filterpole-zeroplot
For a 2nd order Bessel low pass, I have a pole-zero map which has a pole couple of the form: \$p_{1/2} = a \pm jb\$.
Can I deduce the cutoff frequency from the pole location? Does it contain any more information?
EDIT: Since the calculation of the cutoff frequency in the answer depends on the filter type I am wondering, if it is possible to determine the cutoff frequency for any kind of 2nd order low pass solely by the pole location (under the assumption they are complex conjugate)?
EDIT: Thanks to hryghr I see that the starting assumptions were incorrect. The transfer function magnitude can't be found that simply. It is more than ten years since I considered my skills sharp on this topic, and knives don't get sharper in the drawer! But I can't have that I posted something formally incorrect, so here goes attempt #2:
I will derive the transfer function the dirty way .. using Kirchoff's Current Law (KCL) (a very generic method). I call the output node \$V_{o}\$, and the middle node \$V_{x}\$. For the following equations i cut down on writing by writing \$V_{o}\$ instead of the more accurate \$V_{o}(s)\$ :
I: KCL in \$V_{o}\$:
$$ \frac{V_{o}-V_{x}}{R_{2}}+sC_{2}V_{o}=0 $$
$$ V_{x}=V_{o}(1+sR_{2}C_{2}) $$ II: KCL in \$V_{x}\$:
$$ \frac{V_{x}-V_{i}}{R_{1}}+\frac{V_{x}-V_{o}}{R_{2}}+sC_{1}V_{x}=0 $$
Rearranging terms:
$$ R_{2}(V_{x}-V_{i})+R_{1}(V_{x}-V_{o})+sR_{1}R_{2}C_{1}V_{x}=0 $$
Rearranging terms:
$$ V_{x}(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{2}V_{i}-R_{1}V_{o}=0 $$
Substituting \$V_{x}\$ with result of I: $$ V_{o}(1+sR_{2}C_{2})(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{2}V_{i}-R_{1}V_{o}+sR_{1}R_{2}C_{1}V_{o}=0 $$
Collecting terms for \$V_{o}\$
$$ V_{o}((1+sR_{2}C_{2})(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{1})=R_{2}V_{i} $$
Rearranging:
$$ \frac{V_{o}}{V_{i}}=\frac{R_{2}}{(1+sR_{2}C_{2})(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{1}} $$
Expanding terms:
$$ \frac{V_{o}}{V_{i}}=\frac{R_{2}}{R_{1}+R_{2}+sR_{1}R_{2}C_{1}+sR_{1}R_{2}C_{2}+sR_{2}^{2}C_{2}+s^{2}R_{1}R_{2}^{2}C_{1}C_{2}-R_{1}} $$
\$R_{1}\$ cancels, then divide by \$R_{2}\$ top and bottom:
$$ \frac{V_{o}}{V_{i}}=\frac{1}{1+sR_{1}C_{1}+sR_{1}C_{2}+sR_{2}C_{2}+s^{2}R_{1}R_{2}C_{1}C_{2}} $$
Prettified, the transfer function is:
$$ H(s)=\frac{V_{o}(s)}{V_{i}(s)}=\frac{1}{s^{2}R_{1}R_{2}C_{1}C_{2}+s(R_{1}C_{1}+R_{1}C_{2}+R_{2}C_{2})+1} $$
This is probably a nice place to start converting to the standard form that hryghr mentions. It may be that the corner frequency asked for relates to that form. I won't bother to much with that, but move on to find the -3dB point.
The magnitude of the transfer function can for instance be found by calculating:
$$ \left|H(\omega)\right|=\sqrt{H(s\rightarrow j\omega)H(s\rightarrow-j\omega)} $$
Setting \$A=R_{1}R_{2}C_{1}C_{2}\$ and \$B=(R_{1}C_{1}+R_{1}C_{2}+R_{2}C_{2})\$ to simplify this calculation:
$$ \left|H(\omega)\right|=\frac{1}{\sqrt{((j\omega)^{2}A+(j\omega)B+1)((-j\omega)^{2}A+(-j\omega)B+1)}} $$
$$ \left|H(\omega)\right|=\frac{1}{\sqrt{(-\omega{}^{2}A+j\omega B+1)(-\omega{}^{2}A-j\omega B+1)}} $$
$$ \left|H(\omega)\right|=\frac{1}{\sqrt{\omega{}^{4}A^{2}-\omega{}^{2}A(j\omega B-j\omega B+1+1)+\omega^{2}B^{2}+(j\omega B-j\omega B)+1}} $$
$$ \left|H(\omega)\right|=\frac{1}{\sqrt{\omega{}^{4}A^{2}+\omega{}^{2}(B^{2}-2A)+1}} $$
Finding \$B^{2}-2A\$ gives you something like:
$$ R_{1}^{2}(C_{1}+C_{2})^{2}+C_{2}^{2}(2R_{1}R_{2}+R_{2}^{2}) $$
Then to find the -3dB point start at:
$$ \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{\omega{}^{4}A^{2}+\omega{}^{2}(B^{2}-2A)+1}} $$
$$ 2=\omega{}^{4}A^{2}+\omega{}^{2}(B^{2}-2A)+1 $$
So far I have done it all by hand (hopefully no mistakes), but here I call it a day, try mathematica, and get \$\omega\$ for the -3dB frequency as:
$$ w\to\sqrt{\frac{1}{A}-\frac{B^{2}}{2A^{2}}+\frac{\sqrt{8A^{2}-4AB^{2}+B^{4}}}{2A^{2}}} $$
From wiki, the transfer function polynomials for several Bessel filters are: -
If you took two 2nd order filters and multiplied the polynomials you do not end up with the polynomial of a 4th order Bessel filter. You end up with this: -
\$s^4 + 6s^3 + 15s^2 + 18s + 9\$
This informs you that cascading two independent 2nd order Bessel filters does not produce a 4th order Bessel filter or, in other words you have to calculate the values of the components in both 2nd order filters taking into account that you are trying to implement a 4th order system. You might get lucky of course but if you just cascade 2 2nd order Bessel filters it won't be a 4th order Bessel filter.
It's not just Bessel filters that this rule applies to.
Here's a website that might help. The calculator appears to keep a common value for the capacitors used but there are 6 resistors all having different values: -
Best Answer
Yes - you can. There are filter tables for all the well-known lowpass approximations - including the Thomson-Bessel response. These tables list the relavant pole parameters for the various filter orders.
For second order Bessel response the relevant parameters are
$$Q_p=0.5773,\space \Omega_p=\frac{\omega_p}{\omega_0}=1.732$$
Explanation: The pole quality factor \$Q_p\$ is a measure for the angle \$\beta\$ between the pole position and the negative-real axis. We have \$Q_p=\frac12\cos\beta\$ . This angle can also be expressed using the given values for the real part (\$a\$) and the img. part (\$b\$) of the pole.
The normalized angular frequency \$\Omega_p=\frac{\omega_p}{\omega_0}\$ is the ratio of the actual angular pole frequency \$\omega_p\$ divided by the actual cut-off frequency \$\omega_0\$. The actual frequency \$\omega_p\$ is identical to the magnitude of the line which connects the origin and the pole position. This magnitude can also be expressed using the known values for \$a\$ and \$b\$.
In this context, you should note that the above mentioned value \$\Omega_p=1.7320\$ is based on the following definition of cut-off: The cut-off angular frequency \$\omega_0\$ is defined as the inverse of the group delay, hence: \$\omega_0=\frac{1}{\tau}\$.
This commonly used definition makes sense because - in most cases - the Bessel response is applied because of its time behaviour (delay filter). This is in contrast to other responses (Butterworth, Chebyshev, ...) where the frequency response matters primarily.
If the cut-off frequency \$\omega_0\$ is defined using the 3dB criterion for the magnitude the following normalized value applies: \$\Omega_p=\frac{\omega_p}{\omega_0}=1.2723\$.