Question 1 of 8: State-Space Analysis and Pole Placement by State Feedback

Question 1: State-Space Analysis and Pole Placement by State Feedback (20 marks, compulsory)

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 second-order single-input single-output state model with

Find. Whether the pair is controllable and observable; the open-loop transfer function with its poles, zeros and stability verdict; a set of closed-loop pole locations meeting all three specifications; and the state feedback vector $k^T$ together with the calibration gain $K$ that place them.

Approach. Build the controllability and observability matrices and test their determinants, obtain $G_{open}(s) = C(sI-A)^{-1}B + D$ to locate the poles and zeros, then exploit the hint: cancel the open-loop zero with one closed-loop pole so the loop collapses to a first-order response whose time constant is set by the settling-time specification, and finally size $K$ for unity DC gain.

1. Test controllability. For a second-order system $M_c = \begin{bmatrix} B & AB \end{bmatrix}$. With $AB = \begin{bmatrix} -12 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -12 \\ 1 \end{bmatrix}$, $$M_c = \begin{bmatrix} 1 & -12 \\ 0 & 1 \end{bmatrix}, \qquad \det M_c = (1)(1)-(-12)(0) = 1$$ The determinant is non-zero, so $M_c$ has full rank 2 and the system is controllable — every state can be steered by $u$, which is what makes arbitrary pole placement possible in part 4.
2. Test observability. Here $M_o = \begin{bmatrix} C \\ CA \end{bmatrix}$ and $CA = \begin{bmatrix} 1 & 8 \end{bmatrix}\begin{bmatrix} -12 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -4 & 1 \end{bmatrix}$, giving $$M_o = \begin{bmatrix} 1 & 8 \\ -4 & 1 \end{bmatrix}, \qquad \det M_o = (1)(1)-(8)(-4) = 33$$ Again non-zero, so the system is observable: both states can be reconstructed from the measured output.
3. Form the resolvent. The transfer function needs $(sI-A)^{-1}$: $$sI - A = \begin{bmatrix} s+12 & -1 \\ -1 & s \end{bmatrix}, \qquad \det(sI-A) = s(s+12) - 1 = s^2 + 12s - 1$$ so that $$(sI-A)^{-1} = \frac{1}{s^2+12s-1}\begin{bmatrix} s & 1 \\ 1 & s+12 \end{bmatrix}$$
4. Assemble the open-loop transfer function. Multiplying out, $(sI-A)^{-1}B = \dfrac{1}{s^2+12s-1}\begin{bmatrix} s \\ 1 \end{bmatrix}$, and premultiplying by $C$ gives $1\cdot s + 8\cdot 1$ in the numerator. With $D = 0$, $$\boxed{\,G_{open}(s) = \frac{Y(s)}{U(s)} = \frac{s+8}{s^2+12s-1}\,}$$
5. Locate the poles and zeros and judge stability. The single finite zero is at $s = -8$. The poles solve $s^2+12s-1 = 0$: $$s = \frac{-12 \pm \sqrt{144+4}}{2} = -6 \pm \sqrt{37} \;\Rightarrow\; s_1 = +0.0828, \quad s_2 = -12.083$$ One pole lies in the right half plane, therefore the open-loop system is unstable — a slowly diverging mode with time constant $1/0.0828 = 12.1$ s. (The constant term of the characteristic polynomial is negative, which by itself already violates the necessary condition for stability.)
Figure 1.2 — Open-loop pole-zero map. The pole at +0.083 lies in the right half plane, so the uncontrolled system is unstable.
6. Choose the closed-loop pole locations. Zero overshoot rules out any complex pair, so both closed-loop poles must be real. Following the hint, put the first one directly on the open-loop zero, $s = -8$, so that it cancels in the closed-loop transfer function and the input–output response degenerates to first order. For a first-order lag $T_{s(\pm 2\%)} = 4\tau$, so $$\tau \le \frac{0.25}{4} = 0.0625\ \text{s} \;\Rightarrow\; s = -\frac{1}{\tau} \le -16$$ Taking the limiting value gives the pole pair $$\boxed{\,s_{1,2}^{\,cl} = -8 \ \text{(cancels the zero)}, \quad -16\,}$$ whose characteristic polynomial is $(s+8)(s+16) = s^2 + 24s + 128$.
7. Solve for the combined feedback gain. With $u = K(r - k^T x)$ the state equation becomes $\dot{x} = (A - BKk^T)x + BKr$. Writing the combined row vector $g = Kk^T = \begin{bmatrix} g_1 & g_2 \end{bmatrix}$, $$A - Bg = \begin{bmatrix} -12-g_1 & 1-g_2 \\ 1 & 0 \end{bmatrix} \;\Rightarrow\; \det(sI - A + Bg) = s^2 + (12+g_1)s - (1-g_2)$$ Matching this to $s^2+24s+128$ term by term: $$12 + g_1 = 24 \;\Rightarrow\; g_1 = 12, \qquad -(1-g_2) = 128 \;\Rightarrow\; g_2 = 129$$
8. Calibrate $K$ for zero steady-state error. Repeating the resolvent calculation with the closed-loop matrix, the numerator is unchanged, so $$\frac{Y(s)}{R(s)} = \frac{K(s+8)}{(s+8)(s+16)} = \frac{K}{s+16}$$ The zero has cancelled exactly as intended. Zero steady-state error to a unit step demands unity DC gain, $K/16 = 1$, hence $K = 16$ and $$\boxed{\,K = 16, \qquad k^T = \frac{1}{K}\begin{bmatrix} 12 & 129 \end{bmatrix} = \begin{bmatrix} 0.750 & 8.0625 \end{bmatrix}\,}$$
9. Confirm the specifications. The realised loop is the first-order system $16/(s+16)$: its step response is a pure exponential, so $PO = 0\%$ exactly; its time constant is $\tau = 1/16 = 0.0625$ s giving $T_{s(\pm2\%)} = 4\tau = 0.25$ s, exactly at the limit; and its DC gain is unity so $e_{ss} = 0\%$. All three requirements are met simultaneously.