# Control Systems Engineering - Masons Rule Help

1. Feb 1, 2014

### GreenPrint

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

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 [Broken]

2. Relevant equations

3. 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.

Last edited by a moderator: May 6, 2017
2. Feb 1, 2014

### donpacino

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. Feb 1, 2014

### GreenPrint

Is it really just $-\frac{C_{f}s}{1 - C_{f}s}$?

4. Feb 3, 2014

### donpacino

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

so vload/-Idist=1/(CFS)

5. Feb 3, 2014

### GreenPrint

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

6. Feb 4, 2014

### donpacino

Yessir

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