Convert floating point to binary

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Muammar
Muammar le 7 Jan 2012
Commenté : Ian Ono le 19 Oct 2021
Hiii...
I want to ask how to convert floating point to binary in MATLAB
Thank you

Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 7 Jan 2012
e.g.:
a = 1234.57849; % your float point number
n = 16; % number bits for integer part of your number
m = 20; % number bits for fraction part of your number
% binary number
d2b = [ fix(rem(fix(a)*pow2(-(n-1):0),2)), fix(rem( rem(a,1)*pow2(1:m),2))]; %
% the inverse transformation
b2d = d2b*pow2([n-1:-1:0 -(1:m)].');
EDIT [16:32(UTC+4) 08.01.2012]
a = 1234.57849; % your float point number
n = 16; % number bits for integer part of your number
m = 25; % number bits for fraction part of your number
% binary number
d2b = fix(rem(a*pow2(-(n-1):m),2));
% the inverse transformation
b2d = d2b*pow2(n-1:-1:-m).';
  7 commentaires
Walter Roberson
Walter Roberson le 3 Déc 2020
As discussed below, for negative values, you need to be specific about which representation you want to use for the binary fraction.
saba h
saba h le 8 Août 2021
Thanks.

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Plus de réponses (5)

Oliver P
Oliver P le 10 Août 2016
Thank you for the cute and elegant solution! Unfortunately it's only working for positive floats. Negative floats will produce the same result as positive floats, but with negative bits. Which, of course, is not valid. And it's not the proper representation of negative values anyway.
  5 commentaires
Oliver P
Oliver P le 12 Août 2016
Yes, I agree. As far as I'm aware Matlab uses IEEE-754 for all floating-point (single, double and custom) and for unsigned fixed-point calculations. Only for signed fixed-point it's using two's-complement representation.
Walter Roberson
Walter Roberson le 12 Août 2016
Modifié(e) : Walter Roberson le 10 Fév 2017
The "Fixed Point Toolbox" can handle floating point numbers, but are only IEEE 754 if you request very specific formats.
I do not recall that the internal format for floating point number in the Symbolic Toolbox is documented.
The Fixed Point Toolbox offers Separated Sign. I would need to recheck to see if it offers One's Complement.

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Walter Roberson
Walter Roberson le 7 Jan 2012
You need to define the fraction representation in binary.

Muammar
Muammar le 8 Jan 2012
Thank you for all answers...

naghma tabassum
naghma tabassum le 12 Oct 2016
Modifié(e) : Walter Roberson le 12 Oct 2016

Aneesh paulsagin
Aneesh paulsagin le 16 Mar 2018
Modifié(e) : Walter Roberson le 3 Déc 2020
convert complex number to binary number
A = [-0.0040383664156692-0.00294191598222591i, ...
1.00279327279556+0.00768012699728154i, ...
-0.00226521017869135+0.00526418383309796i, ...
0.999498954084202-0.007158248828685i, ...
-0.00549315262581557+0.00808461388120792i, ...
0.998352426774419+0.00927983415466687i, ...
0.00736345881927219+0.00540426830690426i, ...
0.989408434745709-0.0144762821959683i, ...
0.00827899268722473+0.0122398877118786i, ...
0.999298739008971-0.0129949269950415i, ...
-5.47057549608037e-07-0.0130605748664198i, ...
1.01414402334238+0.0131228156923076i, ...
0.000678728159952879-0.00434397278237206i, ...
0.985341332736134+0.0239798712601118i, ...
0.0109818351271128-0.00658607972360998i, ...
1.01709879921672-0.00394256645505557i, ...
0.000335417716939878-0.00461609765687651i, ...
0.996785178287252-3.51718069407279e-05i, ...
-0.0137042758344959+0.00734580139566216i, ...
1.01389851161064+0.00526816880638668i, ...
-0.0143246406043654-0.0173541476823603i, ...
0.984838248467196-0.00274924075252472i, ...
-0.00383017735389232-0.00877400220581385i, ...
0.996013541706753+0.0113592028562242i, ...
-0.00607963966107746-0.00701052911751136i, ...
1.00401827238935-0.0163653626342944i]
  4 commentaires
Walter Roberson
Walter Roberson le 18 Oct 2021
format long g
A = [
1.01709879921672-0.00394256645505557i, ...
0.996013541706753+0.0113592028562242i, ...
-0.00607963966107746-0.00701052911751136i, ...
]
A =
1.01709879921672 - 0.00394256645505557i 0.996013541706753 + 0.0113592028562242i -0.00607963966107746 - 0.00701052911751136i
A_binary_out = reshape(dec2bin(typecast(reshape([real(A(:).'); imag(A(:).')], 1, []),'uint8'),8).',1, []);
tn = tempname(); %temporary file name
fid = fopen(tn, 'w');
fwrite(fid, A_binary_out, 'char');
fclose(fid);
fid = fopen(tn, 'r');
A_binary_in = char(fread(fid, [1 inf], 'uint8'));
fclose(fid);
pairs = typecast(uint8(bin2dec(reshape(A_binary_in, 8, []).')),'double');
A_reconstructed = pairs(1:2:end) + 1i .* pairs(2:2:end)
A_reconstructed =
1.01709879921672 - 0.00394256645505557i 0.996013541706753 + 0.0113592028562242i -0.00607963966107746 - 0.00701052911751136i
If your original A did not happen to be a column vector, then you will need a step to reshape it to the original size.
Ian Ono
Ian Ono le 19 Oct 2021
Great!
Much thanks

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