# Precise Centre frequency of an All-pole digital filter

I'm not an engineer and have essentially taught myself all I know - this present problem is giving me some problems.

I have an all-pole filter (a gammatone filter - impulse response is essentially a damped cosine) implemented as a cascade of identical 2nd-order pole pair sections. The poles are located in the z-domain at $\exp(-\beta + i\theta)$: $\beta$ is the bandwidth in radians/sample and $\theta$ is the 'ringing' frequency in radians/sample.

Given a value for $\theta$, this pole pair does not produce a filter with peak gain at the frequency $\theta$ - I'm trying to find out exactly what the offset is (i.e. at what frequency a filter with a given $\theta$ actually has peak magnitude), so that I can produce filters that peak at precisely a given arbitrary frequency.

I've found an equation for calculating what this offset is given poles located in the s-domain at $-\beta + i\theta$:

$$\mathrm{f_{c,actual}} = \theta \sqrt{1 - \dfrac{\beta^2}{\theta^2}}$$

This formula provides results consistent with observations up to a certain frequency, though I believe it needs to be modified to produce results that work for arbitrary discrete spectrums.. I've tried a number of things and just can't quite figure it out.

Any help is greatly appreciated.

For a filter consisting of a complex conjugate pair of poles, the $$z$$-domain transfer function is: $$H(z) = \frac{a}{\left(1-r(\cos{\theta}-i\sin{\theta})z^{-1}\right)\left(1-r(\cos{\theta}+i\sin{\theta})z^{-1}\right)}\\ = \frac{a}{1 - 2r\cos(\theta)z^{-1} + r^2z^{-2}},$$

where $$a$$ is a constant by which the magnitude frequency response can be normalized, $$r$$ is the pole radius, and $$\theta$$ is the pole angle. The peak of the magnitude frequency response will not be exactly at frequency $$\theta$$, but at a nearby frequency $$\omega_0$$ at which the derivative

$$\frac{|a|2r\sin(\omega)\left(2r\cos\omega - (r^2 + 1)\cos\theta\right)}{\left(4r^2\cos^2\theta - 4r(r^2 + 1)\cos(\theta)\cos(\omega) + 4r^2\cos^2\omega + r^4 - 2r^2 + 1\right)^{3/2}}$$

of the magnitude frequency response $$|H(e^{i\omega})|$$ with respect to $$\omega$$ equals 0. That frequency is:

$$ω_0 = \operatorname{acos}\left(\frac{(r^2 + 1)\cos\theta}{2r}\right)\text{ if }-1 < (r^2 + 1)\cos(\theta)/(2r) < 1$$

Figure 1. The condition for that the magnitude frequency response peaks elsewhere than at $$w_0=0$$ or $$w_0=\pi$$ is that poles stay outside these droplets.

In the following it is assumed that the condition is satisfied as otherwise the peak would be trivially at frequency $$0$$ (if $$\theta < \frac{\pi}{2}$$) or $$\pi$$, whichever is closer to the pole. The normalized magnitude frequency response with $$a$$ chosen so that $$|H(e^{i\omega_0})| = 1$$ reads:

$$|H(e^{i\omega})| = \sqrt{\frac{(r^4 - 2r^2 + 1)\sin^2\theta}{4r^2\cos^2\theta - 4r(r^2 + 1)\cos(\theta)\cos(\omega) + 4r^2\cos^2\omega + r^4 - 2r^2 + 1}}.$$

Given $$\omega_0$$ and $$r$$:

$$\theta = \operatorname{acos}\left(\frac{2r\cos\omega_0}{r^2 + 1}\right)\tag{1}$$

which is something you can use directly to keep the peak of your all-pole bandpass filter where you want it:

$\omega_0$ pole loci">
Figure 2. Pole loci for a selection of constant $$\omega_0$$ traced over $$0 < r < 1$$.

Applying the substitution given by equation (1) gives:

$$H(z) = \frac{a}{1 - \frac{4r^2\cos(\omega_0)}{r^2 + 1}z^{-1} + r^2z^{-2}}$$

Just like $$\theta$$ does not exactly give the peak frequency, $$r$$ does not give the exact bandwidth. The exact -3 dB to -3 dB bandwidth $$B_\text{-3 dB}$$:

Figure 3. -3 dB bandwidth of the filter, with $$\omega_0 = \frac{\pi}{4}$$ and $$r=0.9$$.

is given by:

$$B_\text{-3 dB} = \operatorname{acos}\left(\frac{(r + 1/r)\cos\theta + (r - 1/r)\sin\theta}{2}\right)\\- \operatorname{acos}\left(\frac{(r + 1/r)\cos\theta + (1/r - r)\sin\theta}{2}\right)$$

This was found by solving the values of $$\omega$$ for which $$|H(e^{i\omega})| = \sqrt{2}/2$$. The first $$\operatorname{acos}$$ gives the solution found to the right from the peak and the latter $$\operatorname{acos}$$ the one to the left. More generally, for a bandwidth from magnitude frequency response value $$g$$ to $$g$$:

$$B_g = \operatorname{acos}\left(\frac{(r + 1/r)\cos\theta + \sqrt{1/g^2 - 1}(r - 1/r)\sin\theta}{2}\right)\\-\operatorname{acos}\left(\frac{(r + 1/r)\cos\theta + \sqrt{1/g^2 - 1}(1/r - r)\sin\theta}{2}\right)$$

This is needed when designing a composite filter, for example a gammatone filter, that is a cascade of identical two-pole filters. Usually we want to design the filter to have a given bandwidth, but it is very difficult to solve $$r$$ from the above. So we resort to approximation. Using the value $$D_2$$ of the second derivative of the magnitude frequency response at $$\omega_0$$:

$$D_2 = \frac{(r^2 + 1)^2\cos^2\theta - 4r^2}{(r^2 - 1)^2(1 - \cos^2\theta)},\tag{2}$$

we can approximate the magnitude frequency response with that of a complex one-pole filter that has at $$\omega_0$$ the magnitude frequency response and its first two derivatives matched to those of the two-pole filter:

Figure 4. One-pole approximation with pole angle $$\omega_0$$.

The magnitude frequency response of the one-pole filter is:

$$|\hat H(e^{i\omega})| = \frac{\sqrt{2}\left(\sqrt{1 - 4D_2} - 1\right)}{2\sqrt{\left(2D_2\cos(\omega - \omega_0) - 2D_2 + 1\right)\left(1 - \sqrt{1 - 4D_2} - 2D_2\right)}}.$$

It is symmetrical around $$\omega_0$$ and has $$g$$ to $$g$$ bandwidth:

$$\hat B_g = 2\operatorname{acos}\left(\frac{1/g^2 - 1}{2D_2} + 1\right),$$

or inversely and more usefully:

$$D_2 = \frac{1 - g}{2g^2\left(\cos(\hat B_g/2) - 1\right)}.$$

Equations $$(1)$$ and $$(2)$$ when combined give:

$$D_2 = \frac{2r^2(r^2 + 1)^2\left(1 - \cos(2\omega_0)\right)}{(r^2 - 1)^2(2r^2\cos(2\omega_0) - r^4 - 1)}.$$

Solving that for $$r$$ gives a cumbersome equation with $$\omega_0$$ and $$D_2$$ as arguments that can be used to set the approximate bandwidth of the two-pole filter (c language compatible plaintext with w_0 = $$\omega_0$$ and d_2 = $$D_2$$):

r=sqrt(-sqrt(2)*sqrt(sin(w_0))*sqrt((-sin(w_0)*(pow(d_2+1,2)*pow(cos(w_0),2)
+d_2-1)*sqrt(-pow(d_2+1,2)*pow(cos(w_0),2)+pow(d_2,2)-2*d_2+1)+pow(d_2+1,3)
*pow(cos(w_0),4)-(d_2+1)*(pow(d_2,2)-d_2+2)*pow(cos(w_0),2)+pow(d_2,2)-2*d_2
+1)/sqrt(-pow(d_2+1,2)*pow(cos(w_0),2)+pow(d_2,2)-2*d_2+1))+sin(w_0)*sqrt(-
pow(d_2+1,2)*pow(cos(w_0),2)+pow(d_2,2)-2*d_2+1)-(d_2+1)*pow(cos(w_0),2)+1)
/sqrt(-d_2)


or with alternative symbols:

r = √(- √2·√(SIN(ω_0))·√((- SIN(ω_0)·((d_2 + 1)^2·COS(ω_0)^2 + d_2 - 1)·√(-
(d_2 + 1)^2·COS(ω_0)^2 + d_2^2 - 2·d_2 + 1) + (d_2 + 1)^3·COS(ω_0)^4 - (d_2
+ 1)·(d_2^2 - d_2 + 2)·COS(ω_0)^2 + d_2^2 - 2·d_2 + 1)/√(- (d_2 + 1)^2
·COS(ω_0)^2 + d_2^2 - 2·d_2 + 1)) + SIN(ω_0)·√(- (d_2 + 1)^2·COS(ω_0)^2 +
d_2^2 - 2·d_2 + 1) - (d_2 + 1)·COS(ω_0)^2 + 1)/√(-d_2)


When $$\hat B_\text{-3 dB} = \hat B_{\sqrt{2}/2}$$ is set proportional to $$\omega_0$$ by a constant $$c$$ stepped from 0.1 to 1.0, the upper half-plane pole resides on tracks like these:

Figure 5. Pole loci for bandwidth proportional to peak frequency with proportionality constant $$c$$.