LSB in audio stegnography

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Muhammad fayyaz
Muhammad fayyaz am 28 Apr. 2014
Kommentiert: Walter Roberson am 31 Mär. 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 Kommentar
Walter Roberson
Walter Roberson am 31 Mär. 2016
Note: the proper term is "steganography" not "stegnography"

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Akzeptierte Antwort

Walter Roberson
Walter Roberson am 28 Apr. 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.

Weitere Antworten (2)

Muhammad fayyaz
Muhammad fayyaz am 29 Apr. 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 Kommentare
Walter Roberson
Walter Roberson am 29 Apr. 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 am 30 Apr. 2014
THANK YOU SIR

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Muhammad fayyaz
Muhammad fayyaz am 1 Mai 2014
Bearbeitet: Walter Roberson am 31 Mär. 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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