transfer function - Deriving 2nd order passive low pass filter cutoff frequency

I'm working on a 2nd order passive low pass filter, consisting of two passive low pass filters chained together.

^{simulate this circuit – Schematic created using CircuitLab}

Let $ H(s) = H_1(s)H_2(s) $ where $ H_1(s) $ and $ H_2(s) $ are the transfer functions for each separate filter stage.

Then $ |H(s)| = |H_1(s)||H_2(s)| $

Knowing the magnitude of a passive low pass filter,

$$|H(s)| = \dfrac1{\sqrt{ (\omega R_1C_1)^2 + 1} } \times \dfrac1{\sqrt{ (\omega R_2C_2)^2 + 1} } = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}}$$

Then trying to find the cutoff frequency:

$$\left(\dfrac1{\sqrt{2}}\right)^2 = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}}$$ $$2 = \sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}$$ $$4 = ((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)$$ $$4 = (\omega R_1C_1)^2(\omega R_2C_2)^2 + (\omega R_1C_1)^2 + (\omega R_2C_2)^2 + 1$$

And I'm stuck. Research on the web tells me $ \omega_c = \dfrac1{\sqrt{R_1C_1R_2C_2}} $, but I can't find why? Can anybody show my the derivation to find this?

Answer

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:

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