Control Systems Engineering - Masons Rule Help

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The discussion focuses on using Mason's rule to find the transfer function \(\frac{V_{Load}(s)}{I_{Dist}(s)}\). The user successfully calculated \(\frac{V_{Load}}{I_{Cf}(s)}\) but is uncertain about handling the flow diagram, particularly regarding the loops and the presence of a negative sign in front of \(I_{Dist}\). It is confirmed that there is only one forward path, leading to a gain of \(1 \cdot \frac{1}{C_{f}s} \cdot 1\). The final expression for the transfer function is clarified as \(-\frac{C_{f}s}{1 - C_{f}s}\), with the understanding that the gain is derived from the forward path alone. The discussion concludes with the affirmation that the opposite magnitude would yield a positive transfer function.
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Homework Statement



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

http://imageshack.com/a/img31/4299/zr5.PNG

Homework Equations





The Attempt at a Solution



There's only one forward path, 1*\frac{1}{C_{f}s}*1. Am I supposed to consider the loops to the left of I_{Dist} 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 I_{Dist}. Thanks for any help.
 
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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.
 
Is it really just -\frac{C_{f}s}{1 - C_{f}s}?
 
well there is no loops in the circuits. so the gain would just be the forware gain.

so vload/-Idist=1/(CFS)
 
So if I wanted positive Vload/Idist it would just be the opposite magnitude, -1/(CFS)?
 
Yessir
 

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