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Cascaded low pass filters problem

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Feb23-09, 06:32 PM
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
[tex]\frac{Vout}{Vin}=\frac{1}{(1-\omega^2C^2R^2)+3j\omega CR}[/tex]

Hint: [tex]\frac{Vout}{Vin}=\frac{V1}{Vin}\frac{Vout}{V1}[/tex]
2. Relevant equations
For Capacitor, [tex]Z=\frac{1}{j\omega C}[/tex]
General Potential divider equation [tex]Vout=\frac{Z2}{Z1+Z2}Vin[/tex]

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
[tex]V1=\frac{1}{j\omega C}\frac{1}{R+\frac{1}{j\omega C}}Vin[/tex]
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
[tex]X=\frac{1}{\omega C}[/tex]
while its impedance is
[tex]Z\frac{1}{j\omega C}[/tex]
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
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Feb24-09, 12:19 PM
P: 10
Just like to say I finally got it :)
I just used [tex]V=IZ[/tex] really.
Say that current [tex]I[/tex] flows through the first filter and current [tex]I_{1}[/tex] goes through the second one.
Starting from the right, [tex]I_{1}=j\omega C\times V_{out}[/tex]

[tex]RI_{1}=j\omega CR V_{out}[/tex]

[tex]V_{1}=RI_{1}+V_{out}=(j\omega CR+1) V_{out}[/tex]

[tex]RI=R(I_{1}+j\omega C V_{1})=(2j\omega CR-\omega^2 C^2R^2)V_{out}[/tex]

[tex]V_{in}=RI+V_{1}=(1+3j\omega CR-\omega^2 C^2R^2) V_{out}[/tex]

hence original result is obtained.
Feb24-09, 01:42 PM
P: 220
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.

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