Question 6 of 8: Controllability/Observability vs. Parameter α, and the Transfer Function by Mason's Gain Formula

Question 6: Controllability/Observability vs. Parameter α, and the Transfer Function by 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. $A=\begin{bmatrix}0&0&0\\1&0&0\\0&1-2\alpha&-2\end{bmatrix}$, $B=[1,1,0]^T$, $C=[0,\alpha,1]$.

Find. 1) Controllability and observability, and any $\alpha$-condition. 2) $G(s)=Y(s)/U(s)$, via Mason's Gain Formula on the state diagram.

Approach. Build $M_c=[B\ AB\ A^2B]$ and $M_o=[C;CA;CA^2]$ and test their determinants for singularity as functions of $\alpha$. For the transfer function, lay the state equations out as a signal-flow graph (an integrator chain with the extra direct feedthrough branches that $B=[1,1,0]^T$ creates) and apply Mason's Gain Formula: enumerate every forward path from $U$ to $Y$ and the one feedback loop, then combine with the loop's cofactors.

1. Part 1) — controllability. $$M_c=[B\ \ AB\ \ A^2B]=\begin{bmatrix}1&0&0\\1&1&0\\0&1-2\alpha&2\alpha-1\end{bmatrix},\qquad \det M_c=2\alpha-1.$$ $\boxed{\text{Controllable}\iff\alpha\ne\tfrac12}$ — controllability DOES depend on $\alpha$, failing at exactly $\alpha=0.5$.
2. Part 1) — observability. $$M_o=\begin{bmatrix}C\\CA\\CA^2\end{bmatrix}= \begin{bmatrix}0&\alpha&1\\\alpha&1-2\alpha&-2\\1-2\alpha&4\alpha-2&4\end{bmatrix},\qquad \det M_o=-1.$$ The determinant is a nonzero CONSTANT, independent of $\alpha$: $\boxed{\text{the system is observable for every value of }\alpha}$ (observability does NOT depend on $\alpha$).
3. Part 2) — state diagram and forward paths. The state equations, read directly off $A,B$, give a chain of three integrators with TWO direct input branches (because $B$'s first two entries are both 1): $\dot x_1=u$ (branch $U\to X_1$, gain 1), $\dot x_2=x_1+u$ (branches $X_1\to X_2$ AND $U\to X_2$, both gain 1), and $\dot x_3=(1-2\alpha)x_2-2x_3$ (a branch $X_2\to X_3$ of gain $1-2\alpha$, plus a SELF-LOOP at $X_3$ of gain $-2$ from the $-2x_3$ term). The output taps $y=\alpha x_2+x_3$ (branches $X_2\to Y$ gain $\alpha$, $X_3\to Y$ gain $1$). There are FOUR forward paths from $U$ to $Y$: $$P_1{:}\,U{\to}X_1{\to}X_2{\to}Y=\frac{\alpha}{s^2},\quad P_2{:}\,U{\to}X_1{\to}X_2{\to}X_3{\to}Y=\frac{1-2\alpha}{s^3},$$ $$P_3{:}\,U{\to}X_2{\to}Y=\frac{\alpha}{s},\quad P_4{:}\,U{\to}X_2{\to}X_3{\to}Y=\frac{1-2\alpha}{s^2}.$$ The ONE loop is the self-loop at $X_3$: $L_1=-2/s$, so $\Delta=1-L_1=\dfrac{s+2}{s}$. Paths $P_1,P_3$ never touch $X_3$, so their cofactor keeps the loop: $\Delta_1=\Delta_3=\Delta$. Paths $P_2,P_4$ pass through $X_3$, touching the only loop, so $\Delta_2=\Delta_4=1$.
4. Part 2) — assemble via Mason's Gain Formula. $$G(s)=\frac{P_1\Delta_1+P_2\Delta_2+P_3\Delta_3+P_4\Delta_4}{\Delta} =\frac{\left[\tfrac{\alpha}{s^2}+\tfrac{\alpha}{s}\right]\tfrac{s+2}{s} +\left[\tfrac{1-2\alpha}{s^3}+\tfrac{1-2\alpha}{s^2}\right]}{(s+2)/s}.$$ Both bracketed sums factor as $[\alpha(s+2)+(1-2\alpha)]\cdot\tfrac{s+1}{s^3}=(\alpha s+1)\cdot \tfrac{s+1}{s^3}$ (the $2\alpha$ terms cancel), so $$G(s)=\frac{(\alpha s+1)(s+1)/s^3}{(s+2)/s}=\boxed{\frac{(s+1)(\alpha s+1)}{s^2(s+2)}}.$$ Computing $C(sI-A)^{-1}B$ independently via the matrix formula gives a result IDENTICAL to this Mason's-formula result for symbolic $\alpha$ — a strong cross-check on both methods (which also reveals the numerator always factors with a fixed zero at $s=-1$, regardless of $\alpha$).