Inverse Fourier transforms and partial fractions

In summary, the conversation is about finding the inverse Fourier transform of 1/(iw+3)3 using partial fractions and other properties. However, it seems that there is no specific property for dealing with cubics and the solution may require direct integration. The course is focused on Fourier transforms, not Laplace transforms, and the speakers are struggling to find a solution for their upcoming midterm.
  • #1
Luongo
120
0
1. find the inverse FT of 1/(iw+3)3



2. well partial fractions gave the same thing back... I'm not sure how to transform this as there's no property that deals with cubics.



3. i tried using the differentiation property but it doesn't work as it increases the power of 3 to 4 and so on... how would i go about computing the inverse Fourier transform of this? is there a property involving powers out there? how would you
 
Physics news on Phys.org
  • #2
I'm not familiar with using partial fractions for inverse Fourier transforms but it looks like that would be easy to integrate directly. Do you mean Laplace transform?
 
  • #3
HallsofIvy said:
I'm not familiar with using partial fractions for inverse Fourier transforms but it looks like that would be easy to integrate directly. Do you mean Laplace transform?
the course is called "the Fourier transform and its applications" so yeah I'm pretty sure i don't mean laplace transform. can someone help please my midterms tommorow and i have no clue..
 
  • #4
Hey Luongo, did you ever figure out how to do this problem? I have the exact same question on my problem set and have no idea what to do either, I ended up trying partial fractions just like you and it didn't work so I'm stumped.
 
  • #5
1/(iw+3) = 3/(w2 +9) - iw/(w2 +9) = (1/√(w2 +9))ei arctan(w/3)
 

1. What is an inverse Fourier transform?

An inverse Fourier transform is a mathematical operation that takes a frequency domain representation of a signal and converts it back to its original time domain representation. This is useful in analyzing and manipulating signals in fields such as signal processing, communication systems, and image processing.

2. How is an inverse Fourier transform calculated?

The inverse Fourier transform is calculated using the formula: f(t) = ∫F(ω)e^(iωt) dω, where f(t) is the time domain representation, F(ω) is the frequency domain representation, and i is the imaginary unit. This integral is typically solved using various techniques such as integration by parts or residue calculus.

3. What are partial fractions?

Partial fractions are a method used to decompose a rational function (a function with a polynomial in the numerator and denominator) into simpler fractions. This is useful in performing integration, solving differential equations, and simplifying expressions in mathematics.

4. How are partial fractions used in inverse Fourier transforms?

Partial fractions are used in inverse Fourier transforms to simplify the integral and make it easier to solve. By decomposing a complex function into simpler fractions, the integral can be broken down into smaller, more manageable parts. This allows for a more efficient and accurate calculation of the inverse Fourier transform.

5. What are some applications of inverse Fourier transforms and partial fractions?

Inverse Fourier transforms and partial fractions have many applications in various fields of science and engineering. They are commonly used in signal processing to analyze and manipulate signals, in image processing to remove noise and enhance images, in communication systems to transmit and receive data, and in solving differential equations in physics and engineering problems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
958
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
784
  • Calculus and Beyond Homework Help
Replies
31
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
6K
  • Calculus and Beyond Homework Help
Replies
1
Views
933
Back
Top