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Laplace transforms for transient analysis

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data

    A capacitor of 0.1 F and a resistor of 5 Ω are connected in series; the combination is applied to a step voltage of 20V. Determine the expression for the:

    (a) current that flows in the circuit and

    (b) the voltage across the capacitor in time domain.

    2. Relevant equations

    For question (b)

    Would I be correct to use the following equations?

    Where:
    Vc - is the voltage across the capacitor
    V - is the supply voltage
    t - is the elapsed time since the application of the supply voltage
    RC - is the time constant of the RC charging circuit
    e - is the base of the Natural Logarithm = 2.71828

    Time Constant
    τ = R x C
    τ = 5 x 0.5
    τ = 2.5 seconds

    The voltage formula is given as:
    Vc = V(1-e-t/RC)

    which equals:
    Vc = 20(1-e-t/5)
     
    Last edited: Oct 17, 2013
  2. jcsd
  3. Oct 17, 2013 #2

    gneill

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    Staff: Mentor

    Are you expected to do the problem using Laplace transforms (it's in the thread title), or just write down the result from previous knowledge?

    EDIT: Also, 5 x 0.1 is not 5.
     
    Last edited: Oct 17, 2013
  4. Oct 17, 2013 #3
    Yes. The general title for these question states 'Use Laplace transforms for the transient analysis of networks'.

    Appologies for the typing error.
     
  5. Oct 17, 2013 #4

    gneill

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    So, the equations you've written are really a result you hope to find using Laplace transforms.

    As usual, begin with a circuit equation. Use Laplace domain models for the circuit components and write the appropriate equations...
     
  6. Oct 18, 2013 #5
    I cant find any appropriate equation for this, i dont have any good example for this.

    Can you help?
     
  7. Oct 18, 2013 #6

    gneill

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    You write your usual circuit equations just as you would for an AC circuit, but use the Laplace domain version of the component impedances. Where you were using 'jω' before in the expressions for inductor and capacitor impedances, just use the variable 's'.

    I know it looks silly at first glance, writing AC-type equations for what is apparently a DC circuit, but the Laplace domain's 's' variable takes in all frequencies and can readily model the transient response that happens when a switch closes. You'll see :smile:

    You should have in your text or notes the form that a unit step function takes in the Laplace domain. You'll need that to model the 20V source that's suddenly switched on at time t=0.
     
  8. Oct 18, 2013 #7
    If I(t) is the current at time t, what is the voltage drop across the resistor at time t (in terms of I(t) and R)? What is the voltage drop across the capacitor at time t (in terms of the integral of I(t) with respect to t and C)? Write an equation such that the sum of these two voltage drops is equal to the applied step voltage. Do you know how to take the Laplace Transform of the individual terms in this equation? If so, write the Laplace Transform of the equation.

    Chet
     
  9. Oct 18, 2013 #8
    Z = R + (1 / sC)

    Z = 5 + (1 / s x 0.1)

    Using Ohms Law:

    I(s) = Vin / Z

    I(s) = 20/s / 5+(1 / s x 0.1)

    However:

    Vout(s) = I x (1 / sC)

    Vout(s) = I x (1 / s x 0.1)

    So:

    Vout(s) = [ 20/s / (0.1 + (1 / s x 0.1))] x [1 / s x 0.1]

    Vout(s) = 20 / s(s x 0.1 x 0.1 + 1)

    Vout(s) = 20 / s(s x 0.01 + 1)

    Am i right? What value is s in this example?
     
    Last edited: Oct 18, 2013
  10. Oct 18, 2013 #9

    gneill

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    Something's gone wrong in your algebra when you solved for Vout.

    Start by simplifying your Z a bit. Note that 1/(s x 0.1) = 10/s. Makes the numbers easy!

    You're going to want to solve for I, so you might as well simplify the expression for I(s) before carrying on to the potential across C. Can you reduce the expression for I(s)?
     
  11. Oct 18, 2013 #10
    To solve Z = R + (1 / sC) i need to add (5 + 10/s) = (10+5s/ s)

    for Is= (20s/ 10 +5s) = (20s(10-5s) )/ (10+5s) (10-5s) = (200s - 100s2 ) / (100-25s2)

    and if s2=(-1) then

    Is= (200s+100) / 125
    Is it ok so far?
     
  12. Oct 18, 2013 #11

    gneill

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    Use more parentheses to keep your operations straight.
    $$I(s) = \frac{V(s)}{Z(s)} = \frac{20}{s} \left(\frac{s}{5 s + 10}\right) = ~~?$$
     
  13. Oct 18, 2013 #12
    Is= = 20 / 5s+10 = (100s-200) / (25s2 -100)

    Is=( 4s-8) /-5
     
    Last edited: Oct 18, 2013
  14. Oct 18, 2013 #13

    gneill

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    Egads! Just divide the top and bottom of 20 / (5s+10) by five. Hint: It is desirable to have the denominator have factors of the form (s + n).
     
  15. Oct 18, 2013 #14
    This is the same answer I got, but I used a different method. My starting equation was
    [tex]\frac{\int_0^t{I(t')dt'}}{C}+RI(t)=Vu(t)[/tex]
    where u is the unit step function. Taking the Laplace Transform gives:
    [tex]\frac{I(s)}{sC}+RI(s)=\frac{V}{s}[/tex]
    Solving for I(s):
    [tex]I(s)=\frac{V}{R}\frac{1}{\left(s+\frac{1}{RC}\right)}[/tex]
     
  16. Oct 18, 2013 #15
    Then Is= 4/ (s+2)

    Should i calculate now Vout?
     
    Last edited by a moderator: Oct 18, 2013
  17. Oct 18, 2013 #16

    gneill

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    You could, or you could find I(t) first. You need to provide it as one of the answers...
     
  18. Oct 18, 2013 #17
    It= ( V/R ) ( 1-e (-Rt/L) )

    But i dont have L for this equation. Should i find a different one for It?
     
  19. Oct 18, 2013 #18

    gneill

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    There's no inductor in this circuit.

    Find I(t) by finding the inverse Laplace transform of the I(s) that you found above. You should have a table of Laplace transforms to work with.
     
  20. Oct 18, 2013 #19
    What is the Laplace Transform of the function eat?
     
  21. Oct 18, 2013 #20
    ok then,

    It= 4e -2t

    am i right?
     
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