How the order parameter ($n$) in the Butterworth filter mathematically bridges the gap between the Ideal and Gaussian filters?

To understand this, let's look at the formula for the Butterworth Lowpass Filter (BLPF):

• $D(u,v)$: The current distance of a frequency point from the center of the spectrum.

• $D_0$: The cutoff frequency (the chosen distance where we want to start blocking high frequencies).

• $n$: The order of the filter. This is the magic number.

The magic of the Butterworth filter lies entirely in that exponent, $2n$. By simply changing the value of $n$, we fundamentally alter the geometry of the filter's shape in the frequency domain.

How the Order ($n$) Changes the Shape

Imagine looking at a 1-D cross-section (a side profile) of the filter. Here is what happens as we change $n$:

1. Low Order ($n=1$) - The Gaussian Twin:

   When $n=1$, the transition from the passband (allowing frequencies, where $H\approx 1$) to the stopband (blocking frequencies, where $H\approx 0$) is very slow and gradual. At this low order, the Butterworth curve behaves almost exactly like a Gaussian filter. Because the transition is so smooth, it produces absolutely no "ringing" artifacts in the final image.

2. Moderate Order ($n=2$ or $3$) - The "Sweet Spot":

   As $n$ increases to 2, the "slope" of the cutoff becomes a bit steeper. It holds onto the low frequencies a little better than a Gaussian, but drops off faster once it hits $D_0$. A slight amount of ringing might technically be present, but it is usually completely imperceptible to the human eye. This is why $n=2$ is the standard default for image processing.

3. High Order ($n\to \mathrm{\infty }$) - The Ideal Shape:

   Consider what happens to the math when $n$ becomes very large (e.g., $n=20$ or higher).

   • If $D<D_0$ (inside the cutoff), the fraction $D/D_0$ is less than 1. A number less than 1 raised to a massive power approaches 0. So, $H=1/(1+0)=\mathbf{1}$.

   • If $D>D_0$ (outside the cutoff), the fraction $D/D_0$ is greater than 1. A number greater than 1 raised to a massive power approaches infinity. So, $H=1/(1+\mathrm{\infty })=\mathbf{0}$.

   This creates an immediate, vertical drop-off at $D_0$. The filter abruptly snaps from 1 to 0. This is the exact definition of the Ideal filter, which creates severe ringing artifacts.