Bessel
TBD
Butterworth
Consider the n-th order signal transfer function
$$
\begin{align}
H(j \omega) &= \frac{1}{1 + j( \omega / \omega_0 )^{n}} \\
\end{align}
$$
Signal magnitude can then be determined by
$$
\begin{align}
H(j \omega) H^*(j \omega) &=\frac{1}{1 + j( \omega / \omega_0 )^{n}} \frac{1}{1 – j( \omega / \omega_0 )^{n}} \\
H(j \omega) H^*(j \omega) &=\frac{1}{1 + ( \omega / \omega_0 )^{2n}} \\
|H(j \omega)|^2 &=\frac{1}{1 + ( \omega / \omega_0 )^{2n}} \\
|H(j \omega)| &=\frac{1}{ \sqrt{1 + ( \omega / \omega_0 )^{2n} } } \\
\end{align}
$$
This magnitude response defines the n-th order Butterworth Filter.
Magnitude at Critical Frequency is
$$
\begin{align}
|H(j \omega_0)| &=\frac{1}{ \sqrt{1 + ( \omega_0 / \omega_0 )^{2n} } } \\
|H(j \omega_0)| &=\frac{1}{ \sqrt{ 2 } } \\
\\
|H(j \omega_0)|^2 &=\frac{1}{ 2 } \\
10 \log_{10} |H(j \omega_0)|^2 &\approx -3 dB \\
\end{align}
$$
Maximal Flatness
The goal of the Butterworth filter is to create a filter that is “maximally flat” at some known frequency. This is achieved by maximizing the number of derivatives of the frequency response with respect to frequency that are 0.
To get a better understanding, let’s first consider the equation $X_1(\omega) = a + \omega^4, a \in \mathbb{Z}$. Taking successive derivatives, we have
$$
\begin{align}
X_1(\omega) &= a + \omega^4 \\
\frac{d}{d \omega} X_1(\omega) &= 4 \cdot \omega^3 &&\text{1st Derivative} \\
\frac{d^2}{d \omega^2} X_1(\omega) &= 4 \cdot 3 \cdot \omega^2 &&\text{2nd Derivative} \\
\frac{d^3}{d \omega^3} X_1(\omega) &= 4 \cdot 3 \cdot 2 \cdot \omega &&\text{3rd Derivative} \\
\frac{d^4}{d \omega^4} X_1(\omega) &= 4 \cdot 3 \cdot 2 \cdot 1 &&\text{4th Derivative} \\
\frac{d^5}{d \omega^5} X_1(\omega) &= 0 &&\text{5th Derivative} \\
\end{align}
$$
Generalizing the pattern, we can consider the generalized equation
$$
\begin{align}
X_1(\omega) &= a + \omega^n \\
\end{align}
$$
The kth derivative can then be described as
$$
(k \in \mathbb{Z}) \land (k > 0) \implies
\frac{d^k}{d \omega^k} X_1(\omega) = \cases{
\begin{align}
&\frac{n!}{(n-k)!} \omega^{n-k} &&0 < k < n \\
&n! &&k=n \\
&0 &&k > n \\
\end{align}
}
$$
Note that if we are evaluating the derivatives at $\omega=0$, then the above simplifies to
$$
\left[ \frac{d^k}{d \omega^k} X_1(\omega) \right]_{\omega=0} = \cases{
\begin{align}
&0 &&0 < k < n \\
&n! &&k=n \\
&0 &&k > n \\
\end{align}
}
$$
and the derivatives are only nonzero in the nth derivative. This result becomes useful in the achieving maximal flatness in that the first n-1 derivatives are all 0.
Now, if we define a new frequency response $X_2(\omega) = \frac{1}{a + \omega^4} = [a + \omega^4]^{-1}$, we have
$$
\begin{align}
X_2(\omega) =& \frac{1}{a + \omega^4} \\
X_2(\omega) =& [a + \omega^4]^{-1} \\
\\
\frac{d}{d \omega} X_2(\omega) =& (-1) [a + \omega^4]^{-2} \cdot \frac{d}{d \omega} X_1(\omega) &&\text{1st Derivative} \\
\\
\frac{d^2}{d \omega^2} X_2(\omega) =
&(-1)(-2) [a + \omega^4]^{-3} \cdot \frac{d}{d \omega} X_1(\omega) + &&\text{2nd Derivative}\\
&(-1) [a + \omega^4]^{-2} \cdot \frac{d^2}{d \omega^2} X_1(\omega) \\
\\
\frac{d^3}{d \omega^3} X_2(\omega) =
&(-1)(-2)(-3) [a + \omega^4]^{-4} \cdot \frac{d}{d \omega} X_1(\omega) + &&\text{3rd Derivative}\\
&(-1)(-2) [a + \omega^4]^{-3} \cdot \frac{d^2}{d \omega^2} X_1(\omega) + \\
&(-1)(-2) [a + \omega^4]^{-3} \cdot \frac{d^2}{d \omega^2} X_1(\omega) + \\
&(-1) [a + \omega^4]^{-2} \cdot \frac{d^3}{d \omega^3} X_1(\omega) \\
=
&(-1)(-2)(-3) [a + \omega^4]^{-4} \cdot \frac{d}{d \omega} X_1(\omega) + \\
&2(-1)(-2) [a + \omega^4]^{-3} \cdot \frac{d^2}{d \omega^2} X_1(\omega) + \\
&(-1) [a + \omega^4]^{-2} \cdot \frac{d^3}{d \omega^3} X_1(\omega) \\
\\
\frac{d^4}{d \omega^4} X_2(\omega) =
&(-1)(-2)(-3)(-4) [a + \omega^4]^{-5} \cdot \frac{d}{d \omega} X_1(\omega) + &&\text{4th Derivative}\\
&(-1)(-2)(-3) [a + \omega^4]^{-4} \cdot \frac{d^2}{d \omega^2} X_1(\omega) + \\
&2(-1)(-2)(-3) [a + \omega^4]^{-4} \cdot \frac{d^2}{d \omega^2} X_1(\omega) + \\
&2(-1)(-2) [a + \omega^4]^{-3} \cdot \frac{d^3}{d \omega^3} X_1(\omega) + \\
&(-1)(-2) [a + \omega^4]^{-3} \cdot \frac{d^3}{d \omega^3} X_1(\omega) + \\
&(-1) [a + \omega^4]^{-2} \cdot \frac{d^4}{d \omega^4} X_1(\omega) \\
=
&(-1)(-2)(-3)(-4) [a + \omega^4]^{-5} \cdot \frac{d}{d \omega} X_1(\omega) + \\
&3(-1)(-2)(-3) [a + \omega^4]^{-4} \cdot \frac{d^2}{d \omega^2} X_1(\omega) + \\
&3(-1)(-2) [a + \omega^4]^{-3} \cdot \frac{d^3}{d \omega^3} X_1(\omega) + \\
&(-1) [a + \omega^4]^{-2} \cdot \frac{d^4}{d \omega^4} X_1(\omega) \\
\end{align}
$$
Noting the pattern as combinatorial in nature, we may summarize the kth derivative as a sum of k terms where
$$
\frac{d^k}{d \omega^k} \frac{1}{a + \omega^n} = \sum_{l=1}^k \binom{k}{l} (-1)^l (k-l+1)! [a + \omega^n]^{-(k-l+2)} \left[ \frac{d^l}{d \omega^l} (a + \omega^n) \right]
$$
More importantly, note however that from our previous analysis, the evaluation of $\frac{d^l}{d \omega^l} (a + \omega^n)$ at $\omega=0$ will always be 0 except when $l=k=n$. In short, we have a similar result as before where
$$
\left[ \frac{d^k}{d \omega^k} \frac{1}{a + \omega^n} \right]_{\omega=0} = \cases{
\begin{align}
&0 &&0 < k < n \\
&C \neq 0 &&k=n \\
&0 &&k > n \\
\end{align}
}
$$
which achieves our goal to maximize flatness.
Reviewing the previous analysis, one can also readily claim (again without rigorous proof) that the following statement is also true, which we will use in the next section.
$$
\left[ \frac{d^k}{d \omega^k} \frac{1}{1 + \left( \frac{\omega}{\omega_0} \right)^n} \right]_{\omega=0} = \cases{
\begin{align}
&0 &&0 < k < n \\
&C \neq 0 &&k=n \\
&0 &&k > n \\
\end{align}
}
$$
Coefficients
Up until now, we have determined the desired frequency response for maximal flatness. However, this does not necessarily indicate what the transfer function actually is, and where the poles are. Now, consider the transfer function when substituting the previous frequency, $\omega \in \mathbb{R}$, for any complex frequency, $s \in \mathbb{C}$.
$$
\begin{align}
X_3(s) &= \frac{1}{1 + \left( \frac{s}{\omega_0} \right)^n}
\end{align}
$$
To determine the poles, we first examine the denominator to determine when it equates to 0.
$$
\begin{align}
1 + \left( \frac{s_p}{\omega_0} \right)^n &= 0 \\
\left( \frac{s_p}{\omega_0} \right)^n &= -1 \\
s_p^n &= -1 \cdot \omega_0^n \\
\end{align}
$$
For an nth degree polynomial (the rightmost t, the Fundamental Theorem of Algebra dictates that there must be n roots in the complex plane. Additionally, any complex number may be represented using polar coordinates.
$$
\begin{align}
(r_p e^{\theta_p})^n &= -1 \cdot \omega_0^n \\
\end{align}
$$
Note that the right side is real ($\theta = 0 \pm k \pi$), and the magnitude is known, so the poles must satisfy
$$
\begin{align}
\end{align}
$$
TBD
Chebyshev Type I
TBD
Chebyshev Type II
TBD
Elliptic
TBD