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.......