--- title: "Inverse Chebyshev characteristics, order, poles and zeros" subject: "Filter Design" chapter: 2 key: "semester_5/filter_design:inverse-chebyshev-poles-zeros" source: https://www.bctengineer.com/q/semester_5/filter_design/inverse-chebyshev-poles-zeros site: BCT Engineering --- # Inverse Chebyshev characteristics, order, poles and zeros Filter Design · Chapter 2: Approximation Methods - Asked in: 2081 Bhadra, 2+3+5 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/inverse-chebyshev-poles-zeros - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2 ## Question What are the characteristics of inverse Chebyshev response? Derive the expression to calculate the order of inverse Chebyshev low pass filter. Calculate inverse Chebyshev poles and zeros for given specifications: αmin = 18dB, αmax = 0.25 dB, ωs = 1400 rad/sec and ωp = 1000 rad/sec. ## Short answer Inverse Chebyshev: flat passband, equiripple stopband, finite jω-axis zeros. Order n=5 from n ≥ cosh⁻¹√((10^{1.8}–1)/(10^{0.025}–1))/cosh⁻¹(1.4). Denormalized poles: –202.34±j1239.9, –963.6±j1393.8, –2410.4 rad/s; zeros: ±j1472, ±j2382 rad/s. ## Answer #### Characteristics of the inverse Chebyshev response - **Maximally flat-like, monotonic passband:** no ripple in the passband; $|T(0)| = 1$. - **Equiripple stopband:** the response bounces between 0 and $1/\sqrt{1+1/\varepsilon^2}$ above $\omega_s$, so the minimum stopband attenuation is exactly $\alpha_{min}$. - **Finite transmission zeros** on the $j\omega$ axis at $\omega_s/\cos\frac{(2k-1)\pi}{2n}$, so it is not an all-pole function (numerator has $s^2+\omega_z^2$ factors). - **Transition band** as sharp as a Chebyshev filter of the same order (same order formula). - **Better phase and delay** in the passband than Chebyshev (poles are farther from the $j\omega$ axis), so less ringing. - For odd $n$ the response falls to zero at infinity at $-20$ dB/decade; for even $n$ it levels off at the stopband ripple value. ```text |T| 1 |----.__ | \ passband flat | \ | \ .-. .-. equal ripple | \/ \ / \ stopband +-----------+----+-----+---> w ws zeros ``` #### Order of the inverse Chebyshev low-pass filter The inverse Chebyshev (Chebyshev Type II) response, normalized to the stopband edge ($\Omega = \omega/\omega_s$, stopband begins at $\Omega=1$), is $$ |T(j\Omega)|^2 = \frac{\varepsilon^2 C_n^2(1/\Omega)}{1+\varepsilon^2 C_n^2(1/\Omega)} $$ It is obtained by taking a Chebyshev response, subtracting it from 1 (turning it into a high-pass with equiripple stopband) and replacing $\Omega$ by $1/\Omega$. **Step 1: stopband edge.** At $\Omega = 1$, $C_n(1) = 1$: $$ \alpha_{min} = 10\log_{10}\left(1+\frac{1}{\varepsilon^2}\right) \;\Rightarrow\; \varepsilon^2 = \frac{1}{10^{0.1\alpha_{min}}-1} $$ **Step 2: passband edge.** At $\Omega_p = \omega_p/\omega_s$, $1/\Omega_p = \omega_s/\omega_p > 1$, and the attenuation must not exceed $\alpha_{max}$: $$ \alpha_{max} \ge 10\log_{10}\left(1+\frac{1}{\varepsilon^2 C_n^2(\omega_s/\omega_p)}\right) \;\Rightarrow\; C_n^2\!\left(\frac{\omega_s}{\omega_p}\right) \ge \frac{1}{\varepsilon^2(10^{0.1\alpha_{max}}-1)} = \frac{10^{0.1\alpha_{min}}-1}{10^{0.1\alpha_{max}}-1} $$ **Step 3: solve.** With $C_n(x) = \cosh(n\cosh^{-1}x)$ for $x>1$: $$ n \ge \frac{\cosh^{-1}\sqrt{\dfrac{10^{0.1\alpha_{min}}-1}{10^{0.1\alpha_{max}}-1}}}{\cosh^{-1}\left(\dfrac{\omega_s}{\omega_p}\right)} $$ This is the same expression as for the Chebyshev filter: for the same specification both need the same order. #### Poles and zeros for the given specification **Order.** $\alpha_{max} = 0.25$ dB, $\alpha_{min} = 18$ dB, $\omega_s/\omega_p = 1.4$: $$ \begin{aligned} 10^{0.025}-1 &= 0.05925, \quad 10^{1.8}-1 = 62.0957 \\ \sqrt{62.0957/0.05925} &= 32.372, \quad \cosh^{-1}(32.372) = 4.1702 \\ \cosh^{-1}(1.4) &= 0.8670 \\ n &\ge 4.1702/0.8670 = 4.81 \;\Rightarrow\; n = 5 \end{aligned} $$ **Poles.** Compute Chebyshev poles with the stopband $\varepsilon$, invert them, and multiply by $\omega_s$: $$ \begin{aligned} \varepsilon &= \frac{1}{\sqrt{62.0957}} = 0.12690, \quad \frac{1}{\varepsilon} = 7.88008 \\ a &= \tfrac{1}{5}\sinh^{-1}(7.88008) = 0.55230 \\ \sinh a &= 0.58081, \quad \cosh a = 1.15643 \\ p_k &= -\sinh a \sin\theta_k + j\cosh a\cos\theta_k, \quad \theta_k = \tfrac{(2k-1)\pi}{10} \\ s_k &= \omega_s / p_k \end{aligned} $$ | k | $\theta_k$ | Chebyshev $p_k$ | $1/p_k$ | Pole $= 1400/p_k$ (rad/s) | |---|---|---|---|---| | 1, 5 | 18° | $-0.17948 \pm j1.09983$ | $-0.14453 \mp j0.88564$ | $-202.34 \mp j1239.90$ | | 2, 4 | 54° | $-0.46988 \pm j0.67973$ | $-0.68814 \mp j0.99547$ | $-963.40 \mp j1393.66$ | | 3 | 90° | $-0.58081$ | $-1.72174$ | $-2410.44$ | **Zeros.** On the $j\omega$ axis at $\omega_{z,k} = \omega_s/\cos\theta_k$: $$ \begin{aligned} \omega_{z1} &= \frac{1400}{\cos 18^\circ} = \frac{1400}{0.95106} = 1472.05\ \text{rad/s} \\ \omega_{z2} &= \frac{1400}{\cos 54^\circ} = \frac{1400}{0.58779} = 2381.82\ \text{rad/s} \end{aligned} $$ ($\theta = 90°$ gives a zero at infinity, since $n$ is odd.) **Answer:** $n = 5$; poles at $-202.34 \pm j1239.90$, $-963.40 \pm j1393.66$ and $-2410.44$ rad/s; zeros at $\pm j1472.05$, $\pm j2381.82$ rad/s and one at $\infty$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the passband and stopband behaviours of the inverse Chebyshev response? A. Equiripple passband, monotonic stopband B. Monotonic passband, equiripple stopband C. Monotonic in both D. Equiripple in both Answer: B. Inverse Chebyshev is flat/monotonic in the passband and equiripple in the stopband, with minimum attenuation alpha_min. (Key point: Inverse Chebyshev passband and stopband) 2. Where are the transmission zeros of the inverse Chebyshev filter? A. On the j-omega axis at ws/cos((2k-1)pi/2n) B. On the sigma axis C. At the origin D. All at infinity Answer: A. The zeros are on the j-omega axis at ws/cos(theta_k); for odd n one zero is at infinity. (Key point: Finite zeros) 3. For alpha_max = 0.25 dB, alpha_min = 18 dB, ws/wp = 1.4, the formula gives 4.81. The order is: A. 4 B. 6 C. 5 D. 4.81 Answer: C. The next integer above 4.81 is 5; the order formula is the same as for Chebyshev. (Key point: Order) 4. How are the inverse Chebyshev poles obtained? A. Compute Chebyshev poles with the stopband epsilon, invert them, and multiply by ws B. Use the Chebyshev poles directly with wp C. Reflect Chebyshev poles to the right half plane D. Take the zeros and negate them Answer: A. Poles are s_k = ws/p_k where p_k are Chebyshev poles using epsilon = 1/sqrt(10^(0.1 alpha_min)-1). (Key point: Pole construction) ## Close variants set in other exams - [Chebyshev characteristics and order (1500/4500 rad/s)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Achebyshev-characteristics-order-4500) — Filter Design, chapter 2, asked 2× ## Related in Filter Design - [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× - [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× - [Inverse Chebyshev characteristics and order (0.4 dB, 16 dB)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Ainverse-chebyshev-characteristics-order-16db) — Filter Design, chapter 2, asked 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× - [Chebyshev order, poles and TF (1 dB, 18 dB)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Achebyshev-order-poles-tf-1db-18db) — Filter Design, chapter 2 - [Chebyshev order (0.25 dB, 18 dB, 1650 rad/s)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Achebyshev-order-0-25db-1650) — Filter Design, chapter 2, asked 2× ## Related in other subjects - [Impulse response of ideal low pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-impulse-response) — Signals and Systems, chapter 5, asked 5× - [Impulse response of DT ideal low pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-discrete-time) — Signals and Systems, chapter 6