Question 8 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. 17-Phys-B5 Systems and Control, National Exam, May
2019 — a three-hour closed-book examination with one double-sided handwritten formula/notes
sheet permitted and an approved calculator. Questions 1 and 2 are compulsory; candidates choose
three of the remaining six (3–8). Every question is nonetheless answered in full below so
the paper remains a complete study resource. All eight questions carry equal value (20 marks
each).
Reference texts. N. S. Nise, Control Systems Engineering, 7th ed.
(transient-response specifications, root locus, Routh–Hurwitz, frequency-response and lead
compensator design, state-space representation, steady-state errors); K. Ogata, Modern
Control Engineering, 5th ed. (root-locus construction, PID/rate-feedback compensator design,
controllability/observability, Mason's Gain Formula on signal-flow graphs).
Question 8: State-Space Model — Pole Placement by State Feedback
(20 marks)
Approach. Read the ODEs directly off $A,B$; compute $G_{open}(s)=C(sI-A)^{-1}B$
symbolically; form the controllability/observability matrices and check their determinants; solve
the pole-placement + unity-DC-gain system simultaneously for $K,k_1,k_2$.
Part a) — ODEs.
$$\boxed{\dot x_1=-8x_1+x_2+u,\qquad \dot x_2=x_1.}$$
Part b) — open-loop TF.
$(sI-A)^{-1}=\dfrac1{s^2+8s-1}\begin{bmatrix}s&1\\1&s+8\end{bmatrix}$, so
$C(sI-A)^{-1}B=\dfrac1{s^2+8s-1}\begin{bmatrix}1&8\end{bmatrix}\begin{bmatrix}s\\1\end{bmatrix}
=\dfrac{s+8}{s^2+8s-1}$.
$$\boxed{G_{open}(s)=\frac{s+8}{s^2+8s-1}}.$$
The characteristic polynomial $s^2+8s-1$ has roots $s=-4\pm\sqrt{17}=0.1231,\,-8.1231$: ONE pole
in the right half-plane, so the open-loop system is unstable.
Part c) — controllability/observability.
$$M_{co}=[B\ \ AB]=\begin{bmatrix}1&-8\\0&1\end{bmatrix},\ \det=1\ne0
\ \Rightarrow\ \boxed{\text{controllable}};$$
$$M_{ob}=\begin{bmatrix}C\\CA\end{bmatrix}=\begin{bmatrix}1&8\\0&1\end{bmatrix},\ \det=1\ne0
\ \Rightarrow\ \boxed{\text{observable}}.$$
Part d) — state-feedback pole placement. With
$u=K(r-\mathbf k^T\mathbf x)$, $A_{cl}=A-BK\mathbf k^T$, $B_{cl}=BK$. Requiring
$\det(sI-A_{cl})=(s+10)(s+30)=s^2+40s+300$ AND $G_{cl}(0)=C(-A_{cl})^{-1}B_{cl}=1$ (zero
steady-state error to a step) gives three equations in $K,k_1,k_2$:
$$\boxed{K=\frac{75}{2}=37.5,\qquad k_1=\frac{64}{75}=0.8533,\qquad k_2=\frac{602}{75}=8.0267.}$$
Check: $A_{cl}=\begin{bmatrix}-40&-300\\1&0\end{bmatrix}$ has eigenvalues EXACTLY $-10,-30$.
Part e) — closed-loop TF.
$$\boxed{G_{cl}(s)=\frac{75(s+8)}{2(s+10)(s+30)}=\frac{37.5(s+8)}{(s+10)(s+30)}}$$
Poles: $-10,-30$ (as designed). Zero: $-8$ (inherited unchanged from the open-loop numerator,
since state feedback through $B$ alone does not move $C$'s numerator zero). DC gain:
$G_{cl}(0)=37.5(8)/(10\times30)=300/300=\boxed{1}$ (confirms zero steady-state error to a unit
step, as required).
