insert text as watermark in video( one frame)

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LEKHCHINE Somia
LEKHCHINE Somia el 11 de Ag. de 2019
Comentada: LEKHCHINE Somia el 24 de Ag. de 2019
Hi every one
I use matlab R2018b, I try to insert a text (converting to binary) in one frame from my video using dct method but I get some problem so I need help
this is my code:
% Button pushed function: insrerdataButton
function insrerdataButtonPushed(app, event)
Msg_bin = app.EditField2.Value;
couvert = getimage(app.UIAxes2);
%couvert = rgb2ycbcr(couvert);
taille_msg = size(Msg_bin,1)*size(Msg_bin,2);
n = size(couvert,1);
m = size(couvert,2);
%définit le cle secret
%init de la matrice cle que cle(1,1)=1 et cle(2,1)=1
cle(1,1)=1;
cle(2,1)=1;
a = 1;
b = 1;
%pour obtenir les donneés de la cle cad les positions de l'image
for h=2:taille_msg % on commencer par h=2 puisque on fixer que la cle(1,1)=1 et cle(2,1)=1 pour 2'ieme colonne
a=a+14;%pour le block est de taille 8 et on besoin de 15 pixels pour calculer les block
if (a+10>n)
a=1;
b=b+14;
end
cle(1,h)=a;
cle(2,h)=b;
end
for i=1:taille_msg
% cle(1,i):cle(1,i)+7 =(1:8) cle(2,i):cle(2,i)+7=(1:8)...etc
block(:,:,i)=couvert(cle(1,i):cle(1,i)+7,cle(2,i):cle(2,i)+7);%pour donner les blocks de taille 8*8
end
%Calcule le DCT de chaque Block,;
block = zeros(taille_msg);
dct_image = zeros(taille_msg);
for i=1:taille_msg
dct_image(:,:,i) = dct(block(:,:,i));
% error dans cette instruction
end
% parcour de msg
a=1;
b=1;
for i=1:taille_msg
%pour vérifier que B1(u1,v1)=B1(u2,v2)
% 3 et 2 sont des m et n (Amn)
if (dct_image(3,2,i)==dct_image(2,3,i))
dct_image(3,2,i)=dct_image(3,2,i)+1;
end
% vérifier l'opération de la partie 4 de l'algo
if dct_image(3,2,i)<dct_image(2,3,i) && Msg_bin(b,a)=='1'
valeur= dct_image(3,2,i);
dct_image(3,2,i)=dct_image(2,3,i);
dct_image(2,3,i)=valeur;
else
if dct_image(3,2,i)>dct_image(2,3,i) && Msg_bin(b,a)=='0'
valeur= dct_image(3,2,i);
dct_image(3,2,i)=dct_image(2,3,i);
dct_image(2,3,i)=valeur;
end
end
end
%pour revenir a la ligne 2 de message c a d le parcour de tt le msg
if a==7
a=1;
b=b+1;
else
a=a+1;
end
%le idct
image_marque=couvert;
%calcule le idct
% on utilise le idct du matlab
for i = 1:taille_msg
image_marque(cle(1,i):cle(1,i)+7,cle(2,i):cle(2,i)+7)=uint8(idct(dct_image(:,:,i)));
end
setappdata(0,'cle_kz',cle);
setappdata(0,'image_kz',image_marque);
imshow(image_marque,'Parent',app.UIAxes3);
end
When I push on this button I got that error:
Index in position 3 exceeds array bounds (must not exceed 1).
Error in app1/insrerdataButtonPushed (line 154)
dct_image(:,:,i) = dct(block(:,:,i));
Error using matlab.ui.control.internal.controller.ComponentController/executeUserCallback (line 335)
Error while evaluating Button PrivateButtonPushedFcn.
please help and than you
  4 comentarios
Adam Danz
Adam Danz el 24 de Ag. de 2019
LEKHCHINE Somia's answer moved here as a comment.
Hi
I use matlab R2018b
I insert text as watermark in specific frame from video ( using DCT ) but I have some problem
1- I want to replace the original frame by the watermarked frame
2- the text that I can insert just one char
LEKHCHINE Somia
LEKHCHINE Somia el 24 de Ag. de 2019
2- the text that I can insert just one character (8 binary) how can I insert more

Iniciar sesión para comentar.

Respuestas (1)

Adam Danz
Adam Danz el 12 de Ag. de 2019
Editada: Adam Danz el 12 de Ag. de 2019
Based on the error message, it appears that "block" is an [m x n x 1] array (or simply, [m x n]).
You're accessing "block" in a for-loop that controls the 3rd index value. So I'm guessing you get this error on the 2nd iteration where i equals 2 (below).
for i=1:taille_msg
dct_image(:,:,i) = dct(block(:,:,i)); % ERROR when i==2
Just prior to that for-loop you define "block" as an [m x n] array or zeros. So certainly, the 3rd dimension of "block" cannot be greater than 1.
block = zeros(taille_msg);
Curiously, you defined "block" earlier in your code within a different for-loop and in this definition, "block" does appear to be 3D [m x n x i] but you're overwriting it with the zeros which makes it 2D. Now it no longer has a 3rd dimension.
for i=1:taille_msg
block(:,:,i)=couvert(cle(1,i):cle(1,i)+7,cle(2,i):cle(2,i)+7); % THIS IS 3D
end
% and later...
block = zeros(taille_msg); % THIS IS 2D

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