Pole placement by state feedback with unity DC gain: 22-Elec-B2 Advanced Control Systems, December 2013, Q3

Question 3: Pole placement by state feedback with unity DC gain (25 marks)

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 third-order single-input system in which the whole state is available for feedback, together with a target pole set and a target DC gain.

Given data — Question 3
Quantity	Value
State matrix	$A = \begin{bmatrix}0&1&0\\0&0&-1\\0&0&-2\end{bmatrix}$
Input matrix	$B = [\,0\ \ 1\ \ 1\,]^{\mathsf T}$
Output matrix	$C = [\,1\ \ 0\ \ 0\,]$, $D=0$
Open-loop poles	$s = 0,\ 0,\ -2$ (from the triangular structure of $A$)
Required closed-loop poles	$s = -5$, $s = -3\pm j$
Required DC gain	$y_{ss}/r = 1$
Control law	$u(t) = Kx(t) + Lr(t)$ (note the plus sign on $Kx$)

Find. the row vector $K$ that places the three closed-loop poles, and the scalar $L$ that makes the steady-state output equal to a constant reference.

Figure Q3.1 — the required structure: full state feedback u = Kx places the closed-loop poles, and the scalar reference gain L is then chosen independently so that a constant r produces y = r.

Approach. Confirm controllability, form the desired characteristic polynomial, match it coefficient-by-coefficient against the characteristic polynomial of $A+BK$ to obtain $K$, then choose $L$ from the closed-loop DC gain, which is a separate and independent calculation.

Confirm that arbitrary pole placement is possible. Successive products give $AB = [\,1\ \ -1\ \ -2\,]^{\mathsf T}$ and $A^{2}B = [\,-1\ \ 2\ \ 4\,]^{\mathsf T}$, so $$\mathcal{C} = \begin{bmatrix}B & AB & A^{2}B\end{bmatrix} = \begin{bmatrix} 0 & 1 & -1\\ 1 & -1 & 2\\ 1 & -2 & 4\end{bmatrix},\qquad \det\mathcal{C} = -1 \neq 0 .$$ The pair $(A,B)$ is controllable, so any self-conjugate set of three poles can be assigned.
Write the desired characteristic polynomial. The complex pair contributes $(s+3)^{2}+1 = s^{2}+6s+10$, so $$\alpha_{d}(s) = (s+5)(s^{2}+6s+10) = s^{3}+11s^{2}+40s+50 .$$
Form the closed-loop matrix and its characteristic polynomial. With $K = [\,k_{1}\ \ k_{2}\ \ k_{3}\,]$ and the given sign convention $u = Kx + Lr$, the closed-loop matrix is $A+BK$ (not $A-BK$): $$A + BK = \begin{bmatrix} 0 & 1 & 0\\ k_{1} & k_{2} & k_{3}-1\\ k_{1} & k_{2} & k_{3}-2 \end{bmatrix}.$$ Expanding $\det\!\left(sI-(A+BK)\right)$ gives $$\alpha(s) = s^{3} + (2-k_{2}-k_{3})\,s^{2} + (-k_{1}-k_{2})\,s + (-k_{1}) .$$
Match coefficients. Equating $\alpha(s)$ with $\alpha_{d}(s)$ term by term, working from the constant term upward because it isolates $k_{1}$: $$-k_{1} = 50 \;\Rightarrow\; k_{1} = -50; \qquad -k_{1}-k_{2} = 40 \;\Rightarrow\; k_{2} = 50-40 = 10; $$ $$2-k_{2}-k_{3} = 11 \;\Rightarrow\; k_{3} = 2-10-11 = -19 .$$ $$\boxed{K = \begin{bmatrix} -50 & 10 & -19\end{bmatrix}}$$ Substituting back, the eigenvalues of $A+BK$ are $-5$ and $-3\pm j$ exactly, as required.
Choose $L$ for unity DC gain. In steady state $\dot x = 0$, so $0 = (A+BK)x_{ss} + BLr$ and therefore $$x_{ss} = -(A+BK)^{-1}BL\,r, \qquad \frac{y_{ss}}{r} = -C\,(A+BK)^{-1}B\,L .$$ Evaluating the matrix triple product gives $-C(A+BK)^{-1}B = 0.02$, which is simply $1/50$ — the reciprocal of the constant term of $\alpha_{d}(s)$ divided by the numerator gain of the closed loop. Setting the DC gain to unity, $$0.02\,L = 1 \quad\Longrightarrow\quad \boxed{L = 50}$$ Note that $K$ and $L$ are chosen independently: state feedback fixes the poles and therefore the transient shape, while the reference gain scales the input to fix the steady-state level. $L$ has no effect whatever on stability.

Check: two features of the printed source. The system matrix in the paper has its $(2,1)$ entry missing from the print; reading down the columns against rows 1 and 3, it is taken as $a_{21}=0$, which is the only value consistent with the printed alignment. With that reading $\det\mathcal{C}=-1$ and the design is well posed. Second, the paper writes the control law as $u = Kx+Lr$ with a plus sign, so the closed-loop matrix is $A+BK$ and the gains come out negative; a design written in the more common $u = -Kx+Lr$ convention would report $K = [\,50\ \ -10\ \ 19\,]$ for exactly the same controller.

Final results — Question 3
Quantity	Symbol	Value
Controllability determinant	$\det\mathcal{C}$	$-1$ (controllable)
Desired characteristic polynomial	$\alpha_{d}(s)$	$s^{3}+11s^{2}+40s+50$
State-feedback gain	$K$	$[\,-50\ \ \ 10\ \ -19\,]$
Closed-loop DC transmission	$-C(A+BK)^{-1}B$	0.02
Reference gain	$L$	50
Achieved closed-loop poles	—	$-5$, $-3+j$, $-3-j$

Topic. Full-state-feedback pole placement with reference scaling for unity DC gain.

Key relations. Controllability matrix $\mathcal{C}=[B\ AB\ A^{2}B]$; closed-loop matrix $A+BK$ for the convention $u=Kx+Lr$; coefficient matching of $\det(sI-A-BK)$ against $\prod(s-p_{i})$; DC gain $y_{ss}/r = -C(A+BK)^{-1}BL$.

Why this works. Pole placement and DC-gain trimming are two separate degrees of freedom, and keeping them separate is the whole idea of the $u=Kx+Lr$ structure. The feedback gain $K$ reshapes the characteristic polynomial and hence every mode of the transient; because the pair is controllable, coefficient matching has a unique solution for a single-input system. The scalar $L$ then only rescales the input, which multiplies the closed-loop transfer function by a constant and cannot move a pole. Doing the DC-gain calculation from the steady-state state equation, rather than by forming the closed-loop transfer function symbolically, keeps the arithmetic to one matrix inverse.

Common pitfall. Using $A-BK$ when the question wrote $u=+Kx+Lr$; the pole locations then come out mirrored in sign and every subsequent number is wrong. A second pitfall is computing $L$ as $1/(CB)$ or from the open-loop DC gain — the plant here has poles at the origin, so its own DC gain is infinite, and only the closed-loop DC transmission is meaningful. Finally, remember that $L$ gives an exactly unity DC gain only for the nominal model; it provides no integral action, so a parameter error or a load disturbance leaves a standing offset.

Source. Ogata, Modern Control Engineering, 5th ed., Ch. 10 (pole placement design); Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 7th ed., §7.5 (full state feedback) and §7.9 (reference input scaling).

Question 3 of 6: Pole placement by state feedback with unity DC gain

Question 3: Pole placement by state feedback with unity DC gain (25 marks)