Deadbeat design of a digital PI controller for an integrating plant: 22-Elec-B2 Advanced Control Systems, December 2013, Q4

Question 4: Deadbeat design of a digital PI controller for an integrating plant (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 digital PI controller driving a pure integrator through a zero-order hold, with everything sampled at the same period.

Given data — Question 4
Quantity	Value
Controller	$C(z) = K_{1} + \dfrac{K_{2}}{1-z^{-1}}$ (proportional plus discrete integral)
Continuous plant	$P(s) = 1/s$
Hold	zero-order, output constant over each period
Sample period	$h$ (symbolic; $h=0.5$ s is used for the plotted response)
Requirement	all closed-loop poles at $z=0$ (deadbeat)

Find. (a) $K_{1}$ and $K_{2}$ as functions of $h$; (b) the closed-loop transfer function $T(z)=Y(z)/R(z)$; (c) the unit-step response of $y(t)$, including what happens between the sampling instants.

Figure Q4.1 — the sampled-data loop. Everything inside the digital part advances once per period h; the ZOH holds u constant between samples, so the continuous integrator sees a staircase input.

Approach. Replace the hold-plus-plant combination by its exact discrete equivalent, form the closed-loop characteristic polynomial in $z$, force it to equal $z^{2}$, and then reconstruct the continuous output by integrating the resulting staircase of control values.

Discretise the hold and plant together. The zero-order-hold equivalent is $$G(z) = \left(1-z^{-1}\right)\mathcal{Z}\!\left\{\frac{P(s)}{s}\right\} = \left(1-z^{-1}\right)\mathcal{Z}\!\left\{\frac{1}{s^{2}}\right\} = \frac{z-1}{z}\cdot\frac{hz}{(z-1)^{2}} = \frac{h}{z-1},$$ using the $t \leftrightarrow hz/(z-1)^{2}$ entry of the supplied transform table. Physically this says the obvious thing: an integrator fed a constant $u$ for one period gains exactly $hu$.
Put the controller over a common denominator. $$C(z) = K_{1} + \frac{K_{2}}{1-z^{-1}} = K_{1} + \frac{K_{2}z}{z-1} = \frac{(K_{1}+K_{2})z - K_{1}}{z-1}.$$ The controller pole at $z=1$ is the discrete integrator, which is what will guarantee zero steady-state step error.
Form the closed-loop characteristic polynomial. The open loop is $L(z)=C(z)G(z)$, so $1+L(z)=0$ gives $$(z-1)^{2} + h\big[(K_{1}+K_{2})z - K_{1}\big] = 0,$$ which expands to $$z^{2} + \big[h(K_{1}+K_{2}) - 2\big] z + \big(1 - hK_{1}\big) = 0 .$$
Impose the deadbeat condition. Both poles at the origin means the characteristic polynomial must be exactly $z^{2}$, so both remaining coefficients vanish: $$1 - hK_{1} = 0 \;\Rightarrow\; K_{1} = \frac{1}{h}; \qquad h(K_{1}+K_{2}) - 2 = 0 \;\Rightarrow\; K_{1}+K_{2} = \frac{2}{h} \;\Rightarrow\; K_{2} = \frac{1}{h}.$$ $$\boxed{K_{1}=K_{2}=\frac{1}{h}, \qquad C(z) = \frac{1}{h}\left[1 + \frac{z}{z-1}\right] = \frac{1}{h}\cdot\frac{2z-1}{z-1}}$$ At $h = 0.5$ s this is $K_{1}=K_{2}=2$. The gains scale as $1/h$: a faster sampler needs proportionally more gain to move the integrator the same distance in one period.
Assemble the closed-loop transfer function. With the denominator now equal to $z^{2}$, $$T(z) = \frac{C(z)G(z)}{1+C(z)G(z)} = \frac{h\big[(K_{1}+K_{2})z-K_{1}\big]}{z^{2}} = \frac{h\left[\tfrac{2}{h}z - \tfrac{1}{h}\right]}{z^{2}} \quad\Longrightarrow\quad \boxed{T(z) = \frac{2z-1}{z^{2}} = 2z^{-1}-z^{-2}}$$ Note that $T(z)$ is independent of $h$: the sample period is absorbed entirely into the controller gains. Checking the DC gain, $T(1) = (2-1)/1 = 1$, so the steady-state step error is zero as the integrator promised.
Evaluate the sampled step response. Because $T(z)=2z^{-1}-z^{-2}$ is a two-term finite impulse response, the unit-step response is just the running sum of its coefficients $\{0,\,2,\,-1\}$: $$y(0)=0,\quad y(h)=2,\quad y(2h)=2-1=1,\quad y(kh)=1 \ \ \text{for all } k\ge 2 .$$ The output reaches its final value in two steps and stays there exactly — the defining property of a deadbeat design, and the reason the poles were placed at the origin.
Recover the inter-sample behaviour. The control sequence follows from $U(z)/R(z) = C(z)\big[1-T(z)\big]$ with $R(z)=z/(z-1)$, which reduces to $U(z)=\dfrac{1}{h}\left(2 - z^{-1}\right)$, that is $$u(0)=\frac{2}{h},\qquad u(h)=-\frac{1}{h},\qquad u(kh)=0 \ \ (k\ge2).$$ Since the hold keeps each value constant for a full period and the plant integrates, $y(t)$ is piecewise linear, not a staircase: it ramps up at slope $2/h$ from $0$ to $2$ over $0\le t\lt h$, ramps down at slope $-1/h$ from $2$ to $1$ over $h\le t\lt 2h$, and is then exactly constant at 1. At $h = 0.5$ s the two control values are $+4$ and $-2$. $$\boxed{y(t)\ \text{rises linearly to } 2 \text{ at } t=h,\ \text{falls linearly to } 1 \text{ at } t=2h,\ \text{and stays at } 1}$$

Figure Q4.2 — unit-step response with h = 0.5 s. Between samples y(t) is a straight ramp because P(s) = 1/s integrates the held control: it climbs to 2 at t = h, falls back to 1 at t = 2h, then stays flat. The sampled sequence is deadbeat (final value reached at k = 2) but the continuous output overshoots to 2.0 in between — that overshoot is the price of placing both poles at z = 0.

Check: the sampled response hides a 100 % overshoot. The sampled sequence $\{0,2,1,1,\dots\}$ settles in two steps, but the continuous output passes through 2 on the way — twice the set-point. Any specification written on $y(t)$ rather than on $y(kh)$ would fail this design. This is the standard objection to deadbeat control: it also demands large control amplitudes ($2/h$, unbounded as the sampler is speeded up) and it is exquisitely sensitive to plant gain error, because a pole at $z=0$ has no stability margin to lose. A more conservative pole placement inside the unit circle trades finite settling for a gentler response. The question also asks for $T(z)$ relating $X(z)$ to $R(z)$; $X$ is not defined anywhere in the paper and is read here as the output $Y$.

Final results — Question 4
Quantity	Result
ZOH equivalent of the plant	$G(z) = h/(z-1)$
Characteristic polynomial	$z^{2}+[h(K_{1}+K_{2})-2]z+(1-hK_{1})$
Deadbeat gains	$K_{1}=K_{2}=1/h$ (both $=2$ at $h=0.5$ s)
Controller	$C(z) = \dfrac{2z-1}{h(z-1)}$
Closed-loop transfer function	$T(z) = \dfrac{2z-1}{z^{2}}$, $T(1)=1$
Sampled step response	$y(kh) = 0,\ 2,\ 1,\ 1,\ 1,\dots$
Control sequence	$u(kh) = 2/h,\ -1/h,\ 0,\ 0,\dots$
Continuous output	piecewise linear; peak 2 at $t=h$, constant 1 for $t\ge 2h$

Topic. Zero-order-hold equivalents, deadbeat pole placement for a digital PI controller, and inter-sample behaviour.

Key relations. $G(z)=(1-z^{-1})\mathcal{Z}\{P(s)/s\}$; for $P=1/s$, $G(z)=h/(z-1)$; closed-loop characteristic polynomial $(z-1)^{2}+h[(K_{1}+K_{2})z-K_{1}]$; deadbeat requires that polynomial to be $z^{n}$; $U(z)/R(z)=C(z)[1-T(z)]$; for an integrator plant $y(t)$ is the running integral of the held control, hence piecewise linear.

Why this works. Placing all closed-loop poles at $z=0$ makes the closed-loop transfer function a polynomial in $z^{-1}$, so its impulse response has finite length and its step response reaches the final value in as many steps as there are poles. For a first-order discrete plant with a PI controller there are two poles, hence two steps. The design reduces to matching two coefficients, which is why it can be done in closed form with no iteration. Recovering the inter-sample response then requires nothing beyond the control sequence, because a zero-order hold plus an integrator turns a sequence of numbers into a chain of straight ramps.

Common pitfall. Sketching the step response as a staircase. That is what the sampled data look like, but the physical output of an integrating plant driven by a held control cannot jump — it ramps, and the peak value of 2 lives strictly between samples. A second pitfall is forgetting the $(1-z^{-1})$ factor when discretising, which produces $G(z)=hz/(z-1)^{2}$ and a spurious extra pole. A third is to treat $K_{1}$ and $K_{2}$ as dimensionless constants; both carry a factor $1/h$, and a design transplanted to a different sample rate without rescaling will not be deadbeat.

Source. Franklin, Powell & Workman, Digital Control of Dynamic Systems, 3rd ed., Ch. 4 (discrete equivalents) and §8.3 (deadbeat and finite-settling design); Ogata, Discrete-Time Control Systems, 2nd ed., Ch. 6.

Question 4 of 6: Deadbeat design of a digital PI controller for an integrating plant

Question 4: Deadbeat design of a digital PI controller for an integrating plant (25 marks)