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 thanks SG
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.
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.