k. But the system reflects the input sequence around X=0.

Now to answer, your question about where your analysis is wrong, let Z(n) = X( n-k). Then when this is passed to the system the output Y(n) is Z(-n) which is X(-n-k) and not X(-(n-k)) which is X(-n+k).

It is difficult to wrap one's head around this. Atleast it was so for me. But hope it helps.


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