Hiii...
I want to ask how to convert floating point to binary in MATLAB
Thank you

 Akzeptierte Antwort

Andrei Bobrov
Andrei Bobrov am 7 Jan. 2012

10 Stimmen

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 7 Jan. 2012
Been a while since I saw a 36 bit word ;-)
(Yes, I have used 36 bit computers.)
Julien Cohen
Julien Cohen am 20 Mai 2016
Much thanks (4 years later), this worked perfectly in my implementation of a genetic algorithm function.
OMID Fattemi
OMID Fattemi am 10 Feb. 2017
for implementing GA, I used randint function to generate initial population as
d2b=randint(1,m+n);
then by the following code, decimal floating number is obtained:
b2d = d2b*pow2(n-1:-1:-m)';
Dave Amels
Dave Amels am 20 Mär. 2020
DEC system 10
An Vo
An Vo am 3 Dez. 2020
This code works perfectly for positive number. I run with negative number, the binary sequence contains bit 0, 1 and -1. bit -1 is not right. How can I fix this problem?
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

1 Stimme

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

Walter Roberson
Walter Roberson am 10 Aug. 2016
The proper representation of negative values? As decreed by which King?
Oliver P
Oliver P am 11 Aug. 2016
Bearbeitet: Oliver P am 11 Aug. 2016
I'm not aware of a king here, but an association called IEEE. :-) The IEEE 754 format defines, for example, the leading bit convention. Matlab converts -22.9 properly to binary32 when you use:
dec2bin(typecast(single(-22.9), 'uint32'))
It's also possible to convert to binary64 in a similar way. But not to custom formats. Please do check this webpage for more info on this topic: IEEE 754 Converter
Walter Roberson
Walter Roberson am 11 Aug. 2016
IEEE 754 defines one way to represent single precision numbers as binary, but it is far from being the only valid way.
When people ask about converting negative floating point to binary, the context is most typically the need to transmit quantized signals, which is almost always a fixed-point context, not a floating-point context. IEEE 754 does not deal with fixed point.
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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Muammar
Muammar am 8 Jan. 2012

0 Stimmen

Thank you for all answers...
Aneesh paulsagin
Aneesh paulsagin am 16 Mär. 2018
Bearbeitet: Walter Roberson am 3 Dez. 2020

0 Stimmen

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

A_binary = reshape(dec2bin(typecast(reshape([real(A(:).'); imag(A(:).')], 1, []),'uint64'),64).',1, []);
Ian Ono
Ian Ono am 18 Okt. 2021
Bearbeitet: Ian Ono am 18 Okt. 2021
Walter Robinson, If I use your line to create A_binary and then fwrite it to file. How would I fread it back?
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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