# Time invariance of a System

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.

• probably you have a constraint that $x[n] \neq 0$. Also do you have anything to say about the initial conditions of the solution $y[n]$? Oct 17 '17 at 8:30
• @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. Oct 17 '17 at 8:37
• Can we assume causality? Oct 17 '17 at 14:22
• Why did you tag the question with continuous-signals, as the equation sounds discrete? Jun 25 '19 at 19:30

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.
• 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. Oct 17 '17 at 19:54