6PM lecture. Makeup lecture.
On Monday there will be extra office hrs. Exam on Tuesday.
Example. Lets say we have \(X_{1},X_{2}\) as variables, and \(U\) as input and \(Y\) as output. Then given\begin{align*} X_{1}+\alpha X_{2} & =U\\ \beta X_{1}-3X_{2} & =3U\\ Y & =X_{1}+2X_{2} \end{align*}
The goal is to solve for \(Y\) in terms of \(U\) without all the variables \(X_{1},X_{2}\) involved. This can be solved of course using algebra:
Using Mason method, we first rewrite the equations so that the variables are on the LHS. In the above, this becomes\begin{align*} X_{1} & =U-\alpha X_{2}\\ X_{2} & =\beta X_{1}-2X_{2}-3U \end{align*}
Next, we first set up a signal graph. Each variable becomes a node, like this
Note: At each node, all incoming branch gains are added. We now setup the loop gains. A loop must not visit a node more than once. There are two loops here. The gains on them are \(\left \{ -\alpha \beta ,-2\right \} \). Next, we find all the forward paths from \(U\) to \(Y\). A forward path must not visit same node more than once. There are 4 forward paths. The gains on each are
\begin{align*} M_{1} & =\left ( -3\right ) \left ( 2\right ) =-6\\ M_{2} & =\left ( 1\right ) \left ( \beta \right ) \left ( 2\right ) =2\beta \\ M_{3} & =\left ( 1\right ) \left ( 1\right ) =1\\ M_{4} & =\left ( -3\right ) \left ( -\alpha \right ) \left ( 1\right ) =3\alpha \end{align*}
Now we defined Mason delta \(\Delta \)\[ \Delta =1-\sum \text{loop gains}+\sum \text{loop gains 2 at times}-\sum \text{loop gains 3 at times}\cdots \] In the above, when looking for loop gains 2 at times, the loops must not be sharing a node. Same for loop gains 3 at times and higher sums. In our example, this gives\begin{align*} \Delta & =1-\sum \text{loop gains}\\ & =1-\left ( -\alpha \beta -2\right ) \\ & =3+\alpha \beta \end{align*}
Finally, we define \(\Delta _{i}\), which is Mason \(\Delta \) but with the forward path \(M_{i}\) removed from the graph. There are 4 forward paths in this problem, so there are \(\Delta _{1},\Delta _{2},\Delta _{3},\Delta _{4}\). Each time we remove a forward path, we find \(\Delta \) again using the above Mason rule method. In this problem we see that\begin{align*} \Delta _{1} & =1\\ \Delta _{2} & =1\\ \Delta _{3} & =1-\left ( -2\right ) =3\\ \Delta _{4} & =1 \end{align*}
Finally, we apply the Mason gain formula\begin{align*} \frac{Y}{U} & =\frac{\sum _{i=1}^{4}M_{i}\Delta _{1}}{\Delta }\\ & =\frac{\left ( -6\right ) \left ( 1\right ) +\left ( 2\beta \right ) \left ( 1\right ) +\left ( 1\right ) \left ( 3\right ) +\left ( 3\alpha \right ) \left ( 1\right ) }{3+\alpha \beta }\\ & =\frac{\alpha +2\beta -3}{3+\alpha \beta } \end{align*}
Reader Find \(\frac{Y}{U}\) for this graph
Second example. Here we take a circuit and obtain the equations, then use signal graph in order to use Mason rule to obtain the transfer function
Solving the circuit loops gives (all in Laplace domain)\begin{align*} \left ( R_{1}+sL\right ) I_{1}-I_{2}Ls-V_{in}\left ( s\right ) & =0\\ \left ( R_{2}+\frac{1}{Cs}\right ) I_{2}+LsI_{2}-I_{1}Ls & =0\\ V_{out}\left ( s\right ) & =R_{2}I_{2} \end{align*}
Now the variables are \(I_{1},I_{2}\), so we need to have these on the LHS. To do this, do this trick: Add \(I_{1}\) to each side of the first equation, and add \(I_{2}\) to each side of the second equation, this gives\begin{align*} I_{1} & =\left ( R_{1}+sL\right ) I_{1}-I_{2}Ls-V_{in}\left ( s\right ) +I_{1}\\ I_{2} & =I_{2}+\left ( R_{2}+\frac{1}{Cs}\right ) I_{2}+LsI_{2}-I_{1}Ls \end{align*}
Now set up the signal graph
Reader: Find \(\frac{V_{out}}{V_{in}}\) for the above.\begin{align*} \frac{V_{out}}{V_{in}} & =\frac{\sum _{i=1}^{1}M_{i}\Delta _{i}}{1-\sum \text{one at time}+\sum \text{2 at times}}\\ & =\frac{\left ( -1\right ) \left ( -Ls\right ) \left ( R_{2}\right ) }{1-\sum \left ( R_{1}+Ls+1\right ) +\left ( \frac{1}{Cs}+R_{2}+Ls+1\right ) +\sum \left ( R_{1}+Ls+1\right ) \left ( \frac{1}{Cs}+R_{2}+Ls+1\right ) }\\ & =\frac{LsR_{2}}{1-\left ( R_{1}+R_{2}+\frac{1}{Cs}+2Ls+2\right ) +\left ( R_{1}+Ls+1\right ) \left ( R_{2}+\frac{1}{Cs}+Ls+1\right ) }\\ & =\frac{LsR_{2}}{\frac{1}{Cs}\left ( R_{1}+Ls\right ) \left ( CLs^{2}+CR_{2}s+1\right ) } \end{align*}
Now we take a block diagram and convert to signal graph. Given
We know that \(\frac{Y\left ( s\right ) }{R\left ( s\right ) }=\frac{H\left ( s\right ) G\left ( s\right ) }{1+H\left ( s\right ) G\left ( s\right ) }\) and that \(\frac{E\left ( s\right ) }{R\left ( s\right ) }=\frac{1}{1+H\left ( s\right ) G\left ( s\right ) }\). Use Mason to show the above.
Reader Find \(\frac{Y}{U}\) for this
