i need the verilog code for hyperbolic cordic to implement square root of a given number(either using iterative or pipelined architecture).
Dear Sir,
For this kind of function, I usually start with the McLaurin or Taylor series to calculate it.
Typically, proceed as follows:
To calculate sqrt(x)
1. Count the number of leading zeros of x
2. Take the next 6 bits of the number (you can vary the number of bits a bit.)
Put 5 of these bits through a lookup table.
We have now x ~= (2**-n) * xo *(1 + delta)
(xo is the number you feed to the look-up table. Its the 6-bit approximation of x.)
Now, we have:
sqrt(x) = sqrt(2**-n) * sqrt(xo) * sqrt(1 + delta)
The lookup table need now to produce sqrt(xo), and also (xo^-1). Multiplying x with (xo^-1) will give you (1 + delta).
You need to use a second lookup table to produce sqrt(2**-n) (or whatever), and
sqrt(1 + delta) is approximated by the Mclaurin series of sqrt(1+x) around x=0. You need to 'massage' the whole algorithm so that you get an acceptable precision with 2 or 3 terms of the Mclaurin series.
Best Regards,
Henri
Sorry, there is a second method.
This goes as follows:
1. Calculate the number of leading zeros
2. Take a 6-bit approximation of the number. (see above)
3. Calculate yo = sqrt(xo*2**-n) using a lookup table and some multipliers
4. Use following iterative algoritm to find a better approximation of y.
(You need to do 1-2-3 iterations of this algoritm)
(yo+delta)^2 = x
yo^2 + 2*delta*yo + delta^2 = x
delta = (x -yo^2)/(2*yo) - delta^2/(2*yo)
We discard the second term, and then we get:
delta = (x -yo^2)/(2yo)
y1 = yo + delta
To calculate the division by 1/2yo, you let the lookup table produce (1/yo).
For the second iteration, you calculate
delta = (x - y1^2)/(2yo) (do not calculate (1/yo) again.
y2 = y1 + delta.
You can do more steps if needed.
thank you for the help.........i will see if i can use these in my code.......but i need it using CORDIC Algorithm.........if u have any ideas for that also please tell me.......
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