Control Systems Engineering - Masons Rule Help

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Discussion Overview

The discussion revolves around finding the transfer function \(\frac{V_{Load}(s)}{I_{Dist}(s)}\) using Mason's rule in the context of control systems engineering. Participants are addressing a homework problem that involves analyzing a flow diagram and determining the relationships between various variables.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant notes that there is only one forward path and calculates the gain as \(1 \cdot \frac{1}{C_{f}s} \cdot 1\), questioning whether the loops to the left of \(I_{Dist}\) should be considered non-touching loops.
  • Another participant confirms the presence of only one forward path and agrees with the gain calculation, pointing out a negative sign in the diagram.
  • A participant proposes a potential expression for the transfer function as \(-\frac{C_{f}s}{1 - C_{f}s}\).
  • Another participant asserts that there are no loops in the circuit, suggesting that the gain would simply be the forward gain, leading to the expression \(v_{load}/-I_{dist} = 1/(C_{f}s)\).
  • A follow-up question is raised about obtaining a positive expression for \(\frac{V_{Load}}{I_{Dist}}\), suggesting it would be the opposite magnitude, \(-1/(C_{f}s)\).
  • A participant confirms the last statement with "Yessir".

Areas of Agreement / Disagreement

Participants generally agree on the presence of a single forward path and the calculation of the gain, but there is uncertainty regarding the interpretation of the negative sign and the correct formulation of the transfer function. The discussion remains unresolved regarding the final expression for \(\frac{V_{Load}(s)}{I_{Dist}(s)}\).

Contextual Notes

There are limitations in the discussion regarding the assumptions about the flow diagram and the interpretation of signs, which may affect the final expressions derived. The presence of non-touching loops is also not clearly established.

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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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