VHDL rollover of counter

Can you use up an adder? If so it's as simple as c = a - b. If you take c as 2's compliment, then the magnitude of c is the amount that you need to change a so that it matches b; the sign of c is the direction. It will automagically be the shortest distance when interpreted in this manner.

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Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

I'm not sure a single comparison will do, as the subset if what you want is if A > B then inc(B) elseif A < B then dec(B) but you are also asking for a 'shortest path' test, which adds another test, to decide which way is closer : the maths way, or the wrap-way. and I think that is 4 branches, no matter how you code it ?

Yes, of course, you need the count up test and the count down test, so as to accommodate the case where the input did not change.

Right now I have 6 tests (a pair of 3-way ors depending on the state of the upper two bits of input and counter) and I'm not sure this is covering all cases.

I think my main problem was making the input and counter signed. This takes care of the rollover between -1 and zero both ways work fine, but totally messes up the overflow between largest negative number and largest positive.

Changing to unsigned should help, but I still have to come up with a test for when the < or > comparison give the wrong result.

Thanks, Tim and Jim!

Jon

Well, I changed it over to unsigned, but I still ended up with more or clauses than I think should be needed. Anyway, it seems to be working correctly now.

Thanks,

Jon

2's compliment or unsigned binary is going to behave the same way as far as finding the shortest path. Doing the c = a-b subtraction on the top two (or three) bits will certainly use less logic -- I'm sort of a permanent FPGA Rip Van Winkle, so I don't know if you could to the 3-bit version in a pair of LUTs -- but you certainly could with the 2-bit version, particularly since you only care about the sign bit and equality.

If your synthesizer is at all good with optimization, you should be able to just do the c = a-b with two (or three) bit data, and have the right answer pop out. If not, take the truth table below, and have fun with it.

a b up equal --- --- --- --- 00 00 x 1 00 01 1 0 00 11 0 0 00 10 x x (indeterminate case) 01 10 1 0 01 11 x x " " 01 01 x 1 01 00 0 0 11 00 1 0 11 01 x x " " 11 11 x 1 11 10 0 0 10 10 x 1 10 11 1 0 10 01 0 0 10 00 x x " "

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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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A quadrant test can be efficient in Logic, but I'm not sure it meets your condition of always taking the shortest path ? If the quadrants are opposite, you could take over 128 clocks ?

So I quickly tried some unsigned maths algorithms, which leads to firstly this in Basic Scripts (Vars are all Unsigned Bytes) If uAp =3D uBp Then exit Function Else Diff =3D uAp - uBp if uAp > uBp Then If Diff < 128 Then ' closer than 128, so INC uB uBp =3D uBp+1 Else ' go the other way uBp =3D uBp-1 End If Else ' uAp < uBp , so Diff has wrapped If Diff >=3D 128 Then ' Wrapped, closer than 128, so DEC uB uBp =3D uBp-1 Else ' go the other way uBp =3D uBp+1 End If End If End If

then I spotted that seems to be

If uAp =3D uBp Then exit Function Else Diff =3D uAp - uBp If Diff < 128 Then uBp =3D uBp+1 Else uBp =3D uBp-1 End If End If

and tested this does seem to chose the 'right' direction, and converge with never more than 128 calls.

So your instincts were right.

The magic of unsigned maths ;)