# Stegnaography using DCT

15 views (last 30 days)
Pranoy on 29 Jan 2012
Edited: Fady Samann on 12 Aug 2020
this is the code we are using to implement steganography based on DCT....
I = im2double(I(:,:,1));
subplot(2,2,1),imshow(I);
subplot(2,2,2),imshow(I2);
I2=de2bi(I2(:,:,1));
x=reshape(I2',1,38400);
T = dctmtx(8);
dct = @(x)T * x * T';
B = blkproc(I,[8 8],dct);
k=0;
for i=1:256
for j=1:256
if B(i,j)>0
b=dec2bin(typecast(single(B(i,j)),'uint32'),32);
b(1,7:8)=x(1,k*2+1:k*2+2);
k=k+1;
c=double(typecast(uint32(bin2dec(b)),'single') );
B(i,j)=c;
end
end
end
subplot(2,2,3),imshow(B);
invdct = @(x)T' * x * T;
B3= blkproc(B,[2 2],invdct);
subplot(2,2,4),imshow(B3);
could somebody tell me the problem in this code with respect to the binary and decimal conversion....
kbae on 26 Apr 2013
could you give me the extraction code for this embedding method?thank you

Walter Roberson on 30 Jan 2012
You will find that de2bi() does not return the characters '0' and '1' that are required for bin2dec().
You will also find that the bit ordering is different between de2bi() and dec2bin(). Compare
dec2bin(8,5)
de2bi(8,5)
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 31 Jan 2012
Mostly, be consistent about whether you use de2bi() or dec2bin()

NILANJ on 19 Apr 2013
I2=de2bi(I2(:,:,1)); can any one tell me plz that in this function why (:,:,1) ???
Walter Roberson on 19 Apr 2013
First bit plane -- corresponding to Red in an RGB image.

Maulana Wahid on 9 Jan 2014

Peter Eze on 22 Dec 2016
use transpose(dec2bin(ImageMessage(i,j))) to create the binary string for each pixel of the message to be embedded

juhi patel on 26 Sep 2017
de2bi()is used for you want your data in matrix form.&dec2bin() is used if you want output in string format.you are mixing both function in this code.using de2bi() will solve your problem since image is in matrix form.
Walter Roberson on 26 Sep 2017
I almost always use dec2bin() and subtract '0' to get matrix form. Then I do not have to worry about the fact that de2bi() is part of a communications toolbox rather than part of basic MATLAB.

Fady Samann on 13 Jun 2020
Edited: Fady Samann on 12 Aug 2020
First, you are not doing quantization to remove the samller DCT coefficients after that you can change LSB by masking it and and OR it with the message bit. Use bitwise operations, because dec2bin and bin2dec treat the binary as a string of 0's and 1's.
the process is just like JPG compression but in the middle of quantization and dequantization, you can change the LSB.
Second, when you pick the right DCT coefficient to change its LSB, the coefficient must be greater than one because of changing the LSB of one to zero (from the message bits) will set the coefficient to zero then during the decoding process this coefficient will be skipped because of the if condition (B(i,j)>0).
Note: round the number only after the quantization stage in the encoding and decoding of the message.