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. )
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 revealed by Differential Calculus. If a sine-wave has a peak amplitude of (1), then its instantaneous differential equals (2 π F) , which is also known as (ω) , at zero-crossing. It follows that unit gain will only take place at (F = N / π) . 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.
What follows is also a reason for which by itself, offers poor performance in compressing signals. It usually needs to be combined with other methods of data-reduction, thus possibly resulting in the lossy . And another approach which uses , is , the last of which is a proprietary codec, which minimizes the loss of quality that might otherwise stem from using .
I believe this observation is also relevant to This Earlier Posting of mine, which implied a High-Pass Filter with a cutoff frequency of 1 kHz, that would be part of a Band-Pass Filter. My goal was to obtain a gain of at least 0.5 , over the entire interval, and to simplify the Math.
(Edited 03/21/2017 . )