# Control Systems Engineering - Masons Rule Help

## 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 [Broken]

## 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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donpacino
Gold Member
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 $-\frac{C_{f}s}{1 - C_{f}s}$?

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