Inverse Fourier, can't factor denominator, can't use partial frac.

In summary, the problem is finding the inverse Fourier of a complex fraction with a quadratic denominator. The initial approach of using partial fractions and completing the square did not work, but it is suggested to try factoring the denominator and using the table for a potential solution.
  • #1
fishingspree2
139
0

Homework Statement



Inverse Fourier of:
[ jω+2 ] / [ (jω)2 +5jω+9 ] where j = sqrt(-1)

I tried using partial fractions but the denominator can't be factored...I tried completing the square on the denominator but I get a sum of squares.

What can I try? I am sure I don't have to use the formal definition using the integral, there has to be a way of manipulating it to be able to use the table.

thank you!
 
Physics news on Phys.org
  • #2
fishingspree2 said:

Homework Statement



Inverse Fourier of:
[ jω+2 ] / [ (jω)2 +5jω+9 ] where j = sqrt(-1)

I tried using partial fractions but the denominator can't be factored...I tried completing the square on the denominator but I get a sum of squares.

What can I try? I am sure I don't have to use the formal definition using the integral, there has to be a way of manipulating it to be able to use the table.

thank you!

Why would you say the denominator can't be factored? It is just a quadratic in ω, so certainly has two roots.

RGV
 
  • #3
Alright, I will try to see if it leads me somewhere, do you think I'll be able to solve it this way? it seems like it will get messy
 

FAQ: Inverse Fourier, can't factor denominator, can't use partial frac.

1. What is an inverse Fourier transform?

The inverse Fourier transform is a mathematical operation that converts a frequency domain representation of a function into a time domain representation. It is the opposite of the Fourier transform, which converts a time domain representation into a frequency domain representation.

2. Why can't the denominator be factored in inverse Fourier transforms?

The denominator in inverse Fourier transforms cannot be factored because it represents the frequency variable in the Fourier transform. Factoring it would change the frequency domain representation of the function, making it impossible to accurately convert back to the time domain.

3. What does it mean when we can't use partial fractions in inverse Fourier transforms?

Inverse Fourier transforms cannot use partial fractions because the Fourier transform of a function is unique, meaning it cannot be broken down into simpler components. Therefore, there is no need for partial fractions in the inverse Fourier transform.

4. Is it possible to skip the inverse Fourier transform and just use the Fourier transform?

No, it is not possible to skip the inverse Fourier transform and just use the Fourier transform. The Fourier transform only gives us a frequency domain representation of a function, while the inverse Fourier transform is necessary to convert it back to the original time domain representation.

5. Can inverse Fourier transforms be applied to any type of function?

Yes, inverse Fourier transforms can be applied to any type of function, as long as it is integrable. This means that the function must have a well-defined integral over its entire domain. However, the resulting time domain representation may not always be a simple or easy-to-understand function.

Similar threads

Compute the given Fourier transform by using the given tables
Replies
3
Views
1K
What is the inverse Fourier transform of e^3iωt for solving ut+3ux=0?
Replies
1
Views
2K
Inverse Fourier transform of decaying function
Replies
1
Views
2K
Solving Fourier Inverse: Integrals and Techniques for Exam Revision
Replies
3
Views
1K
Inverse Fourier Transform of cos(4ω + pi/3)
Replies
2
Views
7K
Fourier transform integration using well-known result
Replies
9
Views
2K
Why Can't I Calculate This Inverse Fourier Transform Correctly?
Replies
2
Views
2K
Inverse Fourier transforms and partial fractions
Replies
4
Views
5K
Factoring for partial fraction decompostion
Replies
1
Views
1K
Integral arising from the inverse Fourier Transform
Replies
1
Views
892

Hot Threads

Recent Insights

Back
Top