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Solving Impulse Response of RL Circuit

  1. Feb 2, 2012 #1
    Hi all,

    I'm taking a fourth year time-series analysis course for Physics students.

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

    Let us consider a simple physical system consisting of a resistor (with resistance R) and an inductor (with inductance L) in series. We apply an input voltage a(t) across the pair in series, and measure the output voltage b(t) across the inductor alone.

    We are asked to show analytically what the step and impulse responses would be.

    2. Relevant equations

    Kirchhoff's equations

    H(t) = step or Heaveside function
    δ(t) = delta function

    [itex]\frac{dH(t)}{dt}[/itex] = δ(t)

    Let

    V(t) = source potential
    V_(t) = potential across inductor

    Assume

    I = 0 at t = 0 (no initial current)

    3. The attempt at a solution

    I was able to use the basic circuit equation:

    V(t) = R*I + L*[itex]\frac{dI}{dt}[/itex]

    To solve for the current across the inductor:

    I(t) = [itex]\frac{V(t)}{R}[/itex]-[itex]\frac{V(t)}{R}[/itex]*e[itex]^{(-R/L)*t}[/itex]

    Now setting V(t) = H(t) to find the step response, and knowing that:

    L*[itex]\frac{dI(t)}{dt}[/itex] = V_(t)

    I found that:

    V_(t) = H(t)*e[itex]^{(-R/L)*t}[/itex]

    ...which is the correct step response. However, my problem is in deriving the impulse response. When I try use the current I(t) and substitute in δ(t) for the input potential V(t), my derivation falls apart. I'm not used to working with the δ(t) function, so this is a bit tricky for me. I've since found that the impulse response is:

    V_(t) = δ(t) - (R/L)*e[itex]^{(-R/L)*t}[/itex]*H(t)

    ...but have no idea how to get to this point. Clearly, the impulse response is the time derivative of the step response, but is this just coincidental?

    Any help would be much appreciated. Thanks!

    Riyaad
     
  2. jcsd
  3. Feb 3, 2012 #2

    rude man

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    Have you had the Laplace transform?
     
  4. Feb 3, 2012 #3
    No, we haven't studied Laplace transforms yet, so I was hoping there would be a way to solve without it.
     
  5. Feb 3, 2012 #4

    rude man

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    You know those ads on TV that say, "don't try this at home"? I say, "don't try to solve differential equations with Dirac delta (aka impulse) function excitations without a transform". The logical one here is the Laplace.

    Maybe some math wiz will show us how to do it the classical, or another, way.

    EDIT: I'm surprised that a 4th yr course would not have covered the Laplace or similar (e.g. heaviside) xfrm.

    FYI my answer is vL = k{δ(t) - (1/τ)e-t/τ)} for a unit impulse input kδ(t) where k = 1.0 V-sec and τ = L/R.

    (The units of δ(t) are sec-1.)
     
    Last edited: Feb 3, 2012
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