## Butterworth Filters

There exists a basic type of low-pass filter, called a Butterworth Filter, which is a 2nd-order filter, which therefore has a falloff-rate of -12db /Octave, far above the corner frequency, and this is its general diagram:

Even though it is clear from this diagram that the two capacitors, or the two resistors, are allowed to have different values, the way the design of this filter is mainly taught today, both resistors are made equal, as are both capacitors, thus simplifying the computation of each, once the other has been determined according to what seems practical, applying the same principle as what would be applied for a 1st-order filter.

One basic weakness of this filter, especially in modern applications, is the fact that it will attenuate frequency-components considerably, which are below its corner-frequency. There have historically been two approaches taken to reduce this effect, if any attempt has been made to do so at all:

1. C1 can be given twice the value of C2, but R1 and R2 kept equal. This poses the question of whether the corner-frequency will still be correct. And my estimation is that because of the way Electrical Engineers have defined the corner-frequency, the specific frequency-response at that frequency should remain the square root of 1/2 (or, -3db). But, if C1 is larger than C2, then the frequency-response will not be the same at any other point in the curve. I.e., the curve could be flatter, with response-values closer to unity, at frequencies considerably lower than the corner-frequency.
2. The operational amplifier stage, which in the basic design is just a voltage-follower, can be transformed into a gain-stage, with a gain slightly higher than one. This is done by placing a voltage-divider from the output of an operational amplifier, to yield the feedback voltage, fed to its inverting input. What needs to be stressed here, is that significantly high gain leads to an unstable circuit.

While either approach can be taken, it is important not to apply both at the same time, as the amount of feedback given by C1 would be exaggerated, and would lead to a hot-spot somewhere in the pass-band of this filter. In general, the trend today would be to use approach (2).

## About the Amplitudes of a Discrete Differential

One of the concepts which exist in digital signal processing, is that the difference between two consecutive input samples (in the time-domain) can simply be output, thus resulting in a differential of some sort, even though the samples of data do not represent a continuous function. There is a fact which must be observed to occur at (F = N / 2) – i.e. when the frequency is half the Nyquist Frequency, of (h / 2) , if (h) is the sampling frequency.

The input signal could be aligned with the samples, to give a sequence of [s0 … s3] equal to

0, +1, 0, -1

This set of (s) is equivalent to a sine-wave at (F = N / 2) . Its discrete differentiation [h0 … h3] would be

+1, +1, -1, -1

At first glance we might think, that this output stream has the same amplitude as the input stream. But the problem becomes that the output stream is by same token, not aligned with the samples. There is an implicit peak in amplitudes between (h0) and (h1) which is greater than (+1) , and an implicit peak between (h2) and (h3) more negative than (-1) . Any adequate filtering of this stream, belonging to a D/A conversion, will reproduce a sine-wave with a peak amplitude greater than (1).

(Edit 03/23/2017 : )

In this case we can see, that samples h0 and h1 of the output stream, would be phase-shifted 45⁰ with respect to the zero crossings and to the peak amplitude, that would exist exactly between h0 and h1. Therefore, the amplitude of h0 and h1 will be the sine-function of 45⁰ with respect to this peak value, and the actual peak would be (the square root of 2) times the values of h0 and h1.

(Erratum 11/28/2017 —

And so a logical question which anybody might want an answer to would be, ‘Below what frequency does the gain cross unity gain?’ And the answer to that question is, somewhat obscurely, at (N/3) . This is a darned low frequency in practice. If the sampling rate was 44.1kHz, this is achieved somewhere around 7 kHz, and music, for which that sampling rate was devised, easily contains sound energy above that frequency.

Hence the sequences which result would be:

s = [ +1, +1/2, -1/2, -1, -1/2, +1/2 ]

h = [ +1/2, -1/2, -1, -1/2, +1/2, +1 ]

What follows is also a reason for which by itself, DPCM offers poor performance in compressing signals. It usually needs to be combined with other methods of data-reduction, thus possibly resulting in the lossy ADPCM. And another approach which uses ADPCM, is aptX, the last of which is a proprietary codec, which minimizes the loss of quality that might otherwise stem from using ADPCM.

I believe this observation is also relevant to This Earlier Posting of mine, which implied a High-Pass Filter with a cutoff frequency of 500 Hz, that would be part of a Band-Pass Filter. My goal was to obtain a gain of at most 0.5 , over the entire interval, and to simplify the Math.

— End of Erratum. )

(Posting shortened here on 11/28/2017 . )

Dirk

## A Caveat in Using Vacuuming Robots

I own a “Neato XV Signature” -brand robot vacuum cleaner. I already posted about it here:

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