Deriving 2nd order passive low pass filter cutoff frequency

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:

$$ w\to\sqrt{\frac{1}{A}-\frac{B^{2}}{2A^{2}}+\frac{\sqrt{8A^{2}-4AB^{2}+B^{4}}}{2A^{2}}} $$

A lot of people confuse natural frequency with cut off frequency. The natural frequency is the frequency the system wants to oscillate at. The cut off frequency (or -3dB freq) is just when the transfer function has a magnitude of 0.707

If the two poles of the filter are not close together, the 2nd order canonical terms like the natural frequency and the damping factor start to loose practical meaning. If the poles are close together, the natural frequency will tend to be near the -3dB frequency but not exactly.

The equation you keep seeing $$ f_{n} = \dfrac{1}{(2\pi*\sqrt{R_{1}R_{2}C_{1}C_{2}})} $$

is for the natural frequency. If you solve for the damping factor you'll also see that it's $$ d= \dfrac{\dfrac{(C_{1}R_{1}+C_{2}R_{1}+C_{2}R_{2})}{2}}{\sqrt{R_{1}R_{2}C_{1}C_{2}}} $$

You're attempting to define in an equation for what the -3dB frequency is, so you have to set the transfer function to equal -3dB and just solve for the frequency that results. The problem with that is the math is going to get real ugly really fast. I looked at this once a few years ago and found this relationship. $$ f_{c} = f_{n} * \sqrt{1-2d^2 + \sqrt{4d^4-4d^2+2}} $$

where \$f_{c}\$ is the \$-3dB\$ frequency.

So for an example if you take:

\$ \begin{cases} R_1 = 10k\Omega \\ R_2 = 40k\Omega \\ C_1 = 0.1µF \\ C_2 = 0.01µF \end{cases}\$

You'll get the following numbers:

\$ \begin{cases} f_n = 251.6Hz \\ d = 1.186 \\ f_c = 127.7Hz \end{cases}\$

You can also find the poles which are 458.8Hz and 138.02Hz, so the 3dB frequency is pretty close to the first pole. You'll find if you slide that second pole out further and further the 3db frequency will be pretty close to the first pole.

Hope that helps.

In my opinion this 'cut-off' frequency is not defined as the -3dB point. The real transfer function is: $$ H(s)=\frac{1}{s^2R_1R_2C_1C_2+s(R_1C_1+R_1C_2+R_2C_2)+1}$$

The 'common form' of a second order element in control theory is $$W(s)=\frac{1}{\frac{s^2}{\omega_n^2}+2\frac{\xi}{\omega_n}s+1}$$, where \$\xi\$ is the damping coefficient and \$\omega_n\$ is the natural frequency. If you want to express the natural frequency of \$H(s)\$, you'll find that it is equal to \$\frac{1}{\sqrt{R_1R_2C_1C_2}}\$.

While HKOB's answer really seems reasonable even when evaluating the correct transfer function, MATLAB showed me (using different arbitrary R and C values) that the calculated 'cut-off' frequency is not even close to the -3dB point on the Bode plots.