---
title: "Mapping constant damping ratio line to z-plane"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:constant-damping-ratio-mapping"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/constant-damping-ratio-mapping
site: BCT Engineering
---

# Mapping constant damping ratio line to z-plane

Digital Control System · Chapter 3: Analysis of Discrete Time Control System

- Asked in: 2075 Baisakh, 4 marks
- Video: [Discrete-Time-Systems - Mapping Between s & z planes (Lecture 7 - Part II)](https://www.youtube.com/watch?v=QEPQYPrvjA4) — Technical Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/constant-damping-ratio-mapping
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

Explain with suitable diagram how constant damping ratio line in s-plane is mapped into z-plane.

## Short answer

In the s-plane a constant damping ratio line is a radial line at angle cos⁻¹ζ. Under z=e^{Ts} its magnitude decays exponentially with ω_d while angle grows linearly, so it maps to a logarithmic spiral that starts at z=1 and winds toward the origin. Larger ζ yields a tighter spiral; ζ=0 maps to the unit circle.

## Answer

In the s-plane, a **constant damping ratio line** is a radial line from the origin making angle $\theta = \cos^{-1}\zeta$ with the negative real axis. On it, $s = -\zeta\omega_n + j\omega_n\sqrt{1-\zeta^2}$, i.e. $s = -\omega_d\dfrac{\zeta}{\sqrt{1-\zeta^2}} + j\omega_d$.

#### Mapping with $z = e^{Ts}$
Use $T = 2\pi/\omega_s$:

$$
z = e^{Ts} = \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{\omega_d}{\omega_s}\right)\angle\ 2\pi\frac{\omega_d}{\omega_s}
$$

- $|z| = \exp\left(-\dfrac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\dfrac{\omega_d}{\omega_s}\right)$ decreases exponentially as $\omega_d$ increases.
- $\angle z = 2\pi\omega_d/\omega_s$ increases linearly with $\omega_d$.

Since the radius shrinks exponentially as the angle grows, **a constant-ζ line maps into a logarithmic spiral**. It starts at $z = 1$ (for $\omega_d = 0$) and winds toward the origin.

#### Example: $\zeta = 0.5$
| $\omega_d/\omega_s$ | $\lvert z\rvert$ | $\angle z$ |
|---|---|---|
| 0 | 1 | 0° |
| 1/8 | 0.635 | 45° |
| 1/4 | 0.404 | 90° |
| 1/2 | 0.163 | 180° |

At $\omega_d = \omega_s/2$ the spiral reaches the negative real axis. The part from $\omega_d = 0$ to $\omega_s/2$ fills the upper half of the z-plane. The line for $-\omega_d$ maps to the mirror spiral in the lower half.

```text
                Im z
                 |
                 * B = j0.404  (wd/ws = 1/4)
                 |
                 |     * A = 0.449 + j0.449  (1/8)
                 |
 ------*---------+-----------* O ------ Re z
  C = -0.163     0           1   (wd = 0)
  (wd/ws = 1/2)

 zeta = 0.5 spiral runs O -> A -> B -> C inside
 the unit circle; its mirror image (for -wd) lies
 in the lower half-plane.
```

#### Key points
- $\zeta = 0$ (the $j\omega$ axis) maps onto the unit circle.
- $\zeta = 1$ (the negative real axis) maps onto the segment $0 < z \le 1$ of the positive real axis.
- Larger $\zeta$ gives a spiral that shrinks faster, i.e. tighter toward the origin.
- In design, the desired closed-loop poles lie on the ζ-spiral at angle $\omega_dT = 2\pi/N$, where $N$ is the number of samples per cycle of damped oscillation.

## Check yourself

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

1. In the s-plane, a line of constant damping ratio is
   A. A vertical line at s = -zeta*wn
   B. A radial line from the origin at angle cos^-1(zeta) from the negative real axis
   C. A circle about the origin
   D. A horizontal line at s = j*wd
   Answer: B. Constant zeta means constant angle theta = cos^-1(zeta) with the negative real axis. (Key point: Shape of the s-plane line)

2. A constant-zeta line maps into the z-plane as
   A. A straight radial line
   B. A circle of radius e^(-zeta*wn*T)
   C. A logarithmic spiral winding toward the origin
   D. A vertical line through z = 1
   Answer: C. |z| decays exponentially while the angle grows linearly with wd, giving a logarithmic spiral starting at z = 1. (Key point: Resulting curve in z-plane)

3. On the constant-zeta line, as wd increases, what happens to |z| and the angle of z?
   A. Both increase linearly
   B. |z| decreases exponentially, angle increases linearly
   C. |z| stays 1, angle increases
   D. |z| increases exponentially, angle decreases
   Answer: B. |z| = exp(-2*pi*zeta/sqrt(1-zeta^2) * wd/ws) and angle = 2*pi*wd/ws. (Key point: Magnitude and angle of z)

4. Where does zeta = 0 (the jw axis) map in the z-plane?
   A. The origin
   B. The segment 0 < z <= 1 on the real axis
   C. The unit circle
   D. The negative real axis
   Answer: C. With zeta = 0, |z| = 1 for all frequencies, so the jw axis maps onto the unit circle. (Key point: Special values of zeta)

## Related in Digital Control System

- [Mapping from s-plane to z-plane](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3As-plane-to-z-plane-mapping) — Digital Control System, chapter 3, asked 3×
- [Mapping an s-plane region to the z-plane](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Amap-s-plane-region) — Digital Control System, chapter 3, asked 2×
- [z-plane design with Kv, plant 1/(s(s+2))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Az-plane-design-kv-s2) — Digital Control System, chapter 4

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