1
$\begingroup$

Suppose we have an linear time-invariant (LTI) system which acts on discrete signals. Suppose someone tells us the transfer function is: $$H(z) = \frac{1}{z-2},$$ but doesn't specify the ROC. Now the ROC must be an annulus, so our options are: $$0 < |z| < 2$$ or $$2 < |z| < \infty.$$ Now, my thinking (which is probably wrong since there exist two different impulse responses which give rise to this transfer function), is that we can reconstruct the difference equation. $$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{z-2} \Longrightarrow Y(z)z-2Y(z) = X(z) \Longrightarrow y_{n+1} - 2y_n = x_n,$$ where the last step is due to the property $$Z[x_{n+k}](z) = z^kX(z).$$ The difference equation implies the system is causal and we infer the ROC is $$2 < |z| < \infty.$$ Where is my mistake? Where did I make an implicit assumption?

$\endgroup$

1 Answer 1

1
$\begingroup$

$y_{n+1} - 2y_n = x_n$ still doesn't tell you. You could take it to mean a causal, unstable system: $y_{n+1} = x_n + 2y_n$, or a stable, non-causal system $y_n = \frac 1 2 y_{n+1} - \frac 1 2 x_n$.

You just have to know in advance which one is being talked about*. As a corollary, people are usually going to assume that any pole $|a| > 1$ is an unstable pole of a causal system: it's up to you to tell people when you're deep enough into theory-land that you're taking noncausal systems seriously.

* Hint: if it's a physical system, it's causal. And it's behavior will be strongly affected -- at least eventually -- by whatever nonlinearity it hits when $y_n$ gets huge.

$\endgroup$
3
  • 1
    $\begingroup$ Exactly. And for higher order systems, there are even more possible interpretations of a given difference equation: causal, anti-causal, and generally several non-causal (two-sided) versions. $\endgroup$
    – Matt L.
    Commented Feb 3, 2023 at 11:07
  • $\begingroup$ Thank you for your answer $\endgroup$
    – Algo
    Commented Feb 4, 2023 at 14:16
  • $\begingroup$ "... for higher order systems, there are even more ..." yes. The coin toss came up tails on whether I should include that for completeness, or leave it out for simplicity. $\endgroup$
    – TimWescott
    Commented Feb 4, 2023 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.