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Control Systems Engineering - Masons Rule Help

  • Thread starter GreenPrint
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  • #1
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Homework Statement



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]

Homework Equations





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.
 
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Answers and Replies

  • #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.
 
  • #3
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Is it really just [itex]-\frac{C_{f}s}{1 - C_{f}s}[/itex]?
 
  • #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)
 
  • #5
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So if I wanted positive Vload/Idist it would just be the opposite magnitude, -1/(CFS)?
 
  • #6
donpacino
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Yessir
 

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