Sampled-data loop — discrete model, Jury stability and ramp error: 22-Elec-B2 Advanced Control Systems, May 2015, Q5

Question 5: Sampled-data loop — discrete model, Jury stability and ramp error (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 continuous first-order plant driven through a zero-order hold and closed by a discrete integral controller.

Quantity	Value
Continuous plant	$P(s) = \dfrac{1}{s+1}$, time constant 1 s
Discrete controller	$C(z) = \dfrac{K}{z-1}$, a pure discrete integrator
Sample period	$h = 0.2\ \text{s}$
Hold	zero-order, between the controller and the plant
Sampled signals	the reference, the hold input and the continuous output

Find. The closed-loop pulse transfer function $T(z)$, the range of $K$ for which the loop is stable, the steady-state error to a unit ramp, and a comment on what the output does between the sampling instants.

Question 5: sampled-data loop. The reference, the ZOH input and the continuous output are all sampled with the same period h, so the loop closes entirely in the z-domain.

Approach. Replace the hold-plus-plant cascade by its exact discrete equivalent, close the loop entirely in the z-domain, apply the Jury conditions to the resulting quadratic, and use the discrete velocity constant for the ramp error.

Part (a) — find the zero-order-hold equivalent of the plant. The hold and the plant together are described exactly by $$G(z) = \bigl(1 - z^{-1}\bigr)\,\mathcal{Z}\!\left\{\frac{P(s)}{s}\right\} = \bigl(1 - z^{-1}\bigr)\,\mathcal{Z}\!\left\{\frac{1}{s(s+1)}\right\}.$$ Splitting into partial fractions, $\dfrac{1}{s(s+1)} = \dfrac{1}{s} - \dfrac{1}{s+1}$, and taking the two standard transforms from the appended table, $$\mathcal{Z}\!\left\{\frac{1}{s(s+1)}\right\} = \frac{z}{z-1} - \frac{z}{z - e^{-h}} .$$
Simplify the discrete plant. Multiplying by $(1 - z^{-1}) = (z-1)/z$ collapses the difference to a single first-order term, $$G(z) = \frac{1 - e^{-h}}{z - e^{-h}}, \qquad e^{-0.2} = 0.81873,$$ $$G(z) = \frac{0.18127}{z - 0.81873}.$$ The discrete pole is the continuous pole mapped through $z = e^{sh}$, and the numerator is what is left of the DC gain after one sample period of charging.
Close the loop. Because every signal entering the summing junction is sampled at the same instants, the loop closes in the z-domain with no hidden inter-sample path: $$T(z) = \frac{C(z)G(z)}{1 + C(z)G(z)} = \frac{0.18127K}{(z-1)(z - 0.81873) + 0.18127K},$$ $$\boxed{\;T(z) = \frac{0.18127\,K}{z^2 - 1.81873\,z + \bigl(0.81873 + 0.18127K\bigr)}\;}$$
Part (b) — state the Jury conditions. For a second-order characteristic polynomial $z^2 + a_1 z + a_0$ all roots lie strictly inside the unit circle if and only if $$|a_0| \lt 1, \qquad 1 + a_1 + a_0 \gt 0, \qquad 1 - a_1 + a_0 \gt 0 .$$ Here $a_1 = -1.81873$ and $a_0 = 0.81873 + 0.18127K$.
Apply the three tests. Taking them in turn, $$1 + a_1 + a_0 = 1 - 1.81873 + 0.81873 + 0.18127K = 0.18127K \gt 0 \;\Longrightarrow\; K \gt 0,$$ $$1 - a_1 + a_0 = 3.63746 + 0.18127K \gt 0 \quad\text{(automatic for } K \gt 0),$$ $$a_0 \lt 1: \quad 0.81873 + 0.18127K \lt 1 \;\Longrightarrow\; K \lt \frac{1 - 0.81873}{0.18127} = 1 .$$ The upper limit is exactly unity because the numerator $1 - e^{-h}$ and the quantity $1 - e^{-h}$ appearing in the test are the same number — the result is independent of the sample period chosen.
State the stability range. Combining the three conditions, $$\boxed{\;0 \lt K \lt 1\;}$$ At $K = 1$ the roots are $z = 0.9094 \pm j0.4160$, of modulus exactly 1: the loop is marginally stable and rings at $\arg z / h = 0.4290/0.2 = 2.145\ \text{rad/s}$, a period of about 2.93 s.

Question 5: discrete root locus. The two branches leave the open-loop poles at z = 1 and z = 0.8187, meet, and reach the unit circle exactly at K = 1.

Part (c) — form the discrete velocity constant. The loop has a pole at $z = 1$ supplied by the controller, so it is Type 1 and a ramp produces a finite error. The discrete velocity constant is $$K_v = \frac{1}{h}\lim_{z \to 1}(z-1)\,C(z)G(z) = \frac{1}{h}\lim_{z \to 1}(z-1)\frac{K}{z-1}\cdot\frac{1 - e^{-h}}{z - e^{-h}} = \frac{1}{h}\cdot\frac{K\,(1 - e^{-h})}{1 - e^{-h}} = \frac{K}{h}.$$
Evaluate the ramp error. With $K_v = K/h = 5K$, $$\boxed{\;e_{ss} = \frac{1}{K_v} = \frac{h}{K} = \frac{0.2}{K}\;}$$ Since stability caps $K$ below 1, the smallest error the loop can ever achieve is $e_{ss} \gt 0.2$ — one sample period's worth of ramp travel. At the practical setting $K = 0.5$ the error is 0.4. This bound, not the algebra, is the point of the part: no choice of gain within the stable range makes the loop track a ramp closely, and shortening $h$ is the only remedy.
Comment on the inter-sample behaviour. Everything above describes $y$ only at $t = kh$. Between samples the hold freezes the control at its last value while the plant, a first-order lag, relaxes towards the corresponding final value, so $y(t)$ is a concatenation of exponential arcs of time constant 1 s — not the staircase the hold produces, and not a straight line either. Two consequences follow. First, the reference ramp keeps climbing during each interval while $u$ is frozen, so the true continuous error is larger than the sampled value for most of every period and the boxed result is a best-case reading. Second, as $K$ approaches its stability limit the closed-loop poles approach the unit circle at $\pm 24.6^\circ$, so the sampled sequence rings at about $2.1\ \text{rad/s}$; because that is well below the sampling rate of $2\pi/h \approx 31\ \text{rad/s}$ the hold does not conceal any faster oscillation, and the sampled picture is a faithful summary of the continuous one. Had the ringing frequency approached the Nyquist rate, hidden oscillation between samples would have to be checked explicitly.

Question 5 — results
Quantity	Symbol	Value
Discrete plant pole	$e^{-h}$	0.81873
Zero-order-hold equivalent	$G(z)$	$\dfrac{0.18127}{z - 0.81873}$
(a) Closed-loop pulse transfer function	$T(z)$	$\dfrac{0.18127K}{z^2 - 1.81873z + 0.81873 + 0.18127K}$
(b) Stability range	$K$	$0 \lt K \lt 1$
Roots at the stability limit	$z$	$0.9094 \pm j0.4160$, $\|z\| = 1$
(c) Discrete velocity constant	$K_v$	$K/h = 5K$
(c) Steady-state ramp error	$e_{ss}$	$0.2/K$ (0.4 at $K = 0.5$)
(c) Best achievable ramp error	$\min e_{ss}$	$\gt 0.2$

Topic. Discrete equivalents of continuous plants, the Jury stability test, and steady-state error constants in the z-domain.

Key relations. $G(z) = (1 - z^{-1})\mathcal{Z}\{P(s)/s\}$; for a first-order plant $\dfrac{a}{s+a} \to \dfrac{1 - e^{-ah}}{z - e^{-ah}}$; Jury for $z^2 + a_1z + a_0$: $|a_0| \lt 1$, $1 \pm a_1 + a_0 \gt 0$; $K_v = \frac{1}{h}\lim_{z\to1}(z-1)L(z)$ and $e_{ss} = 1/K_v$ for a unit ramp.

Why this works. The zero-order hold makes the cascade exactly representable in discrete time — no approximation is involved — because the input to the plant really is piecewise constant. Once every signal at the summing junction is sampled synchronously, ordinary single-loop algebra applies with $z$ in place of $s$, and the unit circle replaces the imaginary axis as the stability boundary.

Common pitfall. Taking $G(z)$ as the z-transform of $P(s)$ alone and omitting the $(1 - z^{-1})$ factor, which misplaces the numerator and shifts the stability limit. A second is dividing by $h$ in the wrong direction when forming $K_v$; the check is dimensional, since $e_{ss} = h/K$ must carry the units of the ramp times a time. A third is answering the inter-sample question with “the output is a staircase” — the hold output is a staircase, but $y$ is the plant's smooth response to it.

Source. Franklin, Powell & Workman, Digital Control of Dynamic Systems, 3rd ed., ch. 4 and ch. 8 (discrete equivalents, the Jury test and discrete error constants); Ogata, Discrete-Time Control Systems, 2nd ed., ch. 3 and ch. 4.

Question 5 of 6: Sampled-data loop — discrete model, Jury stability and ramp error

Question 5: Sampled-data loop — discrete model, Jury stability and ramp error (25 marks)