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

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

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

- Asked in: 2072 Asoj, 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-step
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Consider the digital control system as shown in figure below, where the plant is of the first order and has a dead time of 5 second. By choosing reasonable sampling time 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 oscillation is 10. After the controller is designed, determine the response of the system to a unit step input. [Figure: unity-feedback loop: r(t)/R(z) → summing junction → sampler (δT = T) → digital PI controller → ZOH → e⁻⁵ˢ/(s+0.4) → c(t)/C(z)]

## Short answer

Sampling period T=2.5 s. PI controller GD(z)=0.2028(z-0.5873)/(z-1) with Kp=0.1191 and Ki=0.0837. The step response settles to 1 with zero steady-state error.

## Answer

A digital PI controller $G_D(z)=K_p+\dfrac{K_i}{1-z^{-1}}=K\dfrac{z-a}{z-1}$ is designed by the root-locus (angle and magnitude) method in the z-plane. The plant is $G_p(s)=\dfrac{e^{-5s}}{s+0.4}$ (time constant 2.5 s, dead time 5 s).

#### Choice of sampling period
- The dead time should be a whole number of sampling periods, so $T = 5/N$.
- Ten samples per cycle fixes the angle of the desired pole at $\omega_d T = 2\pi/10 = 36^\circ$, whatever T is. Each sample of delay ($z^{-1}$) then adds $36^\circ$ of phase lag at that pole.
- With $T = 1$ s ($N = 5$) the delay alone gives $5\times36^\circ = 180^\circ$. The poles at $z=1$ and $z=e^{-0.4}$ add more than $180^\circ$ again, and one PI zero cannot make up the difference, so no PI controller exists.
- **Choose $T = 2.5$ s**: this equals the plant time constant, the dead time is $2T$, and the angle condition can be met.

#### Pulse transfer function of the plant (with ZOH)
With $e^{-0.4T}=e^{-1}=0.3679$:
$$
\begin{aligned}
G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+0.4)}\right] = z^{-2}\,\frac{(1-e^{-1})/0.4}{z-e^{-1}}\\
&= \frac{1.5803}{z^{2}(z-0.3679)}
\end{aligned}
$$

#### Desired closed-loop poles
$$
\begin{aligned}
|z| &= \exp\left(-\frac{2\pi\zeta}{10\sqrt{1-\zeta^2}}\right) = \exp\left(-\frac{0.6283\times0.5}{0.8660}\right) = 0.6958\\
z_1 &= 0.6958\angle 36^\circ = 0.5629 + j0.4090
\end{aligned}
$$
(This gives $\omega_d = 0.2513$ rad/s and $\omega_n = 0.2902$ rad/s.)

#### Angle condition
Open loop: $G_D(z)G(z) = \dfrac{1.5803K(z-a)}{z^2(z-0.3679)(z-1)}$. Pole angles at $z_1$:

| Pole | Angle at $z_1$ |
|---|---|
| $z=0$ (double) | $2\times36^\circ = 72.0^\circ$ |
| $z=0.3679$ | $64.5^\circ$ |
| $z=1$ | $136.9^\circ$ |
| Total | $273.4^\circ$ |

The zero must give $273.4^\circ - 180^\circ = 93.4^\circ$:
$$
a = 0.5629 - \frac{0.4090}{\tan 93.4^\circ} = 0.5873
$$

#### Magnitude condition
$$
K = \left|\frac{z_1^2(z_1-0.3679)(z_1-1)}{1.5803(z_1-0.5873)}\right| = 0.2028
$$
$$
G_D(z) = \frac{0.2028(z-0.5873)}{z-1} = 0.1191 + \frac{0.0837}{1-z^{-1}}
$$
So $K_p = Ka = 0.1191$ and $K_i = K(1-a) = 0.0837$.

#### Unit-step response
$$
\frac{C(z)}{R(z)} = \frac{0.3204(z-0.5873)}{z^4-1.3679z^3+0.3679z^2+0.3204z-0.1882}
$$
Closed-loop poles: $0.5629\pm j0.4090$ (designed), $0.7562$ and $-0.5141$. Solving the difference equation with $r(k)=1$:

| k | 0–2 | 3 | 4 | 5 | 6 | 7 | 8 | 10 | 12 | 15 | 20 | 29 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| c(kT) | 0 | 0.320 | 0.571 | 0.795 | 0.907 | 0.958 | 0.962 | 0.947 | 0.957 | 0.987 | 0.997 | 1.000 |

- Output is zero for 3 samples (2 for the dead time, 1 for ZOH plus computation).
- The extra real pole at 0.7562 slows the final approach, so the response rises to about 0.96, dips slightly and creeps up to 1 with almost no overshoot.
- Because of the integral action, steady-state error to a step is zero: $c(\infty)=1$. $K_v = K(1-a)(1.5803)/[T(1-0.3679)] = 0.0837\ \text{s}^{-1}$.

**Answer: $T = 2.5$ s, $G_D(z) = 0.2028(z-0.5873)/(z-1)$ ($K_p = 0.1191$, $K_i = 0.0837$); the step response settles at 1 with zero steady-state error.**

## Check yourself

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

1. Why is T = 1 s rejected for the plant e^(-5s)/(s+0.4) with 10 samples per cycle?
   A. The dead time is not a whole number of samples at T = 1 s
   B. Five samples of delay give 5 x 36 = 180 degrees of lag, and the poles at z = 1 and e^-0.4 add more; one PI zero cannot make up the difference
   C. T = 1 s violates the sampling theorem for this plant
   D. The ZOH cannot be used when T is less than the time constant
   Answer: B. Each delay sample adds 36 degrees at the desired pole, so the delay alone uses 180 degrees and no PI zero can supply enough lead. (Key point: Choice of sampling period)

2. What fixes the angle of the desired closed-loop pole at 36 degrees?
   A. The damping ratio 0.5
   B. The plant time constant of 2.5 s
   C. Ten samples per cycle of damped oscillation, so omega_d T = 2 pi/10
   D. The dead time of 5 s
   Answer: C. One damped cycle is 2 pi radians spread over 10 samples, giving 36 degrees per sample whatever T is. (Key point: Angle of desired pole)

3. The poles at z = 0 (double), 0.3679 and 1 contribute a total of 273.4 degrees at z1. What angle must the PI zero contribute?
   A. 273.4 degrees
   B. 86.6 degrees
   C. 180 degrees
   D. 93.4 degrees
   Answer: D. The angle condition requires zero angle minus pole angles = -180, so the zero contributes 273.4 - 180 = 93.4 degrees. (Key point: Angle condition for PI zero)

4. Why is the steady-state error to a unit step zero for the designed system?
   A. The PI controller adds a pole at z = 1 (integral action)
   B. The dead time is a whole number of samples
   C. The damping ratio is 0.5
   D. The extra real closed-loop pole is at 0.7562
   Answer: A. The integrator pole at z = 1 raises the system type, so c(infinity) = 1 for a step. (Key point: Steady-state behaviour)

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