Derivation of transfer function for parallel load cap. circuit

Discussion in 'Homework Help' started by abbefaria, Nov 20, 2011.

  1. abbefaria

    Thread Starter New Member

    Nov 20, 2011
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    Hello,

    I am studying Analog electronics and am learning about frequency response at the moment. My textbook is good but skips steps sometimes, and I've been trying for too long to fill in the blanks. If someone can show me how the text got from point A to point B and where I'm going wrong, I'd really appreciate it!

    The example involves deriving the transfer function for a simple parallel load capacitor circuit [​IMG]

    (The attached schematic isn't completely accurate (sorry): the one in the example has an additional resistor Rs in series with the source).

    Anyway, the resistor in series with the source is labeled Rs, the resistor in parallel with the capacitor is Rp, and the capacitor is Cp. Vo is taken across the parallel elements.

    The book gave something like: from this schematic it can be seen that Vo/Vi = (Rp/(Rs+Rp))*(1/(1+sCp(RsRp/(Rs+Rp))).

    My attempt to derive it is as follows:

    Node voltage at the output terminal yields
    Vo/Rs - Vi/Rs + Vo/Rp + Vo/(1/sCp) = 0

    Vo/Vi = (1/Rs) / [(1/Rs) + (1/Rp) + sCp]

    since 1/Rs + 1/Rp should end up equaling 1, I got them over a single denominator:

    Vo/Vi = (1/Rs) / [(Rp + Rs)/(Rs*Rp) + sCp]

    Factoring out the constants shows that this will never yield the book's answer (or I can't see it, one or the other):

    Vo/Vi = [(1/Rs^2) + (1/RsRp)][1 / (1 + s((Rp + Rs)/(Rs*Rp)Cp)]


    If someone could show me how the book got from node voltage to their transfer function, or where I started going down the wrong path on my calculations, I would really appreciate it! Thanks in advance for the help.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The TF will be given by applying the voltage divider relationship

    G(s)=\frac{V_o}{V_s}=\frac{R_p||\frac{1}{Cs}}{R_s+R_p||\frac{1}{Cs}}

    where

    R_p||\frac{1}{Cs}=\frac{R_p \frac{1}{Cs}}{R_p+\frac{1}{Cs}}=\frac{R_p}{R_pCs+1}

    Make the substitution & it's then a matter of algebraic manipulation
     
  3. Bushi

    New Member

    Jun 9, 2013
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    Vo/Vi = (1/Rs) / [(Rp + Rs)/(Rs*Rp) + sCp]
    take (Rp + Rs)/(Rs*Rp) common
    (1/Rs) / (Rp + Rs)/(Rs*Rp) [1+ sCp (Rp||Rs)]
    (Rp/Rs+Rp)/ [1+ sCp (Rp||Rs)]
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    No problem up to this point, but your next statement has be baffled.

    Where on Earth did you get this notion? 1/Rs + 1/Rp is the some of two indpendendent conductances and would have units of siemens (or mhos). Is there something in the problem statement saying that the sum of this is 1S?

    Okay so far.

    I certainly don't know how it shows that it will never yield the book's answer, but you can tell by inspection that the answer above is WRONG. How can you do this? By checking the units (i.e., performing a dimensional analysis).

    You have the product of two things:

    Vo/Vi = [A]

    where

    A = [(1/Rs^2) + (1/RsRp)]
    B = [1 / (1 + s((Rp + Rs)/(Rs*Rp)Cp)]

    The first has units of S^2 (or inverse-impedance-squared)
    The second is dimensionless since it is off the form
    B = 1/(1+C)

    So your Vo/Vi has units of inverse-impedance-squared when it should clearly be dimensionless. Hence it is wrong. Period. No point going any further until this is fixed.

    Since you didn't show any work on how you got from the last correct expression to the wrong one, I can't even attempt to point out where you went wrong. And if I were a grader, I could give you partial credit only for what you had done up to this point and none for the next step no matter how tiny the error might have been.


    If someone could show me how the book got from node voltage to their transfer function, or where I started going down the wrong path on my calculations, I would really appreciate it! Thanks in advance for the help.[/QUOTE]
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    Anyone realize that this is a very old thread? The OP has long gone on this one.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    You are right! Thanks for pointing it out. I just am not in the habit of checking the OP date.

    Well, maybe someone else benefited, or will in the future when they do a search and run across it.
     
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