---
title: "Controller design for peak time ≤ 1.2 s (ζ = 0.5)"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:design-peak-time-zeta-0-5"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/design-peak-time-zeta-0-5
site: BCT Engineering
---

# Controller design for peak time ≤ 1.2 s (ζ = 0.5)

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

- Asked in: 2068 Magh, 16 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-peak-time-zeta-0-5
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Design a digital controller for the system shown in figure below so that dominant closed loop poles of the system will have damping ratio of 0.5. It is required that the peak time of the response tₚ≤ 1.2 sec. Assume sampling time T = 0.2 sec. [Figure: unity-feedback loop: R(z) → summing junction → sampler → digital controller → m*(kT) → ZOH → 1/s(s+1) → C(z)]

## Short answer

Using lead compensation with zero cancelling the plant pole at 0.8187, the controller is G_D(z)=7.340(z-0.8187)/(z-0.4177). The resulting closed-loop dominant poles are 0.6401±j0.3696, giving damping ratio 0.5 and peak time 1.2 s.

## Answer

A lead-type digital controller $G_D(z) = K\dfrac{z-\alpha}{z-\beta}$ is designed by the z-plane root-locus method. Its zero cancels the slow plant pole.

#### Plant pulse transfer function ($T = 0.2$ s)
With $e^{-0.2} = 0.8187$:
$$
\begin{aligned}
G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})}\\
&= \frac{0.01873(z+0.9355)}{(z-1)(z-0.8187)}
\end{aligned}
$$

#### Desired dominant poles
$$
\begin{aligned}
t_p &= \frac{\pi}{\omega_d} = 1.2 \Rightarrow \omega_d = 2.618\ \text{rad/s}\\
\omega_n &= \frac{\omega_d}{\sqrt{1-0.5^2}} = 3.023\ \text{rad/s},\quad \zeta\omega_n = 1.511\\
|z| &= e^{-\zeta\omega_nT} = e^{-0.3023} = 0.7391, \quad \angle z = \omega_dT = 0.5236\ \text{rad} = 30^\circ\\
z_1 &= 0.6401 + j0.3696
\end{aligned}
$$
That is 12 samples per cycle of damped oscillation, which is adequate.

#### Angle deficiency
| Factor | Angle at $z_1$ |
|---|---|
| zero at $-0.9355$ | $+13.20^\circ$ |
| pole at $1$ | $-134.24^\circ$ |
| pole at $0.8187$ | $-115.80^\circ$ |
| Total | $-236.84^\circ$ |

The controller must add $+56.84^\circ$.

#### Controller zero and pole
- Zero at $\alpha = 0.8187$ cancels the plant pole and removes its $-115.80^\circ$.
- Then the controller pole must give $13.20^\circ - 134.24^\circ - \theta_\beta = -180^\circ$, so $\theta_\beta = 58.96^\circ$:
$$
\beta = 0.6401 - \frac{0.3696}{\tan 58.96^\circ} = 0.4177
$$

#### Gain
$$
K = \left|\frac{(z_1-1)(z_1-0.4177)}{0.01873(z_1+0.9355)}\right| = 7.340
$$
$$
G_D(z) = 7.340\,\frac{z-0.8187}{z-0.4177}
$$

#### Check
- Characteristic equation: $z^2 - 1.2802z + 0.5463 = 0$, roots $0.6401 \pm j0.3696$, which is the design point.
- Step response: $c(kT) = 0,\ 0.137,\ 0.442,\ 0.757,\ 0.994,\ 1.125,\ 1.163,\ 1.141,\ \ldots$. The peak is 1.163 at $t = 6T = 1.2$ s, so $t_p \le 1.2$ s is met.
- Velocity error constant:
$$
K_v = \frac{1}{T}\lim_{z\to1}(z-1)G_D(z)G(z) = \frac{7.340\times0.03625}{0.2\times(1-0.4177)} = 2.28\ \text{s}^{-1}
$$

**Answer: $G_D(z) = 7.340\,\dfrac{z-0.8187}{z-0.4177}$, giving closed-loop poles $0.6401\pm j0.3696$ ($\zeta=0.5$, $t_p=1.2$ s).**

## Check yourself

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

1. With tp = 1.2 s, what is omega_d?
   A. pi x 1.2 = 3.77 rad/s
   B. 1.2 rad/s
   C. pi/1.2 = 2.618 rad/s
   D. 2 pi/1.2 = 5.24 rad/s
   Answer: C. tp = pi/omega_d, so omega_d = pi/1.2 = 2.618 rad/s. (Key point: Damped frequency from peak time)

2. With T = 0.2 s, zeta = 0.5, how is the desired pole z1 found?
   A. Magnitude e^(-zeta omega_n T) = 0.7391, angle omega_d T = 30 degrees
   B. Magnitude zeta = 0.5, angle 60 degrees
   C. Magnitude e^(omega_n T), angle 90 degrees
   D. Magnitude 1, angle omega_d
   Answer: A. z = e^(-sigma T) at angle omega_d T = 0.5236 rad = 30 degrees, giving 0.6401 + j0.3696. (Key point: Desired pole in z-plane)

3. Where is the controller zero placed and why?
   A. At z = 1 to add an integrator
   B. At the origin to remove delay
   C. At z = -0.9355 to cancel the plant zero
   D. At z = 0.8187 to cancel the slow plant pole at e^-T
   Answer: D. Cancelling the pole removes its angle, leaving the controller pole to supply the remaining angle. (Key point: Zero placement)

4. After cancellation, the controller pole must give 58.96 degrees. Then beta is:
   A. 0.6401 - 0.3696/tan(58.96) = 0.4177
   B. 0.6401 + 0.3696/tan(58.96)
   C. 0.8187
   D. 1
   Answer: A. beta = Re(z1) - Im(z1)/tan(angle) = 0.4177. (Key point: Controller pole from angle)

5. Kv = (1/T) lim (z-1) G_D(z) G(z) for this design is about:
   A. 7.34 per second
   B. 0.2 per second
   C. 2.28 per second
   D. 10 per second
   Answer: C. Evaluating at z = 1 gives Kv = 7.340 x 0.03625 / (0.2 x 0.5823) = 2.28 per second. (Key point: Velocity error constant)

## Close variants set in other exams

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