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Inverse laplace of high pass RC filter

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm trying to work out the time domain response of a first order high pass RC filter, eg. http://en.wikipedia.org/wiki/File:High_pass_filter.svg

    I've got two solutions, one using laplace and one using differential equations, but they are different. One equation must be wrong, can anyone point out which one? And why?

    2. Relevant equations

    The laplace transform, or a first order ODE.

    3. The attempt at a solution

    Using laplace:
    [tex](v_{in}-v_{out})sC + \frac{0-v_{out}}{R}[/tex]
    [tex]v_{in}sCR = v_{out}(1+SCR)[/tex]
    [tex]v_{out} = v_{i} \frac{s}{s+\frac{1}{CR}}[/tex]

    Then I need to take the inverse laplace transform. I can't see any inverse transforms in laplace tables (eg. http://en.wikipedia.org/wiki/Laplace_transform" [Broken])for functions of the form [tex]\frac{s}{s+\alpha}[/tex], but I see that multiplication by s in the frequency domain is differentiation in the time domain, so I tried taking the inverse laplace of [tex]\frac{1}{s+\frac{1}{CR}}[/tex] to obtain [tex]f(t)[/tex] and then differentiating the result to give [tex]f'(t)[/tex], ie:

    [tex]F(s) = \frac{1}{s+\alpha}[/tex]
    [tex]f(t) = e^{-\alpha t}[/tex]

    and I want [tex]f'(t)[/tex], the first derivative of [tex]f(t)[/tex], so:

    [tex]f'(t) = -\alpha e^{-\alpha t}[/tex]

    and substituting in my values:

    [tex]v_{out}(t) = - v_{in} \frac{e^{\frac{-t}{RC}}}{RC}[/tex]

    Next, I try the same thing with differential equations:

    [tex]v_{in} = v_{c} + v_{r}[/tex]
    [tex]v_{in} = \frac{1}{C}\int{i} dt + iR[/tex]

    then using [tex]i = \frac{dq}{dt}[/tex]
    [tex]v_{in} = \frac{q}{C} + \frac{dq}{dt}[/tex]

    And solving this using an integrating factor [tex]\mu = e^{\frac{t}{RC}}[/tex]:
    [tex]\mu q = C v_{in}(e^{\frac{t}{RC}} - 1)[/tex]

    And finally computing [tex]v_{out}(t) = i.R = \frac{dq}{dt}.R[/tex] gives me:

    [tex]v_{out}(t) = v_{in} e^{\frac{-t}{RC}}[/tex]

    I'm not sure why this result differs from the laplace version. If I skip the final differentiation in the laplace formulation of the problem (ie. just entirely ignore the existence of the 's' on the top line), then the two equations give the same result, but I don't understand why. Any ideas?


    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 8, 2009 #2
    I've thought about it some more. Physically it makes sense at t=0 for Vout to be equal to Vin, so the 1/RC term in the laplace version of the equation doesn't make sense. I've also done a SPICE simulation in which the differential equation matches perfectly with the simulated waveform, so the differential equation is right and the laplace equation is wrong.

    But I don't understand why the laplace equation is wrong...
  4. Feb 9, 2009 #3
    Finally sorted this out. For the benefit of any googlers etc..., the problem is that I just kinda left [tex]v_{in}[/tex] alone (didn't fully appreciate what I was doing), but infact it could be written as [tex]v_{in}(s)[/tex] and then I needed to substitute in the Laplace transform for this, which for a unit step input is just [tex]\frac{1}{s}[/tex]. And then the 's' on the top line of the laplace equation cancels with the new 's' on the bottom...
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