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RLC inverse Laplace transform question

  1. Dec 27, 2011 #1
    The link below shows a circuit and the s-domain equation for I1(s) that I have derived; My question is on the portion of the equation containing the "(s+a)" terms. I am trying to arrive at the time-domain equation for I1(s) but instead of arriving at the Laplace transform pair shown in the link below my math ends up with an extra "a" term in the numerator as shown. I cannot find a Laplace transform pair for the s-domain equation that I end up (as shown in the link below). Have I made an error in the derivation? Has anyone derived this solution? Thanks.

    Last edited: Dec 27, 2011
  2. jcsd
  3. Dec 29, 2011 #2
    First of all, let's use some of our intuition regarding circuits - intuition that every electrical engineer should have, that is - and predict the activity of this circuit. Notice that the input voltage is Vi/s; this corresponds to a DC signal of magnitude Vi that is applied to the circuit for all time t. Therefore, we should expect to see some type of resonance characterized by the inductor and capacitor "passing" some measure of voltage back and forth until the resistor causes this oscillation to die down via power dissipation in between passes.

    After a long time t - which we can assume represents current in the time domain since the signal has always and will always be constant - the inductor will effectively look like a short circuit since it no longer opposes any current. Thus, the total input voltage will be dropped across the parallel RC combination. This means that there will be a charge on the capacitor of magnitude Q = CoVin with a voltage of Vin across the resistor. Finding the current is now simply a matter of applying Ohm's Law.

    I1(t) = Vin/R

    Regardless of your arithmetic, you have made a fatal mistake using Ohm's Law in your solution; it seems that you took the sum of the input voltage in the Laplace domain and the equivalent impedance of the circuit. In my solution, I multiplied both of these Laplacian terms together and came up with a fraction that i decomposed using partial fractions. From there, you can compute the inverse Laplace transform of the two resulting terms to obtain the activity of the circuit. As it turns out, we can cancel some algebraic terms to obtain the exact qualitative relationship that I predicted above.

    In other words, the current flows completely through - and only depends on the value of - the resistor. This makes sense because there will be no current flowing through the capacitor at long times t. Conversely, the inductor will not resist any current and will resemble a piece of wire with zero impedance.

    Of course, this all depends on the nature of your input voltage; if there are discontinuities or AC oscillations, the response of this circuit in the time domain can become quite complex.
  4. Dec 29, 2011 #3
    Hello GuiltySparks and thanks for your response and intuition.

    I agree with your comments and was aware of the eventual steady-state currents and voltages that will result after the transient event takes place. This simplified circuit is part of a power systems design where I'm contemplating inserting the inductor Lr so as to slow down the rise of the current when Vi (a DC source switched in via a MOSFET) applies power to the RLC circuit. The idea is to minimize switching loss by delaying the rise of I1. I was trying to derive the time-based equation for I1 so that I could manipulate the ckt values that will control the characteristics of the damped oscillation that will occur as a transient. I am only interested in the transient event.
  5. Dec 29, 2011 #4
    I worked out the solution and attached it. Notice that I assume at time t = 0 a DC step function is applied.

    You will have to minimize the amplitude of the trigonometric function. To do this, compute the partial derivatives of the amplitude A with respect to L, R, and C and set them equal to 0. Solve the system of equations for the component values in order to minimize I.

    Attached Files:

  6. Dec 29, 2011 #5
    Thanks for the effort and for the attached derivation; this is very helpful. I will compare it to mine and see where I made and error. Your result makes sense and is what I would've expected.
  7. Dec 30, 2011 #6
    Doing it was actually pretty informative for me, also. I realized that if you take your source to be Vin/s, then that assumes the voltage on the circuit is always Vin for all time t. You would then need to account for initial conditions when doing the Laplace Transform; incorporating a heaviside function allows us to study just the transient event.
    Last edited: Dec 30, 2011
  8. Jan 10, 2012 #7
    Hello Again, GuiltySparks; I recomputed my original derivation above with your inputs and successfully got the same results as you so thanks again for your input.
    I have expanded my analysis on the circuit we discussed previously. I added additional components which complicated the circuit somewhat. If you look at the link below you will see the circuit and note that I have added resistors Ri and Rci plus a capacitor, Ci, with an initial condition of Vi. Again, my circuit and it's components are shown with impedances in the "s" domain; the input source is a step-function which assumed to begin at time t=0 and is shown as "Vi/s". The input capacitor Ci is assumed charged to the input voltage and has it's initial condition represented with its series voltage source "Vi/s".
    Again, my interest is in the transient event and I was trying to derive the time-domain equation for Ii (and eventually for the current in the capacitor Ci). As you see in the link below my derivation came to a stop when I ran into the s-domain equation for Ii which has a cubic s-term in the denominator. I'm not sure how to do a Partial Fraction expansion on this equation for Ii so that I can do an inverse Laplace Transform on Ii to get an expression in the time domain. Any suggestions? Did I make an error?

  9. Jan 10, 2012 #8
    To clarify, both DC sources in the circuit are step inputs? If so, you need to represent them as a heaviside function in the s-domain (or u_0(t)), like I did in the previous derivation. Though, I do have the sneaking suspicion that you are incorporating these into the circuit to account for the initial conditions that are obligatory for Laplace analysis; if so - don't! This will needlessly convolute the circuit - and your analysis.

    As for the charged capacitor, you will have to represent it as either a current source or a voltage source in your analysis - I'm not sure which. Even though you are investigating the transient response, this will need to be expressed as a non-constant voltage function of time. This will be tough, since the Thevenin or Norton voltage equivalent of the capacitor will have an RLC impedance. When the long weekend comes around, I will try to find the answer for both of us. Also, is the charge on the capacitor representative of the initial conditions or an actual charge accumulation on the plates?

    You'll have to determine the roots of the cubic using the general form of the roots for a cubic equation, as show in the Roots of a Cubic Function section here: http://en.wikipedia.org/wiki/Cubic_function. Since you have variable coefficients, I believe that this is the only way to solve for it. I could be wrong, though. A more practical method is to substitute actual component values into the transfer function and solve.

    With that being said, this method is quite tedious and very time consuming. Moreover, you are extremely likely to make a mistake. Perhaps you should build the circuit and analyze its response yourself? Otherwise, there is a panoply of engineering software out there that can expedite your analysis.

    I would help more, but I have my own Engineering coursework to struggle with!
  10. Jan 11, 2012 #9
    The source Vi is a step-input voltage source; the second source in series with the input capacitor, Ci, is meant to depict an initial voltage condition of Vi on the capacitor. I am representing this also as a step-source at t=0. In previous analysis I have always successfully represented the initial conditions on a capacitor as a voltage-source in series with the capacitor's impedance, (1/Ci*s in this case); is this in error?

    Yes, I was surprised to end-up with a cubic equation. I am using a MathCad-like software (called "LiveMath" on my Mac) to crunch the Algebra so I don't think I have any Algebra errors. By the way what software did you use in the previous derivation that you attached?

    I have also been analyzing this circuit with Spice simulation but I was trying to get a deeper insight through the time-domain equations that I was striving for.

    Don't knock yourself out but any insight in the future will be helpful.
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