---
title: "Chebyshev order (0.5 dB, 20 dB, 3500 rad/s)"
subject: "Filter Design"
chapter: 2
key: "semester_5/filter_design:chebyshev-order-2000-3500"
source: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-order-2000-3500
site: BCT Engineering
---

# Chebyshev order (0.5 dB, 20 dB, 3500 rad/s)

Filter Design · Chapter 2: Approximation Methods

- Asked in: 2079 Baisakh, 4+3 marks
- Video: [Problem #1 - An Analog Type-I Chebyshev LowPass Filter Design - DTSP](https://www.youtube.com/watch?v=8M05FuKXIaw) — Padmasri Naban
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-order-2000-3500
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2

## Question

Derive the expression of order n for Chebyshev Low pass filter. Use this expression to find the order from the given specifications: ωp = 2000 rad/s, ωs = 3500 rad/s, αmax = 0.5 dB, αmin = 20 dB.

## Short answer

Order n ≥ arccosh(√((10^(0.1αmin)-1)/(10^(0.1αmax)-1)))/arccosh(ωs/ωp). Using αmax=0.5dB, αmin=20dB, ωs/ωp=3500/2000=1.75 gives n=4.

## Answer

#### Derivation of the 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: $\cosh^{-1}x = \ln\left(x+\sqrt{x^2-1}\right)$.)

#### Order for the given specification

Given $\alpha_{max} = 0.5$ dB, $\alpha_{min} = 20$ dB, $\frac{\omega_s}{\omega_p} = \frac{3500}{2000} = 1.75$.

$$
\begin{aligned}
10^{0.1\alpha_{max}} - 1 &= 10^{0.05} - 1 = 0.12202 \\
10^{0.1\alpha_{min}} - 1 &= 10^{2} - 1 = 99.0000 \\
\sqrt{\frac{99.0000}{0.12202}} &= 28.4843 \\
\cosh^{-1}(28.4843) &= 4.0422 \\
\cosh^{-1}(1.75) &= 1.1588 \\
n &\ge \frac{4.0422}{1.1588} = 3.488
\end{aligned}
$$

**Answer: n = 4** (next integer above 3.488).

A 4th-order Chebyshev filter with 0.5 dB ripple gives about $10\log[1+0.12202\cosh^2(4\times1.1588)] \approx 25.1$ dB at 3500 rad/s, which satisfies the 20 dB requirement.

## Check yourself

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

1. Which ratio is used for wp = 2000 and ws = 3500?
   A. 1.5
   B. 0.571 inverted to 2000
   C. 3500 - 2000
   D. 1.75
   Answer: D. ws/wp = 3500/2000 = 1.75. (Key point: Normalized ratio)

2. Why is the loss at w = 1 exactly amax for every n?
   A. Cn(1) = n
   B. Cn(1) = 1 for all n
   C. Because epsilon = 1
   D. Cn(1) = 0 for all n
   Answer: B. Cn(1) = cos(n cos^-1 1) = 1, so loss = 10log(1+eps^2). (Key point: Why C_n(1) = 1)

3. The formula gives 3.488 for 0.5 dB / 20 dB. The order is:
   A. 3
   B. 5
   C. 4
   D. 3.488
   Answer: C. Round up to 4; a 4th-order filter gives about 25.1 dB at 3500 rad/s. (Key point: Order result)

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