convolution of two function

Question

parham kianian 2020-2-9

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/504498-convolution-of-two-function

编辑： AKASH KUMAR 2022-1-27

I have two functions as below:

t = 0 : pi/180 : pi;

f = exp(-t/3) .* (sin(2*t) + 2*sin(4*t) + 0.4*sin(2*t).*sin(40*t));

h = 10 * exp(-10*t);

I want to calcluate the convolution of f and h (L = f * h (t)). This means that to filter signal f using h and store results in L. In fact, this is the example 2.21 of "A first course in wavelets with Fourier analysis".

Plot of f is:

Plot of L, that is the filtered signla is:

And this is what MATLAB calculates using conv function

G = conv(f,h,'same');

What is going wrong? G should be same as L as depicted in Figure 15.

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Answer 1

sumanth chinna 2020-10-8

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https://ww2.mathworks.cn/matlabcentral/answers/504498-convolution-of-two-function#answer_508846

在 MATLAB Online 中打开

clc 
clear all  
close all
x=[1 1 1 1 -1 -1 -1 -1]
h=[0 1 2 3 4 3 2 1]
l1=length(x)
l2=length(h)
N=max(l1,l2)
cconv(x,h,N)
disp('without using standard function')
x=[x zeros(1,N-l1)]
h=[h zeros(1,N-l2)]
y=zeros(1,N)
for n=1:N
    for m=1:N
        j=mod(n-m,N)
        j=j+1
        y(n)=y(n)+x(m)*h(j)
    end
end

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AKASH KUMAR 2022-1-27

编辑：AKASH KUMAR 2022-1-27

在 MATLAB Online 中打开

%%%%% 
% algorithm to compute convolution 
clc
clear 
close all
x1 = [5,6,1];
h = [10,6,4,8,9,5];
N = length(x1)+length(h)-1;
y=conv(x1,h) % Inbuilt matlab function
x = linearconvolve(x1,h);
x
%% User defined function to find linear convolution
function cnv = linearconvolve(a,b)
L = length(a)+length(b)-1;
cnv = zeros(1,L);
a1=[a,zeros(1,L-length(a))]; % define a new vector of a
b1=[b,zeros(1,L-length(b))];
for i=1:L
    c = 0;
    for j=1:i
        c = c + a1(j)*b1(i-j+1);
    end
      cnv(i) = c;
end
end

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Answer 2

Image Analyst 2020-2-9

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https://ww2.mathworks.cn/matlabcentral/answers/504498-convolution-of-two-function#answer_414618

Looks like it's an edge effect where zeros outside the signal are getting convolved in when the signals start to not overlap anymore. Instead of 'same', use 'full' to see the full signal.

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parham kianian 2020-2-9

When I use option 'full', the results is as follow:

The length of output signla is 2*length(f)-1. The first length(f) data points of output signal have exactly the same shape as shown in Figure 15, but its maximum amplitude is about 150. It should be 2.3 as depicted in Figure 15.

How can I solve this problem?

Image Analyst 2020-2-9

在 MATLAB Online 中打开

parham, the plots are correct for the filters you gave. You might want to normalize h so that the filtered signal is of the same amplitude as the original signal:

t = 0 : pi/180 : pi;
f = exp(-t/3) .* (sin(2*t) + 2*sin(4*t) + 0.4*sin(2*t).*sin(40*t));
h = 10 * exp(-10*t);
% Normalize h
h = h / sum(h);
plot(t, f, 'b-', 'LineWidth', 2);
grid on;
hold on;
plot(t, h, 'r-', 'LineWidth', 2);
legend('f', 'h');
filtered_f = conv(f, h, 'full');
deltat = t(2)-t(1)
tf = linspace(-length(h)*deltat, deltat * (length(f) + length(h)), length(filtered_f));
plot(tf, filtered_f, 'g-', 'LineWidth', 2);
legend('f', 'h', 'filtered_f', 'Interpreter', 'none', 'Location', 'northwest');

I think that maybe what you aren't realizing is that the convolution flips the filter signal from right ot left, so that the red signal essentially has the blip on the right end as it slides across the blue signal, not the left side like it's plotted. So then the red blip is way off the right end of the blue signal, only the flat, essentially zero, part of the signal is there to multiply by the blue signal, so the output signal is a constant zero on the right.

You'll notice that there are two positive humps and two negative humps, just like your desired output signal.

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