I want to design a divider using cordic function, for high number / low number (30/2) can anyone help me.?

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sanjai
sanjai 2022-5-18
评论： Walter Roberson 2022-5-19
I have already design the divider using cordic, but this is working only for when numerator is low and denomerator is high.
for example : x/y , 6/20 = 0.3. but when i give 20/6 it shows output is 1.
x = 6;
y = 20;
z = 0;
format long
k = zeros(20,3);
for i = 1:20
if x > 0
x = x - y*2^(-i);
z = z + 2^(-i);
else
x = x + y*2^(-i);
z = z - 2^(-i);
end
k(i,:) = [i 6/20 z];
end
Can anyone help ,how to design the divider using the cordic for higher number / Lower number (eg : 30/2).

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Walter Roberson
Walter Roberson 2022-5-18
Do not start with i=1. Start with i being the negative of the number of bits before the decimal place.
I believe that your else is incorrect. I believe you should be working with 2^(-(i-1)) so that you "undo" the previous step
  2 个评论
Walter Roberson
Walter Roberson 2022-5-19
In the case where your numerator and denominator are both non-negative or both negative, then you expect a non-negative output; take the absolute values of numerator and denominator, set a flag indicating that you are expecting non-negative, and proceed.
In the case where either numerator or denominator are negative, but not both, then you expect a negative output; take the absolute values of numerator and demoninator, set a flag indicating that you are expecting negative, and proceed.
You are now certain to not be working with negative.
At each step, calculate
x_new = x - y*2^(-i);
If x_new is positive, then
x = x_new;
z = z + 2^(-i);
and otherwise do not change x or z; either way, continue the loop. Stop when x is 0.

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