I am trying to calculate the poles of an analog Gaussian filter. Its characteristic function, $e^{-\log(2)x^2}$, can be expanded into MacLaurin series:

$$2^{-x^2} = e^{-\log(2)x^2} = \lim_{N \to \infty} \sum^N_{k=0}{\frac{(\log(2))^k}{k!}x^{2k}}$$.

This would go to the denominator of the transfer function, so the roots of the sum are the poles of the filter. The more terms, the better the approximation. So far, so good, classic theory.

But if I calculate the poles for, say $N$=13, I get some strange display (blue dots):


Just by looking at the graph, I thought to rotate them 90 degrees and keep only the left side (red points), while comparing them to the Bessel poles of the same order (green points). The circle is for reference, only, while the Bessel poles have been scaled down by $\sqrt{N}$, for comparison. Interesting enough, if I am to calculate the poles for the normalized transfer function (not the squared one), the series terms would have to be divided by an extra $2^k$, and the resulting transfer function would have to be compared with $\sqrt{|H(s)^2|}$, rather than $|H(s)|$.

I am aware that Bessel isn't Gaussian, or vice-versa, but they are two filters who deal with time-domain, rather than frequency, and their responses are quite similar, hence the comparison.

But now I am puzzled: the blue poles are symmetric on the $X$ axis, and if I were to keep the left hand side, I would end up having two extra, purely imaginary poles for odd orders (as is the case here). Rotating them by 90 degrees seems not only to solve this problem, but also to give the correct transfer function magnitude:


The blue trace is with the original poles, left-hand side poles. One note: keeping the extra, purely imaginary poles would result in a resonance peak (not surprisingly), so I removed those and added an extra, single, real pole on $X$ axis, of magnitude $-2|\max\left({s_k}\right)|$, resulting in the current plot. The red trace is with the rotated poles (also left-hand side), while the green one is the Mathematica expression $\sqrt{e^{-\log(2)*x^2}}$ with a 0.1 offset, for better comparison. $N$=13, as above.

All these seem confusing to me: when calculating the poles from the mathematical expression (the series expansion), do we keep them as they are, or do we rotate them 90 degrees (simple multiplication with $j$), and why?

  • $\begingroup$ I guess the variable $x$ in your equation is 'frequency', so you shouldn't compute the poles in the $x$-plane, but use $s=j\omega$ with $\omega=2\pi f$. This will give you the necessary rotation by 90 degrees. $\endgroup$
    – Matt L.
    Commented Sep 10, 2016 at 12:38
  • $\begingroup$ @MattL. That's just the mathematical expression defining the transfer function, not the transfer function itself (2nd pic, green trace). Similar to $R(x)$ in $\frac{1}{\sqrt{1+\epsilon^2 R(x)^2}}$. Using $j\omega$ in $e$ would make it all 1. The transfer function, otoh, is derived from the complex poles, so plotting those ones (2nd pic, blue and red) was done with abs(H(%i*x)) in wxMaxima. $\endgroup$ Commented Sep 10, 2016 at 12:46
  • $\begingroup$ I realized I am wrong in my previous reply because there's $x^2$, so the magnitude would not be unity, but even so, the gaussian frequency response is defined as $|H(j\omega)|^2=e^{-log(2)*\omega^2}$, which is the sum mentioned before, so the poles would have to be deduced from this series. $\endgroup$ Commented Sep 10, 2016 at 13:01
  • 1
    $\begingroup$ What I mean is the following: if $x$ is frequency (which is what I suppose), then you must replace $x$ by $s/(2\pi j)$, which gives you a new function in the variable $s$, and that's the function of which you must compute the poles. $\endgroup$
    – Matt L.
    Commented Sep 10, 2016 at 13:03
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    $\begingroup$ The series coefficients will still be real valued, even if you use $s$ as a variable. $\endgroup$
    – Matt L.
    Commented Sep 10, 2016 at 15:19

1 Answer 1


First of all, computing the poles of an (ideal) Gaussian filter is an impossible task, because its transfer function is not a rational function, and there are simply no poles. This is in contrast to the classic filter approximations, like Butterworth, Chebyshev, and Cauer filters, all of which have rational transfer functions, and, consequently, all of them have poles.

The frequency response of an ideal Gaussian filter is


The corresponding transfer function is a function of $s=j\omega$:


which is of course not rational, and which doesn't have any finite poles.

What you did is compute a rational approximation by developing $1/G(s)$ into a power series, and using only a finite number of terms. However, since you used $\omega$ as a complex variable instead of $s$, the poles are turned by $90$ degrees. This was already mentioned in the comments. The correct way of doing this is as follows:


Using only a finite number of terms gives the following rational approximation of $G(s)$:


the poles of which are of course the roots of the denominator polynomial.

  • $\begingroup$ I didn't think it necessary to say that the series needs infinite terms for convergence. I made a notice in passing, in one of the comments, though. Still, I'll mark this as the answer, thank you. $\endgroup$ Commented Sep 10, 2016 at 18:33
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    $\begingroup$ @aconcernedcitizen: Yes, but the number of terms is actually not the main point; I thought it important to mention that it is only the approximation that makes the filter have poles, otherwise the ideal Gaussian filter doesn't have poles. Furthermore, it's good that the confusion of $s$ vs. $\omega$ is now formulated in a proper answer, not only in comments. $\endgroup$
    – Matt L.
    Commented Sep 10, 2016 at 19:03
  • $\begingroup$ Oh, don't get me wrong, the answer should have been here, instead of the comments, it's just that the approximation applies not only to gaussian, but to Bessel, too. In fact, now that I think about it, almost everywhere I read was "the ... approximation", be it Butterworth, Chebyshev, Cauer, Legendre, etc. I think that's why I didn't bother much to say anything about the series expansion. But, it's also true that this, like Bessel, needs N as large as possible to come close to the response, while the others can do with fixed polynomials orders. Could this fall under "semantics"? :-) $\endgroup$ Commented Sep 11, 2016 at 6:27

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