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Control Systems Engineering - Masons Rule Help

  1. Feb 1, 2014 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find [itex]\frac{V_{Load}(s)}{I_{Dist}(s)}[/itex] using Mason's rule. I was able to solve for [itex]\frac{V_{Load}}{I_{Cf}(S)}[/itex] using Mason's rule but am not sure how to solve for [itex]\frac{V_{Load}(s)}{I_{Dist}(s)}[/itex].

    http://imageshack.com/a/img31/4299/zr5.PNG [Broken]

    2. Relevant equations



    3. The attempt at a solution

    There's only one forward path, [itex]1*\frac{1}{C_{f}s}*1[/itex]. Am I supposed to consider the loops to the left of [itex]I_{Dist}[/itex] to be non-touching loops? I've never solved a problem like this before were I'm asked to solve for the transfer function for a variable that's in the middle of a flow diagram and am not exactly sure. I'm also unsure if in the diagram that's a negative sign in front of [itex]I_{Dist}[/itex]. Thanks for any help.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 1, 2014 #2

    donpacino

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    As you stated there is only one forward path. That means that the gain will always be 1*1/CFS*1, as you already stated. It looks like there is a negative sign to me.

    I don't want to just give you the answer, but you are VERY close.
     
  4. Feb 1, 2014 #3
    Is it really just [itex]-\frac{C_{f}s}{1 - C_{f}s}[/itex]?
     
  5. Feb 3, 2014 #4

    donpacino

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    well there is no loops in the circuits. so the gain would just be the forware gain.

    so vload/-Idist=1/(CFS)
     
  6. Feb 3, 2014 #5
    So if I wanted positive Vload/Idist it would just be the opposite magnitude, -1/(CFS)?
     
  7. Feb 4, 2014 #6

    donpacino

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    Yessir
     
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