I got this answer from here. But i don't know how this method is correct. Until now to prove shift invariance what i did was :
step 1: delay output and write it in terms of input.
step 2: delay input and write the modified output
step 3: if output from step 1 and 2 are same then system is shift invariant.

what this guy has done is "different". the oppenheim book and proakis manokalis book don't follow this method. which book follows this method?

the method.

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.


The so called example system is nonlinear and recursive and those books generally exclude the treatment of nonlinear & recursive systems. Hence you may not find example solutions.

The method relies on the fact that the relationship between the output $y[n]$ and input $x[n]$ remains the same between $y[n-d]$ and $x[n-d]$.

As an absurd example, think of the following. Consider the equation of a circle at origin: $x^2 + y^2 = r^2$. Now, shift this circle to the point $(a,b)$ and re-write its equation: $(x-a)^2 + (y-b)^2 = r^2$. Finally, redefine the variables $x' = x-a$ and $y' = y-b$ and rewrite the equation: $(x')^2 + (y')^2 = r^2$.

It's seen that the circle is shift invariant (not in the sense of the signal processing context though)

  • $\begingroup$ do you know a book that follows this method? $\endgroup$ – abhishek Jan 26 at 12:23
  • $\begingroup$ you should look at every question and answer... no I don't remember. $\endgroup$ – Fat32 Jan 26 at 12:59

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