---
title: "Chebyshev characteristics and order (0.25 dB, 40 dB)"
subject: "Filter Design"
chapter: 2
key: "semester_5/filter_design:chebyshev-characteristics-order-40db"
source: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-characteristics-order-40db
site: BCT Engineering
---

# Chebyshev characteristics and order (0.25 dB, 40 dB)

Filter Design · Chapter 2: Approximation Methods

- Asked in: 2076 Asoj, 2+5+3 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-characteristics-order-40db
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2

## Question

What are the characteristics of Chebyshev filter? Derive an expression to calculate the order of given low pass specifications using Chebyshev approximation. Using your expression calculate the order of Chebyshev low pass filter for following specifications: Passband extending from ω = 0 rad/s to ω = 1000 rad/s, the attenuation should not exceed 0.25 dB. Stopband extending from ω = 2500 rad/s to ω = ∞, the attenuation should not be less than 40 dB.

## Short answer

Chebyshev filter has equiripple passband, monotonic stopband, -20n dB/dec roll-off, and lower order than Butterworth. The order is n ≥ arcosh(√[(10^(0.1αmin)-1)/(10^(0.1αmax)-1)]) / arcosh(ωs/ωp). For αmax=0.25 dB, αmin=40 dB, ωs/ωp=2.5, n=5.

## Answer

#### Characteristics of the Chebyshev filter

- **Equiripple passband:** $|T|$ oscillates between $1$ and $1/\sqrt{1+\varepsilon^2}$ in $0 \le \Omega \le 1$; the ripple height is $\alpha_{max}$ dB and is set by $\varepsilon$.
- **Number of ripples:** the passband contains $n$ half-cycles of ripple between $\Omega = 0$ and $\Omega = 1$, so a higher order gives more ripples (of the same height).
- **DC value:** $|T(0)| = 1$ for odd $n$ and $1/\sqrt{1+\varepsilon^2}$ for even $n$.
- **Monotonic stopband:** beyond $\Omega = 1$ the response falls monotonically; the asymptotic roll-off is $-20n$ dB/decade, but the transition is much sharper than Butterworth of the same order (about $6(n-1)$ dB more stopband attenuation for the same $\varepsilon$).
- **All-pole function:** no finite transmission zeros; poles lie on an **ellipse** in the s-plane.
- **Phase/delay:** phase is more non-linear and group delay has a larger peak near the band edge than Butterworth (poorer transient response, more ringing).
- For a given specification it needs a **lower order** than Butterworth, so fewer components.

#### Derivation of the order

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

Passband $0$ to $1000$ rad/s with $\alpha_{max} = 0.25$ dB; stopband from $2500$ rad/s with $\alpha_{min} = 40$ dB.

$$
\begin{aligned}
\varepsilon^2 &= 10^{0.025}-1 = 0.05925 \\
10^{4}-1 &= 9999 \\
\sqrt{9999/0.05925} &= 410.79, \quad \cosh^{-1}(410.79) = 6.7112 \\
\cosh^{-1}(2.5) &= 1.5668 \\
n &\ge \frac{6.7112}{1.5668} = 4.283
\end{aligned}
$$

**Answer: n = 5.**

## Check yourself

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

1. How does the Chebyshev response behave beyond the passband edge?
   A. It ripples with equal height
   B. It rises again after a dip
   C. It falls monotonically, sharper than Butterworth of the same order
   D. It stays constant at 1/sqrt(1+eps^2)
   Answer: C. Beyond Omega = 1 the response falls monotonically and faster than Butterworth of the same order. (Key point: Stopband behaviour and roll-off)

2. What condition on C_n(Omega_s) gives the order formula?
   A. C_n^2(Omega_s) <= (10^(0.1 alpha_min) - 1)/eps^2
   B. C_n^2(Omega_s) >= (10^(0.1 alpha_min) - 1)/eps^2
   C. C_n(Omega_s) = eps
   D. C_n^2(Omega_s) >= 10^(0.1 alpha_max)
   Answer: B. Attenuation at Omega_s must be at least alpha_min, so 1+eps^2 C_n^2 >= 10^(0.1 alpha_min). (Key point: Stopband condition)

3. For alpha_max = 0.25 dB, which expression gives eps^2?
   A. 10^0.25 - 1
   B. 10^0.025 - 1 = 0.05925
   C. 10^(2.5) - 1
   D. 0.25/10
   Answer: B. eps^2 = 10^(0.1*0.25) - 1 = 10^0.025 - 1 = 0.05925. (Key point: Ripple factor for the spec)

4. With the ratio 6.7112/1.5668 = 4.283, the order for the 40 dB spec is:
   A. 4
   B. 6
   C. 5
   D. 4.283
   Answer: C. The next integer above 4.283 is 5. (Key point: Final order)

## Close variants set in other exams

- [Butterworth order (0.1 dB to 200 rad/s, 30 dB)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abutterworth-order-200-2000) — Filter Design, chapter 2, asked 2×

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