what on earth is this?
they're in all kinds of gadgets, that produce 2 signals, which, if you were to treat them as X, Y coordinates, the signal would draw a little square for you. thus the "quad" in "quadrature".
these are great rotary encoders because it's about the simplest way to tell which direction the knob is getting rotated.
they can be produced mechanically but often they're a little LED, paired with two light sensors, and a slotted wheel between.
my friend @smiffy calls this grey code.. .which… IT TECHNICALLY IS. the special case of grey code reduced to just 2 bits.
a thing I've been thinking about lately is how it just so happens that the quadrature signal coming out of a rotary encoder is pretty close to the signal you need to drive a stepper motor. If I can get enough attention and energyy together I might test this theory out.
QPSK is, so it is written, a very popular modultion/demodulation scheme that builds on thr quadrature encoding, to do a neat magic trick. in an audio/analog signal, QPSK (quadrature phase shift keying) can transmit 2 bits at a time
it works like so:
1. two bits map to one of four corners in a square, we’ll call one of these corners/bitpairs a “symbol”
2. each of those corners corresponds to an angle sine and cosine of a circumscribed circle
3. then for each symbol, cut out a peice of sin pi(e)
now, interestingly, with QAM, corresponding your symbols to a 4 x 4 grid on the phase/amplitude plane is not the only possibility. if you feel confident in your transmission mediujm’s
fidelity, you can slice it up even finer.
or, carefully map out the characteristics of your medium to find an optimal quantisation.
maybe it could be dynamically negotiated.
maybe the optimal arrangement of
points is not a neat grid. fuck around and find out.
okay so, i could try and get a cassette recorder, shitty variety, ans try to
map out an optimal encoding of QAM.
(look how much the “mo-dem” diagram for QAM looks like a frequing esoteric sigil)
the adaptable nature of QAM means the symbol encodings can avoid regions of Phase/Amplitude space that decode poorly and best. each frequency/symbol rate as well. appeently high end of casette with tascam deck and metal tape is 16000khz. the safe or typical is 8000-11000khz
so,, let’s say, optimistically, i can get 11000khz on audio tape, which would the. translate to 5500 baud (symbols per second), then, REALLY optisitically, assume i can get 4 bits into each symbol. that’s 22kbits per second, or 2750 bytes per second. beats the pants off of the c64 standard 300/bits per second. (37 bytes per second)
@zens I literally just finished a course on exactly this at the university a few days ago (and I got a 3/5 grade, so I guess I'm the most authoritative expert on the subject that exists).
Every configuration of signal to noise ratio (which here is determined by your tape), modulation method (e.g. M-ary QAM), and symbol rate, leads to a specific bit error rate (that can be calculated), and you need to balance those choices out to get an error rate that you're OK with.
@zens I don't remember the formulas by heart, but basically if you know the signal to noise ratio of your tape, which modulation you've chosen to use, and at what rate you're putting symbols on the tape, you can calculate at what probability any given bit read off the tape will be read incorrectly. Then you can for example lower the symbol rate, use an error correcting code, use a different modulation (or why not all of these?) to get the probability of errors arbitrarily low.
@zens I'm sincerely sorry if I'm just here explaining something you already knew, I only wanted to bring up this specific relevant part of the subject (the error rate/probability) that you hadn't mentioned yet in the thread
@vurpo it’s interesting! would be good to get the formula, though maybe it’s just multiplying those varables together
@zens the probability of *symbol* error is just the probability that noise in the channel will cause a transmitted symbol (e.g. one of the 16 points in your QAM constellation) to be received as a different one of those points, and using your code (e.g. Gray code) you can estimate how many bits inside the symbol (since one symbol in that case contains 4 bits) would be changed by that error
(and channel=tape, transmitted=recorded, received=read back)
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