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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
source | link

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
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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.")

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