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.
Answer
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.
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