---
title: "Digital PI design for plant with 2 s dead time"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:pi-design-dead-time-2s"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/pi-design-dead-time-2s
site: BCT Engineering
---

# Digital PI design for plant with 2 s dead time

Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System

- Asked in: 2081 Chaitra, 12 marks; 2074 Bhadra, 12 marks; 2082 Kartik, 12 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/pi-design-dead-time-2s
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

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]

## 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.

## Answer

#### Step 1: Plant pulse transfer function ($T = 1$ s)
The dead time of 2 s is exactly $2T$, so it becomes $z^{-2}$:

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

#### Step 2: Desired dominant poles
With 10 samples per cycle, $\omega_dT = 2\pi/10 = 36^\circ$, and

$$
\begin{aligned}
|z| &= \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{\omega_d}{\omega_s}\right) = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac{1}{10}\right) = 0.6958 \\
z_1 &= 0.6958\angle 36^\circ = 0.5629 + j0.4090
\end{aligned}
$$

#### Step 3: PI controller and angle condition
$$
G_D(z) = K_P + \frac{K_I}{1-z^{-1}} = K\frac{z-\alpha}{z-1}, \qquad K = K_P + K_I,\; \alpha = \frac{K_P}{K_P+K_I}
$$

Angles at $z_1$:

| Factor | Angle |
|---|---|
| double pole at origin ($z^2$) | $-2(36^\circ) = -72^\circ$ |
| pole at $0.3679$ | $-64.51^\circ$ |
| controller pole at $1$ | $-136.91^\circ$ |
| **Sum of pole angles** | $-273.41^\circ$ |

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

$$
\alpha = 0.5629 - \frac{0.4090}{\tan 93.41^\circ} = 0.5629 + 0.0244 = 0.5873
$$

#### Step 4: Gain $K$
$$
K = \frac{|z_1|^2\,|z_1-0.3679|\,|z_1-1|}{0.6321\,|z_1-0.5873|} = \frac{0.4841 \times 0.4531 \times 0.5986}{0.6321 \times 0.4097} = 0.5070
$$

$$
G_D(z) = 0.5070\,\frac{z - 0.5873}{z - 1}
$$

$$
K_P = K\alpha = 0.2977, \qquad K_I = K(1-\alpha) = 0.2092
$$

#### Step 5: $K_v$ and steady-state error for a ramp
$$
\begin{aligned}
K_v &= \lim_{z\to1}\frac{(1-z^{-1})G_D(z)G(z)}{T} = \frac{K(1-\alpha)}{T}\cdot\frac{0.6321}{1-0.3679} = 0.2092 \times 1 = 0.2092\ \text{s}^{-1} \\
e_{ss} &= \frac{1}{K_v} = 4.78
\end{aligned}
$$

**Check:** the closed-loop characteristic equation $z^3(z-1) - 0.3679z^2(z-1) + 0.3205(z-0.5873) = 0$ has roots $0.5629 \pm j0.4090$, $0.7562$ and $-0.5141$. 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:** $G_D(z) = 0.5070\dfrac{z-0.5873}{z-1}$ (that is, $K_P = 0.2977$, $K_I = 0.2092$), with $K_v = 0.2092\ \text{s}^{-1}$ and $e_{ss} = 4.78$ for a unit ramp.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With T = 1 s, the dead time e^(-2s) in the plant becomes
   A. z^-1
   B. z^-2
   C. z^2
   D. e^-2
   Answer: B. A delay of 2 s is exactly 2T, i.e. a factor z^-2. (Key point: Dead time in z domain)

2. For 10 samples per cycle (wd*T = 36 degrees) and zeta = 0.5, the desired pole is
   A. 0.6958 at angle 36 degrees (0.5629 + j0.4090)
   B. 0.5 at angle 36 degrees
   C. 1.0 at angle 36 degrees
   D. 0.6958 at angle 18 degrees
   Answer: A. |z| = exp(-2*pi*0.5/0.866 * 0.1) = 0.6958; angle = 36 degrees. (Key point: Desired pole)

3. The PI controller K(z - alpha)/(z - 1) has
   A. A pole at z = 1 and a zero at alpha = Kp/(Kp+Ki)
   B. A zero at z = 1 and a pole at alpha
   C. Two poles at the origin
   D. A pole at z = 0
   Answer: A. The integral action gives a pole at z = 1; the zero is at alpha = KP/(KP+KI). (Key point: PI controller structure)

4. The sum of pole angles is -273.41 degrees. The PI zero must contribute
   A. +93.41 degrees
   B. -93.41 degrees
   C. +180 degrees
   D. +273.41 degrees
   Answer: A. -273.41 + 93.41 = -180 degrees, satisfying the angle condition. (Key point: Zero angle contribution)

5. For this loop, Kv and the ramp steady-state error are
   A. Kv = 0.2093, ess = 4.78
   B. Kv = 0.5070, ess = 1.97
   C. Kv = 4.78, ess = 0.2093
   D. Kv = 0.2977, ess = 3.36
   Answer: A. Kv = K(1-alpha)/T times plant DC gain 1 = 0.2093, and ess = 1/Kv = 4.78. (Key point: Velocity error constant)

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