Question 8 of 8: Canonical forms, signal-flow graph, and transfer function by inspection
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 07-Elec-A2 Systems & Control, December 2015, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula/notes sheet; a Laplace-transform table and the standard ζ–overshoot / ζ–resonant-peak design plots are supplied). Questions 1 and 2 are compulsory; a complete paper is five questions, so candidates choose 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) — time response and second-order specifications (Ch. 4), block/signal-flow reduction and Mason’s rule (Ch. 5), Routh–Hurwitz stability (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID / lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain and phase margins (Ch. 10–11), state space, controllability/observability and canonical forms (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and Mason’s gain formula; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci, the Nyquist plot, the network schematic and the signal-flow graph below are redrawn as inline figures.
Reading the exam figures. Every transfer function in this paper is stated exactly, so every boxed result is confirmed analytically and the exact model governs. Q2 supplies a Bode plot ($G_m=32.2$ dB at $3.9$ rad/s, $P_m=-145^\circ$ at $0.324$ rad/s) and a root locus, and Q7 supplies the uncompensated Bode plot; values read from those curves are flagged as reads and cross-checked against the Routh/analytic results.
Question 8 — Canonical forms, signal-flow graph, and transfer function by inspection [10 + 10]
Part A — Signal-flow graph and state equations [10]
Approach. The three cascaded integrators produce the phase variables; the denominator coefficients become feedback gains into the first summing node and the numerator coefficients become feedforward taps to the output — the controllable (phase-variable) canonical form.
Figure Q8.1 completed — forward chain of three $1/s$ integrators ($x_3\to x_2\to x_1$); feedback gains $-2,-4,-1$ (denominator $s^3+2s^2+4s+1$) into the input node; feedforward taps $5,3,1$ (numerator $5s^2+3s+1$) to $Y$.
Assign phase variables. Take $x_1$ as the output of the last integrator, $x_2$ the middle, $x_3$ the first: $\dot x_1=x_2$, $\dot x_2=x_3$.
Feedback into the first node. The denominator $s^3+2s^2+4s+1$ gives $\dot x_3=-1\,x_1-4\,x_2-2\,x_3+u$ (feedback gains $-a_0,-a_1,-a_2=-1,-4,-2$).
Output feedforward. The numerator $5s^2+3s+1$ gives $y=1\,x_1+3\,x_2+5\,x_3$ (taps $b_0,b_1,b_2=1,3,5$).
State-space matrices. $$A=\begin{bmatrix}0&1&0\\0&0&1\\-1&-4&-2\end{bmatrix},\ B=\begin{bmatrix}0\\0\\1\end{bmatrix},\ C=[\,1\ 3\ 5\,],\ D=0.$$ Check: $C(sI-A)^{-1}B=\dfrac{5s^2+3s+1}{s^3+2s^2+4s+1}=G(s)$. This is the $\boxed{\text{controllable (phase-variable) canonical form}}$.
Part B — Transfer function by inspection [10]
Approach. In observable canonical form the last column of $A$ holds the negated denominator coefficients, $B$ holds the numerator coefficients, and $G(s)=C(sI-A)^{-1}B+D$ reads off directly.
Denominator. The last column $(-4,-6,-8)^\top$ gives $a_0=4,a_1=6,a_2=8$, so the characteristic polynomial is $s^3+8s^2+6s+4$.
Strictly-proper numerator. $B=(8,-5,-3)^\top$ maps to $b_0=8,b_1=-5,b_2=-3$, giving the strictly-proper part $\dfrac{-3s^2-5s+8}{s^3+8s^2+6s+4}$.
Add the direct term. With $D=1$, $$G(s)=1+\frac{-3s^2-5s+8}{s^3+8s^2+6s+4}=\boxed{\frac{s^3+5s^2+s+12}{s^3+8s^2+6s+4}}.$$ Direct evaluation of $C(sI-A)^{-1}B+D$ confirms this exactly.
