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Digital PI design for plant with 2 s dead time

Short answer

Dead time 2T gives z⁻²; G(z) = 0.6321/[z²(z−0.3679)]. Desired dominant poles: 0.6958∠36°. PI controller G_D(z) = K(z−α)/(z−1) has zero at α = 0.5873 to satisfy angle condition. Gain K = 0.5070 yields G_D(z) = 0.5070(z−0.5873)/(z−1) with K_P = 0.2977, K_I = 0.2092. Velocity constant K_v = 0.2092 s⁻¹ and ramp steady‑state error e_ss = 4.78.

Design a digital PI controller such that the dominant closed loop poles have damping ratio ζ=0.5, sampling period T = 1 and ωd/ωₛ=1/10 (10 samples per cycle of sinusoidal oscillation) and dead time of 2 sec. Also find Kᵥ and eₛₛ in response to unit ramp input. [Figure: unity-feedback loop: sampler → GD(z) → ZOH (1-e⁻ᵀˢ)/s → e⁻²ˢ/(s+1) → output]

Step 1: Plant pulse transfer function (T=1T = 1 s)

The dead time of 2 s is exactly 2T2T, so it becomes z−2z^{-2}:

G(z)=z−2(1−z−1) Z[1s(s+1)]=z−2(1−e−1)z−11−e−1z−1=0.6321z2(z−0.3679)\begin{aligned} G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+1)}\right] = z^{-2}\frac{(1-e^{-1})z^{-1}}{1-e^{-1}z^{-1}} \\ &= \frac{0.6321}{z^2(z - 0.3679)} \end{aligned}

Step 2: Desired dominant poles

With 10 samples per cycle, ωdT=2π/10=36∘\omega_dT = 2\pi/10 = 36^\circ, and

∣z∣=exp⁡(−2πζ1−ζ2⋅ωdωs)=exp⁡(−2π(0.5)0.866⋅110)=0.6958z1=0.6958∠36∘=0.5629+j0.4090\begin{aligned} |z| &= \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{\omega_d}{\omega_s}\right) = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac{1}{10}\right) = 0.6958 \\ z_1 &= 0.6958\angle 36^\circ = 0.5629 + j0.4090 \end{aligned}

Step 3: PI controller and angle condition

GD(z)=KP+KI1−z−1=Kz−αz−1,K=KP+KI,  α=KPKP+KIG_D(z) = K_P + \frac{K_I}{1-z^{-1}} = K\frac{z-\alpha}{z-1}, \qquad K = K_P + K_I,\; \alpha = \frac{K_P}{K_P+K_I}

Angles at z1z_1:

FactorAngle
double pole at origin (z2z^2)−2(36∘)=−72∘-2(36^\circ) = -72^\circ
pole at 0.36790.3679−64.51∘-64.51^\circ
controller pole at 11−136.91∘-136.91^\circ
Sum of pole angles−273.41∘-273.41^\circ

For −180∘-180^\circ, the controller zero must contribute +93.41∘+93.41^\circ:

α=0.5629−0.4090tan⁡93.41∘=0.5629+0.0244=0.5873\alpha = 0.5629 - \frac{0.4090}{\tan 93.41^\circ} = 0.5629 + 0.0244 = 0.5873

Step 4: Gain KK

K=∣z1∣2 ∣z1−0.3679∣ ∣z1−1∣0.6321 ∣z1−0.5873∣=0.4841×0.4531×0.59860.6321×0.4097=0.5070K = \frac{|z_1|^2\,|z_1-0.3679|\,|z_1-1|}{0.6321\,|z_1-0.5873|} = \frac{0.4841 \times 0.4531 \times 0.5986}{0.6321 \times 0.4097} = 0.5070 GD(z)=0.5070 z−0.5873z−1G_D(z) = 0.5070\,\frac{z - 0.5873}{z - 1} KP=Kα=0.2977,KI=K(1−α)=0.2092K_P = K\alpha = 0.2977, \qquad K_I = K(1-\alpha) = 0.2092

Step 5: KvK_v and steady-state error for a ramp

Kv=lim⁡z→1(1−z−1)GD(z)G(z)T=K(1−α)T⋅0.63211−0.3679=0.2092×1=0.2092 s−1ess=1Kv=4.78\begin{aligned} K_v &= \lim_{z\to1}\frac{(1-z^{-1})G_D(z)G(z)}{T} = \frac{K(1-\alpha)}{T}\cdot\frac{0.6321}{1-0.3679} = 0.2092 \times 1 = 0.2092\ \text{s}^{-1} \\ e_{ss} &= \frac{1}{K_v} = 4.78 \end{aligned}

Check: the closed-loop characteristic equation z3(z−1)−0.3679z2(z−1)+0.3205(z−0.5873)=0z^3(z-1) - 0.3679z^2(z-1) + 0.3205(z-0.5873) = 0 has roots 0.5629±j0.40900.5629 \pm j0.4090, 0.75620.7562 and −0.5141-0.5141. The desired pair is obtained. The real pole at 0.7562 lies farther from the origin than the desired pair, so it is not negligible. The PI zero at 0.5873 partly offsets it, but the response will be somewhat slower than the pair alone suggests.

Answer: GD(z)=0.5070z−0.5873z−1G_D(z) = 0.5070\dfrac{z-0.5873}{z-1} (that is, KP=0.2977K_P = 0.2977, KI=0.2092K_I = 0.2092), with Kv=0.2092 s−1K_v = 0.2092\ \text{s}^{-1} and ess=4.78e_{ss} = 4.78 for a unit ramp.

✓ Answer checked by Claude Opus 5.5 (high effort)

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