how to do convolution without commands

47 vues (au cours des 30 derniers jours)
omar chavez
omar chavez le 27 Nov 2011
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

Réponse acceptée

Chandra Kurniawan
Chandra Kurniawan le 27 Nov 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 commentaire
David Young
David Young le 27 Nov 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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Plus de réponses (5)

Wayne King
Wayne King le 27 Nov 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 commentaire
omar chavez
omar chavez le 27 Nov 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 le 16 Mar 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 commentaires
Savannah Quinn
Savannah Quinn le 13 Sep 2020
I am getting an index out of bounds due to h(i,j)
Enrique de José María Flores Rodríguez
Modifié(e) : Enrique de José María Flores Rodríguez le 16 Mar 2021
Seem works good on a DICOM image on 2021, thank you !

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SALVADOR CANAS
SALVADOR CANAS le 15 Jan 2018
Those convolutions are convolutions with padding=1, how do you do for padding=0?

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

VIGNESH B S
VIGNESH B S le 28 Juil 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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