Simplified Bass and Treble Control



This feedback loop assumes that if the amplification factor before positive feedback is even slightly less than (1.0), the resulting infinite series will converge, on some finite amplification factor. However, as the reader can see, ‘VR1′ and ‘VR2′ can be turned all the way to the right, thus connecting the feedback summations directly to the Source of the two MOSFETs.

I am assuming that the gain there is slightly less than unity, as following from changes in Gate voltage. However, because there are tolerances in all the resistors, the feedback gain may reach unity in theory. (:1)  If this should happen, the user of the appliance should have a tremendous experience and learn not to do that again.




The resistor-capacitor combinations of (R11,C4) and (R12,C5) produce a rudimentary first-order, low pass and high pass filter each. The MOSFETs achieve that their Source voltages will follow their Gate voltages simply, while small changes to their Drain voltage should be the inverse of those to Source voltage, times (9), as long as the current flowing through ‘R2′ and ‘R3′ can be neglected.

Two resistors were removed from an earlier version of this circuit, that were meant as an enhancement, but that would have prevented the circuit from working.

The way the inverse-logarithmic-taper Variable Resistors (=Potentiometers) work is, that with their wiper at centre-position, its resistance to the right electrode should be 10kΩ, while its resistance to the left electrode should be 90kΩ. This leads to a design barrier because the resistances of ‘R8′ and ‘R10′ automatically need to be 9x ‘R7′ and ‘R9′, so that the inverted signals cancel out on the wipers.


An added complication to this arrangement is that, due to the phase-shifts of the first-order filters, (α) is some multiple of a complex number, that multiple being determined by the setting of the Bass and Treble controls, the complex number determined by the frequency, which will appear normally in the denominator, and thus result in an attenuation, which my notes above are only a gross simplification of. However, well below and above their corner-frequencies, the low-pass and high-pass filters will exhibit little phase-shifting, so that the final result below is still correct:

This also achieves that the maximum attenuation can become:

1 / (( (200kΩ / 220kΩ) * (90kΩ | 220kΩ) / 10kΩ) + 1)

In other words, with the wiper of each potentiometer turned all the way to the left, ‘R2′ and ‘R3′ will act as though in parallel with ‘R8′ and ‘R10′, thus diminishing the achievable inverted gain somewhat.

‘R1′ … ‘R6′, together with the operational amplifier, form a summation circuit, with the input to this circuit as one of its inputs.

(Updated 10/21/2019, 19h35 … )

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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 . )