NivaarExam PrepOfficial exam papers ↗

22-Elec-A2 Systems and Control · May 2016

Question 3 of 8: State space vs. transfer function; controllability, observability, steady-state error

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 2016, 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 ζ–overshoot and 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 specs (Ch. 4), block reduction (Ch. 5), Routh–Hurwitz (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID and lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain/phase margins (Ch. 10–11), state space, controllability/observability (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling. All step responses, block diagrams, pole–zero maps, root loci, the Bode sketches and the Nyquist plot 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 $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$; the figures on the exam paper (step responses, root loci, Bode plots) are read graphically, so the values below carry normal chart-reading tolerance.

Question 3 — State space vs. transfer function; controllability, observability, steady-state error [5 + 5 + 10]

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$$D$$H(s)$
$\begin{bmatrix}0&1\\-1&-1\end{bmatrix}$$\begin{bmatrix}0\\1\end{bmatrix}$$\begin{bmatrix}1&1\end{bmatrix}$$1$$\dfrac{1}{s+1}$

Find. $L(s)=K_pG_p(s)H(s)$; controllability/observability; $K_p$ range.

R(s)−+KpProcessx=Ax+Bu, y=Cx+DuY(s)H(s) = 1/(s+1)Figure Q3.1 — closed loop with dynamic (1/(s+1)) feedback
Figure Q3.1: closed loop with $K_p$, state-space process and dynamic feedback $1/(s+1)$.

Part (1) — open-loop transfer function

  1. Process transfer function. $G_p(s)=C(sI-A)^{-1}B+D$. With $\det(sI-A)=s^2+s+1$, $C(sI-A)^{-1}B=\dfrac{1+s}{s^2+s+1}$, so $G_p(s)=\dfrac{1+s}{s^2+s+1}+1=\boxed{\dfrac{s^2+2s+2}{s^2+s+1}}$ (note $D=1$ makes it non-strictly-proper).
  2. Open loop of the system shown. Multiply by the controller and the feedback element: $L(s)=K_pG_p(s)H(s)=\boxed{\dfrac{K_p\,(s^2+2s+2)}{(s^2+s+1)(s+1)}}$.

Part (2) — controllability and observability

  1. Controllability. $\mathcal{C}=[B\ AB]=\begin{bmatrix}0&1\\1&-1\end{bmatrix}$, $\det\mathcal{C}=-1\neq0\Rightarrow$ controllable.
  2. Observability. $\mathcal{O}=\begin{bmatrix}C\\CA\end{bmatrix}=\begin{bmatrix}1&1\\-1&0\end{bmatrix}$, $\det\mathcal{O}=1\neq0\Rightarrow$ observable.

Part (3) — gain range

  1. Characteristic equation. $1+L(s)=0\Rightarrow(s^2+s+1)(s+1)+K_p(s^2+2s+2)=0$, i.e. $s^3+(2+K_p)s^2+(2+2K_p)s+(1+2K_p)=0$.
  2. (a) Stability. The Routh $s^1$ element numerator is $2K_p^2+4K_p+3\gt0$ for all $K_p$, and the $s^0$ term needs $K_p\gt-0.5$; thus the loop is $\boxed{\text{stable for all }K_p\gt0}$.
  3. (b) Steady-state error. Type-0 loop with $H(0)=1$, so $K_{pos}=K_p\,G_p(0)\,H(0)=K_p\cdot2\cdot1=2K_p$ and $e_{ss(step)}=\dfrac{1}{1+2K_p}$. Requiring $\le0.05$: $1+2K_p\ge20\Rightarrow\boxed{K_p\ge9.5}$.
QuantityValue
$G_p(s)$$(s^2+2s+2)/(s^2+s+1)$
$L(s)$$K_p(s^2+2s+2)/[(s^2+s+1)(s+1)]$
Controllable / observableyes / yes
Stable rangeall $K_p\gt0$
$e_{ss}\le5\%$$K_p\ge9.5$