Problem in Convolution integral by fourier transformation

In summary, the individual Fourier transforms of the Gaussian functions are correct, but when multiplied together to find the convolution, the resulting values have alternate signs and differ in numerical value from the expected result. The cause of this issue is unclear and may require further fine-tuning of the spacing and number of points used in the Fourier transform. It is recommended to first fix any errors in the Fourier transform before attempting to debug the convolution.
  • #1
praban
13
0
Hello,

I am trying to numerically evaluate a convolution integral of two functions (f*g) using Fourier transform (FT) i.e using

FT(f*g) = FT(f) multiplied by FT(g) (1)

I am testing for a known case first. I have taken the gaussian functions (eq. 5, 6 and 7) as given in
http://mathworld.wolfram.com/Convolution.html

I am not using FFT but a naive implementation of FT. The problem I am facing is that the FTs are correct but convolution is not.

When I do inverse FT I do get back the 3 gaussians as given in the wolfram link. However, eq. (1) is not satisfied.

Is there any trick in eq. (1)? I would appreciate any suggestion.

thanks,

Praban
 
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  • #2
You need to supply details of what is wrong. The concept is perfectly correct.
 
  • #3
I am giving the details below (equation numbers are from my original posts, except the equation I have written below).

The sigma1 and sigma2 of eq. (5) and (6) are taken as 2.0 and 2.5

I am using 128 points in my discrete FT (values -63 to +64). Inverse FT of the FT of these gaussians give back those functions (although there are some error - about 10 to 20% - e.g I am getting 0.102 when the actual value is 0.117).

When I check FT(f*g) i.e. FT (gaussian of eq. (7)) = FT(f) multiplied by FT(g) (1)

the following happens

(1) the numerical values are same - however, FT(f*g) has alternate + and - signs for the numbers, while FT(f) mul FT(g) values are all positive

(2) If I change the limits - from -20 to +20 (still 128 points), alternate signs remain but now the numerical values also differ between two sides of equ. (1) and (2).

(3) I did the same with 512 points - results same as point (2) above.

I guess, that even FT of gaussian functions requires some fine tuning of spacing and number of points to choose (i.e. why 10-20% error is coming). However, in eq. (1), sign alternation part if not clear to me.

When I tried a function like exp(-kx)/x, my FT routine works fine (error less than 1%).

I would appreciate any help.

Praban
 
  • #4
praban said:
I am using 128 points in my discrete FT (values -63 to +64). Inverse FT of the FT of these gaussians give back those functions (although there are some error - about 10 to 20% - e.g I am getting 0.102 when the actual value is 0.117).

It's a waste of time thinking about your other problems, until you have fixed that one.

Try some simple functions where you know what the FFT should be. For example
a constant function
sine and cosine functions one period in the 128 points
an alternating sequence of values +1, -1. +1, -1., ...,
all the values zero except for one point.

When those you can do an FFT and an inverse FFT of those and the answers are correct to 5 or 6 significant figures, not 10% or 20%, then you can try debugging your convolutions.
 

1. What is the Convolution integral by Fourier transformation?

The Convolution integral by Fourier transformation is a mathematical operation that combines two functions to create a third function. It is used to analyze signals and systems in various fields such as signal processing, image processing, and communication systems.

2. What is the purpose of using Convolution integral by Fourier transformation?

The purpose of using Convolution integral by Fourier transformation is to simplify the analysis of signals and systems. It allows us to break down complex signals into simpler components, making it easier to understand and manipulate them.

3. What are the steps involved in performing Convolution integral by Fourier transformation?

The steps involved in performing Convolution integral by Fourier transformation are as follows:
1. Take the Fourier transform of the two functions.
2. Multiply the two Fourier transforms.
3. Take the inverse Fourier transform of the product to get the convolution of the two functions.

4. What are some common applications of Convolution integral by Fourier transformation?

Convolution integral by Fourier transformation has a wide range of applications in various fields. Some common applications include image and signal filtering, noise reduction, pattern recognition, and solving differential equations in physics and engineering.

5. What are some common challenges faced in solving problems related to Convolution integral by Fourier transformation?

Some common challenges faced in solving problems related to Convolution integral by Fourier transformation include understanding the concept, choosing the correct functions to convolve, and correctly applying the steps involved. Other challenges may include dealing with complex functions, handling discontinuities, and interpreting the results accurately.

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