0 个投票

% 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 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Walter Roberson
Walter Roberson 2016-3-31
Note: the proper term is "steganography" not "stegnography"

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Walter Roberson
Walter Roberson 2014-4-28

1 个投票

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.

更多回答（2 个）

Muhammad fayyaz
Muhammad fayyaz 2014-4-29

0 个投票

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 个评论
显示 无 隐藏 无

Walter Roberson
Walter Roberson 2014-4-29
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.

请先登录，再进行评论。

Muhammad fayyaz
Muhammad fayyaz 2014-5-1
编辑：Walter Roberson 2016-3-31

0 个投票

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.

类别

在 帮助中心 和 File Exchange 中查找有关 Numbers and Precision 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by