Circuit analysis problem - Laplace

Discussion in 'Homework Help' started by ineedmunchies, Apr 3, 2008.

  1. ineedmunchies

    Thread Starter New Member

    Apr 3, 2008
    1
    0
    The question is as shown in the first picture. Question1.jpg

    It asks that initial condition current generators are used, what I believe models the circuit for t>0 is shown in the second picture with the switch open, and a current generator added. (I am not sure if this is correct.)

    I then used KCL to write equations at node 1 and node 2.

    Node 1:
    \frac{5}{s} - \frac{2}{s}=\frac{V_{1}}{1}+\frac{V_{1}-V_{2}}{2s}


    Node 2:
    \frac{2}{s}+\frac{2}{s}=\frac{V_{2}}{1}-(\frac{V_{1}-V_{2}}{2s})


    These can then be rearraged to give

    \frac{3}{s}=V_{1}(1+\frac{1}{2s})-V_{2}(\frac{1}{2s})

    and

    \frac{4}{s}=V_{1}(\frac{-1}{2s})+V_{2}(1+\frac{1}{2s})

    Which I then put into a matrix and solved for V_{1} and V_{2}

    Giving
    V_{1} = \frac{3}{s} - \frac{2}{1+\frac{1}{2s}}
    which can be simplified to
    V_{1} = \frac{3}{s} - \frac{4}{s+2}

    and
    V_{2} = (\frac{\frac{-3}{2}}{1+\frac{1}{2s}})+\frac{4}{s}

    (*Note the 4/s is not part of the denominator, I couldn't get the brackets to work properly.)

    Which can be simplified to
    V_{2} = \frac{4}{s}-\frac{3}{s+2}

    Then convert these back to the time domain to give:
    V_{1}(t) = 3-4e^{-2t}

    and V_{2}(t) = 4-3e^{-2t}

    Can anyone tell if there is a mistake here or not?
    I don't feel confident this is the correct answer. I think the 5 ohm resistor and 5A current should effect the circuit somehow but do not know how to encorporate it into my equations.
     
  2. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    I think your algebra is off in your solutions for V1 and V2. Starting with your original equations for nodes 1 and 2, I get:
    V1 = (3s + 7/2)/(s(s + 1))
    V2 = (4s + 7/2)/(s(s + 1))

    I've checked, and these values satisfy your original equations, so I'm pretty confident about them.
    To do the inverse Laplace transforms, you'll need to break each of these apart into a sum of two rational expressions. The denominators will be s and s + 1, not s and s + 2 that you show.
    Mark
     
  3. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    In case the bit about breaking up the expressions for V1 and V2 is not clear enough, you'll need to do a partial fraction decomposition for each of them.

    For V1, you need to solve for A and B in this equation:
    (3s + 7/2)/(s(s+1)) = A/s + B/(s + 1)

    When you do the inverse Laplace transform, you'll end up with v1(t) = A + B*e^(-t)

    Do the same thing to find v2(t).
     
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