how to do convolution without commands

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omar chavez
omar chavez el 27 de Nov. de 2011
Comentada: Sai Jothsna Sri el 7 de Sept. de 2023
Hi, im trying to do convolution without any of the commands in matlab.
Just plain for, this is because im trying to use a code that can also be implemented on C.
So far I have this
x=[1 2 3]
h=[-1 0 3]
Z=[0]
[M,N]=size(x)
[L,O]=size(h)
A=N+O-1
for n=1:1:N
for o=1:1:O
y(n,o)=x(n)*h(o)
end
end

Respuesta aceptada

Chandra Kurniawan
Chandra Kurniawan el 27 de Nov. de 2011
Hi, Omar. Do you seeking for 2D convolution code without Matlab toolbox command?? I have the code below
function B = convolve(A, k);
[r c] = size(A);
[m n] = size(k);
h = rot90(k, 2);
center = floor((size(h)+1)/2);
left = center(2) - 1;
right = n - center(2);
top = center(1) - 1;
bottom = m - center(1);
Rep = zeros(r + top + bottom, c + left + right);
for x = 1 + top : r + top
for y = 1 + left : c + left
Rep(x,y) = A(x - top, y - left);
end
end
B = zeros(r , c);
for x = 1 : r
for y = 1 : c
for i = 1 : m
for j = 1 : n
q = x - 1;
w = y -1;
B(x, y) = B(x, y) + (Rep(i + q, j + w) * h(i, j));
end
end
end
end
Save the following code with filename 'convolve.m'. And then create a new m-file and type this code :
clear; clc;
I = [4 4 3 5 4;
6 6 5 5 2;
5 6 6 6 2;
6 7 5 5 3;
3 5 2 4 4];
k = [0 -1 0;
-1 4 -1;
0 -1 0];
Hsl = convolve(I, k)
Watch at the result! You can also compare the result with the matlab toolbox command 'conv2'
Bnd = conv2(I,k,'same')
The both results are same :
Hsl =
6 3 -2 8 9
9 3 0 2 -3
2 0 2 6 -3
9 6 0 2 1
1 8 -6 5 9
Bnd =
6 3 -2 8 9
9 3 0 2 -3
2 0 2 6 -3
9 6 0 2 1
1 8 -6 5 9
>>
  1 comentario
David Young
David Young el 27 de Nov. de 2011
This is a little more complex than necessary - you don't need the first loop that reflects A, just change the index computation in the second loop to reflect the mathematical definition of convolution.

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Más respuestas (5)

Wayne King
Wayne King el 27 de Nov. de 2011
I take it when you say "without commands", you really are just saying without conv(). It appears to me you have 1-D vectors from your initial post. Specifically, you give the example:
x=[1 2 3];
h=[-1 0 3];
You can exploit the relationship between linear convolution, circular convolution, and the DFT by extending the length of your input vectors with zero-padding, multiplying their DFTs, and then taking the inverse DFT.
x = [1 2 3]';
h = [-1 0 3]';
N = length(x)+length(h)-1;
x1 = [x; zeros(N-length(x),1)];
h1 = [h; zeros(N-length(h),1)];
convxh = ifft(fft(x1).*fft(h1));
Compare convxh with
conv(x,h)
  1 comentario
omar chavez
omar chavez el 27 de Nov. de 2011
yes you are right. The thing is that, since the code should be able to apply in several platforms I cant use ifft or fft. Im trying to prove that a code can be put in several languages, like matlab, C,etc

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DI
DI el 16 de Mzo. de 2015
The first answer is not actually full size.
Full size will be like this:
% Written by Dizeng 3/15/2015. Full convolution.
function B = convolve(A, k)
[r,c] = size(A);
[m,n] = size(k);
h = rot90(k, 2);
center = floor((size(h)+1)/2);
Rep = zeros(r + m*2-2, c + n*2-2);
for x = m : m+r-1
for y = n : n+r-1
Rep(x,y) = A(x-m+1, y-n+1);
end
end
B = zeros(r+m-1,n+c-1);
for x = 1 : r+m-1
for y = 1 : n+c-1
for i = 1 : m
for j = 1 : n
B(x, y) = B(x, y) + (Rep(x+i-1, y+j-1) * h(i, j));
end
end
end
end
Hope it will be helpful to others...
  3 comentarios
Savannah Quinn
Savannah Quinn el 13 de Sept. de 2020
I am getting an index out of bounds due to h(i,j)
Enrique de José María Flores Rodríguez
Editada: Enrique de José María Flores Rodríguez el 16 de Mzo. de 2021
Seem works good on a DICOM image on 2021, thank you !

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SALVADOR CANAS
SALVADOR CANAS el 15 de En. de 2018
Those convolutions are convolutions with padding=1, how do you do for padding=0?

Sk Group
Sk Group el 25 de Oct. de 2021
Convolution without conv function in MATLAB | Complete CODE | Explanation | Example And Output

VIGNESH B S
VIGNESH B S el 28 de Jul. de 2022
% The code below is for convolution without conv command.
% Idea behind it is multiplying a element of x with every element in h and
% adding them with a shift.
% Eg: x = [1,2] and h = [4,5,6] say, h is transformed to [3,4,5,0] ; no.of
% zeros to add after h is given by min(len(x) ,len(h)) -1). als zero is
% added so problems with vector addn is removed. (ouput of conv is of same
% length as transformed h).
%then you perform x(1).*h = [4,5,6,0] and x(2).*h = [8,10,12,0] when x(i)
%if i>2 -> you add zeros to the result of x(i).*h in at index 1 and shift
%it. y = conv of x and h = [4,5,6,0] + [0,8,10,12].
clc
clear
x = input('Enter x [..] '); %getting input x and h
Unable to run the 'fevalJSON' function because it calls the 'input' function, which is not supported for this product offering.
h = input('Enter h [..] ');
if length(x)>length(h)
temp_var = x;
x = h;
h = temp_var;
end
y = zeros(1,(length(x)+length(h)-1));
min_val = [length(x),length(h)];
r = min(min_val)-1; ;%minimum of x and h -1
hi = [h,zeros(1,r)];
for i = 1:length(x)
temp = x(i).*hi;
% disp(temp)
if i>=2
tempi = [zeros(1,i-1),temp(1:end-i+1)];
else
tempi = temp;
end
y = y + tempi;
end
disp('The convolution of x and h is:' );
disp(y)

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