Question 6 of 8: Controllability, observability, transfer function by Mason
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 16-Elec-A2 Systems & Control, May 2017, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula sheet; a Laplace-transform table and the standard $\zeta$–overshoot and $\zeta$–resonant-peak design plots are supplied on pages 2–3). Questions 1 and 2 are compulsory; a complete paper is five questions, so a candidate chooses three of Q3–Q8. Each question is worth 20 marks. All eight are worked below so the set is a complete study resource.
Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — Routh–Hurwitz (Ch. 6), root locus (Ch. 8), steady-state error and static error constants (Ch. 7), frequency response, Nyquist, gain/phase margins (Ch. 10), PID and compensator design (Ch. 9–11), state space, controllability/observability and Mason’s rule (Ch. 3, 5, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling, second-order correlations and pole-placement. All block diagrams, root loci, pole–zero maps, Nyquist plots, step responses and the state-realization diagrams below are redrawn as inline figures.
Reading the supplied design charts. The percent-overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ and the second-order model is $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Every result here is derived analytically in the $s$-domain and cross-checked; the design charts are used only to corroborate $\zeta$ from $PO$.
Question 6 — Controllability, observability, transfer function by Mason [20]
Given. $A,B,C$ above with parameter $\alpha$. Find. $M_c,M_o$ and their rank conditions.
Controllability matrix. $AB=[0,\,1,\,1-2\alpha]^{T}$, $A^2B=[0,\,0,\,2\alpha-1]^{T}$, so $$M_c=[B\ AB\ A^2B]=\begin{bmatrix}1&0&0\\1&1&0\\0&1-2\alpha&2\alpha-1\end{bmatrix},\quad\det M_c=2\alpha-1.$$ Controllable iff $\det M_c\ne0$, i.e. $\boxed{\alpha\ne\tfrac12}$ — controllability does depend on $\alpha$.
Observability matrix. $CA=[\alpha,\,1-2\alpha,\,-2]$, $CA^2=[1-2\alpha,\,-2+4\alpha,\,4]$, so $$M_o=\begin{bmatrix}C\\CA\\CA^2\end{bmatrix}=\begin{bmatrix}0&\alpha&1\\ \alpha&1-2\alpha&-2\\ 1-2\alpha&-2+4\alpha&4\end{bmatrix},\quad\det M_o=-1.$$ Since $\det M_o=-1\ne0$ for every $\alpha$, the system is $\boxed{\text{observable for all }\alpha}$ — observability does not depend on $\alpha$.
Interpretation. At $\alpha=\tfrac12$ the mode associated with the eigenvalue $-2$ becomes uncontrollable (it drops out of $M_c$); the loss shows up in the transfer function of Item 2 as a pole–zero cancellation.
Figure Q6.1 — state diagram (two chained integrators feeding a third with self-loop $-2$). Mason’s rule on the forward paths $u\to y$ gives $G(s)$ directly.
Item (2) — transfer function via Mason
Given. the state diagram above. Find. $G(s)$. Approach. solve $(sI-A)x=B$ by forward substitution (the structure is a chain), then $y=Cx$ — identical to summing Mason forward paths over the loop determinant.
Output. $y=\alpha x_2+x_3=\dfrac{s+1}{s^2}\left[\alpha+\dfrac{1-2\alpha}{s+2}\right]=\dfrac{(s+1)\,[\alpha(s+2)+(1-2\alpha)]}{s^2(s+2)}.$ The bracket simplifies to $\alpha s+1$.
Transfer function. $$\boxed{G(s)=\frac{(s+1)(\alpha s+1)}{s^2(s+2)}=\frac{\alpha s^2+(1+\alpha)s+1}{s^3+2s^2}.}$$ At $\alpha=\tfrac12$ the numerator is $\tfrac12(s+1)(s+2)$, cancelling the pole at $-2$ (the uncontrollable mode) and leaving $G=\dfrac{0.5(s+1)}{s^2}$ — consistent with Item 1.
Q6 transfer-function map (shown for $\alpha=0.4$): double pole at the origin, real pole $-2$, and zeros at $-1$ and $-1/\alpha$; at $\alpha=\tfrac12$ a zero lands on $-2$ and cancels that pole.
