Discrete-time model and deadbeat state feedback for a sampled-data loop: 22-Elec-B2 Advanced Control Systems, May 2013, Q5

Question 5: Discrete-time model and deadbeat state feedback for a sampled-data loop (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 first-order continuous plant driven through a zero-order hold at a uniform period $h$, preceded by a purely discrete block that accumulates the error through a unit delay.

Given data
Continuous plant	$P(s)=\dfrac{1}{s+1}$
Hold	zero-order, uniform period $h$
Discrete block	unit delay $z^{-1}$ with unity positive feedback (an accumulator)
Error definition	$e(k)=r(k)-x_{1}(k)$, with $r=0$
States / input / output	$x=\begin{bmatrix}x_{1} & x_{2}\end{bmatrix}^{\mathsf T}$, input $u(k)$, output $x_{1}(k)$
Design target	all closed-loop poles at $z=0$ (deadbeat)

Find. The zero-order-hold equivalent relating $u(k)$ to $x_{1}(k)$; the two-state discrete state-space model; and the row vector $K$ that places both closed-loop eigenvalues at the origin.

Figure Q5.1 - Discrete accumulator driven by e, cascaded with a sampler, zero-order hold, continuous plant P(s) and a second sampler, all at the uniform period h.

Approach. Discretise the continuous branch exactly with the zero-order-hold formula, write the accumulator as a second difference equation, assemble the pair into state-space form, and then choose $K$ so that the closed-loop characteristic polynomial is $z^{2}$.

(a) Exact zero-order-hold equivalent. Between samples the hold keeps $u$ constant, so integrating $\dot{x}_{1}=-x_{1}+u$ over one period gives the exact recursion$$x_{1}(k+1)=e^{-h}x_{1}(k)+\left(\int_{0}^{h}e^{-\tau}\,d\tau\right)u(k)=e^{-h}x_{1}(k)+\bigl(1-e^{-h}\bigr)u(k).$$Writing $a=e^{-h}$ and $b=1-e^{-h}$ for brevity,$$\boxed{\,x_{1}(k+1)=a\,x_{1}(k)+b\,u(k),\qquad \frac{X_{1}(z)}{U(z)}=\frac{1-e^{-h}}{z-e^{-h}}\,}$$The same result follows from the standard hold-equivalent formula $P_{d}(z)=(1-z^{-1})\,\mathcal{Z}\{P(s)/s\}$. Note $a+b=1$, so the discrete model inherits the unit DC gain of the continuous plant.
(b) Write the accumulator. In the discrete block the input to the delay is the sum of $e$ and the delay output, and the delay output is $x_{2}$, so$$x_{2}(k+1)=x_{2}(k)+e(k).$$With $r=0$ the error is $e(k)=-x_{1}(k)$, so$$x_{2}(k+1)=x_{2}(k)-x_{1}(k).$$The block is a discrete integrator: it is what would give the loop zero steady-state error to a constant reference once $u$ is driven from it.
Assemble the state-space model. Stacking the two difference equations with $x=\begin{bmatrix}x_{1} & x_{2}\end{bmatrix}^{\mathsf T}$,$$\boxed{\,x(k+1)=\begin{bmatrix}e^{-h} & 0\\ -1 & 1\end{bmatrix}x(k)+\begin{bmatrix}1-e^{-h}\\ 0\end{bmatrix}u(k),\qquad y(k)=\begin{bmatrix}1 & 0\end{bmatrix}x(k)\,}$$with $D=0$. The matrix is lower triangular, so its open-loop eigenvalues are $z=e^{-h}$ (the sampled plant pole) and $z=1$ (the accumulator), the latter on the unit circle.
Check that the design is possible. The controllability matrix is$$\mathcal{W}_{c}=\begin{bmatrix}B & AB\end{bmatrix}=\begin{bmatrix}b & ab\\ 0 & -b\end{bmatrix},\qquad \det\mathcal{W}_{c}=-b^{2}=-\bigl(1-e^{-h}\bigr)^{2}\neq 0$$for every $h\gt 0$, so both eigenvalues can be placed arbitrarily and a deadbeat solution exists.
(c) Form the closed-loop matrix. With $K=\begin{bmatrix}k_{1} & k_{2}\end{bmatrix}$ and $u(k)=-Kx(k)$,$$A-BK=\begin{bmatrix}a-bk_{1} & -bk_{2}\\ -1 & 1\end{bmatrix}.$$Its characteristic polynomial is $z^{2}-\operatorname{tr}(A-BK)\,z+\det(A-BK)$, and "all poles at zero" means that polynomial must equal $z^{2}$, i.e. both the trace and the determinant must vanish.
Solve the two conditions. The trace condition gives$$a-bk_{1}+1=0\;\Longrightarrow\;k_{1}=\frac{1+a}{b},$$and substituting it into the determinant condition $\det(A-BK)=(a-bk_{1})-bk_{2}=0$ gives $bk_{2}=a-bk_{1}=a-(1+a)=-1$, so$$\boxed{\,K=\begin{bmatrix}\dfrac{1+e^{-h}}{1-e^{-h}} & \dfrac{-1}{1-e^{-h}}\end{bmatrix}\,}$$Both entries grow without bound as $h\to 0$, which is the familiar warning that deadbeat control demands ever larger actuator effort as the sampling rate is raised.
Numerical illustration and interpretation. Taking $h=0.2$ s as a representative period, $a=0.8187$ and $b=0.1813$, so $k_{1}=10.033$ and $k_{2}=-5.5167$; the eigenvalues of $A-BK$ are then both zero to machine precision. Because $A-BK$ is nilpotent with $(A-BK)^{2}=0$, any initial state is driven exactly to the origin in at most two samples, that is in $2h=0.4$ s — the defining property of a deadbeat design, and something no continuous-time controller can achieve.

Question 5 — final results
Quantity	Value
(a) Hold equivalent	$x_{1}(k+1)=e^{-h}x_{1}(k)+(1-e^{-h})u(k)$
(a) Pulse transfer function	$\dfrac{X_{1}(z)}{U(z)}=\dfrac{1-e^{-h}}{z-e^{-h}}$
(b) $A$, $B$	$\begin{bmatrix}e^{-h} & 0\\ -1 & 1\end{bmatrix}$, $\begin{bmatrix}1-e^{-h}\\ 0\end{bmatrix}$
(b) $C$, $D$	$\begin{bmatrix}1 & 0\end{bmatrix}$, $0$
(b) Open-loop eigenvalues	$e^{-h}$ and $1$
(c) Deadbeat gain	$K=\begin{bmatrix}\dfrac{1+e^{-h}}{1-e^{-h}} & \dfrac{-1}{1-e^{-h}}\end{bmatrix}$
(c) Example, $h=0.2$ s	$K=\begin{bmatrix}10.033 & -5.5167\end{bmatrix}$
(c) Settling	exact, in 2 samples

Topic. Exact zero-order-hold discretisation of a continuous plant, sampled-data state-space assembly, and deadbeat pole placement.

Key relations. $A_{d}=e^{A_{c}h}$, $B_{d}=\left(\int_{0}^{h}e^{A_{c}\tau}d\tau\right)B_{c}$; $P_{d}(z)=(1-z^{-1})\mathcal{Z}\{P(s)/s\}$; deadbeat means $\det(zI-A+BK)=z^{n}$; controllability requires $\det\begin{bmatrix}B & AB\end{bmatrix}\neq 0$.

Why this works. The hold makes the input constant between samples, so the continuous solution can be integrated in closed form and the sampled behaviour is exact, not an approximation — this is why the hold equivalent is preferred to a Tustin or Euler substitution when the sample period is not small. Placing every eigenvalue at the origin makes the closed-loop matrix nilpotent, and the Cayley–Hamilton theorem then guarantees that its $n$-th power is exactly zero, so the state reaches the origin in at most $n$ steps rather than merely converging to it. The price is visible in the gain expressions: a shorter sample period gives the controller less time to act, so the demanded control effort grows like $1/h$.

Common pitfall. Discretising with the substitution $z=e^{sh}\approx 1+sh$, which gives $b=h$ instead of $1-e^{-h}$ and silently misplaces the poles at any realistic sample period. The second pitfall is a sign error in the accumulator: the diagram shows a positive feedback path around the delay, which is what makes it an integrator, and the minus sign in $x_{2}(k+1)=x_{2}(k)-x_{1}(k)$ comes from $e=-x_{1}$ with $r=0$, not from the feedback around the delay.

Source. Franklin, Powell & Workman, Digital Control of Dynamic Systems, 3rd ed., §4.3 (hold equivalents) and §8.2–8.3 (pole placement and deadbeat control); Åström & Wittenmark, Computer-Controlled Systems, 3rd ed., §3 and §4.

Question 5 of 6: Discrete-time model and deadbeat state feedback for a sampled-data loop

Question 5: Discrete-time model and deadbeat state feedback for a sampled-data loop (25 marks)