--- title: "Chebyshev response and order (0.5 dB, 22 dB)" subject: "Filter Design" chapter: 2 key: "semester_5/filter_design:chebyshev-response-order-22db" source: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-response-order-22db site: BCT Engineering --- # Chebyshev response and order (0.5 dB, 22 dB) Filter Design · Chapter 2: Approximation Methods - Asked in: 2079 Bhadra, 7+3 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-response-order-22db - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2 ## Question Derive the expression for the responses and order of chebyshev approximation method. Use Chebyshev approximation formula to estimate the order of Chebyshev filter for the following specifications: ωp = 2000 rad/sec; αmax = 0.5 dB; ωs = 2000 rad/sec [as printed]; αmin = 22 dB ## Short answer The Chebyshev response is |T|²=1/[1+ε²C_n²(ω/ωp)] with equal-ripple passband and monotonic stopband. The order is n ≥ arcosh(√[(10^(0.1αmin)-1)/(10^(0.1αmax)-1)]) / arcosh(ωs/ωp). With printed ωs=ωp=2000 rad/s, transition is zero, so assuming the companion problem's ωs=3000 rad/s gives n=5. ## Answer #### Chebyshev response The Chebyshev approximation chooses $$ |T(j\Omega)|^2 = \frac{1}{1+\varepsilon^2 C_n^2(\Omega)},\qquad \Omega = \frac{\omega}{\omega_p} $$ where $C_n$ is the Chebyshev polynomial, defined by $C_n(\Omega) = \cos(n\cos^{-1}\Omega)$ for $|\Omega|\le1$ and $\cosh(n\cosh^{-1}\Omega)$ for $|\Omega|>1$. It obeys the recursion $$ C_0 = 1,\quad C_1 = \Omega,\quad C_{n+1}(\Omega) = 2\Omega C_n(\Omega) - C_{n-1}(\Omega) $$ so $C_2 = 2\Omega^2-1$, $C_3 = 4\Omega^3-3\Omega$, $C_4 = 8\Omega^4-8\Omega^2+1$. Behaviour of the response: - In the passband ($|\Omega|\le1$), $C_n$ swings between $-1$ and $+1$, so $C_n^2$ is between 0 and 1 and $|T|$ ripples between $1$ and $1/\sqrt{1+\varepsilon^2}$ (equal ripple). - In the stopband ($|\Omega|>1$), $C_n$ grows like $2^{n-1}\Omega^n$, so $|T|$ falls monotonically and quickly. - At $\Omega=0$: $|T|=1$ for odd $n$, $1/\sqrt{1+\varepsilon^2}$ for even $n$. At $\Omega=1$: $|T| = 1/\sqrt{1+\varepsilon^2}$ for all $n$. ```text |T| 1 |\ /\ /\ | \/ \/ \ | \ | \__ +-------------+--------> W 1 ``` #### Order of the Chebyshev 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 As printed, $\omega_p = \omega_s = 2000$ rad/s. With $\omega_s/\omega_p = 1$, $\cosh^{-1}(1) = 0$ and $n \to \infty$: a zero-width transition band cannot be met by any finite filter, so the value of $\omega_s$ is clearly a misprint. The companion Butterworth question with the same $\omega_p$, $\alpha_{max}$ and $\alpha_{min}$ uses $\omega_s = 3000$ rad/s, so **assume $\omega_s = 3000$ rad/s** ($\omega_s/\omega_p = 1.5$). $$ \begin{aligned} \varepsilon^2 &= 10^{0.05}-1 = 0.12202 \\ 10^{2.2}-1 &= 157.4893 \\ \sqrt{157.4893/0.12202} &= 35.9263, \quad \cosh^{-1}(35.9263) = 4.2744 \\ \cosh^{-1}(1.5) &= 0.9624 \\ n &\ge \frac{4.2744}{0.9624} = 4.441 \end{aligned} $$ **Answer: n = 5** (next integer above 4.441). Check: with $n = 5$ the attenuation at $\omega_s$ is $10\log_{10}[1+0.12202\cosh^2(5\cosh^{-1}1.5)] = 26.65$ dB $\ge 22$ dB, while $n = 4$ gives only 18.35 dB, which is not enough. The same steps apply to whatever $\omega_s$ is intended; only $\cosh^{-1}(\omega_s/\omega_p)$ changes. For example, $\omega_s = 4000$ rad/s ($\omega_s/\omega_p = 2$) gives $n \ge 4.2744/1.3170 = 3.246$, so $n = 4$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Using C_0 = 1, C_1 = Omega and C_(n+1) = 2*Omega*C_n - C_(n-1), what is C_2? A. 2*Omega^2 - 1 B. 2*Omega^2 + 1 C. Omega^2 - 1 D. 4*Omega^3 - 3*Omega Answer: A. C_2 = 2*Omega*Omega - 1 = 2*Omega^2 - 1. (Key point: Chebyshev polynomial recursion) 2. What is |T| at Omega = 1 for any order n? A. 1 B. 0 C. 1/sqrt(1+eps^2) D. 1/(1+eps^2) Answer: C. C_n(1) = 1 for all n, so |T(1)| = 1/sqrt(1+eps^2). (Key point: Behaviour at Omega = 1) 3. Why can the order not be computed with ws = wp = 2000 rad/s as printed? A. Because cosh^-1(1) = 0 makes n infinite, so a nonzero transition band is needed B. Because alpha_min would be zero C. Because epsilon becomes negative D. Because cosh^-1(1) = 1 gives n = 1 which is wrong Answer: A. With ws/wp = 1 the denominator cosh^-1(1) = 0, so n tends to infinity; a transition band is needed, hence ws = 4000 assumed. (Key point: Misprinted spec handling) ## Close variants set in other exams - [Butterworth order (0.5 dB, 22 dB)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abutterworth-order-2000-3000-22db) — Filter Design, chapter 2, asked 4× ## Related in Filter Design - [Chebyshev order (0.5 dB, 20 dB, 2000 rad/s)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Achebyshev-order-0-5db-2000) — Filter Design, chapter 2 - [Inverse Chebyshev order (0.5 dB, 18 dB, 1800 rad/s)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Ainverse-chebyshev-order-1800-18db) — Filter Design, chapter 2, asked 3× - [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× - [Chebyshev order (0.5 dB, 20 dB, 3500 rad/s)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Achebyshev-order-2000-3500) — Filter Design, chapter 2 - [Inverse Chebyshev order (0.25 dB, 18 dB, 1400 rad/s)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Ainverse-chebyshev-order-1400) — Filter Design, chapter 2, asked 5× - [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× ## Related in other subjects - [S-parameters and Butterworth/Chebyshev filter responses](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3As-parameters-filter-responses) — RF and Microwave Engineering, chapter 3