RGB Optical, RGB Optical Interface (Infinite 1 bit):
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If we used colour and colour to bit-map (code, switch), the total number of possible positions would go up to 16.7 million or more, PAM 16.7M+.
In the bit to bit-switch sense (voltage, voltage steps), optical analogue, that would be 16.7 million bits, and PCM 16.7 million.
If we include lumen (colour and lumen sensor), we can use lumen to multiply the total number of positions.
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1 bit time can represent [X] number of positions, so 1 sample 100 channels would equal 100 bits.
100 (C) x 48000 (S) = 4,800,000 / 1,000,000 = 4.8 Mbits/s.
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Sample interval and sample rate is an interesting one, reducing the interval will increase total fidelity (if enough), but at the same time add frequency outside of hearing range.
If we calculated 48k as the base of 2 bits (2 intervals), between time is the [bit time, interval], sample rate is number of bits in a horizontal line, per second.
48000 /1000 = 48, /1000 = 0.048, x [20.833~ us] = 1 (1 interval). If we increased to 96k, 96000 /1000 = 96, /1000 = 0.096, x [10.416~ us] = 1.
Total bits would look like [48: 1__1] [96: 1_1_1], as you can see we just went from 2 bits, to 3 bits in [20.833~ us] time. This is not enough to notice much in fidelity.
To get similar to 24 bits in [20.833~ us] time, 1152 sample rate is needed (per channel, per second) (868.055~ nano second interval).
24 x 48 =1152, at the same time, there should be a digital filter attenuating past 20-22k to 0, as we can not hear it.
100 (C) x 1152000 (S) = 115,200,000 / 1,000,000 = 115.2 Mbits/s. Position, positions per second.
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Principle of PAM: (1) original signal, (2) PAM signal, (a) amplitude of signal, (b) time.
Return to nil (RTN), off state (no bits).