---
title: "Bode design: PM 55°, GM 10 dB, Kv = 5"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:bode-design-phase-margin"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/bode-design-phase-margin
site: BCT Engineering
---

# Bode design: PM 55°, GM 10 dB, Kv = 5

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

- Asked in: 2072 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/bode-design-phase-margin
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Design a digital controller for the system shown below. Use the Bode diagram approach in s plane. The design specifications are that the phase margin be 55°, the gain margin be at least 10 dB and the static velocity error constant be 5 sec⁻¹. The sampling period is specified as 0.1 sec. [Figure: unity-feedback loop: R(z) → summing junction → sampler → digital controller → ZOH → plant 1/s(s+2) → C(z)]

## Short answer

GD(z) = 40.78(z-0.8187)/(z-0.2608). This lead compensator achieves a static velocity error constant Kv = 5 s⁻¹, phase margin ≈55°, and gain margin ≈12.3 dB.

## Answer

The design is done in the w-plane (bilinear transformation), where the Bode diagram can be used as in the s-plane. A lead compensator $G_D(w) = K\dfrac{1+\tau w}{1+\alpha\tau w}$ is used.

#### Step 1: Pulse transfer function of the plant
With $T = 0.1$ s and $e^{-0.2} = 0.8187$:
$$
\begin{aligned}
G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right]\\
&= \frac{0.004683z + 0.004381}{(z-1)(z-0.8187)} = \frac{0.004683(z+0.9355)}{(z-1)(z-0.8187)}
\end{aligned}
$$

#### Step 2: Transform to the w-plane
Substitute $z = \dfrac{1+0.05w}{1-0.05w}$:
$$
G(w) = \frac{0.5\left(1-\frac{w}{20}\right)\left(1+\frac{w}{600.4}\right)}{w\,(1+0.5017w)}
$$

#### Step 3: Gain for $K_v = 5$
$$
K_v = \lim_{w\to0} w\,G_D(w)G(w) = K\times0.5 = 5 \;\Rightarrow\; K = 10
$$

#### Step 4: Uncompensated Bode data ($10\,G(j\nu)$)
| $\nu$ (rad/s) | 1 | 2 | 4 | 6 | 10 | 20 |
|---|---|---|---|---|---|---|
| Magnitude (dB) | 13.0 | 5.0 | −4.9 | −11.2 | −19.2 | −29.1 |
| Phase (deg) | −119.4 | −140.6 | −164.4 | −177.8 | −194.3 | −217.4 |

Gain crossover is at $\nu = 2.88$ rad/s and the phase margin is only **26.8°**. About 28° or more of lead is needed. The w-plane zero at $w=20$ adds extra lag at high frequency, so a larger lead is needed than in a continuous design.

#### Step 5: Lead compensator
Choose the compensator zero to cancel the plant pole at $w = -1.993$, so $\tau = 0.5017$. Then reduce $\alpha$ until PM = 55°:

| $\alpha$ | New crossover $\nu_c$ | PM | GM |
|---|---|---|---|
| 0.25 | 4.47 | 48.6° | 12.3 dB |
| 0.20 | 4.65 | 52.3° | 12.3 dB |
| **0.17** | **4.76** | **55.0°** | **12.3 dB** |

$$
G_D(w) = 10\,\frac{1 + 0.5017w}{1 + 0.0853w}
$$
Compensated loop: $G_D(w)G(w) = \dfrac{5(1-w/20)(1+w/600.4)}{w(1+0.0853w)}$. Phase margin = 55°, phase crossover at $\nu = 15.7$ rad/s, **gain margin = 12.3 dB (≥ 10 dB)**.

#### Step 6: Back to the z-plane
Substitute $w = \dfrac{2}{T}\dfrac{z-1}{z+1} = 20\dfrac{z-1}{z+1}$:
$$
\begin{aligned}
G_D(z) &= 10\,\frac{1+10.033\frac{z-1}{z+1}}{1+1.706\frac{z-1}{z+1}} = 40.78\,\frac{z-0.8187}{z-0.2608}
\end{aligned}
$$

#### Check of $K_v$
$$
K_v = \frac{1}{T}\lim_{z\to1}(z-1)G_D(z)G(z) = \frac{1}{0.1}\times40.78\times\frac{0.1813}{0.7392}\times\frac{0.009063}{0.1813} = 5.0\ \text{s}^{-1}
$$

**Answer: $G_D(z) = 40.78\,\dfrac{z-0.8187}{z-0.2608}$, giving PM ≈ 55°, GM ≈ 12.3 dB and $K_v = 5\ \text{s}^{-1}$.**

## Check yourself

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

1. G(w) has a 0.5/w term at low frequency. For Kv = 5, what is K?
   A. 2.5
   B. 5
   C. 0.1
   D. 10
   Answer: D. Kv = lim w G_D(w) G(w) = 0.5 K = 5, so K = 10. (Key point: Gain from Kv)

2. What is the phase margin of 10 G(j nu) at its crossover (nu = 2.88 rad/s)?
   A. 55 degrees
   B. 26.8 degrees
   C. 12.3 degrees
   D. 90 degrees
   Answer: B. The phase there is about -153 degrees, giving PM = 26.8 degrees, far below 55 degrees. (Key point: Uncompensated margin)

3. How is the lead zero chosen in this design?
   A. At w = 20, the non-minimum-phase zero
   B. At the origin
   C. To cancel the plant pole at w = -1.993, so tau = 0.5017
   D. At the new crossover frequency
   Answer: C. Cancelling the plant pole leaves a simple first-order roll-off, and only alpha is varied. (Key point: Zero placement)

4. Why is more lead needed than in a continuous design?
   A. The sampler reduces the gain
   B. Kv is higher
   C. The compensator is digital
   D. The w-plane zero at w = 20 adds extra phase lag at high frequency
   Answer: D. The right-half-plane zero (1 - w/20) adds lag near crossover, so more lead is required. (Key point: Extra lead due to w-plane zero)

5. With alpha = 0.17, the compensated system gives:
   A. PM = 55 degrees and GM = 12.3 dB, meeting GM of at least 10 dB
   B. PM = 55 degrees but GM = 5 dB
   C. PM = 26.8 degrees and GM = 12.3 dB
   D. PM = 48.6 degrees and GM = 10 dB
   Answer: A. alpha = 0.17 gives PM 55.0 degrees and GM 12.3 dB, satisfying all specifications. (Key point: Result check)

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