# Cascaded low pass filters problem

 P: 10 1. The problem statement, all variables and given/known data Given the coupled RC network shown below (see attachment), show that the voltage transfer function is $$\frac{Vout}{Vin}=\frac{1}{(1-\omega^2C^2R^2)+3j\omega CR}$$ Hint: $$\frac{Vout}{Vin}=\frac{V1}{Vin}\frac{Vout}{V1}$$ 2. Relevant equations For Capacitor, $$Z=\frac{1}{j\omega C}$$ General Potential divider equation $$Vout=\frac{Z2}{Z1+Z2}Vin$$ 3. The attempt at a solution I find this all a bit confusing :( I know that the second filter is acting as a load for the first filter, so I know I just can't write $$V1=\frac{1}{j\omega C}\frac{1}{R+\frac{1}{j\omega C}}Vin$$ So would I need to combine the total impedance of the second filter with the impedance of the first capacitor? Actually, could someone clear up impedance and reactance? For a capacitor reactance is $$X=\frac{1}{\omega C}$$ while its impedance is $$Z\frac{1}{j\omega C}$$ So if a resitor is connected in series, how/what would I combine to obtain the total resistance/impedance? Any help would be appreciated, I think you can tell that my ideas are a bit muddled :s thanks SG Attached Thumbnails
 P: 10 Just like to say I finally got it :) I just used $$V=IZ$$ really. Say that current $$I$$ flows through the first filter and current $$I_{1}$$ goes through the second one. Starting from the right, $$I_{1}=j\omega C\times V_{out}$$ $$RI_{1}=j\omega CR V_{out}$$ $$V_{1}=RI_{1}+V_{out}=(j\omega CR+1) V_{out}$$ $$RI=R(I_{1}+j\omega C V_{1})=(2j\omega CR-\omega^2 C^2R^2)V_{out}$$ $$V_{in}=RI+V_{1}=(1+3j\omega CR-\omega^2 C^2R^2) V_{out}$$ hence original result is obtained.