how to calculate the convolution of two signal without using CONV() ?

Hi everyone, i was wondering how to calculate the convolution of two sign without Conv();. I need to do that in order to show on a plot the process. i know that i must use a for loop and a sleep time, but i dont know what should be inside the loop, since function will come from a pop-up menu from two guides.(guide' code are just ready);
option = get(handles.popupmenu1,'value');
option2 = get(handles.popupmenu2,'value');
// something that switch the func.
conv(x,h);
//plot the conv.

4 Commenti

For 1-D convolution, you need two for loops, one inside the other. What does "show on a plot the process" mean? What exactly are you going to plot? Why do you need to plot anything anyway? Is this homework? Why not simply use conv() and not do all this complicated, unnecessary stuff? (Actually the convolution is not that hard if you'd just try it.) What do you mean by "two guides"? Do you have 1 fig file, or two?
it is homework. i have to use two fig, and i need to show how the convolution is done, like a delay inside the for loop.
Not sure what that means. Why do you need a delay? Do you want to show a set of boxes representing the kernel moving along over the top of another set of boxes representing the signal? Doing the convolution is one thing (a really easy thing), but providing the animation, if that is what is required will be a lot harder.
Lets leave out that part, I just really need to know how to set the for loop for two different signal.

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 Risposta accettata

close all
clearvars
%x=input('Enter x: ')
x=sin(2*pi*0.1.*(1:1:11));
%h=input('Enter h: ')
h=[1 2 3 4 5 3 1 -1];
% convolution
m=length(x);
n=length(h);
X=[x,zeros(1,n)];
H=[h,zeros(1,m)];
for i=1:n+m-1
Y(i)=0;
for j=1:m
if(i-j+1>0)
Y(i)=Y(i)+X(j)*H(i-j+1);
else
end
end
end
% plot results
figure;
subplot(3,1,1); stem(x, '-b^'); xlabel('n');
ylabel('x[n]'); grid on;
subplot(3,1,2); stem(h, '-ms');
xlabel('n'); ylabel('h[n]'); grid on;
subplot(3,1,3); stem(Y, '-ro');
ylabel('Y[n]'); xlabel('----->n'); grid on;
title('Convolution of Two Signals without conv function');

3 Commenti

what exactly is '-b^', '-ro' and '-ms' ?
the first characters represent the colour and the second characters represent the shape:
'-b^': blue triangle
'-ro': red circle
'-ms': magenta square
How to make the graph starts from zero not one?

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Più risposte (5)

close all
clearvars
%x=input('Enter x: ')
x=sin(2*pi*0.1.*(1:1:11));
%h=input('Enter h: ')
h=[1 2 3 4 5 3 1 -1];
% convolution
m=length(x);
n=length(h);
X=[x,zeros(1,n)];
H=[h,zeros(1,m)];
for i=1:n+m-1
Y(i)=0;
for j=1:m
if(i-j+1>0)
Y(i)=Y(i)+X(j)*H(i-j+1);
else
end
end
end
% plot results
figure;
subplot(3,1,1); stem(x, '-b^'); xlabel('n');
ylabel('x[n]'); grid on;
subplot(3,1,2); stem(h, '-ms');
xlabel('n'); ylabel('h[n]'); grid on;
subplot(3,1,3); stem(Y, '-ro');
ylabel('Y[n]'); xlabel('----->n'); grid on;
title('Convolution of Two Signals without conv function');
Convolution without conv function in MATLAB | Complete CODE | Explanation | Example And Output

clear all;

clc;

x = input ('Enter the first signal/sequence ');

h = input('Enter the second signal/sequence ');

Find the length of the individual signals, and thus the output

[~,L]= size (x);

[~,M] = size (h);

len y = L + M - 1

y = zeros (1, len_y);

len = 1:1en_y;

for i = 0 / l * eny

for j = 0/1 * eny

if ((i-j+1)>0 && (i-j+1)<=M && (j+1)<=L) y(i+1)=y(i+1)+x(j+1),^ * h(i-j+1);

end

end

end

disp (y)

stem (len, y)

1 Commento

what do you mean by y(i+1)=y(i+1)+x(j+1),^ * h(i-j+1); ? what does ,^ mean?

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Hi guys i need to find a matlab code which can take 2 inputs of 2 different signals and generates and y(t) signal. This y(t) signal must be a another signal. So I'm basicaly looking for a signal for output. This is soooo urgent

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