LCR Circuit Transfer Function & Diif. equations

Discussion in 'Homework Help' started by kvothe, Jul 24, 2011.

  1. kvothe

    Thread Starter New Member

    Jul 24, 2011
    8
    0
    could someone please show & explain the steps involved in deriving both equations for a series LCR circuit as shown in the attached image.
     
  2. kvothe

    Thread Starter New Member

    Jul 24, 2011
    8
    0
    does anyone know how i can use matlab/simulink to derive the time solution for a simulated series LCR circuit? (i have the simulation of the step responce from the transfer function)
     
  3. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    For both cases you can perform KVL, making sure that you have the components in the correct domain. Frequency for the first case, and time domain for the differential equations.

    The first one is pretty trivial as you convert each element into its frequency domain counterpart and then solve accordingly.

    The time domain equation requires you to know the relationships between current and voltage of capacitors and inductors.

    Since it is a single series loop, it would be easy enough to perform a KVL on the circuit. Knowing the following:

    V_{l}(t) = L\frac{di(t)}{dt}

    V_{c}(t) = \frac{1}{C}\int_0^t \! i(x)dx

    V_{r}(t) = Ri(t)

    Therefore after performing KVL you get,

    V(t) = Ri(t) + L\frac{di(t)}{dt} + \frac{1}{C}\int_0^t \! i(x)dx

    Differentiating both sides to get rid of the integration,

    \frac{dV(t)}{dt} = R\frac{di(t)}{dt} + L\frac{d^{2}i(t)}{dt^{2}} + \frac{1}{C}i(t)

    After normalizing the differential equation you get,

    \frac{d^{2}i(t)}{dt^{2}} + \frac{R}{L}\frac{di(t)}{dt} + \frac{1}{LC}i(t) = \frac{1}{L}\frac{dV(t)}{dt}

    Also regarding that question, remember that these relationships for current hold for the inductor and capacitor as well, and since this is a series circuit, each of these is equal to the current flowing through the entire circuit at all times.

    I_{l}(t) = \frac{1}{L}\int_0^t \! v(x)dx

    I_{c}(t) = C\frac{dV_{c}(t)}{dt}
     
    Last edited: Jul 24, 2011
  4. kvothe

    Thread Starter New Member

    Jul 24, 2011
    8
    0
    thanks, i didn't realise how thick i was being! somehow made it far more complicated in my head than it actually was!
     
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