# Cascaded low pass filters problem

1. Feb 23, 2009

### Stef42

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 Files:

• ###### circuit.jpg
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2. Feb 24, 2009

### 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.

3. Feb 24, 2009

### 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.