State Space Model from a Transfer Function, Pole Placement by State Feedback: 17-Phys-B5 Systems and Control, December 2017, Q5

Question 5: State Space Model from a Transfer Function, 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. $Y(s)/U(s)=(2s^2+6s+20)/(s^3+10s^2+2s+30)$; desired closed-loop poles $-20,\ -2\pm j2$; zero steady-state error to a unit step.

Find. 1) $\{A,B,C\}$ in Controller Canonical Form (CCF). 2) $K$ and $\mathbf{k}$.

Approach. The CCF reads its $A,B,C$ matrices directly off the transfer function's coefficients. For the feedback design, first find the COMBINED gain vector $K\mathbf{k}^T$ that places the poles (a direct coefficient match, since $A-B(K\mathbf{k}^T)$ stays in companion form), THEN choose the scalar $K$ separately so the closed loop's DC gain is exactly 1 (zero step error) — only after that does $\mathbf{k}=(K\mathbf{k}^T)/K$ come out uniquely.

Part 1) — Controller Canonical Form. For a strictly-proper TF $\dfrac{b_2s^2+b_1s+b_0}{s^3+a_2s^2+a_1s+a_0}$, CCF reads the denominator coefficients into the bottom row of $A$ and the numerator coefficients directly into $C$: $$\boxed{A=\begin{bmatrix}0&1&0\\0&0&1\\-30&-2&-10\end{bmatrix},\quad B=\begin{bmatrix}0\\0\\1\end{bmatrix},\quad C=\begin{bmatrix}20&6&2\end{bmatrix}.}$$ (Indeed $C(sI-A)^{-1}B$ reproduces the given transfer function exactly.)
Part 2) — combined feedback gain from pole placement. With $u=K(r-\mathbf{k}^Tx)$, the closed-loop matrix is $A_{cl}=A-B(K\mathbf{k}^T)$; writing $f^T=K\mathbf{k}^T=[f_1,f_2,f_3]$, $A_{cl}$ stays in companion form with bottom row $[-30{-}f_1,\,-2{-}f_2,\,-10{-}f_3]$, so its characteristic polynomial is $$s^3+(10{+}f_3)s^2+(2{+}f_2)s+(30{+}f_1).$$ The desired poles give $(s+20)(s^2+4s+8)=s^3+24s^2+88s+160$. Matching coefficients: $$10+f_3=24,\quad2+f_2=88,\quad30+f_1=160\ \Rightarrow\ \boxed{f^T=[130,\ 86,\ 14]}.$$
Part 2, cont'd — DC gain condition fixes $K$, then $\mathbf{k}$. With $A_{cl}=A-Bf^T$, the reference-to-output DC gain (per unit $K$) is $C(-A_{cl})^{-1}B=\dfrac{1}{8}$; requiring the FULL closed loop's DC gain to equal $1$ (zero steady-state error to a step) fixes $$\boxed{K=8}\qquad\Rightarrow\qquad \boxed{\mathbf{k}=\frac{f^T}{K}=\left[\frac{65}{4},\ \frac{43}{4},\ \frac{7}{4}\right]=[16.25,\ 10.75,\ 1.75]}.$$ With these exact numbers, $A_{cl}=A-B(K\mathbf{k}^T)$ has eigenvalues exactly $-20,\,-2\pm j2$, and $C(-A_{cl})^{-1}(BK)=1$.

Final results — Question 5
Item	Result
CCF	$A=\begin{bmatrix}0&1&0\\0&0&1\\-30&-2&-10\end{bmatrix}$, $B=[0,0,1]^T$, $C=[20,6,2]$
Combined gain $K\mathbf{k}^T$	$[130,\ 86,\ 14]$
$K$	$8$
$\mathbf{k}$	$[16.25,\ 10.75,\ 1.75]$

Topic: Controller Canonical Form and pole placement by state feedback with a DC-gain (zero steady-state error) constraint.

Key relations: CCF reads $A,B,C$ directly off a transfer function's coefficients; $A_{cl}=A-B(K\mathbf{k}^T)$; for a step input, zero steady-state error requires $C(-A_{cl})^{-1}(BK)=1$.

Why this works: because $A$ is already companion form, subtracting $Bf^T$ only touches the BOTTOM row, so pole placement in CCF is a direct coefficient match with no matrix inversion needed — the harder Ackermann's-formula route is only required when the model is NOT already in companion form. The gain $K$ and the vector $\mathbf{k}$ are only separable once an extra condition (here, DC gain $=1$) is imposed, since $u=K(r-\mathbf{k}^Tx)$ depends on their PRODUCT for the pole locations but on $K$ alone for the reference scaling.

Common pitfall: placing the poles and stopping, without checking the resulting DC gain (the "$K$" in $u=K(r-\mathbf{k}^Tx)$ is a genuinely free design parameter that pole placement alone does NOT fix); and forgetting that $\mathbf{k}$, not $K$, is what actually multiplies each state in the feedback path.

Source: Nise, Control Systems Engineering, Ch. 3 (state-space representation) and Ch. 12 (pole placement by state feedback); Ogata, Modern Control Engineering, Ch. 12.

Question 5 of 8: State Space Model from a Transfer Function, Pole Placement by State Feedback

Question 5: State Space Model from a Transfer Function, Pole Placement by State Feedback (20 marks)