---
title: "Inverse Chebyshev order (0.25 dB, 18 dB, 1400 rad/s)"
subject: "Filter Design"
chapter: 2
key: "semester_5/filter_design:inverse-chebyshev-order-1400"
source: https://www.bctengineer.com/q/semester_5/filter_design/inverse-chebyshev-order-1400
site: BCT Engineering
---

# Inverse Chebyshev order (0.25 dB, 18 dB, 1400 rad/s)

Filter Design · Chapter 2: Approximation Methods

- Asked in: 2072 Chaitra, 5+3 marks; 2071 Chaitra, 5+3 marks (asked 2 times)
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/inverse-chebyshev-order-1400
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2

## Question

Derive an expression to calculate the order of Inverse Chebyshev low pass filter. Use this formula to estimate the order of Inverse Chebyshev low pass filter having following specification: αmax = 0.25 dB, ωp = 1000 rad/s, αmin = 18 dB, ωs = 1400 rad/s

## Short answer

The order is n ≥ arccosh(√((10^(0.1αmin)−1)/(10^(0.1αmax)−1))) / arccosh(ωs/ωp). For αmax=0.25 dB, αmin=18 dB, ωp=1000 rad/s, ωs=1400 rad/s, n=5.

## Answer

#### Order of an inverse Chebyshev low-pass filter
The inverse Chebyshev response is flat (monotonic) in the passband and equiripple in the stopband. Normalized to the **stopband edge** $\omega_s$ = 1 rad/s:
$$
|T(j\omega)|^2 = \frac{\varepsilon^2 C_n^2(1/\omega)}{1 + \varepsilon^2 C_n^2(1/\omega)}, \qquad
\alpha(\omega) = 10\log_{10}\left[1 + \frac{1}{\varepsilon^2 C_n^2(1/\omega)}\right]
$$
**Stopband ($\omega \ge 1$):** $1/\omega \le 1$, so $C_n^2(1/\omega)$ swings between 0 and 1. The least attenuation (where $C_n^2 = 1$) must be $\alpha_{min}$:
$$
\alpha_{min} = 10\log_{10}\left(1 + \frac{1}{\varepsilon^2}\right) \;\Rightarrow\; \varepsilon^2 = \frac{1}{10^{0.1\alpha_{min}} - 1}
$$
**Passband:** the normalized passband edge is $\omega_p/\omega_s < 1$, where the loss must not exceed $\alpha_{max}$:
$$
\begin{aligned}
10\log_{10}\left[1 + \frac{1}{\varepsilon^2 C_n^2(\omega_s/\omega_p)}\right] &\le \alpha_{max} \\
C_n^2\left(\frac{\omega_s}{\omega_p}\right) &\ge \frac{1}{\varepsilon^2(10^{0.1\alpha_{max}} - 1)} = \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}
\end{aligned}
$$
Since $\omega_s/\omega_p > 1$, $C_n = \cosh(n\cosh^{-1}x)$, so
$$
n \ge \frac{\cosh^{-1}\sqrt{\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}}}{\cosh^{-1}(\omega_s/\omega_p)}
$$
This is the same expression as for the Chebyshev filter.

#### Order for the given specifications
$\alpha_{max}$ = 0.25 dB, $\alpha_{min}$ = 18 dB, $\omega_p$ = 1000 rad/s, $\omega_s$ = 1400 rad/s.
$$
\begin{aligned}
10^{0.1\alpha_{min}} - 1 &= 10^{1.8} - 1 = 62.0957 \\
10^{0.1\alpha_{max}} - 1 &= 10^{0.025} - 1 = 0.05925 \\
\sqrt{62.0957/0.05925} &= 32.3723, \qquad \frac{\omega_s}{\omega_p} = \frac{1400}{1000} = 1.4 \\
n &\ge \frac{\cosh^{-1}(32.3723)}{\cosh^{-1}(1.4)} = \frac{4.1702}{0.8670} = 4.810
\end{aligned}
$$
So **n = 5**.

**Answer:** a 5th-order inverse Chebyshev filter is required, with $\varepsilon^2 = 1/62.096 = 0.01610$.

## Check yourself

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

1. Which describes the inverse Chebyshev response?
   A. Flat monotonic passband and equiripple stopband
   B. Equiripple passband and monotonic stopband
   C. Equiripple in both bands
   D. Monotonic in both bands with no zeros
   Answer: A. Finite zeros in the stopband give the stopband ripple. (Key point: Inverse Chebyshev response)

2. How is ε^2 found from αmin?
   A. ε^2 = 10^(0.1αmin) - 1
   B. ε^2 = 1/(10^(0.1αmin) - 1)
   C. ε^2 = 1/(10^(0.1αmax) - 1)
   D. ε^2 = αmin/20
   Answer: B. The least stopband attenuation αmin = 10 log(1 + 1/ε^2). (Key point: Epsilon)

3. What is the order formula?
   A. n ≥ log[...] / (2 log(ωs/ωp))
   B. n ≥ cos^-1(ωs/ωp) / cosh^-1(...)
   C. n ≥ cosh^-1 sqrt[(10^(0.1αmin)-1)/(10^(0.1αmax)-1)] / cosh^-1(ωs/ωp)
   D. n ≥ sinh^-1(...) / sinh^-1(ωs/ωp)
   Answer: C. Same as for Chebyshev. (Key point: Order formula)

4. For αmax 0.25 dB, αmin 18 dB, ωs/ωp = 1.4, n computes to 4.81. The order is?
   A. 5
   B. 4
   C. 6
   D. 3
   Answer: A. Round up to the next integer. (Key point: Result)

## Close variants set in other exams

- [Chebyshev order (0.1 dB, 20 dB, 2500 rad/s)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Achebyshev-order-0-1db-2500) — Filter Design, chapter 2, asked 5×
- [Chebyshev order (0.25 dB, 20 dB, 1500 rad/s)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Achebyshev-order-0-25db-1500) — Filter Design, chapter 2, asked 6×
- [Inverse Chebyshev order (0.5 dB, 25 dB, 1800 rad/s)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Ainverse-chebyshev-order-1800-25db) — Filter Design, chapter 2, asked 5×

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