---
title: "Controller design with 8 samples per cycle, step response and Kv"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:design-8-samples-step-kv"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/design-8-samples-step-kv
site: BCT Engineering
---

# Controller design with 8 samples per cycle, step response and Kv

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

- Asked in: 2082 Chaitra (new course), 9 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/design-8-samples-step-kv
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Consider a digital control system shown in figure below. The plant is a second order system described as G(s)=1/s(s+1). The sampling period of the system is taken to be T = 1 s. Design a digital controller such that the dominant closed loop poles of the system will have damping ratio ζ=0.5 and number of samples per cycle of damped sinusoidal oscillation to be 8. Following the design, obtain the unit step response of the designed system and static velocity error constant Kᵥ. [Figure: unity-feedback loop: R(s) → summing junction → sampler (T = 1 s) → digital controller → ZOH → G(s) → Y(s)]

## Short answer

Digital controller: GD(z)=0.7991(z-0.3679)/(z-0.1926). The step response overshoots by 16.3% and settles to 1. Static velocity error constant Kv=0.626 s^-1.

## Answer

#### Step 1: Plant pulse transfer function ($T = 1$ s)
$$
G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)} = \frac{0.3679(z + 0.7183)}{(z-1)(z-0.3679)}
$$

#### Step 2: Desired poles
With 8 samples per cycle, $\omega_dT = 2\pi/8 = 45^\circ$, and

$$
\begin{aligned}
|z| &= \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{1}{8}\right) = e^{-0.4534} = 0.6354 \\
z_{1,2} &= 0.6354\angle \pm 45^\circ = 0.4493 \pm j0.4493
\end{aligned}
$$

#### Step 3: Angle condition
| Factor | Angle at $z_1$ |
|---|---|
| zero at $-0.7183$ | $+21.05^\circ$ |
| pole at $1$ | $-140.79^\circ$ |
| pole at $0.3679$ | $-79.73^\circ$ |
| **Total** | $-199.47^\circ$ |

The controller must add $+19.47^\circ$, so a lead compensator is needed.

#### Step 4: Lead controller $G_D(z) = K\dfrac{z-0.3679}{z-\beta}$
The zero cancels the plant pole at 0.3679 and adds $+79.73^\circ$. The controller pole must contribute $79.73^\circ - 19.47^\circ = 60.26^\circ$:

$$
\beta = 0.4493 - \frac{0.4493}{\tan 60.26^\circ} = 0.1926
$$

#### Step 5: Gain
$$
K = \frac{|z_1-1|\,|z_1-0.1926|}{0.3679\,|z_1+0.7183|} = \frac{0.7107 \times 0.5175}{0.3679 \times 1.2511} = 0.7991
$$

$$
G_D(z) = 0.7991\,\frac{z - 0.3679}{z - 0.1926}
$$

#### Step 6: Closed-loop pulse transfer function
$$
\begin{aligned}
G_DG &= \frac{0.2940(z + 0.7183)}{(z-1)(z-0.1926)} \\
\frac{C(z)}{R(z)} &= \frac{0.2940z + 0.2112}{z^2 - 0.8986z + 0.4038}
\end{aligned}
$$

The poles are $0.4493 \pm j0.4493$, as desired. ✔

#### Step 7: Unit step response
With $R(z) = z/(z-1)$, the difference equation is
$c(k) = 0.8986c(k-1) - 0.4038c(k-2) + 0.2940r(k-1) + 0.2112r(k-2)$, where $r(k) = 1$:

| $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| $c(k)$ | 0 | 0.294 | 0.769 | 1.078 | 1.163 | 1.115 | 1.038 | 0.987 | 0.973 | 0.981 | 0.994 |

The maximum overshoot is about **16.3%** at $k = 4$, which matches $\zeta = 0.5$. One damped cycle takes about 8 samples, and $c(\infty) = 1$ (type-1 system).

#### Step 8: Static velocity error constant
$$
K_v = \lim_{z\to1}\frac{(1-z^{-1})G_DG}{T} = \frac{0.2940 \times 1.7183}{1 \times (1 - 0.1926)} = 0.626\ \text{s}^{-1}
$$

**Answer:** $G_D(z) = 0.7991\dfrac{z-0.3679}{z-0.1926}$. The step response overshoots by 16.3% and settles to 1, and $K_v = 0.626\ \text{s}^{-1}$.

## Check yourself

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

1. For 8 samples per damped cycle and zeta = 0.5, the desired pole angle and magnitude are:
   A. 45 degrees and 0.5
   B. 60 degrees and 0.6354
   C. 22.5 degrees and 0.8
   D. 45 degrees and 0.6354
   Answer: D. wd T = 2 pi/8 = 45 degrees; |z| = exp(-2 pi zeta/sqrt(1-zeta^2) / 8) = 0.6354. (Key point: Pole location from samples per cycle)

2. Why is a lead controller with zero at 0.3679 chosen?
   A. The plant contributes -199.47 degrees, so +19.47 is needed; the zero cancels the plant pole at 0.3679
   B. The plant has too much lead
   C. To add an integrator
   D. To reduce gain
   Answer: A. Total angle -199.47 degrees means +19.47 degrees deficiency (a lead). The zero cancels the pole at 0.3679, and the controller pole supplies the remaining angle. (Key point: Controller structure)

3. What is the unit step response behaviour of the designed system?
   A. No overshoot, final value 0.5
   B. About 16.3% overshoot at k = 4 and settles to 1
   C. About 50% overshoot
   D. Unstable oscillation
   Answer: B. c(k) peaks at 1.163 at k = 4 and tends to 1 because the system is type 1. (Key point: Step response)

4. Kv for the designed system is:
   A. 3.16 per second
   B. 1.0 per second
   C. 0.626 per second
   D. infinite
   Answer: C. Kv = (1/T) x 0.2940 x 1.7183/(1 - 0.1926) = 0.626 for T = 1. (Key point: Kv value)

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