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?


1 Answer 1


$y_{n+1} - 2y_n = x_n$ 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.

  • 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

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