Question 3 of 8: Signal-Flow Graph — Mason's Gain Formula

Question 3: Signal-Flow Graph — 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. Six forward-chain gains $1,\,1/s,\,10,\,1/s,\,8,\,1/s$ from $R(s)$ to $Y(s)$ through nodes $n_1,\dots,n_5$; six labelled branches closing loops: $-3$ ($n_2\to n_1$), $-2$ ($n_4\to n_3$), $-2$ ($Y\to n_5$), $-1$ ($n_3\to n_1$), $-1$ ($Y\to n_1$), and $+2$ ($n_2\to n_5$, a FORWARD-direction shortcut, not a backward loop).

Find. Loop/path counts; Mason's formula in general form; the simplified $G(s)=Y(s)/R(s)$.

Approach. Identify every closed directed cycle in the graph (a "loop" must return to its own starting node), classify touching vs non-touching by shared nodes, identify every forward path from $R$ to $Y$, then apply Mason's Gain Formula.

1. Part 1) — loops, paths, and touching. Because the $+2$ branch runs $n_2\to n_5$ in the SAME (forward) direction as the main chain, it does not close a loop by itself — it creates a SECOND forward path, and it also closes ONE new loop when combined with the $Y\to n_1$ feedback branch. Enumerating every directed cycle: $$L_a\!:n_1{\to}n_2{\to}n_1\ (-3/s),\quad L_b\!:n_3{\to}n_4{\to}n_3\ (-2/s),\quad L_c\!:n_5{\to}Y{\to}n_5\ (-2/s),$$ $$L_d\!:n_1{\to}n_2{\to}n_3{\to}n_1\ (-10/s),\quad L_e\!:n_1{\to}\cdots{\to}Y{\to}n_1\ (-80/s^3),$$ $$L_f\!:n_1{\to}n_2{\to}n_5{\to}Y{\to}n_1\ (\text{via the }+2\text{ shortcut},\ -2/s^2).$$ That is 6 loops. Checking every pair by shared nodes, the non-touching pairs are $(L_a,L_b)$, $(L_a,L_c)$, $(L_b,L_c)$, $(L_b,L_f)$, $(L_c,L_d)$ — 5 non-touching pairs. The only mutually non-touching TRIPLE is $(L_a,L_b,L_c)$ — 1 non-touching triple. Forward paths: the main chain $P_1$ ($R\to n_1\to\cdots\to Y$, gain $80/s^3$) and the shortcut path $P_2$ using the $+2$ branch ($R\to n_1\to n_2\to n_5\to Y$, gain $2/s^2$) — 2 paths.

   Loops	Non-touching (2 at a time)	Non-touching (3 at a time)	Paths
   6	5	1	2

2. Part 2) — Mason's Gain Formula, general form. $$\boxed{G(s)=\frac{Y(s)}{R(s)}=\frac{P_1\Delta_1+P_2\Delta_2} {1-\left(\sum L_i\right)+\left(\sum L_jL_k\right)-\left(\sum L_lL_nL_m\right)}}$$ with $\Delta_k$ = the value of $\Delta$ with every loop touching path $k$ struck out.
3. Part 3) — substituting gains. $\sum L_i=L_a+\cdots+L_f =-\dfrac{17}{s}-\dfrac{2}{s^2}-\dfrac{80}{s^3}$; the non-touching-pair sum is $L_aL_b+L_aL_c+L_bL_c+L_bL_f+L_cL_d=\dfrac{36}{s^2}+\dfrac{4}{s^3}$; the lone triple is $L_aL_bL_c=-\dfrac{12}{s^3}$. So $$\Delta=1+\frac{17}{s}+\frac{38}{s^2}+\frac{96}{s^3}=\frac{s^3+17s^2+38s+96}{s^3}.$$ Path $P_1$ ($80/s^3$) touches every node in the graph, so $\Delta_1=1$. Path $P_2$ ($2/s^2$) touches nodes $\{n_1,n_2,n_5,Y\}$ only — the ONLY loop it doesn't touch is $L_b$ (which lives entirely on $\{n_3,n_4\}$) — so $\Delta_2=1-L_b=1+2/s=(s+2)/s$. Then $$P_1\Delta_1+P_2\Delta_2=\frac{80}{s^3}+\frac{2}{s^2}\cdot\frac{s+2}{s} =\frac{80+2(s+2)}{s^3}=\frac{2(s+42)}{s^3}.$$ Dividing by $\Delta$: $$\boxed{G(s)=\frac{2(s+42)}{s^3+17s^2+38s+96}}.$$ This EXACTLY matches the numerator $K(s+42)$ ($K=2$) that Question 4 quotes as "the transfer function obtained from the signal-flow graph in Question 3" — strong independent confirmation that the branch-tracing above is correct (the source's own denominator constants in Question 4 are rounded/illustrative for that question's step- and frequency-response figures; see the Question 4 Verify note).