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This is probably a very stupid question. In many places (e.g. here), the Butterworth filters, e.g. lowpass, are described as being "allpole" filters, that have all of these poles on the unit circle.

But shouldn't poles on the unit circle lead to a pole in the amplitude response of the filter at that particular frequency, in the same way as a zero attenuates that particular frequency? Instead, the Butterworth lowpass has strong attenuation in the upper frequency spectrum, despite having all of its poles on the left-hand side of the unit circle (the high frequency side).

Could anyone please clear my confusion ?

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You've made an understandable mistake. You are probably looking at this picture:

Picture of Butterworth poles, in a circle on the s plane

That is not the unit circle, and it isn't even in the $z$ domain.

What you are looking at is the locations of the poles for a 4-pole Butterworth filter in the Laplace domain. These are values of $s$, not $z$, and the circle is not a unit circle -- it's radius is defined by the cutoff frequency of the filter (which is why the radius is indicated as being $\omega_0$).

The Butterworth is one of the "old modern" filters, invented before we could just start with a desired frequency-domain response and synthesize the optimal filter. All of these (Butterworth, Tchebychev, eliptic, Gaussian) were originally designed as continuous-time filters, and the canonical representations of them are in the Laplace ($s$) domain. Implementations of these as IIR filters in the $z$ domain are sometimes-useful approximations.

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    $\begingroup$ The unit circle in the z-plane maps to the imaginary axis of the s-plane. $\endgroup$
    – Hilmar
    May 17 at 22:31
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    $\begingroup$ "... that said explicitly that the Butterworth poles are on the "unit circle" though". You can edit your question with that comment. Especially if you can cite the reference. It's pretty common to design filters in the $s$ domain that are normalized to $\omega_0 = 1$ (no units). In that case, the Butterworth poles would be on that unit circle -- but it's not the z Transform's unit circle. $\endgroup$
    – TimWescott
    May 18 at 0:31
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    $\begingroup$ @tobalt Perhaps the material at the following web page would be helpful to you, dsprelated.com/showarticle/994.php $\endgroup$ May 18 at 8:14
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    $\begingroup$ "It looks like the s domain is a whole different world to the z domain"... Well, yes and no. Both are about turning "difficult" time-domain descriptions of linear time- (or shift-) invariant systems into much easier to handle frequency-domain descriptions. Both involve reducing those descriptions to ratios of polynomials, both have regions of stability (or instability) -- much of the underlying math applies. It's just the next layer up that doesn't -- it's like the difference between building a good solid house out of wood, vs. one out of stone or straw bales or whatever. $\endgroup$
    – TimWescott
    May 18 at 20:54
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    $\begingroup$ I don't know! "Canonical continuous-time filters" maybe. Or just "canonical filters". $\endgroup$
    – TimWescott
    May 21 at 16:52

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