Fourier Transforms: F[Rect] and F[sinc] Relationship Explanation

  • Thread starter Thread starter spaghetti3451
  • Start date Start date
  • Tags Tags
    Fourier
Click For Summary
SUMMARY

The discussion centers on the relationship between the Fourier Transform of the rectangular function, F[Rect], and the sinc function, F[sinc]. It establishes that F[Rect] = sinc implies F[sinc] = Rect with minor constants. The duality property of the continuous Fourier transform is crucial for understanding this relationship, as it allows for the transformation between these two functions. The integral formula for the Fourier transform, 2πf(-w) = ∫[−∞, ∞] F(t)e^(-iwt) dt, is also referenced as a foundational equation in this context.

PREREQUISITES
  • Understanding of Fourier Transform concepts
  • Familiarity with the sinc function and its properties
  • Knowledge of the rectangular function (Rect)
  • Basic calculus, particularly integration techniques
NEXT STEPS
  • Study the duality property of the continuous Fourier transform
  • Learn about the inverse Fourier transform and its applications
  • Explore the properties and applications of the sinc function in signal processing
  • Investigate the mathematical derivation of the Fourier Transform for Rect and sinc functions
USEFUL FOR

Students and professionals in mathematics, electrical engineering, and signal processing who are looking to deepen their understanding of Fourier Transforms and their applications in analyzing signals.

spaghetti3451
Messages
1,311
Reaction score
31

Homework Statement



Explain how F[Rect] = sinc implies F[sinc] = REct +/- a few constants.


Homework Equations



2\pi f(-w) = \int^{\infty}_{-\infty} F(t) e^{-iwt} dt

The Attempt at a Solution



I have no idea!
 
Physics news on Phys.org
What's the formula for the inverse transform?
 
of which function exactly?
 
Of any function. What is the defining equation for getting the inverse transform?
 
I assume rect is short for "rectangle". The question asks you either to prove duality property of continuous Fourier transform or to use it. Search for the duality, you'll see what i mean.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K