# How does a time delay affect the difference equation of a LTI-system?

I'm right now working on my digital signal processing homework and among the exercises is a tough LTI-system in a canonical-like form, however right before the output is a time delay.

Suppose the following linear time invariant system:

This LTI-system has a difference equation of $$y(n)=y(n-1)+x(n)$$

But what happens to the difference equation if we add a time delay at the very end of the system?

Do we then get the difference equation $$y(n)=y(n-2)+x(n-1)$$ or just $$y(n)=y(n-1)+x(n-1)$$? The example above is a mimicking my homework, but is of course much smaller and simpler - I just want to make sure I understand the principle.

• Hint: Define a quantity $w[n]$ just before that last time delay and write the equation for $w[n]$ in terms of $x[n]$. Write $y[n]$ in terms of $w[n]$. Now you have 2 difference equations. Can you combine them to eliminate $w[n]$? Commented May 16 at 12:22
• @AndyWalls Hmm I don't quite get your hint. I could write w[n] = w[n-1]+x[n] and then say y[n]=w[n-1], but then I get that y[n]=w[n]-x[n] and then I say y[n]=w[n-1], so y[n+1]=w[n] and I obtain y[n]=y[n+1]-x[n] which can't be right Commented May 16 at 17:27

Let's call the output of the first system $$y_1[n]$$ and that of the second $$y_2[n]$$. It's trivial to see that $$y_2[n]$$ is just the delayed version of $$y_1[n]$$, i.e.

$$y_2[n] = y_1[n-1] \tag{1}$$

We can delay to the first difference equations by simply subtracting 1 from ALL indices.

$$y_1[n-1] = y_1[n-2] - x[n-1] \tag{2}$$

Popping this into eq (1) we get

$$y_2[n] = y_1[n-2] - x[n-1] \tag{3}$$

In the Z-domain that simply turns into a multiplication with $$z^{-1}$$, i.e.

$$Y_2[z] = z^{-1} \cdot Y_1[z] \tag{4}$$

• Thank you very much for the neat answer! I suspected this might be the correct Ansatz, but I wasn't sure. Thank you very very much :D Commented May 16 at 13:48