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I am looking for an algorithm to calculate a floating point divide. There are a number of options, but the one that is most clear to me and easiest to implement on the hardware I am designing seems to be the NewtonâRaphson iterative method. I\'m trying to understand how many iterations it will take. The formula for that in the Wikipedia article assumes an initial estimate for the reciprocal of 48/17 - D * 32/17. This would require 2 iterations to gain 16 bits of accuracy or 3 iterations for IEEE 32 bit floating point numbers. My needs are for something between that range, but would like to minimize the number of iterations... ideally one.
I believe most implementations use a table lookup for the seed value of the reciprocal of the divisor. In the FPGA I am using, a convenient size would be 2k entries (11 bit address) of 18 bits. I\'m wondering if this would be sufficiently more accurate than the above estimate so that one iteration would be sufficient. I\'m not clear on how to calculate this. I know in general the formula converges rapidly with the error being squared, so I\'m thinking an initial value that is good to some 11 bits would produce more than 20 bits for one turn of the Newton-Raphson crank (20 bits being a level of accuracy called \"good enough\" for this application). But I\'d like to have something that verifies this rather than using a rule of thumb. At some point I will probably have to justify the correctness of the calculations.
Because of the hardware facilities available the calculations have intermediate mantissas of 18 or 36 bits with 28 bits stored when in memory/stack other than ToS which is the register in the ALU (>36 bits). I don\'t think 11 bits is quite as good as required, but I\'m thinking one crank of the NR machine and I\'m \"good\". However, I need to be able to show how good. Anyone know of good documentation on this matter?
Oh, I guess I should mention later on I might be repeating this process for a square root calculation. A measurement using a sensor that is at least temporarily deprecated requires a square root. I managed to show the sensor was not capable of providing an accuracy at the low end that would produce meaningful results. But if they find a better sensor they may want to add that sensor as an option and the square root will be back. No! The HORROR!
Whatever... It\'s just an algorithm...
--
Rick C.
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- Tesla referral code - https://ts.la/richard11209
I believe most implementations use a table lookup for the seed value of the reciprocal of the divisor. In the FPGA I am using, a convenient size would be 2k entries (11 bit address) of 18 bits. I\'m wondering if this would be sufficiently more accurate than the above estimate so that one iteration would be sufficient. I\'m not clear on how to calculate this. I know in general the formula converges rapidly with the error being squared, so I\'m thinking an initial value that is good to some 11 bits would produce more than 20 bits for one turn of the Newton-Raphson crank (20 bits being a level of accuracy called \"good enough\" for this application). But I\'d like to have something that verifies this rather than using a rule of thumb. At some point I will probably have to justify the correctness of the calculations.
Because of the hardware facilities available the calculations have intermediate mantissas of 18 or 36 bits with 28 bits stored when in memory/stack other than ToS which is the register in the ALU (>36 bits). I don\'t think 11 bits is quite as good as required, but I\'m thinking one crank of the NR machine and I\'m \"good\". However, I need to be able to show how good. Anyone know of good documentation on this matter?
Oh, I guess I should mention later on I might be repeating this process for a square root calculation. A measurement using a sensor that is at least temporarily deprecated requires a square root. I managed to show the sensor was not capable of providing an accuracy at the low end that would produce meaningful results. But if they find a better sensor they may want to add that sensor as an option and the square root will be back. No! The HORROR!
Whatever... It\'s just an algorithm...
--
Rick C.
- Get 1,000 miles of free Supercharging
- Tesla referral code - https://ts.la/richard11209