Generate a 32-bit binary stream
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Steve Rogerson
on 21 Feb 2020
Commented: Steve Rogerson
on 26 Feb 2020
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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Accepted Answer
Walter Roberson
on 22 Feb 2020
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.
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