Why different answers are coming during convolution of two functions using different number of data points ?

I am trying to calculate convolution of two Gaussian function using 'conv' function in MATLAB.e.g.

h = @(x) gaussmf(x,[4,0]);
f = @(x) gaussmf(x,[3,0]);
x = -5:.1:5;
y = conv(h(x),f(x));
plot(y)

Now if in the second case if I use x = -5:0.05:5; here x-spacing is 0.05 instead of 0.1 as in the first case, so number of data points become twice. Convolution, y in the second case is coming twice to that of 1st case( i.e. peak hight is increasing twice). But this should not be the case. If we increase the number of points in a function to make it more smoother it should not affect the convolution results. Thanks

Answer 1

[Image Analyst]

Who told you this incorrect statement: "If we increase the number of points in a function to make it more smoother it should not affect the convolution results."? The length of the convolution is the length of the two arrays you convolve summed together. So if you double the number of elements in x, then y will have double the number of elements, just as you observed. And the peak value will be about twice as high because you're multiplying and summing twice as many elements as before.

[Image Analyst]

Debu's "Answer" moved here since it's not an "Answer" to the original question:

Thanks for your answer.

I am new in doing convolution, so may be my comment is wrong.

I agree that number of elements in y will be doubled as length of y =length(h)+length(f)-1 and this is ok. But as we all know that changing the number of x to twice or thrice or more will not change the peak value of the function f(x) and h(x) but you are saying that peak value of y will be changed accordingly.

So if I want to calculate convolution of any two functions, there will have different convolution results depending on number of elements then what will be the acceptable answer ?. How do I know my answer is correct or not ?, because if anyone try to reproduce my results using same function but taking different number of points he will not get same answer. The reason I am saying this is because when we talk about functions, we don't care about number of points(elements) if its not too small or not too large. And convolution of two functions is also a function, so I was thinking that if we increase number of points in f(x) and h(x) convolution function y(x) should have more points rather increasing its peak hight.

Thanks again. Deb

[Image Analyst]

Let's take a very simple case of a rect function convolved with a rect function. Let's say the function is [0,1,1,0] and you convolve that with itself. Now the peak value will be when it overlaps itself 100%. The value will be 0*0+1*1+1*1+0*0 = 2. Now let's say that you have twice as many elements so that it's [0,0,1,1,1,1,0,0] so the peak will be 4.

It's well known that to not change the mean of the function, you need to divide the kernel by the sum of the kernel so that the sum of the normalized kernel is 1. Whatever it sums to is what the image will be multiplied by. So if your kernel is ones(3) (9 elements of 1 in a 3x3 array), and you do

blurredImage = conv2(double(grayImage), kernel);

The mean of blurredImage will be 9 times the mean of grayImage. So to get an output in the same range as the input you have to divide by the sum of the kernel:

kernel = ones(3,3);
kernel = kernel / sum(kernel);
blurredImage = conv2(double(grayImage), kernel);

If you don't the uint8 image could go past 255, which it can't so it gets clipped at 255 and your image looks saturated.

I hope that explains it better.

[Debdas]

Ok I got the answer. So you want to say that I have to scale the output to get the same peak hight. As convolution is basically: do the ffts of two array an then multiply both ffts and then do inverse fft of that arrays. Even if i want to change the number of elements of a function at the end I just had to scale with output with dt(sampling interval). So this will not affect the output. Also I found some useful answer on the following link.

http://in.mathworks.com/matlabcentral/answers/15770-scaling-the-fft-and-the-ifft

Thanks a lot

[Matt J]

Answer Accepted on Debu's behalf.