Controller Canonical Form and Pole Placement by State Feedback: 17-Phys-B5 Systems and Control, May 2017, Q5

Question 5: Controller Canonical Form and Pole Placement by State Feedback (20 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. $Y(s)/U(s)=(2s^2+5s+2)/(s^3+8s^2+2s+30)$; desired closed-loop poles $-10,-2\pm j3$; state feedback law $u=K(r-k^Tx)$; zero SSE to a unit step.

Find. 1) The CCF matrices $A,B,C$. 2) The scalar gain $K$ and feedback vector $k=[k_1,k_2,k_3]^T$.

Approach. Read the CCF matrices directly off the transfer function's coefficients (standard form). For pole placement, note that in CCF the last row of $A-BKk^T$ is exactly the negative of the desired characteristic polynomial's coefficients, giving three linear equations in the three products $Kk_1,Kk_2,Kk_3$; a fourth equation comes from forcing the closed-loop DC gain to 1 (the zero-SSE-to-step condition), which pins down $K$ itself and hence each $k_i$.

Part 1) — Controller Canonical Form. For $a_2=8,a_1=2,a_0=30$ and $b_2=2,b_1=5,b_0=2$, the CCF is $$A=\begin{bmatrix}0&1&0\\0&0&1\\-30&-2&-8\end{bmatrix},\quad B=\begin{bmatrix}0\\0\\1\end{bmatrix},\quad C=\begin{bmatrix}2&5&2\end{bmatrix},\quad D=0.$$ (Indeed $C(sI-A)^{-1}B$ reproduces the given transfer function exactly.)
Part 2) — desired characteristic polynomial. $$(s+10)(s+2-j3)(s+2+j3)=(s+10)(s^2+4s+13)=\boxed{s^3+14s^2+53s+130}.$$
Part 3) — match coefficients of $A-BKk^T$. Since $B=[0,0,1]^T$, the feedback $u=K(r-k^Tx)$ only modifies the LAST row of $A$: new last row $=[-30-Kk_1,\,-2-Kk_2,\,-8-Kk_3]$, giving char. poly. $s^3+(8+Kk_3)s^2+(2+Kk_2)s+(30+Kk_1)$. Matching to $s^3+14s^2+53s+130$: $$Kk_3=6,\qquad Kk_2=51,\qquad Kk_1=100.$$
Part 4) — zero-SSE-to-step condition fixes $K$. The closed-loop numerator is unchanged in form ($K$ scales the input), so $G_{cl}(0)=K\cdot b_0/130=2K/130$; setting this to 1 (zero SSE to a unit step needs unity DC gain) gives $\boxed{K=65}$. Then $$k_1=\frac{100}{65}=\frac{20}{13}\approx1.538,\quad k_2=\frac{51}{65}\approx0.785,\quad k_3=\frac{6}{65}\approx0.092.$$ With these values, $A-BKk^T$ has eigenvalues exactly $-10,\,-2\pm j3$, and the resulting $r\to y$ transfer function has $G_{cl}(0)=1$.

Final results — Question 5
Item	Result
CCF	$A=\begin{bmatrix}0&1&0\\0&0&1\\-30&-2&-8\end{bmatrix}$, $B=[0,0,1]^T$, $C=[2,5,2]$
Desired char. poly.	$s^3+14s^2+53s+130$
$K$	$\boxed{65}$
$k=[k_1,k_2,k_3]^T$	$[20/13,\ 51/65,\ 6/65]\approx[1.538,\ 0.785,\ 0.092]$

Topic: Controller Canonical Form and pole placement by full state feedback.

Key relations: CCF built directly from a transfer function's coefficients; in CCF, $A-BKk^T$'s last row gives the closed-loop characteristic polynomial by inspection (no determinant expansion needed); zero SSE to a step under $u=K(r-k^Tx)$ requires $G_{cl}(0)=1$.

Why this works: the Controller Canonical Form is constructed precisely so that state feedback through $B=[0,\dots,0,1]^T$ only perturbs the bottom row of $A$, turning an otherwise nonlinear pole-placement search into simple linear coefficient matching — this is the main practical reason CCF is the preferred realization for this design method.

Common pitfall: stopping after solving the three $Kk_i$ products (which is one equation short of four unknowns) without using the steady-state-error condition to pin down $K$ itself; and forgetting that the closed-loop numerator is scaled by $K$, not by the plant's own $b_0$ alone, when computing the DC gain.

Source: Nise, Control Systems Engineering, Ch. 3 (state-space) and Ch. 12 (state-feedback design); Ogata, Modern Control Engineering, Ch. 10.

Question 5 of 8: Controller Canonical Form and Pole Placement by State Feedback

Question 5: Controller Canonical Form and Pole Placement by State Feedback (20 marks)