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Cascaded low pass filters problem

  1. Feb 23, 2009 #1
    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

    Attached Files:

  2. jcsd
  3. Feb 24, 2009 #2
    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.
  4. Feb 24, 2009 #3
    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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