Electrical – When eqvivalent transfer function of two systems in series is a product of two transfer functions that describe these systems

I can redraw the block diagram as follows:

Where $$H_3(s) = \frac{R_3}{R_2}$$ and $$H_4(s) = \frac{R_3}{R_1}$$

From this, $$\frac{V_o}{V_{in}} = \frac{-H_1(s)\times H_4(s)}{1+H_1(s)H_2(s)H_3(s)}$$

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:

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