--- title: "Chebyshev poles lie on an ellipse" subject: "Filter Design" chapter: 2 key: "semester_5/filter_design:chebyshev-poles-ellipse" source: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-poles-ellipse site: BCT Engineering --- # Chebyshev poles lie on an ellipse Filter Design · Chapter 2: Approximation Methods - Asked in: 2073 Chaitra, 7 marks - Video: [chebyshev filter poles formation on ellipse explanation](https://www.youtube.com/watch?v=7p8w8T4-098) — harsha mangipudi - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/chebyshev-poles-ellipse - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2 ## Question Show that the poles of chebyshev filter lie on an ellipse. Also show the major and minor axes. ## Short answer Poles are σ_k = –sinh a sinθ_k, ω_k = cosh a cosθ_k, θ_k=(2k–1)π/(2n), a=(1/n)sinh⁻¹(1/ε). Eliminating θ_k gives σ²/sinh²a + ω²/cosh²a = 1, an ellipse; semi-minor axis = sinh a (σ-axis), semi-major = cosh a (jω-axis), foci at ±j1. ## Answer The poles of a Chebyshev low-pass filter lie on an ellipse whose semi-minor axis is $\sinh a$ (along the real axis) and whose semi-major axis is $\cosh a$ (along the imaginary axis), where $a = \frac{1}{n}\sinh^{-1}(1/\varepsilon)$. #### Finding the poles The normalized Chebyshev response is $|T(j\omega)|^2 = \dfrac{1}{1+\varepsilon^2 C_n^2(\omega)}$ with $C_n(\omega) = \cos(n\cos^{-1}\omega)$. Replacing $\omega$ by $s/j$, the poles satisfy $$ 1 + \varepsilon^2 C_n^2(s/j) = 0 \;\Rightarrow\; \cos\left[n\cos^{-1}(s/j)\right] = \pm\frac{j}{\varepsilon} $$ Let $\cos^{-1}(s/j) = u + jv$. Then $$ \cos(nu)\cosh(nv) - j\sin(nu)\sinh(nv) = \pm\frac{j}{\varepsilon} $$ - Real part: $\cos(nu)\cosh(nv) = 0$. Since $\cosh(nv) \ge 1$, $\cos(nu) = 0$, so $u_k = \dfrac{(2k-1)\pi}{2n}$, $k = 1,\dots,2n$. - Imaginary part: with $\sin(nu_k) = \pm1$, $\sinh(nv) = 1/\varepsilon$, so $v = a = \dfrac{1}{n}\sinh^{-1}\dfrac{1}{\varepsilon}$. Now $s = j\cos(u_k + ja) = j[\cos u_k\cosh a - j\sin u_k\sinh a]$: $$ s_k = \sigma_k + j\omega_k = \sin u_k \sinh a + j\cos u_k \cosh a $$ The left-half-plane poles (for a stable filter) are $$ \sigma_k = -\sinh a\,\sin u_k, \qquad \omega_k = \cosh a\,\cos u_k, \qquad k = 1,\dots,n $$ #### Showing the locus is an ellipse $$ \left(\frac{\sigma_k}{\sinh a}\right)^2 + \left(\frac{\omega_k}{\cosh a}\right)^2 = \sin^2u_k + \cos^2u_k = 1 $$ This is an ellipse centred at the origin. #### Major and minor axes | Axis | Direction | Half-length | Full length | |---|---|---|---| | Major | $j\omega$ axis | $\cosh a$ | $2\cosh a$ | | Minor | $\sigma$ axis | $\sinh a$ | $2\sinh a$ | Since $\cosh^2 a - \sinh^2 a = 1$, the foci are at $s = \pm j1$. ```text jw | cosh a x . . + . . o . | . ----x--------+--------o---- sigma -sinh a | sinh a . | . x . . + . . o | x = LHP poles used, o = RHP mirror ``` **Geometric construction:** draw two circles of radius $\sinh a$ and $\cosh a$; draw radial lines at angles $u_k$ from the $j\omega$ axis. Each pole takes its real part from the small circle and its imaginary part from the large circle (compare: Butterworth poles lie on the unit circle). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. From the imaginary part with sin(nu_k) = ±1, what is found? A. sinh(nv) = 1/eps, so v = a = (1/n) sinh^-1(1/eps) B. cosh(nv) = 1/eps C. v = pi/2n D. v = eps Answer: A. The imaginary part gives sinh(nv) = 1/eps, hence a = (1/n) sinh^-1(1/eps). (Key point: Imaginary-part equation gives a) 2. Which is the major axis of the pole ellipse? A. Along sigma with half-length cosh a B. Along sigma with half-length sinh a C. Along j-omega with half-length sinh a D. Along j-omega with half-length cosh a Answer: D. Since cosh a > sinh a, the longer axis lies on the j-omega axis with half-length cosh a; minor is sinh a on sigma. (Key point: Major and minor axes) 3. Where are the foci of the pole ellipse? A. At ±1 on the sigma axis B. At the origin C. At s = ±j1, the passband edge, because cosh^2 a - sinh^2 a = 1 D. At s = ±j cosh a Answer: C. Focal distance squared is cosh^2 a - sinh^2 a = 1, so foci are at ±j1. 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