Butterworth Filter: Derivation and Maximally Flat Response

1. The Butterworth Magnitude Response

As we discussed, Stephen Butterworth's goal was a filter with a frequency response that was as flat as possible in the passband. He stated that "an ideal electrical filter should not only completely reject the unwanted frequencies but should also have uniform sensitivity for the wanted frequencies." The approximation he developed mathematically formalizes this concept.

The magnitude squared of the frequency response for an Nth-order low-pass Butterworth filter is defined as:

$|H(j\omega ){|}^2=\frac{G_0^2}{1+{\left(\frac{\omega }{{\omega }_c}\right)}^{2N}}$

Let's break down these parameters:

• $G_0$ is the DC gain (the gain at $\omega =0$). We will assume $G_0=1$ for simplicity in our derivations.
• ${\omega }_c$ is the cutoff frequency. This is the frequency at which the magnitude is attenuated by a factor of $1/\sqrt{2}$, or -3 dB, regardless of the filter order.
• $N$ is the order of the filter. As $N$ increases, the filter's response gets closer to an ideal "brick-wall" filter.

To get a feel for this response, let's watch a short video.

As the video illustrates, for any order $N$:

• At $\omega =0$, $|H(j0){|}^2=1$, so the DC gain is unity.
• At $\omega ={\omega }_c$, $|H(j{\omega }_c){|}^2=1/(1+1)=1/2$. The magnitude is $|H(j{\omega }_c)|=1/\sqrt{2}\approx 0.707$, which corresponds to the -3 dB point.
• As $\omega \to \infty$, $|H(j\omega ){|}^2\to 0$. The response rolls off towards zero.

The image below summarizes this behavior for various filter orders, showing the response on both linear and logarithmic (decibel) scales.

2. Deriving the Butterworth Transfer Function, H(s)

Now for the central task: how do we get from the magnitude response $|H(j\omega )|$ to a realizable transfer function $H(s)$ that we can build as a circuit? This involves moving from the frequency domain ($j\omega$) to the complex frequency domain ($s$) and is a beautiful application of complex analysis.

The derivation involves three main steps:

1. Relate the magnitude-squared response $|H(j\omega ){|}^2$ to a function in the s-plane, $H(s)H(-s)$.
2. Find the poles of this combined function, $H(s)H(-s)$.
3. Select the poles that correspond to a stable filter to construct $H(s)$.

The following reading provides the complete mathematical details for this derivation.

Let's consolidate and explain the key steps from that reading.

Step 1: From $|H(j\omega ){|}^2$ to $H(s)H(-s)$

From your study of Laplace transforms, you know that for a stable system with a real impulse response, the magnitude-squared frequency response is given by:
$|H(j\omega ){|}^2=H(s)H(-s){|}_{s=j\omega }$
We can use this to move from the frequency domain to the s-domain. Let's start with our normalized Butterworth response (${\omega }_c=1,G_0=1$):
$|H(j\omega ){|}^2=\frac{1}{1+{\omega }^{2N}}$
To find $H(s)H(-s)$, we substitute $\omega =s/j$. This gives:
$H(s)H(-s)=\frac{1}{1+(s/j{)}^{2N}}=\frac{1}{1+(s^2/j^2{)}^N}=\frac{1}{1+(s^2/(-1){)}^N}=\frac{1}{1+(-1{)}^N{s}^{2N}}$

Step 2: Finding the Poles

The poles of $H(s)H(-s)$ are the roots of the denominator polynomial, i.e., the values of $s$ for which:
$1+(-1{)}^N{s}^{2N}=0⟹s^{2N}=-(-1{)}^{-N}$
This looks a bit complicated. A more direct way is to note that $(s/j{)}^{2N}=-1$. The roots of $-1$ in the complex plane are $e^{j(\pi +2k\pi )}$ for any integer $k$.
So, we are looking for $s$ such that:
$(s/j{)}^{2N}=e^{j\pi (2k+1)}$
Taking the $2N$-th root of both sides gives:
$s_k/j=e^{j\frac{\pi (2k+1)}{2N}}\text{for }k=0,1,\dots ,2N-1$
$s_k=j\cdot e^{j\frac{\pi (2k+1)}{2N}}=e^{j\pi /2}\cdot e^{j\frac{\pi (2k+1)}{2N}}=e^{j\pi \left(\frac{1}{2}+\frac{2k+1}{2N}\right)}$
These $2N$ poles are all located on a circle of radius 1 in the s-plane, spaced at equal angles. For a general cutoff frequency ${\omega }_c$, the poles lie on a circle of radius ${\omega }_c$.

Step 3: Selecting the Stable Poles

For a filter to be stable and causal, all of its poles must lie in the left half of the s-plane (LHP), where the real part is negative.

The $2N$ poles we found for $H(s)H(-s)$ are symmetrically distributed about the imaginary axis. For every pole $s_k$ in the LHP, there is a corresponding pole $-s_k$ in the right-half plane (RHP). The $N$ poles in the LHP belong to $H(s)$, and the $N$ poles in the RHP belong to $H(-s)$.

Therefore, to construct our stable filter $H(s)$, we simply select the $N$ poles from the set $\{s_k\}$ that have a negative real part. The following video provides a great visual explanation of this process.

Once we have identified the $N$ stable poles $p_1,p_2,\dots ,p_N$, we can write the transfer function as the product of their factors:
$H(s)=\frac{1}{(s-p_1)(s-p_2)\cdots (s-p_N)}$
The numerator is 1 to ensure the DC gain is 1. The denominator is known as a Butterworth polynomial.

3. The "Maximally Flat" Property Explained

We've now derived the filter, but why is it called "maximally flat"? Let's analyze this property formally, as it's the defining feature of the Butterworth approximation.

As the reading shows, the "maximally flat" property can be seen by performing a Maclaurin series expansion of the magnitude-squared function $|H(j\omega ){|}^2=G^2(\omega )$ around $\omega =0$. Let's use the normalized response again:
$G^2(\omega )=\frac{1}{1+{\omega }^{2N}}=(1+{\omega }^{2N}{)}^{-1}$
The general form of a Maclaurin series for $(1+x{)}^{\alpha }$ is $1+\alpha x+\frac{\alpha (\alpha -1)}{2!}{x}^2+\dots$. In our case, $x={\omega }^{2N}$ and $\alpha =-1$.
$G^2(\omega )=1-({\omega }^{2N})+\frac{(-1)(-2)}{2!}({\omega }^{2N}{)}^2-\cdots =1-{\omega }^{2N}+{\omega }^{4N}-\dots$
The key insight here is that the lowest power of $\omega$ in the series after the constant term is $2N$.

A function's k-th derivative at zero is non-zero only if its Maclaurin series contains an ${\omega }^k$ term (or can be related to it). Since all terms between ${\omega }^0$ and ${\omega }^{2N}$ are missing, it means that the first $2N-1$ derivatives of $G^2(\omega )$ with respect to $\omega$ are zero at $\omega =0$:
$\frac{d^k{G}^2(\omega )}{d{\omega }^k}{|}_{\omega =0}=0\text{for }k=1,2,\dots ,2N-1$
This is the mathematical definition of being maximally flat. For a given filter order $N$, the Butterworth response is the "flattest" possible function near DC that can be realized by a polynomial of that order. It hugs the value 1 as closely as possible before it starts to roll off.

Conclusion

In this lesson, we moved from the concept of a "maximally flat" filter to its complete mathematical description. You now understand not only what a Butterworth filter is, but how to derive it from first principles.

Key Takeaways:

• The Butterworth filter's magnitude-squared response is given by $|H(j\omega ){|}^2=\frac{1}{1+(\omega /{\omega }_c{)}^{2N}}$, which is defined by its order $N$ and cutoff frequency ${\omega }_c$.
• The transfer function $H(s)$ is derived by finding the $2N$ poles of $H(s)H(-s)$, which lie on a circle in the s-plane, and then selecting the $N$ poles in the left-half plane to ensure stability.
• The "maximally flat" property is a direct consequence of the mathematical form of the response. The first $2N-1$ derivatives of the magnitude-squared response are zero at $\omega =0$, making the passband extremely flat near DC.

Preview of the Next Lesson:
We have now mastered the maximally flat approach. But what if we're willing to sacrifice this perfect passband flatness for a faster roll-off? In our next lesson, we will derive the Chebyshev Type I filter approximation and analyze its equiripple properties. This will introduce a new set of trade-offs and a powerful new class of polynomials to our filter design toolkit.