Question 8 of 8: Signal-flow state model (Mason) and Nyquist stability

Question 8 — Signal-flow state model (Mason) and Nyquist stability [12 + 8]

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.

Part A — State equations & transfer function by Mason [12]

[Figure not reproduced: Figure Q8.1 redrawn — forward chain of three $1/s$ integrators; self-loops $-2,-1,-1$; an output feedforward of gain $2$ from $x_1$ to $Y$; and a long feedback of gain $-1$ from $x_3$ to $\dot x_1$. See the official exam paper.]

Approach. Read each $\dot x_i$ node as the sum of its incoming branch gains; then apply Mason’s rule with the integrator loops.

1. State equations (from the $\dot x_i$ nodes). $\dot x_1=-2x_1-x_3+u$; $\dot x_2=2x_1-x_2$; $\dot x_3=x_2-x_3$; output $y=2x_1+x_3$. Hence $$A=\begin{bmatrix}-2&0&-1\\2&-1&0\\0&1&-1\end{bmatrix},\ B=\begin{bmatrix}1\\0\\0\end{bmatrix},\ C=\begin{bmatrix}2&0&1\end{bmatrix}.$$
2. Loops (Mason). Three self-loops $L_1=-\tfrac2s$, $L_2=-\tfrac1s$, $L_3=-\tfrac1s$ (mutually non-touching) and the long feedback $L_4=(\tfrac1s)(2)(\tfrac1s)(1)(\tfrac1s)(-1)=-\tfrac{2}{s^3}$ (touches all).
3. Determinant. $\Delta=1-(L_1+L_2+L_3+L_4)+(L_1L_2+L_1L_3+L_2L_3)-L_1L_2L_3=1+\tfrac4s+\tfrac5{s^2}+\tfrac4{s^3}$; $\times s^3\Rightarrow s^3+4s^2+5s+4$ (the characteristic polynomial).
4. Forward paths. $P_1$: $U\!\to\!\dot x_1\!\to\!\cdots\!\to\!x_3\!\to\!Y=\tfrac{2}{s^3}$, $\Delta_1=1$ (touches all). $P_2$: $U\!\to\!\dot x_1\!\to\!x_1\!\to\!Y=\tfrac2s$, touching only $L_1,L_4$, so $\Delta_2=1-(L_2+L_3)+L_2L_3=1+\tfrac2s+\tfrac1{s^2}$.
5. Transfer function. $G=\dfrac{P_1\Delta_1+P_2\Delta_2}{\Delta}$; multiplying through by $s^3$,$$\boxed{G(s)=\frac{2s^2+4s+4}{s^3+4s^2+5s+4}=\frac{2(s^2+2s+2)}{s^3+4s^2+5s+4}.}$$ (Identical to $C(sI-A)^{-1}B$, confirming the reading.)

Part B — Nyquist stability [8]

[Figure not reproduced: Figure Q8.2 redrawn — figure-eight Nyquist. The critical point $-1$ lies inside the right lobe, which is traced clockwise ; the negative-real-axis crossing (waist) is at $\approx-1.7$. See the official exam paper.]

1. Count the encirclements. Only the right lobe encloses $-1$, and its arrows (top moving right, bottom moving left) show it is traced clockwise: $N=+1$ CW encirclement of $-1$.
2. Apply the criterion. For a clockwise $\Gamma$-path, $Z=N+P$ with $P=1$ open-loop RHP pole: $Z=1+1=\boxed{2}$. Since $Z\ne0$, the closed loop is $\boxed{\text{unstable}}$.
3. Closed-loop pole count. The system is 4th order (3 LHP + 1 RHP open-loop poles $\Rightarrow$ 4 closed-loop poles). With $Z=2$ in the RHP, there are $\boxed{2\text{ in the LHP and }2\text{ in the RHP}}$.
4. Gain margin. The plot crosses the negative real axis (phase crossover) at $\approx-1.7$, so $|L(j\omega_{pc})|=1.7$ and $\text{GM}=\dfrac{1}{1.7}=\boxed{0.59\ \text{V/V}=-4.6\ \text{dB}}$. The $\lt1$ (negative-dB) margin is the signature of a conditionally stable, open-loop-unstable loop: reducing the gain to $\approx59\%$ would push $-1$ out of the lobe and change the encirclement count.