Question 5 of 6: Sampled-data loop — closed-loop pulse transfer function, root locus and inter-sample response

Question 5: Sampled-data loop — closed-loop pulse transfer function, root locus and inter-sample response (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 control loop in which the reference, the controller output and the plant output are all sampled with period $h$, and a zero-order hold reconstructs the control signal.

Find. The closed-loop pulse transfer function $T(z) = Y(z)/R(z)$, the root locus as $K$ runs from zero to infinity, whether stability holds for every $K$ (and the limiting value if not), and the continuous output $y(t)$ over $0 \le t \le 4h$ when $Kh = 1.5$.

Approach. Replace the hold-plus-plant chain by its zero-order-hold equivalent in $z$, close the loop in the discrete domain, analyse the resulting second-order characteristic polynomial with the root-locus rules and the Jury conditions, and finally propagate the difference equations sample by sample — then reconstruct the continuous signal by remembering that a held input into an integrator produces straight ramps.

1. Part (a) — find the zero-order-hold equivalent of the plant. The hold and the plant must be discretised together, never separately: $$G(z) = \left(1 - z^{-1}\right)\mathcal{Z}\!\left\{\frac{P(s)}{s}\right\} = \left(1 - z^{-1}\right)\mathcal{Z}\!\left\{\frac{1}{3s^{2}}\right\}.$$ From the transform table appended to the paper, $1/s^{2}$ maps to $hz/(z-1)^{2}$, so $$G(z) = \frac{z-1}{z}\cdot\frac{1}{3}\cdot\frac{hz}{(z-1)^{2}} = \frac{h}{3(z-1)} .$$ The continuous integrator becomes a discrete integrator with gain $h/3$, which is exactly what one expects: over one sample the held input $u$ moves the output by $uh/3$.
2. Close the loop. The open-loop pulse transfer function is $$L(z) = C(z)G(z) = \frac{K}{z}\cdot\frac{h}{3(z-1)} = \frac{Kh/3}{z(z-1)},$$ and with unity feedback $$\boxed{\;T(z) = \frac{L}{1+L} = \frac{Kh/3}{z^{2} - z + Kh/3}\;}$$ It is convenient to name the single dimensionless design parameter $\beta \equiv Kh/3$; the whole of parts (b) and (c) depends on $K$ and $h$ only through it. The controller's one-sample delay is what makes the loop second order — it supplies the open-loop pole at $z = 0$ that the plant does not have.
3. Part (b) — set up the root locus. The characteristic equation is $z^{2} - z + \beta = 0$, i.e. $1 + \beta/[z(z-1)] = 0$. There are two open-loop poles, at $z = 0$ and $z = 1$, and no finite zeros, so both branches run to infinity. The real-axis segment between the poles, $0 \le z \le 1$, lies to the left of an odd number of poles and therefore belongs to the locus; the two asymptotes are at $\pm 90^\circ$ about the centroid $(0+1)/2 = 0.5$.
4. Locate the breakaway point and the vertical branch. The roots are $$z = \frac{1 \pm \sqrt{1 - 4\beta}}{2},$$ which are real and distinct for $\beta \lt 1/4$, and coincide at $$\boxed{\;z = 0.5 \ \text{ when }\ \beta = \tfrac14,\ \text{ i.e. } Kh = 0.75\;}$$ For $\beta \gt 1/4$ the pair becomes complex with constant real part $1/2$, so the locus leaves the real axis at right angles and travels straight up and down the line $\operatorname{Re}z = 0.5$. Since the product of the roots equals the constant term, $|z|^{2} = \beta$, so the radius grows as $\sqrt{\beta}$ along that vertical line.
5. Part (c) — find where the locus leaves the unit circle. The system is not stable for all $K$: the branches march steadily outward. Setting $|z| = \sqrt{\beta} = 1$ gives $\beta = 1$, at which $z = 0.5 \pm j\sqrt{3}/2 = 1\angle{\pm}60^\circ$, exactly on the unit circle. Hence $$\boxed{\;\text{stable for } 0 \lt K \lt K_{crit} = \frac{3}{h}, \qquad \text{marginal at } K = 3/h\;}$$ The same conclusion follows from the Jury test applied to $z^{2} + a_1 z + a_0$ with $a_1 = -1$, $a_0 = \beta$: the three conditions $|a_0| \lt 1$, $1 + a_1 + a_0 \gt 0$ and $1 - a_1 + a_0 \gt 0$ reduce to $0 \lt \beta \lt 1$. Note the limit is inversely proportional to $h$ — sampling more slowly reduces the gain the loop can tolerate, which is the discrete-time face of the phase lag that the hold introduces.

6. Part (d) — propagate the sampled loop. With $Kh = 1.5$, $\beta = 0.5$, so $T(z) = 0.5/(z^{2} - z + 0.5)$ and the difference equation is $$y(k) = y(k-1) - 0.5\,y(k-2) + 0.5\,r(k-2).$$ The reference is the single pulse $r = \{1, 0, 0, 0, 0\}$. Stepping forward from rest, $$y = \{\,0,\; 0,\; 0.5,\; 0.5,\; 0.25\,\} \ \text{ at }\ t = 0,\,h,\,2h,\,3h,\,4h .$$ The two leading zeros are the two open-loop delays — one from the controller, one inherent in the hold — so nothing can appear at the output before $t = 2h$.
7. Recover the control sequence. The error is $e(k) = r(k) - y(k)$ and the controller gives $u(k) = K\,e(k-1)$: $$e = \{\,1,\ 0,\ -0.5,\ -0.5,\ -0.25\,\}, \qquad u\,h = \{\,0,\ 1.5,\ 0,\ -0.75,\ -0.75\,\}.$$ Quoting $uh$ rather than $u$ keeps the answer free of the unspecified sample period; the control itself is $u(h) = 1.5/h$, which grows without bound as the sampling is made faster, the usual signature of a deadbeat-style pulse response.
8. Reconstruct the continuous output. This is the part the sampled sequence alone cannot tell you. Between samples the hold keeps $u$ constant while the plant integrates it, so $$\dot{y} = \frac{u}{3} \;\Longrightarrow\; \Delta y = \frac{u\,h}{3} \ \text{ over each interval},$$ giving increments $\{0,\ +0.5,\ 0,\ -0.25\}$ across the four intervals. The continuous output is therefore piecewise linear: $$\boxed{\;y(t) = \begin{cases} 0, & 0 \le t \lt h\\[2pt] 0.5\,(t/h - 1), & h \le t \lt 2h\\[2pt] 0.5, & 2h \le t \lt 3h\\[2pt] 0.5 - 0.25\,(t/h - 3), & 3h \le t \le 4h \end{cases}\;}$$ with $y(4h) = 0.25$. It is flat over the first interval (no control yet), ramps linearly up to $0.5$ over the second, holds while $u = 0$, then ramps down to $0.25$.
9. Annotate the sketch. Three features earn the marks: the response is made of straight line segments, never a staircase, because a zero-order hold feeding an integrator produces ramps; the peak value $0.5$ is reached exactly at $t = 2h$ and the continuous signal never exceeds its sampled values, so unlike many sampled-data problems there is no hidden inter-sample overshoot here; and the ringing that follows ($0.5$, $0.25$, then negative values beyond $4h$) reflects the complex pole pair at $0.5 \pm j0.5 = 0.707\angle{\pm}45^\circ$, whose modulus $0.707$ means each oscillation decays by about 30 % per sample.