Fourier Transform (Numerical Analysis)

Click For Summary
SUMMARY

The forum discussion focuses on calculating the finite Fourier transform of order m for various sequences using the formula Uk = (1/N)∑uke^(-2πi*k*j/N) from j=0 to N-1, where 0≤k≤N-1. The participants analyze three specific sequences: a constant sequence (uk = 1), an alternating sequence (uk = (-1)^k for N even), and a linear sequence (uk = k). Key insights include the realization that the sum for the constant sequence yields 1 for k=0 and 0 for k≠0, while the alternating sequence's transform mirrors the first due to symmetry.

PREREQUISITES
  • Understanding of Fourier transforms and their applications in numerical analysis.
  • Familiarity with complex exponentials and their properties.
  • Basic knowledge of summation notation and series convergence.
  • Experience with mathematical software or programming languages for numerical computation.
NEXT STEPS
  • Study the properties of the Discrete Fourier Transform (DFT) and its applications.
  • Learn about Fast Fourier Transform (FFT) algorithms for efficient computation.
  • Explore the implications of Fourier analysis in signal processing and data analysis.
  • Investigate the convergence criteria for Fourier series and transforms.
USEFUL FOR

Mathematicians, engineers, data scientists, and students involved in numerical analysis or signal processing who seek to deepen their understanding of Fourier transforms and their computational methods.

Scootertaj
Messages
97
Reaction score
0
1. Calculate the finite Fourier transform of order m of the following sequences:

a) uk = 1, 0[itex]\leq[/itex]k[itex]\leq[/itex]N-1
b) uk = (-1)k, 0[itex]\leq[/itex]k[itex]\leq[/itex]N-1 N even
c) uk = k, 0[itex]\leq[/itex]k[itex]\leq[/itex]N-1

2. Homework Equations
Uk = (1/N)[itex]\sum[/itex]uke-2pi*i*k*j/N from j=0 to N-1 ; 0<=k<=N-1

Attempt:
a) First thing that I tried is that [itex]\sum[/itex]x^k = [itex]\frac{1}{1-x}[/itex] but that doesn't seem to get where I want. For example, I know that for a), we get 1 for k=0 and 0 for k[itex]\neq[/itex]0 , but this would give 1/N (1) for k = 0 and I don't know what for any other k.
So, I found a formula that says:
[itex]\sum[/itex]ei*k*j from j = 0 to N-1
={ 0 if e[itex]\neq[/itex]1,
{ N-1 else.

b) wouldn't we get (1/N)[itex]\sum[/itex]-e-2pi*i*k*j/N from j=0 to N-1 ; 0<=k<=N-1 which is the same as part a) ?
c) I think I need to use the idea that [itex]\sum[/itex]k*x^k = [itex]\frac{x}{(1-x)<sup>2</sup>}[/itex]

Obviously, if this formula is valid (I have no idea), then it would give 1 for 0 and 0 for other k for part a) which is correct

Any ideas?
 
Last edited:
Physics news on Phys.org
Just a bump
 
Nevermind I got it
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K