How Do You Apply Convolution to Solve Integral Equations in Fourier Transforms?

In summary, the conversation discussed finding the Fourier transform of a given equation involving a second derivative and an integral with an exponential function. It was suggested to use the property of the Dirac delta function and Parseval's Theorem. However, it was later corrected that the integral can be written as a convolution and the Fourier transform can be calculated using the property of the exponential function. The final result is the same as the previous method, but quicker.
  • #1
MathematicalPhysicist
Gold Member
4,699
372
I need to find the Fourier transform of f(x) which is given by the equation:
[tex]-\frac{d^2f(x)}{dx^2}+\frac{1}{a^3}\int_{-\infty}^{\infty}dx'exp(-\lambda|x-x'|)f(x')=\frac{b}{a^2}exp(-\lambda|x|)[/tex]

ofcourse Iv'e taken the Fourier tarnsform of both sides, but I don't see how to calcualte the Fourier tranform of the integral in the above equation, I feel I need to use the definition of dirac's delta function, but don't see how to do this, any ideas, hints?

thanks in advance.
 
Last edited:
Physics news on Phys.org
  • #2
Without doing the Fourier transform, it looks to me like you'll need to use the property that [itex]\delta(x'-x)f(x')=f(x)[/itex]. Just note which of x and x' is actually a variable for integration and which is the "constant" inside the integration. Use Parseval's Theorem.
 
Last edited:
  • #3
I'm not sure it's correct I got that the F.T of the integral without the constant 1/a^3 is:
2pi*f(0)*e^(a-\lambda|x|), is this correct?
 
  • #4
Well actually disregard my previous advice, sorry! Note that the integral is equivalent to [itex]e^{-\lambda |x|} \ast f(x)[/itex] where the operator is convolution. In the frequency domain, this becomes multiplication. Now you simply need to know [itex]e^{-\lambda |x|} \Leftrightarrow \frac{2\lambda}{\lambda^2 + \omega^2} [/itex] which is actually the same answer as you'd arrive by from what I previously said, but in much less time.
 
Last edited:
  • #5
so using convolution you say, ok I'll try it, thanks.
 

Related to How Do You Apply Convolution to Solve Integral Equations in Fourier Transforms?

1. What is a Fourier Transform?

A Fourier Transform is a mathematical technique used to decompose a complex signal into its individual frequency components. It is commonly used in signal processing and data analysis to better understand the underlying frequencies present in a signal.

2. How does a Fourier Transform work?

A Fourier Transform works by converting a signal from its original domain (e.g. time or space) to the frequency domain, where it can be represented as a combination of sine and cosine waves of different frequencies. This allows us to isolate and analyze specific frequencies within a signal.

3. What is the difference between a Fourier Transform and a Fourier Series?

A Fourier Transform is used for continuous signals, while a Fourier Series is used for discrete signals. A Fourier Series decomposes a signal into a sum of discrete frequency components, while a Fourier Transform can handle continuous signals with infinite frequency components.

4. What are some practical applications of Fourier Transform?

Fourier Transform has a wide range of applications in various fields such as signal processing, image processing, data analysis, and quantum mechanics. It is used for tasks such as noise filtering, compression, and spectral analysis of signals.

5. Are there any limitations to Fourier Transform?

Yes, there are some limitations to Fourier Transform. For example, it assumes that the signal is periodic, which may not always be the case. It also assumes that the signal is infinite, which is not always true in real-world scenarios. Additionally, Fourier Transform is not suitable for non-linear signals or signals with sudden changes.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
635
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
683
  • Calculus and Beyond Homework Help
Replies
3
Views
895
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
910
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
Back
Top