RC Transfer function

Discussion in 'Homework Help' started by alexmath, Apr 4, 2015.

  1. alexmath

    Thread Starter New Member

    May 2, 2014
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    Suppose R = C = 1 then the transfer function from the input voltage to the voltage across the capacitor is 1/ (s+1). So Vc(S) / V(S) = 1/(s+1). Getting back to time-domain: Vc(t) = V(t) * e^-t. What's wrong here? Thank you!
     
    Last edited: Apr 4, 2015
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,386
    496
    C is complex, R isn't.
     
  3. MrAl

    Well-Known Member

    Jun 17, 2014
    2,425
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    Hello there,

    Constant voltage sources V(s) transform into V/s they do not stay the same when there is a step input. So V(s) => V/s so your Laplace is not V/(s+1) it is (V/s)/(s+1)=V/(s*(s+1)). If the voltage source is 1v then this is simply 1/(s^2+s).

    Find the inverse of that and you'll be very happy :)
     
  4. alexmath

    Thread Starter New Member

    May 2, 2014
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    Thank you! I'm very happy right now! Have a good day.
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,720
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    The statement R=C=1 is nonsensical.

    That's like saying that the height of a car and the weight of a building are both equal to 1. Totally meaningless.
     
  6. Papabravo

    Expert

    Feb 24, 2006
    10,137
    1,786
    If C is capacitance measured in farads, then it is a real number. The reactance of C, Xc, with units of ohms is also a real number. The impedance or admitaance of a capacitor C is a complex number. In the ideal case it would have a zero real part. For a real component there will be some equivalent series resistance.
     
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