# Natural frequencies of a bandpass filter

I have the following frequency transform used to design a low-pass prototype out of a bandpass specification mask: $$\omega=\frac{\omega_0}{B}\left(\frac{\omega'}{\omega_0}-\frac{\omega_0}{\omega'}\right)$$ where $\omega_0$ and $B$ are constants, $\omega$ represents the angular frequency of the low-pass filter and $\omega'$ is the angular frequency of the bandpass filter. Since the transfer functions of the filters must be hermitic the function can be reduced to:$$\omega=\frac{\omega_0}{B}\left|\frac{\omega'}{\omega_0}-\frac{\omega_0}{\omega'}\right|$$

Given that $\omega_0/B=2.45$, I wonder if by solving for $\omega'$ the equation can be rewriten like this:$$\omega'^2-\frac{\omega_0\omega}{2.45}\omega'-\omega_0^2=0$$ and from here just substitude the natural frecuencies of the low-pass prototype and get the natural frequencies of the bandpass filter since I have ignored the absolute value to solve for $\omega'$ and I don't know if that can be important when computing the result.

• Yes - it is correct. Please note that B is the bandwidth in rad/sec. Hence: wo/B=Q).
– LvW
Apr 30, 2018 at 11:31

When in doubt, plot! Matlab code :

w_array = -10:1:10;
w0_array = -10:1:10;
wl_array_1 = zeros(length(w_array), length(w0_array));
wl_array_2 = zeros(length(w_array), length(w0_array));
syms wl

for i=1:length(w_array)
w = w_array(i);
for j=1:length(w0_array)
w0 = w0_array(j);
s1 = solve(wl^2 - w0*w*wl/2.45 - w0^2 == 0, wl, 'PrincipalValue', true);
if isempty(s1)
s1 = NaN;
end
wl_array_1(i,j) = double(s1);
s2 = solve(w - 2.45*abs(wl/w0 - w0/wl) == 0, wl, 'PrincipalValue', true);
if isempty(s2)
s2 = NaN;
end
wl_array_2(i,j) = double(s2);
end
end

figure(1)
for j=1:length(w0_array)
plot(w_array,wl_array_1(:,j)), hold on
end
xlabel('\omega'), ylabel('\omega´'), xlim([-10,10]), ylim([-45,5])
title('Not considering the absolute in equation'), hold off,

figure(2)
for j=1:length(w0_array)
plot(w_array,wl_array_2(:,j)), hold on
end
xlabel('\omega'), ylabel('\omega´'), xlim([-10,10]), ylim([-45,5])
title('Considering the absolute in equation'), hold off


As can be seen, the absolute ensures that solutions only exist for positive values of $\omega$ and $\omega_0$. Removing the absolute, solutions exist for negative $\omega$ and $\omega_0$. Now it's up to you to decide what it means and if this is important.

For my part, I find it easier to do it in Laplace:

$$s=\frac{s^2+\omega_0^2}{BW s}$$

where BW is the bandwidth, and $\omega_0$ is the center frequency. For a real root, $\Re$:

$$\frac{s^2+\omega_0^2}{BW s}+\Re=0 => s^2+BW \Re s + \omega_0^2=0$$

Which is easy to solve. If the root is pure imaginary:

$$\frac{s^2+w_0^2}{BW s}+j \Im=0=>s^2+j BW \Im s + \omega_0^2=0$$

and if it's complex, $p=\Re+j \Im$:

$$\frac{s^2+w_0^2}{BW s}+p=0=>s^2+BW p s + \omega_0^2=0$$

This one is a bit more complicated to solve, not impossible. The bare, complex roots for the equation come out as:

$$s=-\frac{BW p}{2}\pm\sqrt{\frac{BW^2p^2-4\omega_0^2}{2}}$$

and solving for complex, you get four posibilities, out of which choose, like before, only the positive frequencies:

$$Real=BW^2(\Re^2-\Im^2)-4\omega^2$$ $$Imag=2BW^2\Re\Im$$ $$Mag=\sqrt{Real^2+Imag^2}$$ $$\Phi=\frac{1}{2}\sqrt{Mag}\frac{\arctan2(Imag, Real)}{2}$$ $$s=\frac{1}{2}\left[-BW\Re\pm\sqrt{Mag}\cos(\frac{\Phi}{2})\pm j\left(\sqrt{Mag}\sin\frac{\Phi}{2}\pm BW\Im\right)\right]$$

Looks monstruous, but it works.