| Electronics > Beginners |
| Why binary is represented by two bits 0 and 1 and not three bits? |
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| alexanderbrevig:
Trits? Qbits it where it's at ;) |
| Vtile:
--- Quote from: BrianHG on October 28, 2019, 10:17:38 am --- --- Quote from: Vtile on October 27, 2019, 11:07:45 pm ---Why binary not ternary.. I suppose one interesting reason is that all widely used number systems are even based, 2,8,10,16,60. --- End quote --- That's only after the invention of '0'. --- End quote --- Do you refer zero as logical or numerical operator or both. :wtf: |
| tggzzz:
--- Quote from: TimFox on October 28, 2019, 08:00:07 pm --- --- Quote from: tggzzz on October 28, 2019, 08:44:23 am --- --- Quote from: paulca on October 28, 2019, 08:26:09 am ---Although, be careful, when you look at the analogue side of digital circuits they are very often tri-state. 1, 0 and floating. The later is sometimes a burden, but sometimes can be used to your advantage. You might have a chip which has a 0 or a 1 (high or low) on it's outputs, but when you pull the "chip enable" low they disconnect the the outputs which gives you a third state which you can pull high or low with a resistor for your purposes. --- End quote --- All "digital" circuits are actually analogue; that's most obvious with ECL and derivatives. Some CMOS logic gates can be used as linear amplifiers. Logic gates interpret input voltages/currents as digital signals. When those inputs are within defined limits, the gate's outputs will (eventually) be within limits that other gates can interpret as a digital signal. The few digital circuits that you might encounter include photon counting devices and femtoamp circuits. --- End quote --- Relay logic uses truly discrete binary signals --- End quote --- No. You are halfway there but your are conflating/equating digital signals with the analogue variables (voltage/current) used to represent digital signals. Consider, for example, that if the current is marginally less than the latch in (or hold) currents, the relay will not reliably be in the expected state. |
| T3sl4co1l:
Or even worse, what if a relay starts chattering, or a contact starts arcing at a rate faster than subsequent coils can respond? How could those conditions possibly be expressed purely in relay logic states? :) Everything is fundamentally analog, because the flow of charge and electromagnetic fields are continuous and not quantized (at least, not to any degree we care about, and even then, not in nearly the same way). I'll add this just because it's a common retort -- shot noise proves the electric charge is quantized as electrons, but it doesn't mean charge or current is quantized in bulk. If I move a charged comb around sufficiently slowly, I will affect the field in its vicinity by less than an electron's worth of charge -- when charges are able to move freely, there is no shot noise (or not as much), and so it is perfectly meaningful to speak of continuum charge. (I would make similar arguments regarding quantum mechanics as well, but I doubt it would be very productive to do so.) Tim |
| tggzzz:
--- Quote from: T3sl4co1l on October 29, 2019, 01:11:57 am ---Or even worse, what if a relay starts chattering, or a contact starts arcing at a rate faster than subsequent coils can respond? How could those conditions possibly be expressed purely in relay logic states? :) --- End quote --- Indeed. I considered mentioning that, but thought it an unnecessarily "complex" version of the simpler point. I also considered mentioning metastability :) --- Quote ---Everything is fundamentally analog, because the flow of charge and electromagnetic fields are continuous and not quantized (at least, not to any degree we care about, and even then, not in nearly the same way). --- End quote --- Photon counting applications with, say, an APD? Sure the pulse out of an APD is analogue, but the photon signal is digital. --- Quote ---I'll add this just because it's a common retort -- shot noise proves the electric charge is quantized as electrons, but it doesn't mean charge or current is quantized in bulk. If I move a charged comb around sufficiently slowly, I will affect the field in its vicinity by less than an electron's worth of charge -- when charges are able to move freely, there is no shot noise (or not as much), and so it is perfectly meaningful to speak of continuum charge. --- End quote --- It looks like you are conflating "electric field" with "charge". Analogy: if I have a 1kg weight on a balanced lever, moving the pivot changes the counterbalancing force by less than (or more than) 1kg. |
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