Pole placement by state feedback with reference scaling: 22-Elec-B2 Advanced Control Systems, May 2015, Q4

Question 4: Pole placement by state feedback with reference scaling (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 single-output state model and three required closed-loop pole locations.

Quantity	Value
State matrix	$A = \begin{pmatrix}-1 & -2 & -2\\ 0 & -1 & 1\\ 1 & 0 & -1\end{pmatrix}$
Input matrix	$B = \begin{pmatrix}2 & 0 & 1\end{pmatrix}^{T}$
Output matrix	$C = \begin{pmatrix}1 & 0 & 1\end{pmatrix}$, $D = 0$
Control law	$u = L\,r - K x$, with $K = \begin{pmatrix}k_1 & k_2 & k_3\end{pmatrix}$
Required closed-loop poles	$s = -10$, $s = -3 \pm j4$
Steady-state requirement	$e = r - y \to 0$ for a step reference

Find. The feedback row vector $K$ that places the three poles, and the scalar reference gain $L$ that removes the steady-state tracking error.

Question 4: the three open-loop poles (grey) and the three placed closed-loop poles (red) at s = -10 and s = -3 +/- j4.

Approach. Confirm the pair $(A,B)$ is controllable, expand the closed-loop characteristic polynomial symbolically in the three unknown gains, match it term by term against the desired polynomial, and finally choose $L$ so that the closed-loop DC gain from $r$ to $y$ is exactly one.

Verify that arbitrary pole placement is possible. Successive multiplications give $$AB = \begin{pmatrix}-4\\ 1\\ 1\end{pmatrix}, \qquad A^2B = \begin{pmatrix}0\\ 0\\ -5\end{pmatrix}, \qquad \mathcal{C} = \begin{pmatrix}2 & -4 & 0\\ 0 & 1 & 0\\ 1 & 1 & -5\end{pmatrix},$$ and expanding along the second row, $\det\mathcal{C} = -10 \neq 0$. The pair is controllable, so any self-conjugate set of three poles is reachable by state feedback and the design is guaranteed to have a solution.
Write the desired characteristic polynomial. Multiplying out the required factors, $$\Delta_{d}(s) = (s+10)\bigl[(s+3)^2 + 16\bigr] = (s+10)(s^2 + 6s + 25),$$ $$\Delta_{d}(s) = s^3 + 16s^2 + 85s + 250 .$$
Expand the closed-loop polynomial in the unknown gains. With $u = -Kx$ the state matrix becomes $A - BK$, and expanding $\det\bigl(sI - A + BK\bigr)$ symbolically gives $$\Delta_{cl}(s) = s^3 + \bigl(2k_1 + k_3 + 3\bigr)s^2 + \bigl(2k_1 + k_2 + 4k_3 + 5\bigr)s + \bigl(-2k_1 + 3k_2 + 3k_3 + 5\bigr).$$ Note that the open-loop polynomial $s^3 + 3s^2 + 5s + 5$ is recovered when $k_1 = k_2 = k_3 = 0$, which is a useful check on the expansion before going further.
Match the coefficients. Equating like powers of $s$ produces three linear equations in three unknowns: $$\begin{aligned} s^2: &\quad 2k_1 + k_3 + 3 = 16\\ s^1: &\quad 2k_1 + k_2 + 4k_3 + 5 = 85\\ s^0: &\quad -2k_1 + 3k_2 + 3k_3 + 5 = 250 \end{aligned}$$
Solve the linear system. From the first equation $k_3 = 13 - 2k_1$; substituting into the second gives $k_2 = 80 - 2k_1 - 4k_3 = 28 + 6k_1$, and the third then reduces to $-2k_1 + 3(28 + 6k_1) + 3(13 - 2k_1) = 245$, i.e. $10k_1 = 122$. Hence $$\boxed{\;K = \begin{pmatrix}k_1 & k_2 & k_3\end{pmatrix} = \begin{pmatrix}12.2 & 101.2 & -11.4\end{pmatrix} = \tfrac{1}{5}\begin{pmatrix}61 & 506 & -57\end{pmatrix}\;}$$ The third gain is negative, which is perfectly admissible — state feedback gains carry no sign restriction, and here the third state must be fed back positively to pull the lightly damped open-loop pair leftwards.
Confirm the placement. Substituting $K$ back and factoring, $$\det\bigl(sI - A + BK\bigr) = s^3 + 16s^2 + 85s + 250 = (s+10)(s^2 + 6s + 25),$$ whose roots are $s = -10$ and $s = -3 \pm j4$ as required. The dominant pair has $\zeta = 3/5 = 0.6$ and $\omega_n = 5\ \text{rad/s}$, with the real pole placed twice as far out so that the pair genuinely dominates.
Set up the reference gain. The feedback gain $K$ fixes the poles but also changes the DC gain, so a step reference would otherwise be tracked with an offset. The closed-loop system is $$\dot x = (A - BK)x + BL\,r, \qquad y = Cx,$$ and at steady state $\dot x = 0$ gives $x_{ss} = -(A-BK)^{-1}BL\,r$, so $$y_{ss} = -C\,(A-BK)^{-1}B\,L\,r \stackrel{!}{=} r \;\Longrightarrow\; L = \frac{1}{-C\,(A-BK)^{-1}B}.$$
Evaluate the reference gain. Inverting the closed-loop matrix and forming the scalar product gives $-C(A-BK)^{-1}B = 1/250$, so $$\boxed{\;L = 250\;}$$ and the complete control law is $u(t) = 250\,r(t) - \begin{pmatrix}12.2 & 101.2 & -11.4\end{pmatrix}x(t)$.
Check the finished design. The closed-loop transfer function is $$\frac{Y(s)}{R(s)} = C\bigl(sI - A + BK\bigr)^{-1}BL = \frac{250\,(3s^2 + 6s + 1)}{s^3 + 16s^2 + 85s + 250},$$ whose value at $s = 0$ is $250/250 = 1$. The step response therefore settles on the reference exactly, and $e = r - y \to 0$ as required.

Check: the zero steady-state error here is a calibration, not an integrator. The gain $L = 250$ cancels the DC error of the nominal model only. Any drift in $A$, $B$ or $C$ reintroduces an offset, because there is no integral action to absorb it, and a load disturbance entering at the plant input is not rejected at all. If robust zero error were required the design would have to be extended with an integrator state and the poles re-placed for the augmented fourth-order system. The question as set asks only for the nominal condition, which is what is delivered above.

Question 4 — results
Quantity	Symbol	Value
Controllability determinant	$\det\mathcal{C}$	$-10$ (controllable)
Open-loop characteristic polynomial	$\Delta_{ol}$	$s^3 + 3s^2 + 5s + 5$
Desired characteristic polynomial	$\Delta_{d}$	$s^3 + 16s^2 + 85s + 250$
Feedback gains	$K$	$\begin{pmatrix}12.2 & 101.2 & -11.4\end{pmatrix}$
Reference gain	$L$	250
Dominant pair damping and frequency	$\zeta,\ \omega_n$	0.6, 5 rad/s
Closed-loop DC gain	$Y(0)/R(0)$	1 (zero step error)

Topic. Pole placement by full state feedback, and reference scaling to remove the resulting steady-state offset.

Key relations. $\det(sI - A + BK) = \Delta_{d}(s)$ solved by coefficient matching; placement is possible for arbitrary poles if and only if $\text{rank}[\,B\ AB\ \dots\ A^{n-1}B\,] = n$; $L = \bigl[-C(A-BK)^{-1}B\bigr]^{-1}$; equivalently $L = 1/G_{cl}(0)$ where $G_{cl}$ is the unscaled closed-loop transfer function.

Why this works. State feedback can move the eigenvalues anywhere but cannot change the input or output maps, so the closed-loop DC gain is whatever the placement leaves behind. Scaling the reference by its reciprocal is the cheapest possible fix: one multiplication, no extra state and no effect on the poles just placed.

Common pitfall. Reaching for Ackermann's formula or a controllable canonical transformation when direct coefficient matching is faster for a third-order problem; and forgetting $L$ altogether, which leaves all three poles correctly placed but a step response settling at $1/250$ of the reference. A third pitfall is treating the negative gain $k_3$ as an algebra error and changing its sign.

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

Question 4 of 6: Pole placement by state feedback with reference scaling

Question 4: Pole placement by state feedback with reference scaling (25 marks)