Generate a 32-bit binary stream

Question

0 个投票

I am trying to genrate a binary stream using this function and have searched a lot but came out emty.

The main problem I am facing is mainly the binary addition on every addition and the size of the increasing stream.

The Equation is below:

I tried to use the rand fucntion but it isnt right. I realised the Binary addition was not happening.

The final result should be a float number within [0,1).

pold = [];
pf=0;
for i=1:32
    param = round(rand(1,1));
    pf = param + pf;
    pold(i) = pf*2^(-i);
    pold(i);
end

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Answer 1

Walter Roberson 2020-2-22

0 个投票

sum(randi([0,1],1,32).*2.^-(0:31))

Or

randi([0,2^32-1])/2^32

Or you can use rnd() or related to tell rand to use the shr3cong generator, which has the required results.

Or you could rand() and throw away bits 33 to 52.

Steve Rogerson 2020-2-26

Thank you kind Sir!