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The system is described with the following recursive differences equation:

$$y[n]-4y[n-1]+4y[n-2]=20x[n]+10x[n-1]$$

now lets say the input is delayed by k, then:

$$y[n]-4y[n-1]+4y[n-2]=20x[n-k]+10x[n-1-k]$$

and now the output by the same k:

$$y[n-k]-4y[n-1-k]+4y[n-2-k]=20x[n-k]+10x[n-1-k]$$

here is the problem , I cannot see how same expression will be acquired . I've tried performing simple substitution of n-k=m , which leads to:

$$y[m]-4y[m-1]+4y[m-2]=20x[m]+10x[m-1]$$

which is still not quite the same. Obviously the equation is linear differences equation with constant coefficients , therefore I suppose the system indeed TI but how to mathematically show this specific case.

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You've actually proved time-invariance already. It's just a matter of clean notation to see this. Let's use $y_1[n]$ to denote the response to a delayed input $x[n-k]$. The sequence $y_1[n]$ satisfies the following difference equation:

$$y_1[n]-4y_1[n-1]+4y_1[n-2]=20x[n-k]+10x[n-k-1]\tag{1}$$

The delayed response to the input $x[n]$ satisfies

$$y[n-k]-4y[n-k-1]+4y[n-k-2]=20x[n-k]+10x[n-k-1]\tag{2}$$

Now we have to check if $y_1[n]$ and $y[n-k]$ satisfy the same difference equation. Comparing $(1)$ and $(2)$ we see that this is indeed the case.

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  • $\begingroup$ Thanks , was just a matter of perspective after all $\endgroup$ – bertington313 May 30 at 19:51

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