Question 7 of 8: State space: transfer function, controllability/observability, pole placement
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 2013, 3 hours, closed book (approved calculator + one handwritten formula sheet, and a printed short table of Laplace transforms). Eight questions of 20 marks each; Questions 1 and 2 are compulsory and the candidate chooses three of the remaining six (100 marks total). Every question is worked in full below, since the whole set is a study resource.
Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — root locus (Ch. 8), frequency-response / Bode and Nyquist (Ch. 10), stability and Routh–Hurwitz (Ch. 6), PID and cascade compensation (Ch. 9), steady-state error (Ch. 7), state space and pole placement (Ch. 11–12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — companion treatment of the s-domain, dominant-poles modelling and controllability/observability; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, root loci, Bode/Nyquist plots and pole–zero maps below are redrawn as inline figures.
Reading the exam figures. Two questions (Q2 and Q6) require reading a second-order model off a printed step-response curve; the values read from those curves are stated explicitly in the Given block so a reader can reproduce every subsequent number.
Question 7 — State space: transfer function, controllability/observability, pole placement [10 + 10]
Part A — Transfer function, eigenvalues, controllability/observability [10]
Approach. The matrices are in controllable canonical (companion) form, so the transfer function and characteristic polynomial can be read off directly.
Characteristic polynomial / eigenvalues. The bottom row $[\,0,-2,-2,-1\,]$ gives $\det(sI-A)=s^4+s^3+2s^2+2s=s(s^3+s^2+2s+2)=s(s+1)(s^2+2)$. Eigenvalues: $$\boxed{s=0,\ -1,\ \pm j\sqrt{2}}.$$ Two eigenvalues sit on the imaginary axis ($\pm j1.414$) and one at the origin, so the open-loop system is not asymptotically stable.
Transfer function by inspection. For controllable canonical form with $C=[c_0\ c_1\ c_2\ c_3]$, the numerator is $c_0+c_1 s+c_2 s^2+c_3 s^3=1+s+s^2+s^3$: $$G(s)=\frac{s^3+s^2+s+1}{s^4+s^3+2s^2+2s}=\frac{(s+1)(s^2+1)}{s(s+1)(s^2+2)}=\frac{s^2+1}{s(s^2+2)}.$$ The factor $(s+1)$ cancels — the eigenvalue at $s=-1$ is a hidden mode.
Controllability. In controllable canonical form the controllability matrix $[\,B\ AB\ A^2B\ A^3B\,]$ is always full rank; here rank $=4$, so the system is controllable.
Observability. The observability matrix has rank $3\lt 4$: the system is not observable. The $(s+1)$ pole–zero cancellation identifies the mode at $s=-1$ as the unobservable mode (it is invisible at the output $y$). Since $-1$ is in the left half-plane, the system is still detectable.
Figure Q7.1 — Open-loop eigenvalues ($\times$) at $0,-1,\pm j\sqrt2$; the transfer-function zero at $-1$ ($\circ$) cancels the pole, marking the unobservable mode.
Part B — Pole placement by state feedback [10]
Approach. In companion form, state feedback shifts the last-row coefficients directly, so $K$ is the coefficient-by-coefficient difference between the desired and open-loop characteristic polynomials.
Open-loop coefficients. $s^4+1s^3+2s^2+2s+0$, i.e. $[a_3,a_2,a_1,a_0]=[1,2,2,0]$.
Desired at $-1,-2,-3,-4$. $(s+1)(s+2)(s+3)(s+4)=s^4+10s^3+35s^2+50s+24$, so $[a_3',a_2',a_1',a_0']=[10,35,50,24]$.
Gain vector. $K=[a_0'-a_0,\ a_1'-a_1,\ a_2'-a_2,\ a_3'-a_3]=\boxed{[\,24,\ 48,\ 33,\ 9\,]}$.
Alternative target $-2,-3,-4,-5$. $(s+2)(s+3)(s+4)(s+5)=s^4+14s^3+71s^2+154s+120\Rightarrow K=[\,120,\ 152,\ 69,\ 13\,]$ — much larger gains.
Implementation issues and concerns. (1) State feedback assumes all four states are measured. They are not outputs here, so in practice an observer would be needed — but the system is unobservable (the $s=-1$ mode is hidden), so a full-state observer cannot reconstruct that mode from $y$. Because that mode is stable ($-1$), the pair is detectable and an observer-based controller is still workable, but the $-1$ mode will run open-loop-uncontrolled through the estimator. (2) Placing a closed-loop pole exactly at $-1$ (first target) coincides with the unobservable eigenvalue; feasible via true full-state feedback (the system is controllable) but awkward with an observer. (3) Moving the poles further left to $-2,-3,-4,-5$ requires markedly larger feedback gains ($K$ roughly $5\times$ bigger), which means larger control effort, higher bandwidth, greater actuator-saturation risk and amplified measurement noise — a worse practical design even though it is mathematically just as achievable. The faster set also avoids the $-1$/unobservable-mode coincidence.
