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Convert floating point to binary

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

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Andrei Bobrov
Andrei Bobrov am 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 Kommentare
Walter Roberson
Walter Roberson am 3 Dez. 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 am 8 Aug. 2021
Thanks.

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Weitere Antworten (5)

Oliver P
Oliver P am 10 Aug. 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 Kommentare
Oliver P
Oliver P am 12 Aug. 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 am 12 Aug. 2016
Bearbeitet: Walter Roberson am 10 Feb. 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 am 7 Jan. 2012
You need to define the fraction representation in binary.

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

naghma tabassum
naghma tabassum am 12 Okt. 2016
Bearbeitet: Walter Roberson am 12 Okt. 2016

Aneesh paulsagin
Aneesh paulsagin am 16 Mär. 2018
Bearbeitet: Walter Roberson am 3 Dez. 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 Kommentare
Walter Roberson
Walter Roberson am 18 Okt. 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 am 19 Okt. 2021
Great!
Much thanks

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