# Trouble showing Time Invariance of recursive system

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.

## 1 Answer

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.

• Thanks , was just a matter of perspective after all – bertington313 May 30 '20 at 19:51