Question 3 of 8: Signal Flow Diagrams — Mason’s Gain Formula (20 marks)

Question 3: Signal Flow Diagrams — Mason’s Gain Formula (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The multi-feedback signal flow graph of Figure Q3.1, redrawn below. Reading the graph left to right, the forward chain is $R\to n_1\to n_2\to n_3\to n_4\to n_5\to Y$ with branch gains $1,\ 1/s,\ 10,\ 1/s,\ 8,\ 1/s$. A feedforward branch of gain 2 runs from $n_2$ to $n_5$. Five feedback branches close the graph: $-3$ from $n_2$ to $n_1$, $-2$ from $n_4$ to $n_3$, $-2$ from $Y$ to $n_5$, $-1$ from $n_4$ to $n_1$, and $-1$ from $Y$ to $n_1$.

Find. (1) the loop and path counts; (2) Mason’s formula in general symbols; (3) the transfer function reduced to a ratio of polynomials.

[Figure not reproduced: Figure S3.1 - Redrawn Figure Q3.1 signal flow graph. Blue = feedforward gain 2 (node 2 to node 5); the five lower arcs are the feedback branches. See the official exam paper.]

Approach. Enumerate every closed loop and every forward path, work out which loops are mutually non-touching, build the graph determinant $\Delta$ and each cofactor $\Delta_j$, then apply Mason’s gain formula and clear the powers of $s$.

1. Identify the two forward paths. The direct chain and the branch that jumps the middle integrator via the feedforward gain 2:$$P_1=1\cdot\frac{1}{s}\cdot 10\cdot\frac{1}{s}\cdot 8\cdot\frac{1}{s}=\frac{80}{s^3},\qquad P_2=1\cdot\frac{1}{s}\cdot 2\cdot\frac{1}{s}=\frac{2}{s^2}.$$
2. Enumerate the closed loops. Tracing every cycle gives six:$$L_1=\frac{-2}{s}\ (Y,n_5),\quad L_2=\frac{-80}{s^3}\ (\text{full chain}),\quad L_3=\frac{-2}{s^2}\ (n_1,n_2,n_5,Y),$$$$L_4=\frac{-10}{s^2}\ (n_1..n_4),\quad L_5=\frac{-3}{s}\ (n_1,n_2),\quad L_6=\frac{-2}{s}\ (n_3,n_4).$$
3. Find the non-touching combinations. Two loops are non-touching when they share no node. Checking all pairs leaves five: $(L_1,L_4)$, $(L_1,L_5)$, $(L_1,L_6)$, $(L_3,L_6)$ and $(L_5,L_6)$. Only one triple is mutually non-touching, $(L_1,L_5,L_6)$. This answers the table:

   How many loops?	Non-touching, 2 at a time?	Non-touching, 3 at a time?	How many paths?
   6	5	1	2

4. Write Mason’s formula in general terms (part 2, no values yet):$$G(s)=\frac{P_1\Delta_1+P_2\Delta_2}{\Delta},\qquad \Delta=1-\sum_i L_i+\sum_{j,k}L_jL_k-\sum_{l,n,m}L_lL_nL_m,$$with $\Delta_j$ the determinant of the sub-graph left after deleting every loop that touches path $j$.
5. Assemble the determinant. Summing the loops, the non-touching pairs and the single triple,$$\sum L_i=-\frac{7}{s}-\frac{12}{s^2}-\frac{80}{s^3},\qquad \sum L_jL_k=\frac{16}{s^2}+\frac{24}{s^3},\qquad \sum L_lL_nL_m=-\frac{12}{s^3},$$$$\Delta=1+\frac{7}{s}+\frac{28}{s^2}+\frac{116}{s^3}.$$
6. Evaluate the path cofactors. $P_1$ touches every node, so $\Delta_1=1$. $P_2$ misses $n_3$ and $n_4$, and the only loop confined to those nodes is $L_6$, hence $\Delta_2=1-L_6=1+\dfrac{2}{s}$.
7. Apply Mason and clear the denominators.$$G_{cl}(s)=\frac{\dfrac{80}{s^3}+\dfrac{2}{s^2}\left(1+\dfrac{2}{s}\right)}{1+\dfrac{7}{s}+\dfrac{28}{s^2}+\dfrac{116}{s^3}}=\frac{\dfrac{2}{s^2}+\dfrac{84}{s^3}}{\Delta}.$$Multiplying numerator and denominator by $s^3$,$$\boxed{G_{cl}(s)=\frac{2s+84}{s^3+7s^2+28s+116}=\frac{2(s+42)}{s^3+7s^2+28s+116}}$$This is exactly the transfer function Question 4 quotes, $2(s+42)/[(s+5.67)(s^2+1.33s+20.46)]$ — expanding that product gives $s^3+7s^2+28.0011s+116.0082$, matching to the rounding of the printed factors. The agreement is a complete check on the graph reading.