Question 3 of 6: Full state feedback with pole placement and unity DC gain

Question 3: Full state feedback with pole placement and 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 state-space plant whose input vector carries an unspecified scalar gain $B$, with the output taken as the third state.

Find. The state-feedback row vector $K$ that places the three closed-loop poles as specified, and the scalar reference gain $L$ that makes the steady-state output equal the steady-state reference.

Approach. Expand $\det\!\left(sI-(A-bK)\right)$ symbolically, match it coefficient by coefficient against the desired characteristic polynomial to get $K$, then evaluate the resulting closed-loop transfer function at $s=0$ to get $L$.

1. Part 1 — find the open-loop characteristic polynomial. The plant matrix is block lower-triangular in its first column, so expanding along the top row$$\det(sI-A)=(s+1)\left[s(s+2)+1\right]=(s+1)(s+1)^{2}=(s+1)^{3}=s^{3}+3s^{2}+3s+1$$All three open-loop poles sit at $s=-1$. The plant transfer function is $c(sI-A)^{-1}b=B/(s+1)^{3}$ — a triple lag with no finite zero, which is worth knowing before designing anything.
2. Part 2 — form the closed-loop characteristic polynomial symbolically. With $u=Lr-Kx$ and $K=[k_{1}\;k_{2}\;k_{3}]$, the feedback term $bK$ only populates the first row, because $b=B[1\;0\;0]^{T}$. Expanding $\det\!\left(sI-(A-bK)\right)$ gives a tidy result:$$\det\!\left(sI-(A-bK)\right)=(s+1)^{3}+Bk_{1}(s+1)^{2}+Bk_{2}s+Bk_{3}$$Each gain therefore enters exactly one place in the polynomial, which is what makes this a three-equation linear match rather than an Ackermann computation.
3. Part 3 — expand and match against the desired polynomial. The requested poles give$$\alpha_{d}(s)=(s+5)\left(s^{2}+6s+10\right)=s^{3}+11s^{2}+40s+50$$while the expansion above is $s^{3}+(3+Bk_{1})s^{2}+(3+2Bk_{1}+Bk_{2})s+(1+Bk_{1}+Bk_{3})$. Equating coefficients in descending order and back-substituting each result into the next:
4. Part 4 — solve the three equations in order. $$3+Bk_{1}=11\Rightarrow Bk_{1}=8;\qquad 3+16+Bk_{2}=40\Rightarrow Bk_{2}=21;\qquad 1+8+Bk_{3}=50\Rightarrow Bk_{3}=41$$$$\boxed{\;K=\frac{1}{B}\begin{bmatrix}8&21&41\end{bmatrix}\;}$$Substituting $B=1$ and computing the eigenvalues of $A-bK$ numerically returns $-5$ and $-3\pm j$, confirming the placement.
5. Part 5 — choose $L$ from the closed-loop DC gain. The closed-loop transfer function is $Y(s)/R(s)=L\,c\,\left(sI-(A-bK)\right)^{-1}b$. Because $c$ picks the third state and $b$ drives only the first, the numerator is the $(3,1)$ entry of the adjugate, which evaluates to the constant $B$ with no $s$ dependence:$$\frac{Y(s)}{R(s)}=\frac{BL}{s^{3}+11s^{2}+40s+50}$$State feedback moves poles but leaves the numerator alone, and here that numerator is a pure gain.
6. Part 6 — impose unity DC gain. Setting $s=0$,$$\frac{Y(0)}{R(0)}=\frac{BL}{50}=1\quad\Longrightarrow\quad\boxed{\;L=\frac{50}{B}\;}$$so the reference gain is exactly the constant term of the desired characteristic polynomial divided by the input gain. Evaluating $c\left(-(A-bK)\right)^{-1}bL$ numerically at $B=1$ returns $1.000$, which closes the design. Note that this feed-forward trick gives zero steady-state error only if $B$ and the plant model are exact; there is no integral action in the loop, so a modelling error in $B$ shows up directly as a DC offset.