--- title: "Chebyshev characteristics and minimum order" subject: "Filter Design" chapter: 2 key: "semester_5/filter_design:chebyshev-characteristics-order-25db" source: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-characteristics-order-25db site: BCT Engineering --- # Chebyshev characteristics and minimum order Filter Design · Chapter 2: Approximation Methods - Asked in: 2080 Baisakh, 3+4+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-25db - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2 ## Question What are the characteristics of chebyshev magnitude response? Derive an expression to calculate the order (n) of a Chebyshev filter for given lowpass specifications. Determine the minimum order n of chebyshev filter for following specifications. αp = 1 dB, αs = 25 dB and (ωs/ωp) = 1.5, where the symbols have their usual meanings. ## Short answer Chebyshev response has equiripple passband, monotonic stopband, and steeper transition than Butterworth. Order n ≥ arccosh(√((10^(0.1αs)-1)/(10^(0.1αp)-1)))/arccosh(ωs/ωp). For αp=1dB, αs=25dB, ωs/ωp=1.5, minimum n=5. ## Answer The Chebyshev (Type I) response is an all-pole approximation that allows equal ripple in the passband to get a much steeper transition into the stopband. #### Characteristics of the Chebyshev magnitude response - **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. ```text |T| 1 |~\ /~\ /~\ | \/ \/ \ <- equal ripple (alpha_max) | \ | \_ | \___ monotonic stopband +-----------------+-------> w wp ``` #### 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: $\cosh^{-1}x = \ln\left(x+\sqrt{x^2-1}\right)$.) #### Minimum order for the given specification Given $\alpha_p = 1$ dB, $\alpha_s = 25$ dB, $\omega_s/\omega_p = 1.5$. $$ \begin{aligned} \varepsilon^2 &= 10^{0.1} - 1 = 0.25893 \\ 10^{2.5} - 1 &= 315.2278 \\ \sqrt{\frac{315.2278}{0.25893}} &= 34.8919 \\ \cosh^{-1}(34.8919) &= 4.2452 \\ \cosh^{-1}(1.5) &= 0.9624 \\ n &\ge \frac{4.2452}{0.9624} = 4.411 \end{aligned} $$ **Answer: minimum order n = 5.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the passband behaviour of a Chebyshev type I filter? A. Ripple only in the stopband B. Equal ripple between 1 and 1/sqrt(1+eps^2), with n half-cycles in 0 to 1 C. Monotonic and maximally flat D. Ripple growing with frequency Answer: B. The passband has equal-height ripples set by epsilon, with n half-cycles up to w = 1. (Key point: Equiripple passband) 2. What is |T(0)| for a Chebyshev filter? A. 0 B. 1 for odd n and 1/sqrt(1+eps^2) for even n C. 1/sqrt(1+eps^2) for all n D. 1 for all n Answer: B. Cn(0) is +/-1 for even n and 0 for odd n, so gain at DC is lower for even n. (Key point: DC gain) 3. For 1 dB, 25 dB and ws/wp = 1.5 the formula gives 4.411. n is: A. 5 B. 6 C. 4 D. 4.411 Answer: A. Round up to n = 5. (Key point: Order result) 4. Which statement is correct for Chebyshev versus Butterworth? A. Chebyshev poles lie on the jw axis B. Chebyshev poles lie on an ellipse and it needs a lower order for a given spec C. Chebyshev poles lie on a circle and it needs a higher order D. Chebyshev has linear phase and needs a higher order Answer: B. Elliptical pole locus gives steeper transition so fewer components, at the cost of poorer phase/delay. 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