3 ms·
With this primitive type of system, pseudorandom is your best chance to get accurate correlation. You want each burst to differ in as many of the positions for
by hackcasual 4y ago
With this primitive type of system, pseudorandom is your best chance to get accurate correlation. You want each burst to differ in as many of the positions for maximum discrimination. If you're just incrementing, every other pair of messages will differ by just 1 bit, meaning it's difficult to tell them apart in a noisy situation.
Edit: Also this circuit is simpler than an adder, since there's no carry
- CamperBob2 4y agoEdit: Also this circuit is simpler than an adder, since there's no carry Which is interesting in itself on occasion. If you need to count to a particular binary power on an FPGA, an LFSR for that word width is the most parsimonious way to do it AFAIK.
- jacquesm 4y ago> If you're just incrementing, every other pair of messages will differ by just 1 bit That's not true. 000 001 010 011 100 101 110 111 Plenty of multi-bit changes in just that short sequence. It just isn't guaranteed that more than one bit will change, in about half that will be the case, in the remainder there will be multi-bit changes.
- adrian_b 4y agoThe parent poster said every other pair, which is perfectly true (in English that means for a half of the transitions, i.e. those from even numbers to odd numbers), so your correction is false.
- jacquesm 4y agoAh I see now what he means. Ok, that could have been worded a bit clearer but fine, yes, if you only consider the pairs then it makes sense. 0->1 1 bit, 1->2 2 bits, then 2->3 one bit again, 3 to 4 three bits, and so on. So the better way to phrase it would have been: 'on even to odd transitions the difference will only be one bit'.