Question 5 of 6: Proportional control of a discrete plant and its continuous-time origin

Question 5: Proportional control of a discrete plant and its continuous-time origin (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 discrete-time plant with poles at $z = 0$ and $z = 0.5$ and a zero outside the unit circle, in a unity-feedback loop with a pure gain.

Find. The stabilising range of $K$, the shape of the root locus, and the continuous-time plant whose zero-order-hold equivalent is $P(z)$.

Approach. Stability in discrete time means all roots inside the unit circle, so the Jury conditions on the quadratic characteristic polynomial replace Routh; the same polynomial yields the locus; and part (c) is the inverse of the standard step-invariant transformation $P(z) = (1 - z^{-1})\,\mathcal{Z}\{P_c(s)/s\}$.

1. Part (a) — form the characteristic polynomial. With unity feedback and $C = K$, $$1 + K P(z) = 0 \;\Longrightarrow\; z(z-0.5) + K(z - 1.5) = 0,$$ $$\Delta(z) = z^{2} + (K - 0.5)\,z - 1.5K = z^{2} + a_1 z + a_0 .$$
2. Apply the Jury test for a monic quadratic. All roots lie inside the unit circle if and only if $$|a_0| \lt 1, \qquad \Delta(1) = 1 + a_1 + a_0 \gt 0, \qquad \Delta(-1) = 1 - a_1 + a_0 \gt 0 .$$ Substituting $a_1 = K - 0.5$ and $a_0 = -1.5K$: $$|1.5K| \lt 1 \Rightarrow -\tfrac{2}{3} \lt K \lt \tfrac{2}{3}; \qquad 0.5 - 0.5K \gt 0 \Rightarrow K \lt 1; \qquad 1.5 - 2.5K \gt 0 \Rightarrow K \lt 0.6 .$$
3. Intersect the three conditions. The binding constraint is $\Delta(-1) \gt 0$, so $$\boxed{-\tfrac{2}{3} \lt K \lt 0.6},\qquad\text{and for a positive gain } \;0 \lt K \lt 0.6 .$$ At $K = 0.6$ the roots are $z = 0.9$ and $z = -1.0$: one root sits exactly on the unit circle at $z=-1$, which in discrete time is a sustained oscillation that alternates sign every sample (period $2h$), not the smooth ringing a continuous system would show.
4. Part (b) — establish the real-axis segments of the locus. For $K \gt 0$ a real point is on the locus when the number of real poles and zeros to its right is odd. Counting: to the right of $z = 1.5$ there is nothing (even, not on the locus); between $0.5$ and $1.5$ there is the zero (odd, on the locus); between $0$ and $0.5$ there are the zero and one pole (even, not on the locus); to the left of the origin there are the zero and both poles (odd, on the locus). The locus therefore occupies $[0.5,\,1.5]$ and $(-\infty,\,0]$.
5. Show the locus never leaves the real axis. The discriminant of $\Delta(z)$ is $$(K-0.5)^{2} + 6K = K^{2} + 5K + 0.25,$$ which is strictly positive for every $K \gt 0$. Both branches are therefore real at all gains: there is no breakaway pair, no complex arc, and no damped-oscillation region. One branch runs from the pole at $z = 0.5$ rightwards into the zero at $z = 1.5$; the other leaves the pole at $z = 0$ and travels left along the negative real axis towards $-\infty$, which is the single asymptote ($n - m = 1$, angle $180^{\circ}$).
6. Mark where the branches cross the unit circle. Setting $\Delta(-1) = 0$ gives $K = 0.6$ (the left-going branch leaving at $z = -1$); setting $\Delta(1) = 0$ gives $K = 1$ (the right-going branch leaving at $z = +1$). The smaller of the two governs, confirming the Jury answer: the loop is already unstable before the right-hand branch reaches the boundary.
7. Part (c) — set up the inverse zero-order-hold problem. The step-invariant (ZOH) transformation is $$P(z) = \left(1 - z^{-1}\right)\mathcal{Z}\!\left\{\frac{P_c(s)}{s}\right\} \;\Longrightarrow\; \mathcal{Z}\!\left\{\frac{P_c(s)}{s}\right\} = \frac{z}{z-1}P(z) = \frac{z - 1.5}{(z-0.5)(z-1)} .$$ The left-hand side is the z-transform of the sampled unit-step response of $P_c$, so recovering that sequence recovers the plant.
8. Expand into sampled step-response values. Partial fractions in the $z/(z-a)$ basis give $$\frac{z-1.5}{(z-0.5)(z-1)} = -\frac{z}{z-1} + \frac{4z}{z-0.5} - 3,$$ so the sampled step response is $$g[n] = -1 + 4(0.5)^{n} - 3\delta[n] \;\Longrightarrow\; g = \{0,\,1,\,0,\,-0.5,\,-0.75,\,-0.875,\ldots\} \to -1 .$$ It starts at zero as it must, jumps to $+1$ after one sample and then decays to $-1$ — another inverse response, consistent with the zero outside the unit circle.
9. Recognise the pole at $z = 0$ as one sample of delay. Sampling maps a continuous pole $s = p$ to $z = e^{ph}$, and no finite $p$ gives $z = 0$; equivalently, the $-3\delta[n]$ term above cannot come from any exponential. The pole at the origin is therefore a pure transport delay of one sample, and the factorisation to invert is $$P(z) = z^{-1}\cdot\frac{z - 1.5}{z - 0.5} .$$
10. Invert the remaining factor. Take the trial form $P_{c0}(s) = (s-p)/(s+p)$, whose step response is $-1 + 2e^{-pt}$. Sampling that at period $h$ and applying the ZOH transformation gives $$\left(1-z^{-1}\right)\left[-\frac{z}{z-1} + \frac{2z}{z-r}\right] = \frac{z - 1.5}{z - 0.5} \quad\text{provided } r = e^{-ph} = 0.5 .$$ Both the sample-by-sample values and the transfer function match exactly, so $$\boxed{P_c(s) = e^{-hs}\,\frac{s - p}{s + p}, \qquad p = \frac{\ln 2}{h}} .$$ With the natural choice $h = 1$ s this is $P_c(s) = e^{-s}\,(s - 0.6931)/(s + 0.6931)$.
11. Sanity-check the result. The DC gain is $P_c(0) = -p/p = -1$, matching $P(1) = -1$ exactly; the right-half-plane zero at $s = +p$ is what produces the sampled zero at $z = 1.5$ outside the unit circle; and the negative DC gain explains why the stabilising gain range is so narrow and why $K$ must stay below 0.6 rather than being generously large.