Transfer function

Discussion in 'Homework Help' started by MilK, Apr 10, 2010.

  1. MilK

    Thread Starter Member

    Mar 1, 2008
    25
    0
    Hi,

    is the transfer function of this circuit

    vo/vin = R / ( { [2sL parallel 1/sC] + R } parallel 1/sC )

    ?

    Thks
     
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    No, inductors are like resistors when you combine them - paralleling reduces the value.

    If Rs = 0 then that capacitor to ground doesn't factor into the transfer function.

    It's just:

    <br />
\frac{Vo}{Vin} = \frac{R}{0.5sL || \frac{1}{sC} + R}<br />
     
  3. MilK

    Thread Starter Member

    Mar 1, 2008
    25
    0
    "If Rs = 0 then that capacitor to ground doesn't factor into the transfer function."

    how come?

    Rs will be replace by short circuit if = 0?
     
  4. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Yes, a 0 ohm resistor is a short circuit.
    Since the voltage source is right across that capacitor to ground you know the voltage across it - it's exactly Vin at all times.
    It has no effect on the other components.
     
    MilK likes this.
  5. MilK

    Thread Starter Member

    Mar 1, 2008
    25
    0
    am i right to say, since cap across Vin is always Vin, doesn't change w.r.t circuit, so it doesn't affect the transfer function as transfer function is a measure of how o/p varies with i/p across the affect components?

    if Rs not = 0, the cap to gnd will be included in the transfer function?
     
    Last edited: Apr 11, 2010
  6. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    The transfer function is the relationship between input and output.
    You're looking at voltages and the only way that capacitor could affect the output voltage is if its current caused a voltage drop somewhere between the input and output.

    In this case the capacitor would increase the current coming from the voltage source but since there's no resistance it doesn't do anything else. The current flows only through the capacitor and the voltage source because it's a 0 resistance path, it doesn't go through the output resistor at all.

    If Rs wasn't 0 the capacitor current would add a voltage drop between the input and output, or equivalently the current would split between the voltage source and the output resistor's branch.
     
  7. MilK

    Thread Starter Member

    Mar 1, 2008
    25
    0
    "The current flows only through the capacitor and the voltage source because it's a 0 resistance path, it doesn't go through the output resistor at all."

    so when you say transfer function is:

    <br />
\frac{Vo}{Vin} = \frac{R}{0.5sL || \frac{1}{sC} + R}<br />

    the Vin you are referring to is the voltage across the cap. and not directly from source( although they are the same ) ?
     
  8. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    You're right, that transfer function is from the capacitor to the output but with Rs = 0 that's identical to Vin/Vout.
     
    MilK likes this.
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