---
title: "Chebyshev order and elliptical pole locus"
subject: "Filter Design"
chapter: 2
key: "semester_5/filter_design:chebyshev-order-pole-ellipse"
source: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-order-pole-ellipse
site: BCT Engineering
---

# Chebyshev order and elliptical pole locus

Filter Design · Chapter 2: Approximation Methods

- Asked in: 2081 Bhadra, 3+4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-order-pole-ellipse
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2

## Question

Derive the expression of the order of a low pass Chebyshev approximation and then prove that locus of its pole is an ellipse centered at origin.

## Short answer

n ≥ cosh⁻¹√((10^{0.1α_min}–1)/(10^{0.1α_max}–1)) / cosh⁻¹(ω_s/ω_p). Normalized poles satisfy σ²/sinh²a + ω²/cosh²a = 1, with a=(1/n)sinh⁻¹(1/ε), an ellipse centered at origin; semi-minor axis sinh a on real axis, semi-major cosh a on jω axis.

## Answer

#### Order of a Chebyshev low-pass filter

The Chebyshev low-pass magnitude response (normalized so that the passband edge is $\Omega = \omega/\omega_p = 1$) is

$$
|T(j\Omega)|^2 = \frac{1}{1+\varepsilon^2 C_n^2(\Omega)}
$$

where $C_n(\Omega)$ is the Chebyshev polynomial of order $n$:

$$
C_n(\Omega) = \begin{cases} \cos(n\cos^{-1}\Omega), & |\Omega| \le 1 \\ \cosh(n\cosh^{-1}\Omega), & |\Omega| > 1 \end{cases}
$$

The attenuation in dB is $\alpha(\Omega) = 10\log_{10}\left[1+\varepsilon^2 C_n^2(\Omega)\right]$.

**Step 1: ripple factor from the passband edge.** At $\Omega = 1$, $C_n(1) = 1$ for every $n$, so the attenuation is exactly $\alpha_{max}$:

$$
\alpha_{max} = 10\log_{10}(1+\varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1
$$

**Step 2: stopband condition.** At $\Omega_s = \omega_s/\omega_p\ (>1)$ the attenuation must be at least $\alpha_{min}$:

$$
10\log_{10}\left[1+\varepsilon^2 C_n^2(\Omega_s)\right] \ge \alpha_{min}
\;\Rightarrow\; C_n^2(\Omega_s) \ge \frac{10^{0.1\alpha_{min}}-1}{\varepsilon^2}
$$

**Step 3: solve for n.** Since $\Omega_s > 1$, $C_n(\Omega_s) = \cosh(n\cosh^{-1}\Omega_s)$:

$$
\cosh(n\cosh^{-1}\Omega_s) \ge \sqrt{\frac{10^{0.1\alpha_{min}}-1}{10^{0.1\alpha_{max}}-1}}
$$

$$
n \ge \frac{\cosh^{-1}\sqrt{\dfrac{10^{0.1\alpha_{min}}-1}{10^{0.1\alpha_{max}}-1}}}{\cosh^{-1}\left(\dfrac{\omega_s}{\omega_p}\right)}
$$

The order is the next whole number above this value. (Useful identity: $\cosh^{-1}x = \ln\left(x+\sqrt{x^2-1}\right)$.)

#### Locus of the poles is an ellipse

The poles are the left-half-plane roots of $1+\varepsilon^2 C_n^2(s/j) = 0$, i.e. $C_n = \pm j/\varepsilon$. Putting $\cos^{-1}(s/j) = u + jv$:

$$
\cos(nu)\cosh(nv) - j\sin(nu)\sinh(nv) = \pm \frac{j}{\varepsilon}
$$

Equating real parts: $\cos(nu)=0 \Rightarrow u_k = \dfrac{(2k-1)\pi}{2n}$. Equating imaginary parts: $\sin(nu_k)=\pm1 \Rightarrow v = a = \dfrac{1}{n}\sinh^{-1}\dfrac{1}{\varepsilon}$. Then $s = j\cos(u_k + ja)$ gives (taking the left-half-plane roots)

$$
s_k = \sigma_k + j\omega_k,\quad \sigma_k = -\sinh a\,\sin\frac{(2k-1)\pi}{2n},\quad \omega_k = \cosh a\,\cos\frac{(2k-1)\pi}{2n}
$$

for $k = 1, 2, \dots, n$ (normalized to $\omega_p = 1$; multiply by $\omega_p$ to denormalize).

Now eliminate $\theta_k = (2k-1)\pi/2n$. From the pole expressions:

$$
\frac{\sigma_k}{\sinh a} = -\sin\theta_k, \qquad \frac{\omega_k}{\cosh a} = \cos\theta_k
$$

Squaring and adding, and using $\sin^2\theta_k + \cos^2\theta_k = 1$:

$$
\frac{\sigma_k^2}{\sinh^2 a} + \frac{\omega_k^2}{\cosh^2 a} = 1
$$

This is the equation of an **ellipse centred at the origin** of the s-plane. Since $\cosh a > \sinh a$:
- semi-major axis $= \cosh a$, along the $j\omega$ axis;
- semi-minor axis $= \sinh a$, along the $\sigma$ axis;
- foci at $\pm j1$ (because $\cosh^2 a - \sinh^2 a = 1$), i.e. at the passband edge.

```text
            jw
      .-----+-----.   cosh a
     /  x   |   .  \
    |       |       |
 ---x-------+-------+--- sigma
    |  sinh a       |
     \  x   |   .  /
      '-----+-----'
  (only left-half poles x are used)
```

As $\varepsilon \to 0$ (smaller ripple), $a$ grows and $\sinh a \approx \cosh a$, so the ellipse approaches a circle. Relative to its size the response then approaches the Butterworth response.

## Check yourself

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

1. How are the poles of the Chebyshev function found?
   A. By solving C_n(s/j) = 0
   B. By solving 1 + eps^2 C_n^2(s/j) = 0, i.e. C_n = ±j/eps
   C. By solving C_n(s) = eps
   D. By setting |T| = 1
   Answer: B. Poles are the roots of the denominator 1 + eps^2 C_n^2 with omega replaced by s/j, giving C_n = ±j/eps. (Key point: Pole equation)

2. With cos^-1(s/j) = u + jv, the real-part equation cos(nu) cosh(nv) = 0 gives:
   A. cosh(nv) = 0
   B. cos(nu) = 0, so u_k = (2k-1)pi/2n
   C. sin(nu) = 0
   D. v = 0
   Answer: B. Since cosh(nv) >= 1, cos(nu) must be zero, so u_k = (2k-1)pi/(2n). (Key point: Real part gives angle)

3. Squaring and adding sigma/sinh a = -sin(theta) and omega/cosh a = cos(theta) gives:
   A. sigma^2 + omega^2 = 1
   B. sigma^2/cosh^2 a + omega^2/sinh^2 a = 1
   C. sigma^2/sinh^2 a + omega^2/cosh^2 a = 1
   D. sigma^2 - omega^2 = 1
   Answer: C. sin^2 + cos^2 = 1 gives an ellipse with semi-axis sinh a on sigma and cosh a on j-omega. (Key point: Ellipse equation)

4. As the ripple (epsilon) becomes very small, what happens to the ellipse?
   A. It collapses onto the jw axis
   B. It approaches a circle (Butterworth-like)
   C. It becomes a hyperbola
   D. It shrinks to the origin
   Answer: B. Small epsilon makes a large, so sinh a ≈ cosh a and the ellipse approaches a circle. (Key point: Meaning of small epsilon)

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