---
title: "Digital PI design, plant e^(−5s)/(s+0.4), Kv"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:pi-design-dead-time-5s-kv"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/pi-design-dead-time-5s-kv
site: BCT Engineering
---

# Digital PI design, plant e^(−5s)/(s+0.4), Kv

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

- Asked in: 2077 Chaitra, 12 marks
- 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-5s-kv
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Consider a digital control system shown below. By choosing a reasonable sampling period T, design a digital PI controller such that the dominant closed loop poles have a damping ratio ζ of 0.5 and the number of samples per cycle of damped sinusoidal oscillation is 10. Also find the static velocity error constant. [Figure: unity-feedback loop: R(z) → summing junction → sampler (T) → digital PI controller → (1-e⁻ᵀˢ)/s → e⁻⁵ˢ/(s+0.4) → C(z)]

## Short answer

Sampling period T=2.5 s. PI controller GD(z)=0.2028(z-0.5873)/(z-1). Static velocity error constant Kv=0.0837 s^-1.

## Answer

#### Step 1: Choice of sampling period
The plant has time constant $1/0.4 = 2.5$ s and dead time 5 s. Choose **$T = 2.5$ s**:
- The dead time is then exactly $2T$, giving a clean $z^{-2}$.
- $e^{-0.4T} = e^{-1}$, which gives simple numbers.
- With 10 samples per cycle, the oscillation period is 25 s. This is reasonable for a process with 5 s dead time.

A smaller $T$ such as 1 s is not suitable. The dead time becomes $z^{-5}$, the phase lag at the desired point exceeds what a single PI zero can make up, and the angle condition cannot be met.

$$
\begin{aligned}
G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+0.4)}\right] = z^{-2}\cdot\frac{(1-e^{-1})}{0.4}\cdot\frac{z^{-1}}{1-e^{-1}z^{-1}} \\
&= \frac{1.5803}{z^2(z - 0.3679)}
\end{aligned}
$$

#### Step 2: Desired poles (ζ = 0.5, 10 samples per cycle)
$$
|z| = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac{1}{10}\right) = 0.6958, \qquad \angle z = 36^\circ, \qquad z_1 = 0.5629 + j0.4090
$$

#### Step 3: PI controller $G_D(z) = K\dfrac{z-\alpha}{z-1}$
Angles at $z_1$:
- double pole at origin: $-72^\circ$
- pole at 0.3679: $-64.51^\circ$
- pole at 1: $-136.91^\circ$

The sum is $-273.41^\circ$, so the PI zero must add $+93.41^\circ$:

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

#### Step 4: Gain
$$
K = \frac{|z_1|^2\,|z_1-0.3679|\,|z_1-1|}{1.5803\,|z_1-0.5873|} = \frac{0.4841 \times 0.4531 \times 0.5986}{1.5803 \times 0.4097} = 0.2028
$$

$$
G_D(z) = 0.2028\,\frac{z - 0.5873}{z - 1}
$$

so $K_P = K\alpha = 0.1191$ and $K_I = K(1-\alpha) = 0.0837$.

#### Step 5: Static velocity error constant
$$
\begin{aligned}
K_v &= \lim_{z\to1}\frac{(1-z^{-1})G_DG}{T} = \frac{1}{2.5}\cdot K(1-\alpha)\cdot\frac{1.5803}{1-0.3679} \\
&= \frac{0.0837 \times 2.5}{2.5} = 0.0837\ \text{s}^{-1}
\end{aligned}
$$

The steady-state error for a unit ramp is $1/K_v = 11.9$.

**Check:** the closed-loop roots are $0.5629 \pm j0.4090$ (as designed), $0.7562$ and $-0.5141$. The extra real pole at 0.7562 is not cancelled by the PI zero (0.5873), so it adds a slow tail to the response.

**Answer** (with $T = 2.5$ s): $G_D(z) = 0.2028\dfrac{z-0.5873}{z-1}$ and $K_v = 0.0837\ \text{s}^{-1}$.

## Check yourself

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

1. Why choose T = 2.5 s for the plant e^-5s/(s+0.4)?
   A. It is the smallest possible T
   B. Larger T always gives better performance
   C. It makes the plant type 2
   D. The dead time is exactly 2T giving z^-2, and e^-0.4T = e^-1 is simple
   Answer: D. 5 s = 2T gives a clean z^-2 delay, with pole e^-1 = 0.3679. (Key point: Sampling period choice)

2. With zeta = 0.5 and 10 samples per cycle, the desired pole has angle and magnitude:
   A. 36 degrees and 0.6958
   B. 45 degrees and 0.6354
   C. 36 degrees and 0.5
   D. 18 degrees and 0.8
   Answer: A. 360/10 = 36 degrees; |z| = exp(-2 pi (0.5)/0.866 x 1/10) = 0.6958. (Key point: Pole angle)

3. The plant and PI pole at 1 give -273.41 degrees total. What does the PI zero do?
   A. Adds -93.41 degrees
   B. Adds +93.41 degrees, placing alpha = 0.5873
   C. Cancels the pole at 1
   D. Adds +180 degrees
   Answer: B. To reach -180 the zero must add 93.41 degrees, so alpha = 0.5629 - 0.4090/tan 93.41 degrees = 0.5873. (Key point: PI zero)

4. Kv of the designed system is:
   A. 2.80 per second
   B. 0.2028 per second
   C. 0.0837 per second
   D. 11.9 per second
   Answer: C. Kv = (1/T) K(1-alpha) x 1.5803/(1-0.3679) = 0.0837 per second; the ramp error is 1/Kv = 11.9. (Key point: Kv)

## Close variants set in other exams

- [Digital PI design, plant e^(−2s)/(s+2)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Api-design-dead-time-s2) — Digital Control System, chapter 4, asked 2×

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