I have been out of touch with DSP for a while. I was trying to refresh my basics and ended up with this doubt on IIR filter stability condition. I understood that for an iir filter to be stable, the poles should be inside the unit circle which was derived from its analog counterpart (i.e poles should be on Left hand of s plane). But then is there any extra condition that poles should be only real? Could it be complex also?

  • $\begingroup$ If poles are only real you're limited to the horizontal axis on your unit circle. Of course you can have stable filters with complex poles, but if you want your output to be real you need your pole to be at least complex conjugate $\endgroup$
    – Florent
    Commented May 16, 2018 at 6:56

1 Answer 1


For a causal system to be stable, all the poles have to be inside the unit circle. In general, they may be complex, although they always come in complex conjugate pairs in order to keep the output real for real inputs.

So, the answer is no, there are no extra requirements for the poles to be on the real axis. However, they must be complex conjugates for the output to be real for a real input.

I would like to point out that the region of convergence of the Z Transform may include the unit circle, and yet the system may be non causal (but stable). For example, consider the impulse response: $h[n] = -a^{n}u[-n-1]$, for $|a| > 1$, which corresponds to a system having a pole outside the unit circle, and yet it is a stable (although non-causal) system.

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    $\begingroup$ i have no idea why this answer was downvoted. i upvoted to bring it back to 0. $\endgroup$ Commented May 16, 2018 at 20:54
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    $\begingroup$ An exception is the case of all pass filters where poles can be found outside the unit circle and the filter will still be stable (indeed, having exactly unity gain) so long as there are reflections inside the unit circle---i.e. for every pole outside the unit circle there exists a zero with the same angle that is equidistant from the unit circle on the inside. Real output still requires complex conjugate pairing for all complex pole zero pairs. $\endgroup$
    – dmedine
    Commented Sep 5, 2022 at 2:22

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