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General Second Order Circuit

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Refer Attachment.
    I am trying to derive the second order equation using the natural response of the circuit


    2. Relevant equations
    Refer Attachment


    3. The attempt at a solution
    Nodal Analysis:
    v1+[itex]\frac{1}{2}[/itex][itex]\frac{dv1}{dt}[/itex]+[itex]\frac{1}{3}[/itex][itex]\frac{dv2}{dt}[/itex]=0
    Mesh Analysis LHS:
    i+v1=0

    I can't see how to get everything in terms of v1 or v2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Nov 23, 2011 #2

    rude man

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    Homework Helper
    Gold Member

    First I would label the components from left to right: R1, R2, C1, C2.

    Then I would write equations summing currents at v1 and v2.

    Then solve the diff. eq's (zero initial conditions), which gets you v1 and v2. Then your answer is v1 - v2.

    The reason you should label the components is otherwise you get all mixed up as to which resistor is which, for example. And you wind up with equations like your
    i + v1 = 0 which seemingly mixes volts and amps in terms within the same equation, making dimensional checking impossible. The greatest tool for error checking of equations is dimensional checking!
     
  4. Nov 23, 2011 #3
    Thanks for getting the ball rolling rude man. The most challenging part is coming up with the differential equation.

    [tex]\begin{array}{l}
    {\rm{Find natural response:}} \\
    \frac{{{v_1}}}{1} + \frac{1}{2}\frac{{d{v_1}}}{{dt}} + \frac{{{v_1} - {v_2}}}{1} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\frac{1}{2}\frac{{d{v_1}}}{{dt}} + 2{v_1} - {v_2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\
    \frac{{{v_2} - {v_1}}}{1} + \frac{1}{3}\frac{{d{v_2}}}{{dt}} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to {v_1} = {v_2} + \frac{1}{3}\frac{{d{v_2}}}{{dt}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \\
    {\rm{Substitute }}{v_1}{\rm{ into equation 1:}} \\
    \frac{1}{2}\left[ {\frac{{d{v_2}}}{{dt}} + \frac{1}{3}\frac{{{d^2}{v_2}}}{{d{t^2}}}} \right] + 2{v_2} + \frac{2}{3}\frac{{d{v_2}}}{{dt}} - {v_2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \frac{1}{6}\frac{{{d^2}{v_2}}}{{d{t^2}}} + \frac{1}{2}\frac{{d{v_2}}}{{dt}} + \frac{2}{3}\frac{{d{v_2}}}{{dt}} + {v_2} = 0 \\
    \frac{{{d^2}{v_2}}}{{d{t^2}}} + 7\frac{{d{v_2}}}{{dt}} + 6{v_2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to {\rm{let }}{v_2} = {e^{st}}\,\,\,\,\,\,\,\,\,\,\,\,\, \to {s^2}{e^{st}} + 7s{e^{st}} + 6{e^{st}} = 0 \\
    {\rm{characteristic equation = }}{s^2} + 7s + 6 \to \,{\rm{ roots }} - 6{\rm{ and }} - 1 \\
    {v_{2{\rm{natural}}}}(t) = A{e^{ - 6t}} + B{e^{ - t}} \\
    {v_2}(t) = 5 + A{e^{ - 6t}} + B{e^{ - t}} \\
    @t = 0\,\,{v_2}(0) = 0 = 5 + A + B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to A + B = - 5 \\
    \frac{{d{v_2}}}{{dt}} = - 6A{e^{ - 6t}} - B{e^{ - t}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to @t = 0\,\,\frac{{d{v_2}(0)}}{{dt}} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 0 = - 6A - B \\
    - 6A - B = 0 \\
    A + B = - 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to A = 1,\,\,\,B = - 6 \\
    {v_2}(t) = 5 + {e^{ - 6t}} - 6{e^{ - t}} \\
    {\rm{ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - }} \\
    {\rm{Find }}{v_0}{\rm{:}} \\
    {{\rm{v}}_0} = {v_1} - {v_2} \\
    {\rm{From equation 2: }}{v_1} = {v_2} + \frac{1}{3}\frac{{d{v_2}}}{{dt}} \\
    {v_0} = {v_2} + \frac{1}{3}\frac{{d{v_2}}}{{dt}} - {v_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to {v_0} = \frac{1}{3}\frac{{d{v_2}}}{{dt}} \\
    {v_0}(t) = \frac{1}{3}\left( { - 6{e^{ - 6t}} + 6{e^{ - t}}} \right)\,\,\,\,\,\, \to {v_0}(t) = 2\left( {{e^{ - t}} - {e^{ - 6t}}} \right)V \\
    \end{array}[/tex]
     
  5. Nov 26, 2011 #4

    rude man

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    You certainly seem to know what you're doing. I don't intend to check your math fully. What I did check was correct. I can help you verify your answer though since I have a 'cheat sheet' which lists the transfer function of your network, and with apologies I must use the Laplace transform method, so if you don't understand much of the following, don't be concerned. I am just trying to give you the answer I got & maybe something for you to look at later when you do hit transform methods.

    V2/Vin = 1/[1 + (R1C1 + R2C2 + R1C2)s + R1C1R2C2s^2]

    so your characteristic equation should be the denominator above set to zero, and it looks like it might just be!

    Then a step input transforms to 5/s so
    V2 = 5/s[1 + (R1C1 + R2C2 + R1C2)s + R1C1R2C2s^2].

    However, you're asked to find V1-V2 which conveniently happens to be the current flowing into C2 multiplied by R2, so i = sC2V2 and

    V1-V2 = iR2 = sC2R2V2 = 5R2C2/[1 + (R1C1 + R2C2 + R1C2)s + R1C1R2C2s^2].

    This inverse-transforms to
    V1-V2 = 5R2C2{1/(T1-T2)}[exp(-t/T1) - exp(-t/T2)] volts.

    where T1 and T2 are the coefficients in the chas. equation:
    (T1+s)(T2+s) = 1 + (R1C1 + R2C2 + R1C2)s + R1C1R2C2s2), T1, T2 > 0 and real. I'm leaving it up to you to solve the appropriate quadratic if you want.

    Your answer is in the right form so other than a math mistake you're looking very good.
     
  6. Nov 26, 2011 #5
    No worries. I typed out the maths for my own benefit more than anything else. Just as a learning tool.
     
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