Question 5 of 6: Sampled-data loop: ZOH equivalent, Jury stability range and a deadbeat redesign

Question 5: Sampled-data loop: ZOH equivalent, Jury stability range and a deadbeat redesign (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 unity-feedback sampled-data loop with a one-sample-delay digital controller driving a first-order continuous plant through a zero-order hold.

Find. (a) $T(z)=Y(z)/R(z)$; (b) the range of $K$ that puts both closed-loop roots inside the unit circle; (c) a stable controller giving zero steady-state step error at the sample instants.

Approach. Replace the hold-plus-plant chain by its exact ZOH equivalent $P(z)$, close the loop algebraically, apply the Jury conditions to the resulting quadratic, and then add a pole at $z=1$ to make the loop Type 1 in the discrete sense.

1. Part (a) — form the ZOH equivalent of the plant. The standard result is $P(z)=\left(1-z^{-1}\right)\mathcal{Z}\left\{P(s)/s\right\}$. Expanding by partial fractions with $a=0.2$,$$\frac{P(s)}{s}=\frac{1}{s\left(s+a\right)}=\frac{1}{a}\left[\frac{1}{s}-\frac{1}{s+a}\right]$$and taking the $z$-transform of each term from the supplied table,$$\mathcal{Z}\left\{\frac{P(s)}{s}\right\}=\frac{1}{a}\left[\frac{z}{z-1}-\frac{z}{z-e^{-ah}}\right]$$
2. Part (a) — evaluate it. Multiplying by $\left(1-z^{-1}\right)=\left(z-1\right)/z$ collapses the bracket:$$P(z)=\frac{1}{a}\left[1-\frac{z-1}{z-e^{-ah}}\right]=\frac{1}{a}\cdot\frac{1-e^{-ah}}{z-e^{-ah}}$$With $a=0.2$ and $h=1$ s, $e^{-0.2}=0.818731$, so$$\boxed{P(z)=\frac{0.906346}{z-0.818731}}$$The check that costs nothing: $P(1)=0.906346/0.181269=5.000$, which is the continuous DC gain $1/0.2$. A ZOH equivalent always preserves the DC gain, and if yours does not, the partial fractions are wrong.
3. Part (a) — close the loop. The forward path is $L(z)=C(z)P(z)=\dfrac{0.906346K}{z\left(z-0.818731\right)}$, and with unity feedback$$T(z)=\frac{L(z)}{1+L(z)}\;\Rightarrow\;\boxed{T(z)=\frac{0.906346K}{z^{2}-0.818731z+0.906346K}}$$The $z^{-1}$ in the controller has become the second closed-loop pole: a one-sample computation delay costs exactly one order.
4. Part (b) — apply the Jury test to the quadratic. For $z^{2}+a_{1}z+a_{0}$ with $a_{1}=-0.818731$ and $a_{0}=0.906346K$, all roots lie inside the unit circle if and only if the three conditions$$\left|a_{0}\right|\lt1,\qquad 1+a_{1}+a_{0}\gt0,\qquad 1-a_{1}+a_{0}\gt0$$hold simultaneously. Taking them in turn:$$\left|0.906346K\right|\lt1\;\Rightarrow\;\left|K\right|\lt1.10333$$$$1-0.818731+0.906346K\gt0\;\Rightarrow\;K\gt-\frac{0.181269}{0.906346}=-0.200$$$$1+0.818731+0.906346K\gt0\;\Rightarrow\;K\gt-2.00666$$
5. Part (b) — intersect them. The binding pair is the first and second conditions, so$$\boxed{-0.200\lt K\lt1.10333}$$Both endpoints have closed forms worth quoting because they are exact rather than numerical:$$K_{\min}=-a=-0.2,\qquad K_{\max}=\frac{a}{1-e^{-ah}}=\frac{0.2}{0.181269}$$At $K=-0.2$ a root leaves through $z=+1$ (the loop loses DC feedback), and at $K=1.10333$ the pair leaves through the unit circle as a complex conjugate pair, i.e. as a sustained oscillation. In practice only $0\lt K\lt1.10333$ is useful, since negative gain gives positive feedback at DC.
6. Part (c) — identify what is missing. With the original controller the loop is Type 0 in the discrete sense: $L(1)=0.906346K/(1-0.818731)=5K$ is finite, so a unit step leaves$$e_{ss}=\frac{1}{1+L(1)}=\frac{1}{1+5K}$$which is $0.153$ even at the largest stable gain. Driving this to zero requires $L(1)\to\infty$, i.e. a discrete integrator — a controller pole at $z=1$. This is the exact discrete analogue of the continuous $1/s$; nothing about the sample rate changes the argument.
7. Part (c) — choose the controller. Take$$\boxed{C(z)=\frac{K\left(z-0.818731\right)}{z-1}}$$which is bi-proper (hence realisable) and cancels the plant pole with its zero, leaving the first-order loop$$L(z)=\frac{0.906346K}{z-1},\qquad T(z)=\frac{0.906346K}{z-1+0.906346K}$$The single closed-loop pole is $z=1-0.906346K$, so the loop is stable for$$0\lt K\lt\frac{2}{0.906346}=2.20666$$and $L(1)=\infty$ gives $e_{ss}=0$ for a step, as required.
8. Part (c) — take the natural gain and get a deadbeat response. Choosing $K$ to put the closed-loop pole at the origin,$$1-0.906346K=0\;\Rightarrow\;\boxed{K=1.10333}\;\Rightarrow\;T(z)=\frac{1}{z}$$The step response is then $Y(z)=T(z)R(z)=1/\left(z-1\right)$, i.e. $y(0)=0$ and $y(k)=1$ for every $k\ge1$: the error is zero after a single sample, not merely in the limit. The control sequence follows from $U(z)/R(z)=C/\left(1+CP\right)=K\left(z-0.818731\right)/z$, giving$$u(0)=1.10333,\qquad u(k)=0.200\ \ (k\ge1)$$and the steady control effort $0.200$ is exactly the value needed to hold the plant output at 1 through its DC gain of 5.