I have this small question about the time invariance of a system. Which is:

If the current output is multiplied by the current input (see both are variables) will the system be time variant or time invariant?

To be precise, the question I am dealing with is: $$ y(n-3) + y(n-1) + y(n)x(n) = x(n-3) $$

I would be glad if anyone could help.

  • $\begingroup$ probably you have a constraint that $x[n] \neq 0$. Also do you have anything to say about the initial conditions of the solution $y[n]$? $\endgroup$
    – Fat32
    Commented Oct 17, 2017 at 8:30
  • $\begingroup$ @Fat32 this is all I have got. It was one of the question in the quiz I had yesterday. I am not able to figure out how to go ahead with this equation. $\endgroup$
    – Copernicus
    Commented Oct 17, 2017 at 8:37
  • $\begingroup$ Can we assume causality? $\endgroup$
    – AnVij
    Commented Oct 17, 2017 at 14:22
  • $\begingroup$ Why did you tag the question with continuous-signals, as the equation sounds discrete? $\endgroup$ Commented Jun 25, 2019 at 19:30

1 Answer 1


Yes, this system is time-invariant assuming $x[n]\neq0 $ for all $n $.

The test for this is $$x_{1}[n]=x[n-n_{0}] $$ $$y_{1}[n]=y[n-n_{0}] $$

So let's first express in terms of just $y[n]$. $$ y[n]=\frac{x[n-3]-y[n-3]-y[n-1]}{x[n]} $$ Next, we go through the test. $$y[n-n_{0}]=\frac{x[n-3-n_{0}]-y[n-3-n_{0}]-y[n-1-n_{0}]}{x[n-n_{0}]} $$ Substituting $y_{1}[n] $ $$y_{1}[n]=\frac{x[n-3-n_{0}]-y[n-3-n_{0}]-y[n-1-n_{0}]}{x[n-n_{0}]} $$ $$y_{1}[n]=\frac{x[n-3-n_{0}]-y_{1}[n-3]-y_{1}[n-1]}{x[n-n_{0}]} $$ And finally substituting $x_{1}[n] $ $$y_{1}[n]=\frac{x_{1}[n-3]-y_{1}[n-3]-y_{1}[n-1]}{x_{1}[n]} $$ Because the shifted sequence has the exact same relationship, it is said to be time-invariant. Most systems that don't alter $n$ or $t$ meet this. Be wary of anything multiplying or otherwise messing with the vector arguments beyond simple delays.

  • $\begingroup$ I think I get it. The point which perplexed me was the x(n)y(n). And we know that when the coefficients are time varying then the system is time varying. $\endgroup$
    – Copernicus
    Commented Oct 17, 2017 at 19:12
  • $\begingroup$ Not necessarily. The coefficients on the indices being time-varying would be a red flag, but the normal impulse coefficients can be time varying and keep the overall system time-invariant. Time-invariant just means that the relationship between input and output is independent with regards to shifting (the same input shifted left or right $n_{0}$ will give the same output shifted the same manner). If I answered your question, please mark the thread as solved. $\endgroup$
    – Keegs
    Commented Oct 17, 2017 at 19:54

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