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LSB in audio stegnography

3 visualizaciones (últimos 30 días)
Muhammad fayyaz
Muhammad fayyaz el 28 de Abr. de 2014
Comentada: Walter Roberson el 31 de Mzo. de 2016
% Embed message length in the first 16 samples
% (I don't understand how. Can any one explain?)
str = dec2bin(length(message),16);
if length(message) < length(y)
for a = 1 : length(str)
y(a, nbits-1) = str(a);
end
else
disp('error')
end
I don't understand how they are embedding here. Please help me.
  1 comentario
Walter Roberson
Walter Roberson el 31 de Mzo. de 2016
Note: the proper term is "steganography" not "stegnography"

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Respuesta aceptada

Walter Roberson
Walter Roberson el 28 de Abr. de 2014
They are converting the message length to 16 text-coded binary locations, and are taking the resulting 16 bits and using them to replace the least significant bits of 16 successive text-coded binary representations.

Más respuestas (2)

Muhammad fayyaz
Muhammad fayyaz el 29 de Abr. de 2014
sir your answer is correct but i don't understand how this statement works, i need little bit more explanation here
y(a, nbits-1) = str(a);
  2 comentarios
Walter Roberson
Walter Roberson el 29 de Abr. de 2014
Assume that the original signal has been converted into binary by using dec2bin(). dec2bin() returns an array of characters, each one being the character '0' or the character '1'. Then y(a,nbits-1) is a straight forward replacement of one element of the array with one element of the character vector "str".
Somewhere after this section of code, you will find a call to bin2dec(y) . That operation will take the character array y and convert the representation back into decimal numbers.
Muhammad fayyaz
Muhammad fayyaz el 30 de Abr. de 2014
THANK YOU SIR

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Muhammad fayyaz
Muhammad fayyaz el 1 de Mayo de 2014
Editada: Walter Roberson el 31 de Mzo. de 2016
clc, clear all;
[x,fs,nbits]=wavread('camera.wav');
y=((2^(nbits-1)*x(:,1)));
for i=1:length(y)
if y(i)<0
z(i)=1;
else
z(i)=0;
end
if y(i)<0
y(i)=-1*y(i);
end
end
y=dec2bin(y);
when i run this command the value of decimal is different from the value of binary i need help for example
value of y= 1011
11
.
.
.
.
.
100
101
1
11
1000
from 1 upto 311 column, i don't about these value from where it come after this 311 column the values of decimal is same with the value of binary
10 2
1101 13
111 7
1101 13
0 0
100 4
11 3
1001 9
10101 21
0 0
111 7
.
.
.
.
till the end
the values are same
i copy of all the value in xls file than i understand that after 311 column the values are matching
please help me sir.

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