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22-Elec-A2 Systems and Control · May 2015

Question 7 of 8: State space: controllability, pole placement, closed-loop transfer function

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, 07-Elec-A2 Systems & Control, May 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 design plots are supplied with the paper). 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 pole placement (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, Bode plots, signal-flow graphs and the Nyquist plot below are redrawn as inline figures.

Reading the exam figures. Where a transfer function is stated exactly (Q1–Q5, Q7, Q8A) every result is confirmed analytically and the exact model governs. Q4 supplies a Bode plot ($G_m=16.1$ dB at $2.05$ rad/s, $P_m=-85.1^\circ$ at $0.589$ rad/s) and a root locus; Q5 Parts A–B and Q6 use printed frequency-response plots, and Q8B a printed Nyquist diagram — values read from those curves are flagged as reads and cross-checked against the analytics wherever a transfer function is available.

Question 7 — State space: controllability, pole placement, closed-loop transfer function [8 + 7 + 5]

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$$B$$C$
$\begin{bmatrix}1&1\\1&1\end{bmatrix}$$\begin{bmatrix}1\\0\end{bmatrix}$$\begin{bmatrix}2&-1\end{bmatrix}$

Find. Controllability/observability; $K_1,K_2$ (and any reference gain for zero error); $G_{cl}(s)$.

Part 1 — Controllability & observability [8]

  1. Controllability. $AB=\begin{bmatrix}1\\1\end{bmatrix}$, so $\mathcal{C}=[B\ AB]=\begin{bmatrix}1&1\\0&1\end{bmatrix}$, $\det\mathcal{C}=1\ne0\Rightarrow\boxed{\text{controllable}}$.
  2. Observability. $CA=\begin{bmatrix}1&1\end{bmatrix}$, so $\mathcal{O}=\begin{bmatrix}C\\CA\end{bmatrix}=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$, $\det\mathcal{O}=3\ne0\Rightarrow\boxed{\text{observable}}$.
  3. Note. Open-loop poles are the eigenvalues of $A$: $\det(sI-A)=s(s-2)$, i.e. $0$ and $+2$ — the plant is unstable and needs feedback.

Part 2 — State-feedback pole placement [7]

  1. Closed-loop matrix. $A_{cl}=A-BK=\begin{bmatrix}1-K_1&1-K_2\\1&1\end{bmatrix}$.
  2. Characteristic polynomial. $\det(sI-A_{cl})=s^2+(K_1-2)s+(K_2-K_1)$.
  3. Match $(s+2)(s+3)=s^2+5s+6$. $K_1-2=5\Rightarrow\boxed{K_1=7}$; $K_2-K_1=6\Rightarrow\boxed{K_2=13}$.
  4. Zero steady-state error. The DC gain of the placed system is $-CA_{cl}^{-1}B=-0.5\ne1$, so a reference scaling is required: $u=\bar Nr-K_1x_1-K_2x_2$ with $\bar N=\dfrac{1}{-CA_{cl}^{-1}B}=\boxed{-2}$. This makes $y_{ss}=r$ exactly.

Part 3 — Closed-loop transfer function [5]

  1. Compute $G_{cl}=C(sI-A_{cl})^{-1}B$. With $A_{cl}=\begin{bmatrix}-6&-12\\1&1\end{bmatrix}$ (from $K=[7,13]$), $\det(sI-A_{cl})=(s+2)(s+3)$ and $$G_{cl}(s)=\frac{2s-3}{(s+2)(s+3)}=\frac{2s-3}{s^2+5s+6}.$$
  2. Interpretation. Poles at $-2,-3$ as designed; a right-half-plane zero at $s=+1.5$ (non-minimum phase) and $G_{cl}(0)=-0.5$ — hence the reference gain $\bar N=-2$ needed above to reach unity DC gain.
Closed loop poles -2,-3; RHP zero at +1.50ReIm
Closed-loop poles $-2,-3$ (placed) with the non-minimum-phase zero at $+1.5$; a reference gain $\bar N=-2$ restores unity DC gain.
QuantityResult
Controllable / observableyes ($\det=1$) / yes ($\det=3$)
$K_1$ / $K_2$$7$ / $13$
Reference gain $\bar N$$-2$ (for zero step error)
$G_{cl}(s)$$\dfrac{2s-3}{s^2+5s+6}$ (zero at $+1.5$)