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Homework Help: Impulse Response of an RLC Circuit

  1. Dec 6, 2017 #1
    I bolded the portions I need help with.

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

    A series circuit consists of a resistor with a resistance of 16 ohms, an inductor with inductance of 2 H, and a capacitor with a capacitance of 0.02 F.
    At time t = 0 there is no charge on the capacitor and no current in the circuit. If a 9 V battery is connected at at t=0 and disconnected at t=2
    (a) The charge, q(t), on the capacitor at all t
    (b) Compute the impulse response of the circuit as a function of time and classify the response as under-damped, over-damped, or critically-damped, explaining your rationale.
    (c) Suppose the resistor were changed to make the circuit response critically-damped. What would be the value of R? Compute the new impulse response with this value B.


    2. Relevant equations
    The equation:
    2q'' + 16q + 50 = 9(1 - U(t-2)). Where U is the unit step function.

    3. The attempt at a solution
    So, I found the q(t) by applying the Laplace Transform on the equation and then the inverse Laplace Transform Q(s). My work for the first part is in the image attached.

    However, I am unsure how to calculate the Impulse Response without delta involved in the equation. I believe the circuit is under-damped because: sqrt(16^2 - 4(2)(50)) is not a real solution. However, I don't know how that would relate to the Impulse Response.
     

    Attached Files:

  2. jcsd
  3. Dec 6, 2017 #2

    donpacino

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  4. Dec 6, 2017 #3
    I still don't get it. There's no delta function, which means there's never an instantaneous impulse. Or is the answer simply the transfer function (e.g. Laplace Transform of the above equation)?
     
  5. Dec 6, 2017 #4

    donpacino

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    You're right for the wrong reasons!

    asking for the impulse response is asking how your system will react when an impulse is put in, just like asking for the "step response" would be asking what would happen if there is a step function at the input.

    that being said, the laplace transform of an impulse at zero is... 1. So in the frequency domain you multiply your function by 1 to get the impulse response, therefore the transfer function is equal to the impulse response. If you wanted to get the step response, in the frequency domain you would multiple your function by 1/s. Does that make sense?
     
  6. Dec 6, 2017 #5

    donpacino

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    As far as the damping, my answer will depend on your level of knowledge in electrical engineering.

    If you are well versed in the frequency domain, you will know that the poles of your system (the denominator in your transfer function) will determine your damping.

    You can also determine damping by looking at a plot of the data. Do you know the difference between over/under/critical damped?
     
  7. Dec 6, 2017 #6
    Okay, so the reason the Laplace Transform of the impulse is 1 at zero is because it occurs at t=0, right? If it occurred at t=1, it'd be e^-s, correct? Then, the Impulse will be a function of s (dependent on what the frequency of the signal is, assumed to be at t=0, because that's when the impulse occurs)?
     
  8. Dec 6, 2017 #7

    donpacino

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    Yes that is correct! in general impulse response just means transfer function unless they specify.
     
  9. Dec 6, 2017 #8

    donpacino

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    How are you doing on parts B and C?
     
  10. Dec 6, 2017 #9
    I know in the time domain, over-damped means a real, non-repeating root, under means imaginary root, and critical means real, repeating root. Underdamped will oscillate, the rest decay (same way as the exponential curve), but they will usually converge to some value.

    However, I'm not really aware of how this translates to the frequency domain.
     
  11. Dec 6, 2017 #10

    donpacino

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    You know everything you need to know! the roots (poles) of a system are the best way to tell
     
  12. Dec 6, 2017 #11
    Based off my formula for Q(s) poles exist at s=0 and two complex roots (-4+3i & -4-3i). The complex roots suggest underdamped. Does the simple pole at 0 impact this?
     
  13. Dec 6, 2017 #12

    donpacino

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    oooo I just found a mistake in your work. unfortunately you are making life difficult on yourself Your actual system olny has poles 2 poles at -4 +-3. The pole at zero is caused by you incorperating the step function (which in an input) into your transfer function. Remember tf = output / input. the 9*(1-u(t-2)) is your input and should not be incorperated into your tf. And yes, it is underdamped. If there was a pole at zero in your system, the system would be marginally stable, ie at steady state it would oscillate.
     
  14. Dec 6, 2017 #13
    2q'' + 16q' + 50q = 9(1 - U(t-2))
    So, we just apply the Laplace Transform to 2q'' + 16q' + 50q = 9? How would we then solve for q(t), to account for U? Or do we only omit the unit step function when determining whether it's damped (i.e. only take the Laplace Transform of the left of the above equation)?
     
  15. Dec 6, 2017 #14

    rude man

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    Bit confusing.
    The L. transform of an impulse input delayed by time T is e-sT. Your problem does not deal with delayed inputs.

    What "frequency of the signal"? It's the same impulse at T as at 0, just time-delayed. The spectrum of an impulse anywhere is constant with respect to frequency, all the way to infinity. But I dont think that is what you meant.

    The input pulse described is an approximation to an impulse input δ(t). It approaches an impulse as the pulse width approaches zero while keeping the time-voltage product constant (so it gets taller and thinner). It would be significantly more cumbersome to have to determine the output with the given pulse than with a Dirac delta input δ(t).
     
  16. Dec 6, 2017 #15
    So, the Laplace Transform is just e^0 = 1. If you meant when I asked if it at t=1, I was more making sure I understood the more general case.

    By "frequency of the signal," I meant that s would be the variable, so the magnitude of the impulse response would be dependent on s. Did I misunderstand something?
     
  17. Dec 6, 2017 #16

    rude man

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    .OK. Just that e-s is dimensionally incorrect. You set T=1, can't argue with that!
    Right, but just to be sure, the magnitude of the impulse response would be dependent on s even if the input is δ(t). Except the whole output would be time-delayed by 1 sec.
     
  18. Dec 6, 2017 #17
    Wait, so why would δ(t) delay one second? Wouldn't δ(t-1), delay a second, because that's when the impulse would occur?
     
  19. Dec 6, 2017 #18

    rude man

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    I didn't mean that. As you say, δ(t) is an impulse that occurs at t=0 and δ(t-1) occurs at t=1 sec.
    What I did mean is that the output transform is a function of s whether your input is δ(t) or δ(t-1).

    If L-1[Y(s)] is the time response for an input of δ(t) then L-1[Y(s)e-sT] is the response for an input of δ(t-T).
     
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