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? Text

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.

  • $\begingroup$ 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 $w[n]$? $\endgroup$
    – Andy Walls
    Commented May 16 at 12:22
  • $\begingroup$ @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 $\endgroup$
    – Luxdragon
    Commented May 16 at 17:27

1 Answer 1


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.

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

  • $\begingroup$ 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 $\endgroup$
    – Luxdragon
    Commented May 16 at 13:48

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