I have a system which has the following transfer function $$H(s)=\frac{\beta + s}{s^{2} + 2\alpha s + \beta^{2}}$$ where $\beta = \sqrt{\omega^{2} + \alpha^{2}}$ and $\alpha>0$.

This system, as already discussed in a dsp question is stable. How can I check the other two properties?


If a random (wide-sense stationary) process $n(t)$ passes through a time-invariant filter $h(\tau)$, then the autocorrelation function is $$R_{n^\prime}(t)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} R_{n}(t-\tau+\tau^\prime) h(\tau) h(\tau^\prime) d \tau d \tau^\prime$$ For white noise input, $R_{n}(t)=\delta(t)$ and $$ R_{n^\prime}(t)=\int_{-\infty}^{\infty} h(\tau-t) h(\tau) d \tau $$ The power spectrum of the output is the Laplace transform of $R_{n^\prime}(t)$ and it is possible to prove that $$ S_{n^\prime}(s)= H(-s) \int_{-\infty}^{\infty} h(\tau) e^{-s \tau}d \tau=H(-s) H(s) $$ In my case I have $H(s)=\frac{\beta + s}{s^{2} + 2\alpha s + \beta^{2}}$. I would like to prove that the filter is stable, causal and minimum phase.

  • 2
    $\begingroup$ What are your thoughts about this? If you know the definition of "minimum phase", what problems did you encounter when trying to find out if the given system satisfies the respective conditions? A good answer to that question should be quite long because given only the expression for the transfer function, you can't say anything about stability or causality. You need more information. $\endgroup$
    – Matt L.
    Jul 10, 2022 at 11:05
  • $\begingroup$ I've added an answer to your other question, clarifying a few points that are also relevant to this question. $\endgroup$
    – Matt L.
    Jul 10, 2022 at 12:05
  • 1
    $\begingroup$ What makes a filter “unique”? Stable, causal, and minimum phase I get, but what is “unique”? $\endgroup$
    – Peter K.
    Jul 10, 2022 at 14:54
  • $\begingroup$ And -- since this looks like homework -- what work have you done so far to answer it on your own? $\endgroup$
    – TimWescott
    Jul 10, 2022 at 22:56
  • 1
    $\begingroup$ Thanks to all. I edited my post $\endgroup$
    – Mark
    Jul 11, 2022 at 20:01

1 Answer 1


In general, the algebraic expression of a transfer function alone doesn't uniquely describe a single system. For the example in your question with two complex conjugate poles in the left-half plane, the given transfer function describes two systems: one is causal and stable, the other one is non-causal and unstable.

Illustration of ROC for each case.

A complete (external) description of an LTI system is given by the transfer function and the corresponding region of convergence (ROC). For a causal system, the ROC is a right half-plane. Since the ROC to the right of the two poles includes the imaginary axis, the corresponding causal system is stable. Since the ROC to the left of the two poles does not include the imaginary axis, the corresponding non-causal system is unstable.

So, assuming that the given transfer function is meant to describe a causal system, BIBO (bounded-input, bounded-output)-stability follows immediately.

For answering the question if that causal system is a minimum-phase system, you just have to ask if it is possible to invert the system by a causal and stable system. If that is the case, the system is a minimum-phase system, otherwise it isn't.

  • 3
    $\begingroup$ Thx @PeterK. for the nice plots! $\endgroup$
    – Matt L.
    Jul 10, 2022 at 18:42
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    $\begingroup$ You're welcome! I like having some pics when ROCs are talked about, so I thought it'd add to your excellent answer. $\endgroup$
    – Peter K.
    Jul 10, 2022 at 19:39

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