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Homework Help: Integration and RC Step response

  1. Feb 22, 2010 #1
    Integration

    1. The problem statement, all variables and given/known data

    Hi everyone,

    I'm new here. This is my first post. I found physicsforums when researching on google a solution for a doubt I've had when trying to solve rc step response. Basically, the RC analyse showed at http://freespace.virgin.net/ljmayes.mal/circuittheory/Rcstep.htm" solves step by step the mathematical behave of RC circuits.

    I understand the whole anylise but the line with the follow part (integrating):

    2. Relevant equations

    [tex]\int[/tex]dt = RC[tex]\int[/tex][tex]\frac{1}{Vin - Vc}[/tex]dVc

    t = - CR ln(Vin - Vc) + const


    Thanks in advance
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Feb 22, 2010 #2

    Mark44

    Staff: Mentor

    Yes, that's correct (or at least close. The right side should be ln |x| + const.

    Similarly,
    [tex]\int\frac{du}{u}~=~ln|u| + C[/tex]

    Using an ordinary substitution in your problem, u = Vin - VC, what will du be?
     
    Last edited by a moderator: Apr 24, 2017
  4. Feb 22, 2010 #3
    du in function of Vc ?

    du = -1 ? is that right?
     
  5. Feb 22, 2010 #4

    Mark44

    Staff: Mentor

    No, if u = Vin - VC, then du = -dVC.
     
  6. Feb 22, 2010 #5
    right!? so how would you solve

    [tex]
    \int
    [/tex][tex]\frac{1}{Vin - Vc}[/tex]dVc ??

    Would you differentiate 1/Vin-Vc in function of Vc and then integrate the result?

    sorry I still don't understand the "-" on the right side of the eq.

    thanks
     
  7. Feb 22, 2010 #6

    Mark44

    Staff: Mentor

    Use and ordinary substitution with u = Vin - VC.

    What is du?
    Make the substitution and do the integration. What do you get?
     
  8. Feb 22, 2010 #7
    - ln (Vin - Vc) + const
    thanks very much Mark44
     
    Last edited: Feb 22, 2010
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