9 added 752 characters in body edited Nov 5 '16 at 16:52 robert bristow-johnson 12.8k33 gold badges2020 silver badges5252 bronze badges i don't think it's particularly remarkable that Butterworth filters, defined as all-pole filters that are maximally flat at $$\omega=0$$ (for LPF prototype, meaning the most possible derivatives of $$|H(j\omega)|$$ are zero at $$\omega=0$$), have s-plane poles that lie equally spaced on the left half-circle of radius $$\omega_0$$. from the "maximally flat" and "no zeros", you can derive $$|H(j\omega)|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2N}}$$ for the $$N$$th-order Butterworth. so $$|H(s)|^2 = \frac{1}{1 + \left(\frac{s}{j \omega_0}\right)^{2N}}$$ $$s=p_n$$ is a pole when the denominator is zero. $$1 + \left(\frac{p_n}{j \omega_0}\right)^{2N} = 0$$ or $$\left(\frac{p_n}{j \omega_0}\right)^{2N} = -1$$ $$p_n^{2N} = - (j \omega_0)^{2N}$$ $$|p_n| = \omega_0$$ $$2N \cdot \arg(p_n) = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$$$2N \cdot \arg\{p_n\} = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$ $$\arg(p_n) = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$$$\arg\{p_n\} = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$ for $$N$$th-order Tchebyshev (Type 1, which is all-pole), it's like this: $$|H(j\omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\omega}{\omega_c}\right)}$$ where $$T_N(x) \triangleq \begin{cases} \cos\big(N \, \arccos(x) \big), & \text{if }|x| \le 1 \\ \cosh\big(N \, \operatorname{arccosh}(x) \big), & \text{if }x \ge 1 \\ (-1)^N \, \cosh\big(N \, \operatorname{arccosh}(-x) \big), & \text{if }x \le -1 \end{cases}$$ are the $$N$$th-order Tchebyshev polynomials and satisfy the recursion: \begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x) \quad \quad \forall n \in \mathbb{Z} \ge 1 \end{align} and $$\omega_c$$ is the "passband cutoff" frequency and not to be confused with the -3 dB frequency $$\omega_0$$. (but the two are related.) the passband ripple parameter is $$\epsilon = \sqrt{10^{\tfrac{dB_\text{ripple}}{10}} - 1}$$ analytic extension again: $$|H(s)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{s}{j \omega_c}\right)}$$ and again $$s=p_n$$ is a pole when the denominator is zero. $$1 + \epsilon^2 T_N^2\left(\frac{p_n}{j \omega_c}\right) = 0$$ or $$T_N\left(\frac{p_n}{j \omega_c}\right) = \pm \frac{j}{\epsilon}$$ (because $$\cos(\theta) = \cosh(j \theta)$$ we can use either $$\cos()$$ or $$\cosh()$$ expression for $$T_N()$$ $$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \pm \frac{j}{\epsilon}$$ $$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\pm \frac{j}{\epsilon}\right)$$ since $$y = \cosh(x) = \tfrac12 ( e^x + e^{-x} )$$ and $$x = \operatorname{arccosh}(y) = \log \left( y \pm \sqrt{y^2-1} \right)$$ then $$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm \frac{j}{\epsilon} \pm \sqrt{\left(\pm\frac{j}{\epsilon}\right)^2-1} \right)$$ $$N \log \left( \frac{\Re(p_n)+j \Im(p_n)}{j \omega_c} \pm \sqrt{\left(\frac{\Re(p_n)+j \Im(p_n)}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ $$N \log \left( \frac{=j \Re(p_n)+\Im(p_n)}{\omega_c} \pm \sqrt{\left(\frac{-j \Re(p_n) + \Im(p_n)}{\omega_c}\right)^2 - 1} \right) = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$$$N \log \left( \frac{-j \Re(p_n)+\Im(p_n)}{\omega_c} \pm \sqrt{\left(\frac{-j \Re(p_n) + \Im(p_n)}{\omega_c}\right)^2 - 1} \right) \\ = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ oh dear i might not get this blasted out in 12 hours i've decided that i am too lazy to grok through this. if anyone wants to pick it up, feel free to. lotsa conversion between rectangular and polar notation of complex values. remember when $$w = \pm \sqrt{\ z \ }$$ then $$|w| = +\sqrt{|z|}$$ and \begin{align} \arg\{w\} &= \frac12 \arg\{z \} + \arg\{ \pm 1\} \\ &= \frac12 \arg\{z \} + \frac{\pi}{2}(1 \pm 1) \end{align} and remember $$\log(z) = \log|z| + j\arg\{z\} + j 2 \pi n \quad n \in \mathbb{Z}$$ you may add any integer multiple of $$2 \pi$$ (say "$$2 \pi n$$") to any $$\arg\{\cdot\}$$ (choose the right-hand $$\log()$$ which is how you can get different poles for $$p_n$$). if you like mathematical masturbation with complex variables, knock yourself out. i don't think it's particularly remarkable that Butterworth filters, defined as all-pole filters that are maximally flat at $$\omega=0$$ (for LPF prototype, meaning the most possible derivatives of $$|H(j\omega)|$$ are zero at $$\omega=0$$), have s-plane poles that lie equally spaced on the left half-circle of radius $$\omega_0$$. from the "maximally flat" and "no zeros", you can derive $$|H(j\omega)|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2N}}$$ for the $$N$$th-order Butterworth. so $$|H(s)|^2 = \frac{1}{1 + \left(\frac{s}{j \omega_0}\right)^{2N}}$$ $$s=p_n$$ is a pole when the denominator is zero. $$1 + \left(\frac{p_n}{j \omega_0}\right)^{2N} = 0$$ or $$\left(\frac{p_n}{j \omega_0}\right)^{2N} = -1$$ $$p_n^{2N} = - (j \omega_0)^{2N}$$ $$|p_n| = \omega_0$$ $$2N \cdot \arg(p_n) = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$ $$\arg(p_n) = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$ for $$N$$th-order Tchebyshev (Type 1, which is all-pole), it's like this: $$|H(j\omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\omega}{\omega_c}\right)}$$ where $$T_N(x) \triangleq \begin{cases} \cos\big(N \, \arccos(x) \big), & \text{if }|x| \le 1 \\ \cosh\big(N \, \operatorname{arccosh}(x) \big), & \text{if }x \ge 1 \\ (-1)^N \, \cosh\big(N \, \operatorname{arccosh}(-x) \big), & \text{if }x \le -1 \end{cases}$$ are the $$N$$th-order Tchebyshev polynomials and satisfy the recursion: \begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x) \quad \quad \forall n \in \mathbb{Z} \ge 1 \end{align} and $$\omega_c$$ is the "passband cutoff" frequency and not to be confused with the -3 dB frequency $$\omega_0$$. (but the two are related.) the passband ripple parameter is $$\epsilon = \sqrt{10^{\tfrac{dB_\text{ripple}}{10}} - 1}$$ analytic extension again: $$|H(s)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{s}{j \omega_c}\right)}$$ and again $$s=p_n$$ is a pole when the denominator is zero. $$1 + \epsilon^2 T_N^2\left(\frac{p_n}{j \omega_c}\right) = 0$$ or $$T_N\left(\frac{p_n}{j \omega_c}\right) = \pm \frac{j}{\epsilon}$$ (because $$\cos(\theta) = \cosh(j \theta)$$ we can use either $$\cos()$$ or $$\cosh()$$ expression for $$T_N()$$ $$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \pm \frac{j}{\epsilon}$$ $$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\pm \frac{j}{\epsilon}\right)$$ since $$y = \cosh(x) = \tfrac12 ( e^x + e^{-x} )$$ and $$x = \operatorname{arccosh}(y) = \log \left( y \pm \sqrt{y^2-1} \right)$$ then $$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm \frac{j}{\epsilon} \pm \sqrt{\left(\pm\frac{j}{\epsilon}\right)^2-1} \right)$$ $$N \log \left( \frac{\Re(p_n)+j \Im(p_n)}{j \omega_c} \pm \sqrt{\left(\frac{\Re(p_n)+j \Im(p_n)}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ $$N \log \left( \frac{=j \Re(p_n)+\Im(p_n)}{\omega_c} \pm \sqrt{\left(\frac{-j \Re(p_n) + \Im(p_n)}{\omega_c}\right)^2 - 1} \right) = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ oh dear i might not get this blasted out in 12 hours i don't think it's particularly remarkable that Butterworth filters, defined as all-pole filters that are maximally flat at $$\omega=0$$ (for LPF prototype, meaning the most possible derivatives of $$|H(j\omega)|$$ are zero at $$\omega=0$$), have s-plane poles that lie equally spaced on the left half-circle of radius $$\omega_0$$. from the "maximally flat" and "no zeros", you can derive $$|H(j\omega)|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2N}}$$ for the $$N$$th-order Butterworth. so $$|H(s)|^2 = \frac{1}{1 + \left(\frac{s}{j \omega_0}\right)^{2N}}$$ $$s=p_n$$ is a pole when the denominator is zero. $$1 + \left(\frac{p_n}{j \omega_0}\right)^{2N} = 0$$ or $$\left(\frac{p_n}{j \omega_0}\right)^{2N} = -1$$ $$p_n^{2N} = - (j \omega_0)^{2N}$$ $$|p_n| = \omega_0$$ $$2N \cdot \arg\{p_n\} = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$ $$\arg\{p_n\} = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$ for $$N$$th-order Tchebyshev (Type 1, which is all-pole), it's like this: $$|H(j\omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\omega}{\omega_c}\right)}$$ where $$T_N(x) \triangleq \begin{cases} \cos\big(N \, \arccos(x) \big), & \text{if }|x| \le 1 \\ \cosh\big(N \, \operatorname{arccosh}(x) \big), & \text{if }x \ge 1 \\ (-1)^N \, \cosh\big(N \, \operatorname{arccosh}(-x) \big), & \text{if }x \le -1 \end{cases}$$ are the $$N$$th-order Tchebyshev polynomials and satisfy the recursion: \begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x) \quad \quad \forall n \in \mathbb{Z} \ge 1 \end{align} and $$\omega_c$$ is the "passband cutoff" frequency and not to be confused with the -3 dB frequency $$\omega_0$$. (but the two are related.) the passband ripple parameter is $$\epsilon = \sqrt{10^{\tfrac{dB_\text{ripple}}{10}} - 1}$$ analytic extension again: $$|H(s)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{s}{j \omega_c}\right)}$$ and again $$s=p_n$$ is a pole when the denominator is zero. $$1 + \epsilon^2 T_N^2\left(\frac{p_n}{j \omega_c}\right) = 0$$ or $$T_N\left(\frac{p_n}{j \omega_c}\right) = \pm \frac{j}{\epsilon}$$ (because $$\cos(\theta) = \cosh(j \theta)$$ we can use either $$\cos()$$ or $$\cosh()$$ expression for $$T_N()$$ $$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \pm \frac{j}{\epsilon}$$ $$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\pm \frac{j}{\epsilon}\right)$$ since $$y = \cosh(x) = \tfrac12 ( e^x + e^{-x} )$$ and $$x = \operatorname{arccosh}(y) = \log \left( y \pm \sqrt{y^2-1} \right)$$ then $$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm \frac{j}{\epsilon} \pm \sqrt{\left(\pm\frac{j}{\epsilon}\right)^2-1} \right)$$ $$N \log \left( \frac{\Re(p_n)+j \Im(p_n)}{j \omega_c} \pm \sqrt{\left(\frac{\Re(p_n)+j \Im(p_n)}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ $$N \log \left( \frac{-j \Re(p_n)+\Im(p_n)}{\omega_c} \pm \sqrt{\left(\frac{-j \Re(p_n) + \Im(p_n)}{\omega_c}\right)^2 - 1} \right) \\ = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ oh dear i might not get this blasted out in 12 hours i've decided that i am too lazy to grok through this. if anyone wants to pick it up, feel free to. lotsa conversion between rectangular and polar notation of complex values. remember when $$w = \pm \sqrt{\ z \ }$$ then $$|w| = +\sqrt{|z|}$$ and \begin{align} \arg\{w\} &= \frac12 \arg\{z \} + \arg\{ \pm 1\} \\ &= \frac12 \arg\{z \} + \frac{\pi}{2}(1 \pm 1) \end{align} and remember $$\log(z) = \log|z| + j\arg\{z\} + j 2 \pi n \quad n \in \mathbb{Z}$$ you may add any integer multiple of $$2 \pi$$ (say "$$2 \pi n$$") to any $$\arg\{\cdot\}$$ (choose the right-hand $$\log()$$ which is how you can get different poles for $$p_n$$). if you like mathematical masturbation with complex variables, knock yourself out. 8 added 221 characters in body edited Nov 5 '16 at 6:25 robert bristow-johnson 12.8k33 gold badges2020 silver badges5252 bronze badges i don't think it's particularly remarkable that Butterworth filters, defined as all-pole filters that are maximally flat at $$\omega=0$$ (for LPF prototype, meaning the most possible derivatives of $$|H(j\omega)|$$ are zero at $$\omega=0$$), have s-plane poles that lie equally spaced on the left half-circle of radius $$\omega_0$$. from the "maximally flat" and "no zeros", you can derive $$|H(j\omega)|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2N}}$$ for the $$N$$th-order Butterworth. so $$|H(s)|^2 = \frac{1}{1 + \left(\frac{s}{j \omega_0}\right)^{2N}}$$ $$s=p_n$$ is a pole when the denominator is zero. $$1 + \left(\frac{p_n}{j \omega_0}\right)^{2N} = 0$$ or $$\left(\frac{p_n}{j \omega_0}\right)^{2N} = -1$$ $$p_n^{2N} = - (j \omega_0)^{2N}$$ $$|p_n| = \omega_0$$ $$2N \cdot \arg(p_n) = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$ $$\arg(p_n) = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$ for $$N$$th-order Tchebyshev (Type 1, which is all-pole), it's like this: $$|H(j\omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\omega}{\omega_c}\right)}$$ where $$T_N(x) \triangleq \begin{cases} \cos\big(N \, \arccos(x) \big), & \text{if }|x| \le 1 \\ \cosh\big(N \, \operatorname{arccosh}(x) \big), & \text{if }x \ge 1 \\ (-1)^N \, \cosh\big(N \, \operatorname{arccosh}(-x) \big), & \text{if }x \le -1 \end{cases}$$ are the $$N$$th-order Tchebyshev polynomials and satisfy the recursion: \begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x) \quad \quad \forall n \in \mathbb{Z} \ge 1 \end{align} and $$\omega_c$$ is the "passband cutoff" frequency and not to be confused with the -3 dB frequency $$\omega_0$$. (but the two are related.) the passband ripple parameter is $$\epsilon = \sqrt{10^{\tfrac{dB_\text{ripple}}{10}} - 1}$$ analytic extension again: $$|H(s)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{s}{j \omega_c}\right)}$$ and again $$s=p_n$$ is a pole when the denominator is zero. $$1 + \epsilon^2 T_N^2\left(\frac{p_n}{j \omega_c}\right) = 0$$ or $$T_N\left(\frac{p_n}{j \omega_c}\right) = \pm \frac{j}{\epsilon}$$ (because $$\cos(\theta) = \cosh(j \theta)$$ we can use either $$\cos()$$ or $$\cosh()$$ expression for $$T_N()$$ $$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \pm \frac{j}{\epsilon}$$ $$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\pm \frac{j}{\epsilon}\right)$$ since $$y = \cosh(x) = \tfrac12 ( e^x + e^{-x} )$$ and $$x = \operatorname{arccosh}(y) = \log \left( y \pm \sqrt{y^2-1} \right)$$ then $$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm \frac{j}{\epsilon} \pm \sqrt{\left(\pm\frac{j}{\epsilon}\right)^2-1} \right)$$ . . (still more left to do) i dunno, that's how i would do it.$$N \log \left( \frac{\Re(p_n)+j \Im(p_n)}{j \omega_c} \pm \sqrt{\left(\frac{\Re(p_n)+j \Im(p_n)}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ (now gimme 500 points because, if there is one thing that i am learning from Cap'n Combover, is that i deserve whatever i claim that i deserve. so don't be a Nasty Woman and gimme 500 points.)$$N \log \left( \frac{=j \Re(p_n)+\Im(p_n)}{\omega_c} \pm \sqrt{\left(\frac{-j \Re(p_n) + \Im(p_n)}{\omega_c}\right)^2 - 1} \right) = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ (or "You're fired.")oh dear i might not get this blasted out in 12 hours i don't think it's particularly remarkable that Butterworth filters, defined as all-pole filters that are maximally flat at $$\omega=0$$ (for LPF prototype, meaning the most possible derivatives of $$|H(j\omega)|$$ are zero at $$\omega=0$$), have s-plane poles that lie equally spaced on the left half-circle of radius $$\omega_0$$. from the "maximally flat" and "no zeros", you can derive $$|H(j\omega)|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2N}}$$ for the $$N$$th-order Butterworth. so $$|H(s)|^2 = \frac{1}{1 + \left(\frac{s}{j \omega_0}\right)^{2N}}$$ $$s=p_n$$ is a pole when the denominator is zero. $$1 + \left(\frac{p_n}{j \omega_0}\right)^{2N} = 0$$ or $$\left(\frac{p_n}{j \omega_0}\right)^{2N} = -1$$ $$p_n^{2N} = - (j \omega_0)^{2N}$$ $$|p_n| = \omega_0$$ $$2N \cdot \arg(p_n) = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$ $$\arg(p_n) = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$ for $$N$$th-order Tchebyshev (Type 1, which is all-pole), it's like this: $$|H(j\omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\omega}{\omega_c}\right)}$$ where $$T_N(x) \triangleq \begin{cases} \cos\big(N \, \arccos(x) \big), & \text{if }|x| \le 1 \\ \cosh\big(N \, \operatorname{arccosh}(x) \big), & \text{if }x \ge 1 \\ (-1)^N \, \cosh\big(N \, \operatorname{arccosh}(-x) \big), & \text{if }x \le -1 \end{cases}$$ are the $$N$$th-order Tchebyshev polynomials and satisfy the recursion: \begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x) \quad \quad \forall n \in \mathbb{Z} \ge 1 \end{align} and $$\omega_c$$ is the "passband cutoff" frequency and not to be confused with the -3 dB frequency $$\omega_0$$. (but the two are related.) the passband ripple parameter is $$\epsilon = \sqrt{10^{\tfrac{dB_\text{ripple}}{10}} - 1}$$ analytic extension again: $$|H(s)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{s}{j \omega_c}\right)}$$ and again $$s=p_n$$ is a pole when the denominator is zero. $$1 + \epsilon^2 T_N^2\left(\frac{p_n}{j \omega_c}\right) = 0$$ or $$T_N\left(\frac{p_n}{j \omega_c}\right) = \pm \frac{j}{\epsilon}$$ (because $$\cos(\theta) = \cosh(j \theta)$$ we can use either $$\cos()$$ or $$\cosh()$$ expression for $$T_N()$$ $$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \pm \frac{j}{\epsilon}$$ $$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\pm \frac{j}{\epsilon}\right)$$ since $$y = \cosh(x) = \tfrac12 ( e^x + e^{-x} )$$ and $$x = \operatorname{arccosh}(y) = \log \left( y \pm \sqrt{y^2-1} \right)$$ then $$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm \frac{j}{\epsilon} \pm \sqrt{\left(\pm\frac{j}{\epsilon}\right)^2-1} \right)$$ . . (still more left to do) i dunno, that's how i would do it. (now gimme 500 points because, if there is one thing that i am learning from Cap'n Combover, is that i deserve whatever i claim that i deserve. so don't be a Nasty Woman and gimme 500 points.) (or "You're fired.") i don't think it's particularly remarkable that Butterworth filters, defined as all-pole filters that are maximally flat at $$\omega=0$$ (for LPF prototype, meaning the most possible derivatives of $$|H(j\omega)|$$ are zero at $$\omega=0$$), have s-plane poles that lie equally spaced on the left half-circle of radius $$\omega_0$$. from the "maximally flat" and "no zeros", you can derive $$|H(j\omega)|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2N}}$$ for the $$N$$th-order Butterworth. so $$|H(s)|^2 = \frac{1}{1 + \left(\frac{s}{j \omega_0}\right)^{2N}}$$ $$s=p_n$$ is a pole when the denominator is zero. $$1 + \left(\frac{p_n}{j \omega_0}\right)^{2N} = 0$$ or $$\left(\frac{p_n}{j \omega_0}\right)^{2N} = -1$$ $$p_n^{2N} = - (j \omega_0)^{2N}$$ $$|p_n| = \omega_0$$ $$2N \cdot \arg(p_n) = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$ $$\arg(p_n) = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$ for $$N$$th-order Tchebyshev (Type 1, which is all-pole), it's like this: $$|H(j\omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\omega}{\omega_c}\right)}$$ where $$T_N(x) \triangleq \begin{cases} \cos\big(N \, \arccos(x) \big), & \text{if }|x| \le 1 \\ \cosh\big(N \, \operatorname{arccosh}(x) \big), & \text{if }x \ge 1 \\ (-1)^N \, \cosh\big(N \, \operatorname{arccosh}(-x) \big), & \text{if }x \le -1 \end{cases}$$ are the $$N$$th-order Tchebyshev polynomials and satisfy the recursion: \begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x) \quad \quad \forall n \in \mathbb{Z} \ge 1 \end{align} and $$\omega_c$$ is the "passband cutoff" frequency and not to be confused with the -3 dB frequency $$\omega_0$$. (but the two are related.) the passband ripple parameter is $$\epsilon = \sqrt{10^{\tfrac{dB_\text{ripple}}{10}} - 1}$$ analytic extension again: $$|H(s)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{s}{j \omega_c}\right)}$$ and again $$s=p_n$$ is a pole when the denominator is zero. $$1 + \epsilon^2 T_N^2\left(\frac{p_n}{j \omega_c}\right) = 0$$ or $$T_N\left(\frac{p_n}{j \omega_c}\right) = \pm \frac{j}{\epsilon}$$ (because $$\cos(\theta) = \cosh(j \theta)$$ we can use either $$\cos()$$ or $$\cosh()$$ expression for $$T_N()$$ $$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \pm \frac{j}{\epsilon}$$ $$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\pm \frac{j}{\epsilon}\right)$$ since $$y = \cosh(x) = \tfrac12 ( e^x + e^{-x} )$$ and $$x = \operatorname{arccosh}(y) = \log \left( y \pm \sqrt{y^2-1} \right)$$ then $$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm \frac{j}{\epsilon} \pm \sqrt{\left(\pm\frac{j}{\epsilon}\right)^2-1} \right)$$ $$N \log \left( \frac{\Re(p_n)+j \Im(p_n)}{j \omega_c} \pm \sqrt{\left(\frac{\Re(p_n)+j \Im(p_n)}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ $$N \log \left( \frac{=j \Re(p_n)+\Im(p_n)}{\omega_c} \pm \sqrt{\left(\frac{-j \Re(p_n) + \Im(p_n)}{\omega_c}\right)^2 - 1} \right) = \log \left( \pm j\left( \frac{1}{\epsilon} \pm \sqrt{\frac{1}{\epsilon^2}+1} \right) \right)$$ oh dear i might not get this blasted out in 12 hours 7 added 19 characters in body edited Nov 1 '16 at 2:40 robert bristow-johnson 12.8k33 gold badges2020 silver badges5252 bronze badges i don't think it's particularly remarkable that Butterworth filters, defined as all-pole filters that are maximally flat at $$\omega=0$$ (for LPF prototype, meaning the most possible derivatives of $$|H(j\omega)|$$ are zero at $$\omega=0$$), have s-plane poles that lie equally spaced on the left half-circle of radius $$\omega_0$$. from the "maximally flat" and "no zeros", you can derive $$|H(j\omega)|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2N}}$$ for the $$N$$th-order Butterworth. so $$|H(s)|^2 = \frac{1}{1 + \left(\frac{s}{j \omega_0}\right)^{2N}}$$ $$s=p_n$$ is a pole when the denominator is zero. $$1 + \left(\frac{p_n}{j \omega_0}\right)^{2N} = 0$$ or $$\left(\frac{p_n}{j \omega_0}\right)^{2N} = -1$$ $$p_n^{2N} = - (j \omega_0)^{2N}$$ $$|p_n| = \omega_0$$ $$2N \cdot \arg(p_n) = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$ $$\arg(p_n) = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$ for $$N$$th-order Tchebyshev (Type 1, which is all-pole), it's like this: $$|H(j\omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\omega}{\omega_c}\right)}$$ where $$T_N(x) \triangleq \begin{cases} \cos\big(N \, \arccos(x) \big), & \text{if }|x| \le 1 \\ \cosh\big(N \, \operatorname{arccosh}(x) \big), & \text{if }x \ge 1 \\ (-1)^N \, \cosh\big(N \, \operatorname{arccosh}(-x) \big), & \text{if }x \le -1 \end{cases}$$ are the $$N$$th-order Tchebyshev polynomials and satisfy the recursion: \begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x) \quad \quad \forall n \in \mathbb{Z} \ge 1 \end{align} and $$\omega_c$$ is the "passband cutoff" frequency and not to be confused with the -3 dB frequency $$\omega_0$$. (but the two are related.) the passband ripple parameter is $$\epsilon = \sqrt{10^{\tfrac{dB_\text{ripple}}{10}} - 1}$$ analytic extension again: $$|H(s)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{s}{j \omega_c}\right)}$$ and again $$s=p_n$$ is a pole when the denominator is zero. $$1 + \epsilon^2 T_N^2\left(\frac{p_n}{j \omega_c}\right) = 0$$ or $$T_N\left(\frac{p_n}{j \omega_c}\right) = \frac{j}{\epsilon}$$$$T_N\left(\frac{p_n}{j \omega_c}\right) = \pm \frac{j}{\epsilon}$$ (because $$\cos(\theta) = \cosh(j \theta)$$ we can use either $$\cos()$$ or $$\cosh()$$ expression for $$T_N()$$ $$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \frac{j}{\epsilon}$$$$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \pm \frac{j}{\epsilon}$$ $$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\frac{j}{\epsilon}\right)$$$$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\pm \frac{j}{\epsilon}\right)$$ since $$y = \cosh(x) = \tfrac12 ( e^x + e^{-x} )$$ and $$x = \operatorname{arccosh}(y) = \log \left( y \pm \sqrt{y^2-1} \right)$$ then $$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \frac{j}{\epsilon} \pm \sqrt{\left(\frac{j}{\epsilon}\right)^2-1} \right)$$$$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm \frac{j}{\epsilon} \pm \sqrt{\left(\pm\frac{j}{\epsilon}\right)^2-1} \right)$$ . . (still more left to do) i dunno, that's how i would do it. (now gimme 500 points because, if there is one thing that i am learning from Cap'n Combover, is that i deserve whatever i claim that i deserve. so don't be a Nasty Woman and gimme 500 points.) (or "You're fired.") i don't think it's particularly remarkable that Butterworth filters, defined as all-pole filters that are maximally flat at $$\omega=0$$ (for LPF prototype, meaning the most possible derivatives of $$|H(j\omega)|$$ are zero at $$\omega=0$$), have s-plane poles that lie equally spaced on the left half-circle of radius $$\omega_0$$. from the "maximally flat" and "no zeros", you can derive $$|H(j\omega)|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2N}}$$ for the $$N$$th-order Butterworth. so $$|H(s)|^2 = \frac{1}{1 + \left(\frac{s}{j \omega_0}\right)^{2N}}$$ $$s=p_n$$ is a pole when the denominator is zero. $$1 + \left(\frac{p_n}{j \omega_0}\right)^{2N} = 0$$ or $$\left(\frac{p_n}{j \omega_0}\right)^{2N} = -1$$ $$p_n^{2N} = - (j \omega_0)^{2N}$$ $$|p_n| = \omega_0$$ $$2N \cdot \arg(p_n) = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$ $$\arg(p_n) = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$ for $$N$$th-order Tchebyshev (Type 1, which is all-pole), it's like this: $$|H(j\omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\omega}{\omega_c}\right)}$$ where $$T_N(x) \triangleq \begin{cases} \cos\big(N \, \arccos(x) \big), & \text{if }|x| \le 1 \\ \cosh\big(N \, \operatorname{arccosh}(x) \big), & \text{if }x \ge 1 \\ (-1)^N \, \cosh\big(N \, \operatorname{arccosh}(-x) \big), & \text{if }x \le -1 \end{cases}$$ are the $$N$$th-order Tchebyshev polynomials and satisfy the recursion: \begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x) \quad \quad \forall n \in \mathbb{Z} \ge 1 \end{align} and $$\omega_c$$ is the "passband cutoff" frequency and not to be confused with the -3 dB frequency $$\omega_0$$. (but the two are related.) the passband ripple parameter is $$\epsilon = \sqrt{10^{\tfrac{dB_\text{ripple}}{10}} - 1}$$ analytic extension again: $$|H(s)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{s}{j \omega_c}\right)}$$ and again $$s=p_n$$ is a pole when the denominator is zero. $$1 + \epsilon^2 T_N^2\left(\frac{p_n}{j \omega_c}\right) = 0$$ or $$T_N\left(\frac{p_n}{j \omega_c}\right) = \frac{j}{\epsilon}$$ (because $$\cos(\theta) = \cosh(j \theta)$$ we can use either $$\cos()$$ or $$\cosh()$$ expression for $$T_N()$$ $$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \frac{j}{\epsilon}$$ $$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\frac{j}{\epsilon}\right)$$ since $$y = \cosh(x) = \tfrac12 ( e^x + e^{-x} )$$ and $$x = \operatorname{arccosh}(y) = \log \left( y \pm \sqrt{y^2-1} \right)$$ then $$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \frac{j}{\epsilon} \pm \sqrt{\left(\frac{j}{\epsilon}\right)^2-1} \right)$$ . . (still more left to do) i dunno, that's how i would do it. (now gimme 500 points because, if there is one thing that i am learning from Cap'n Combover, is that i deserve whatever i claim that i deserve. so don't be a Nasty Woman and gimme 500 points.) (or "You're fired.") i don't think it's particularly remarkable that Butterworth filters, defined as all-pole filters that are maximally flat at $$\omega=0$$ (for LPF prototype, meaning the most possible derivatives of $$|H(j\omega)|$$ are zero at $$\omega=0$$), have s-plane poles that lie equally spaced on the left half-circle of radius $$\omega_0$$. from the "maximally flat" and "no zeros", you can derive $$|H(j\omega)|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2N}}$$ for the $$N$$th-order Butterworth. so $$|H(s)|^2 = \frac{1}{1 + \left(\frac{s}{j \omega_0}\right)^{2N}}$$ $$s=p_n$$ is a pole when the denominator is zero. $$1 + \left(\frac{p_n}{j \omega_0}\right)^{2N} = 0$$ or $$\left(\frac{p_n}{j \omega_0}\right)^{2N} = -1$$ $$p_n^{2N} = - (j \omega_0)^{2N}$$ $$|p_n| = \omega_0$$ $$2N \cdot \arg(p_n) = -\pi + 2N \cdot \frac{\pi}{2} + 2 \pi n$$ $$\arg(p_n) = \frac{\pi}{2} + \frac{\pi}{N}\left( n - \tfrac12 \right)$$ for $$N$$th-order Tchebyshev (Type 1, which is all-pole), it's like this: $$|H(j\omega)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{\omega}{\omega_c}\right)}$$ where $$T_N(x) \triangleq \begin{cases} \cos\big(N \, \arccos(x) \big), & \text{if }|x| \le 1 \\ \cosh\big(N \, \operatorname{arccosh}(x) \big), & \text{if }x \ge 1 \\ (-1)^N \, \cosh\big(N \, \operatorname{arccosh}(-x) \big), & \text{if }x \le -1 \end{cases}$$ are the $$N$$th-order Tchebyshev polynomials and satisfy the recursion: \begin{align} T_0(x) & = 1 \\ T_1(x) & = x \\ T_{n+1}(x) & = 2xT_n(x) - T_{n-1}(x) \quad \quad \forall n \in \mathbb{Z} \ge 1 \end{align} and $$\omega_c$$ is the "passband cutoff" frequency and not to be confused with the -3 dB frequency $$\omega_0$$. (but the two are related.) the passband ripple parameter is $$\epsilon = \sqrt{10^{\tfrac{dB_\text{ripple}}{10}} - 1}$$ analytic extension again: $$|H(s)|^2 = \frac{1}{1 + \epsilon^2 T_N^2\left(\frac{s}{j \omega_c}\right)}$$ and again $$s=p_n$$ is a pole when the denominator is zero. $$1 + \epsilon^2 T_N^2\left(\frac{p_n}{j \omega_c}\right) = 0$$ or $$T_N\left(\frac{p_n}{j \omega_c}\right) = \pm \frac{j}{\epsilon}$$ (because $$\cos(\theta) = \cosh(j \theta)$$ we can use either $$\cos()$$ or $$\cosh()$$ expression for $$T_N()$$ $$\cosh\big(N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) \big) = \pm \frac{j}{\epsilon}$$ $$N \, \operatorname{arccosh}\left(\frac{p_n}{j \omega_c}\right) = \operatorname{arccosh}\left(\pm \frac{j}{\epsilon}\right)$$ since $$y = \cosh(x) = \tfrac12 ( e^x + e^{-x} )$$ and $$x = \operatorname{arccosh}(y) = \log \left( y \pm \sqrt{y^2-1} \right)$$ then $$N \log \left( \frac{p_n}{j \omega_c} \pm \sqrt{\left(\frac{p_n}{j \omega_c}\right)^2 - 1} \right) = \log \left( \pm \frac{j}{\epsilon} \pm \sqrt{\left(\pm\frac{j}{\epsilon}\right)^2-1} \right)$$ . . (still more left to do) i dunno, that's how i would do it. (now gimme 500 points because, if there is one thing that i am learning from Cap'n Combover, is that i deserve whatever i claim that i deserve. so don't be a Nasty Woman and gimme 500 points.) (or "You're fired.") 6 added 684 characters in body edited Nov 1 '16 at 2:33 robert bristow-johnson 12.8k33 gold badges2020 silver badges5252 bronze badges 5 added 1271 characters in body edited Oct 31 '16 at 7:34 robert bristow-johnson 12.8k33 gold badges2020 silver badges5252 bronze badges 4 added 7 characters in body edited Oct 31 '16 at 4:21 robert bristow-johnson 12.8k33 gold badges2020 silver badges5252 bronze badges 3 added 2 characters in body edited Oct 30 '16 at 7:09 robert bristow-johnson 12.8k33 gold badges2020 silver badges5252 bronze badges 2 added 492 characters in body edited Oct 30 '16 at 6:49 robert bristow-johnson 12.8k33 gold badges2020 silver badges5252 bronze badges 1 answered Oct 30 '16 at 4:03 robert bristow-johnson 12.8k33 gold badges2020 silver badges5252 bronze badges