Question 4 of 6: Deadbeat digital control of an integrator behind a zero-order hold

Question 4: Deadbeat digital control of an integrator behind a zero-order hold (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 single-loop sampled-data system: reference, controller output and plant output all sampled at the same period, with a zero-order hold ahead of a continuous integrator.

Find. The two controller gains, the resulting discrete closed-loop transfer function, and the unit-step response including what happens between sampling instants.

Approach. Replace the continuous plant by its ZOH equivalent, form the closed-loop characteristic polynomial in $z$, force it to $z^{2}$ to place both poles at the origin, then invert $T(z)R(z)$ for the sample sequence and integrate the held control between samples for the continuous trace.

1. Part (a) — find the ZOH equivalent of the plant. For an integrator behind a zero-order hold the discrete transfer function follows from the standard pair $1/s^{2}\leftrightarrow hz/(z-1)^{2}$:$$G(z)=(1-z^{-1})\,\mathcal{Z}\!\left\{\frac{1}{s^{2}}\right\}=\frac{z-1}{z}\cdot\frac{hz}{(z-1)^{2}}=\frac{h}{z-1}$$So one sample period of held input produces exactly $h$ units of output — the discrete model of an integrator is an accumulator with gain $h$.
2. Form the closed-loop characteristic polynomial. With $C(z)=(k_{1}z+k_{0})/(z-1)$ the loop gain is $L(z)=h(k_{1}z+k_{0})/(z-1)^{2}$, so$$(z-1)^{2}+h(k_{1}z+k_{0})=z^{2}+\left(hk_{1}-2\right)z+\left(1+hk_{0}\right)$$The controller supplies its own pole at $z=1$ — discrete integral action — which is what will eventually kill the steady-state step error.
3. Impose all poles at the origin. Deadbeat means the characteristic polynomial is $z^{2}$, so both non-leading coefficients vanish:$$hk_{1}-2=0,\qquad 1+hk_{0}=0\quad\Longrightarrow\quad \boxed{\;k_{1}=\frac{2}{h},\qquad k_{0}=-\frac{1}{h}\;}$$$$C(z)=\frac{2z-1}{h\,(z-1)}$$Both gains scale as $1/h$: a faster sampler needs proportionally more gain to move the output the same distance in one step.
4. Part (b) — assemble the closed-loop transfer function. Since the characteristic polynomial is now $z^{2}$, the closed loop is simply the loop numerator over $z^{2}$:$$T(z)=\frac{L(z)}{1+L(z)}=\frac{h\left(k_{1}z+k_{0}\right)}{z^{2}}=\boxed{\;\frac{2z-1}{z^{2}}=2z^{-1}-z^{-2}\;}$$Notice that $h$ has cancelled completely: the sampled closed-loop behaviour is identical for every sample period, a hallmark of deadbeat design on an integrator plant. Checking the DC gain, $T(1)=(2-1)/1=1$, so the step is tracked exactly.
5. Part (c) — get the sampled step sequence. With $R(z)=z/(z-1)$,$$Y(z)=T(z)R(z)=\frac{2z-1}{z(z-1)}=2z^{-1}+z^{-2}+z^{-3}+\cdots$$so the output samples are $y(0)=0$, $y(h)=2$, and $y(kh)=1$ for every $k\ge 2$. The error sequence is $e=\{1,-1,0,0,\dots\}$ and the control sequence is$$u=\left\{\frac{2}{h},\ -\frac{1}{h},\ 0,\ 0,\dots\right\}$$Two moves of the actuator and the loop is finished — that is what deadbeat buys.
6. Reconstruct the inter-sample behaviour, which is the graded part. Between samples the input to the plant is a CONSTANT held value and the plant is an integrator, so $y(t)$ is a straight line on each interval with slope equal to that held value. It is never a staircase. Concretely:$$y(t)=\frac{2}{h}t \ \ \text{on } [0,h);\qquad y(t)=2-\frac{1}{h}(t-h)\ \ \text{on } [h,2h);\qquad y(t)=1\ \ \text{for } t\ge 2h$$The response therefore ramps up to exactly 2 at $t=h$, ramps back down to 1 at $t=2h$, and then sits flat forever.
7. State the design consequence. The peak of 2 is a 100 percent overshoot, and it is reached at a sampling instant, so it is visible in the sampled sequence as well as between samples. Deadbeat control on this plant settles in two samples but pays for it with an actuator excursion of $2/h$ and a doubled output — on a real machine either the actuator saturates or the overshoot is unacceptable, and the usual remedy is to place the closed-loop poles somewhere inside the unit circle rather than at the origin.