Question 5 of 8: State-space model, pole placement by state feedback
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 5 — State-space model, pole placement by state feedback [20]
Phase-variable (CCF) layout. Denominator coefficients (negated) fill the bottom feedback row, numerator coefficients form $C$: $$A=\begin{bmatrix}0&1&0\\0&0&1\\-30&-2&-8\end{bmatrix},\ B=\begin{bmatrix}0\\0\\1\end{bmatrix},\ C=[\,2\ 5\ 2\,],\ D=0.$$
Check. $C(sI-A)^{-1}B=\dfrac{2s^2+5s+2}{s^3+8s^2+2s+30}$ reproduces the given transfer function exactly.
Q5 controllable-canonical realization: three cascaded integrators; feedback gains $-30,-2,-8$ into the input summer, output taps $2,5,2$.
Item (2) — pole placement with zero step error
Given. desired poles $-10,\,-2\pm j3$. Find. $K$ and $\mathbf{k}$. Approach. the feedback $u=Kr-Kk^{T}x$ modifies only the last row of $A$ (CCF), so match coefficients of the desired characteristic polynomial, then set the scalar $K$ from the unity-DC-gain (zero-step-error) condition.
Closed-loop last row. With $A_{cl}=A-BKk^{T}$ the characteristic polynomial is $s^3+(8+Kk_3)s^2+(2+Kk_2)s+(30+Kk_1)$. Matching: $$Kk_1=100,\quad Kk_2=51,\quad Kk_3=6.$$
Zero step error fixes $K$. The closed-loop TF is $\dfrac{K(2s^2+5s+2)}{s^3+14s^2+53s+130}$; unity DC gain requires $\dfrac{2K}{130}=1$, so $\boxed{K=65}$.
Feedback vector. Divide by $K$: $$\boxed{\mathbf{k}=\begin{bmatrix}100/65\\51/65\\6/65\end{bmatrix}=\begin{bmatrix}1.538\\0.785\\0.0923\end{bmatrix}.}$$ A direct eigenvalue check of $A-BKk^{T}$ returns $\{-10,-2\pm j3\}$, and the DC gain is $1$ — both specs met.
Desired closed-loop pole map: dominant pair $-2\pm j3$ ($\zeta=0.55$) plus a real pole at $-10$ well to the left.
