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, December 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, $\zeta$–resonant-peak and phase-margin 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), lead/lag and PID design (Ch. 9–11), state space and controllability/observability (Ch. 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling, second-order correlations, pole placement and frequency-domain compensator design. All block diagrams, root loci, pole–zero maps, Bode and Nyquist plots, closed-loop magnitude curves and step responses below are redrawn as inline figures.
Reading the supplied design charts. The percent-overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$, the phase-margin chart uses $\Phi_m\approx100\,\zeta$, 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 from the exact transfer functions given in the paper; the supplied plots are used only to corroborate $\zeta$, $\Phi_m$ and the crossover frequencies. Where a printed figure’s hand-read tick differs from the exact value, the exact value governs and the gap is flagged.
Question 5 — State-space model, pole placement by state feedback [20]
Phase-variable layout. Denominator coefficients (negated) fill the bottom row, numerator coefficients form $C$: $$A=\begin{bmatrix}0&1&0\\0&0&1\\-30&-2&-10\end{bmatrix},\ B=\begin{bmatrix}0\\0\\1\end{bmatrix},\ C=[\,20\ 6\ 2\,],\ D=0.$$
Check. $C(sI-A)^{-1}B=\dfrac{2s^2+6s+20}{s^3+10s^2+2s+30}$ reproduces the given transfer function exactly.
Q5 controllable-canonical realization: three cascaded integrators; feedback gains $-30,-2,-10$ into the input summer, output taps $20,6,2$.
Item (2) — pole placement with zero step error
Given. desired poles $-20,\,-2\pm j2$. Find. $K$ and $\mathbf{k}$. Approach. the feedback $u=Kr-K\mathbf{k}^{T}\mathbf{x}$ modifies only the last row of $A$ (CCF), so match the desired characteristic coefficients, then set the scalar $K$ from the unity-DC-gain (zero-step-error) condition.
Closed-loop last row. With $A_{cl}=A-BK\mathbf{k}^{T}$ the characteristic polynomial is $s^3+(10+Kk_3)s^2+(2+Kk_2)s+(30+Kk_1)$. Matching: $$Kk_1=130,\quad Kk_2=86,\quad Kk_3=14.$$
Zero step error fixes $K$. The closed-loop TF is $\dfrac{K(2s^2+6s+20)}{s^3+24s^2+88s+160}$; unity DC gain requires $\dfrac{20K}{160}=1$, so $\boxed{K=8}$.
Feedback vector. Divide by $K$: $$\boxed{\mathbf{k}=\begin{bmatrix}130/8\\86/8\\14/8\end{bmatrix}=\begin{bmatrix}16.25\\10.75\\1.75\end{bmatrix}.}$$ A direct eigenvalue check of $A-BK\mathbf{k}^{T}$ returns $\{-20,-2\pm j2\}$ and the DC gain is $1$ — both specs met.
Desired closed-loop pole map: dominant pair $-2\pm j2$ ($\zeta=0.707$) plus a real pole at $-20$ well to the left.
