Generate a 32-bit binary stream
12 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Steve Rogerson
am 21 Feb. 2020
Kommentiert: Steve Rogerson
am 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
0 Kommentare
Akzeptierte Antwort
Walter Roberson
am 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.
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Logical finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!