Cascaded low pass filters problem

[Stef42]

Homework Statement

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}

Homework Equations

For Capacitor, Z=\frac{1}{j\omega C}
General Potential divider equation Vout=\frac{Z2}{Z1+Z2}Vin

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

 

[Stef42]

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.

 

[Kruum]

Glad you figured it out! If you still need to know the difference between impedance and reactance, it's quite simple: reactance is the imaginary part of impedance. Plus resistance is the real part of impedance.