2
$\begingroup$

From my understanding of Fourier transform, a Fourier transform of a signal in time domain will give the different frequency components of the signal in frequency domain, specified by their contributions(energies) in the original.

The Fourier transform of rectangular function is a sinc function

If I take Fourier transform of two different rectangular pulses, with width w1 and w2, with w1>w2, the sinc function of one width w1 is narrow and has large peak as compared to the one with width w2(wider and smaller peak). How do I intuitively understand this behavior?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ Think of the limit cases: a very narrow impulse (ideally a Dirac impulse) contains all frequencies and has a flat spectrum, whereas a constant time domain function has only a DC component, i.e. a Dirac impulse in the frequency domain (at frequency $0$). I.e. the shorter a time domain signal is the wider is its spectrum and vice versa. $\endgroup$ Commented Sep 10, 2014 at 20:40

1 Answer 1

Reset to default
1
$\begingroup$

Note that the peak height of the Fourier transform of a wider rectangular function is proportionally higher. This is necessary to preserve total energy, as per Parseval's theorem. However, the energy required to create the sharp edges of a rectangle in the time domain is still out there on the tails of the Sinc function in the frequency domain, roughly the same (but more wiggly).

But when you scale the vertical axis of 2 plots so that they have the same peak height, the envelope of the FT of a wider rect function looks narrower in the plots of that transform. But that's only in relation to the peak with a scaling applied.

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.