I was trying to solve the below system is a Time variant or Time Invariant System:


Now for a system to be time invariant the system property should be :

y(n,k) = T[x(n-k)] 


y(n,k) = y(n-k)

for all possible values of k.

Coming back to system equation:

$y(n,k) = T[x(n-k)] = nx^{2}(n-k)$ delayed by k sample


$y(n-k) = nx^{2}(n-k)$


$y(n,k) = y(n-k)$

The the system is time invariant, but the solution in book states that the system is time variant. I am not sure where I am going wrong or my understanding is not correct?

  • $\begingroup$ Please define what y(n,k) means. $\endgroup$ – John Aug 12 '14 at 15:56
  • $\begingroup$ As per my understanding it means delay the input sequence by k samples to find the output sequence value $\endgroup$ – Programmer Aug 12 '14 at 16:00
  • $\begingroup$ The selected answer is very good explanation. I also found the following video helpful in understanding time-invariance. However, the video uses an example in continuous domain and not discrete domain. youtube.com/watch?v=j-DEErDE4-U $\endgroup$ – avroshk Jun 27 '17 at 23:22
  • $\begingroup$ check this out:youtube.com/watch?v=BvQIrR2vLZE I also found the following video helpful in understanding time-invariance. However, the video uses an examples to clear doubts $\endgroup$ – helpuneed Feb 6 '19 at 16:26

Here is a simple way of looking at it which avoids any algebra and uses minimal symbolism. Let us calculate the output $y$ for an input $x$ which is a unit pulse function. We have that

  • $y[-2] = (-2)(x[-2])^2 = 0$ since $x[-2] = 0$.

  • $y[-1] = (-1)(x[-1])^2 = 0$ since $x[-1] = 0$.

  • $y[0] = (0)(x[0])^2 = 0$ (regardless of the value of $x[0]$).

  • $y[1] = (1)(x[1])^2 = 0$ since $x[1] = 0$.

  • $y[2] = (2)(x[2])^2 = 0$ since $x[2] = 0$.

and so on. The entire input is blocked and the output $y$ is always $0$.

What about the output $\hat{y}$ when the input $\hat{x}$ is just the previous $x$ delayed by one time unit? The only difference now is that $\hat{x}[1] = 1$ and we have that

  • $\hat{y}[1] = (1)(\hat{x}[1])^2 = 1$ since $\hat{x}[1] = 1$.

It is, I hope, obvious that $\hat{x}[n] = 0$ for all integers $n$ except $n=1$. So, since the input $\hat{x}$ is just a delayed version of input $x$, is is true that the output $\hat{y}$ is just a delayed version of output $y$? What, if anything, does this tell you about the system properties?

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The $y(n-k)=(n-k)x^2(n-k)$ is not equal too $nx^2(n-k)$. Hence it's not a time invariant system

They actually have the exact same example here :) https://ccrma.stanford.edu/~jos/fp/Showing_Linearity_Time_Invariance.html

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delay the input by k samples and denote the output by y(n-k) hence, y(n) = n x^2(n) => y(n,k) = n x^2(n-k) => y(n-k) = (n-k) x^2(n-k) now, after comparing this we can see that y(n,k) is not= y(n-k) so this is time varient. theory says that if y(t) = t x(t) than this is time varient because it depends on t. but if y(t) = 10 x(t) here it doesn't depend on 10 so it is time invarient system. i hope you understand. thanks. if anything is not right than let me know.

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  • $\begingroup$ Thank you for providing an answer. However, it requires more editing from you (esp. formulas) to be usable. $\endgroup$ – Laurent Duval Nov 6 '16 at 11:04



$Y[n-n_0] \ne X[n-n_0]$ because $Y[n-n_0]=n-n_0(x^2[n-n_0])$, and $X[n-n_0]=nx^2[n-n_0]$

Hence it's a time variant system.

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