This answer and the comments in it made me wonder whether the following statement is true:

If a transfer function is improper, then that system cannot be causal and stable at the same time.

I had thought that this was true for a while. But the other day I wondered why. For example, the transfer function


is improper, but the ROC $\{s:\mathrm{Re}(s)>-1\}$ would make it stable (it contains the imaginary axis) and causal (it consists of a left-sided plane).

So how are causality and stability related in improper systems?


1 Answer 1


An improper system cannot be causal and stable. If the order of the numerator is greater than the order of the denominator, you'll always have at least one pole at infinity. Consequently, not all poles are in the left half-plane (or inside the unit circle in the case of discrete-time systems).

The system in your example is clearly unstable:


Part of it is an ideal differentiator ($s$), which is unstable.

Also take a look at this related answer.

  • $\begingroup$ I feel like there's something I still don't get. Why would it be unstable if one can find a ROC that contains the imaginary axis? $\endgroup$
    – Tendero
    Mar 21, 2018 at 20:18
  • $\begingroup$ @Tendero: In continuous time you can't: $H(j\omega)\rightarrow\infty$ for $\omega\rightarrow\infty$. In discrete time, $H(z)=z$ cannot be causal and stable, but it can be anti-causal and stable: $h[n]=\delta[n+1]$. $\endgroup$
    – Matt L.
    Mar 21, 2018 at 20:25
  • $\begingroup$ In the answer I quoted in my OP, you stated that $1-s$ can be causal and unstable. I don't see however how one could pick a ROC so that any improper transfer funcion is causal in continuous time, as there will always be at least one pole at infinity, like you pointed out. So how is this possible? All possible ROCs for improper transfer functions will be limited to both the right and the left, so causality doesn't seem possible to achieve. $\endgroup$
    – Tendero
    Mar 28, 2018 at 11:32

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