spectrum of rectangular pulse

see

It isn't, always. It depends on how it is distributed over 0. For a balanced rectangle:

^ | ,..........A.........., : | : : | : : | : : | : x -t | +t v

I think the transform is 2*a*t*sinc(2*PI*f*t).

If you shift the rectangle to be asymmetric around x=0, then the transform will be different.

Jon

Hi,

Sinc(x) is Sin(x)/x.

So compare the shape of this function with the envelope of the components in the spectrum of a narrow rectangular pulse and all should be clear.

Cheers - Joe

In case you are just wondering what the whole idea is, at all...

A Fourier transform can be just a frequency domain representation of a time domain function. Time and frequency are... kin and conjugate to each other.

The spectral width in one domain times the spectral width in the other domain will be greater than a finite constant (1, usually), which implies that something narrow in terms of time information will be wide in terms of frequency information and the converse. The transform of the rectangle, 2*a*t*sinc(2*PI*f*t), exhibits this spectral area product in the "f*t" unitless term inside the sinc() function. Narrowing 't' spreads out the result over broader 'f'.

Jon

That would change the relative phases of the frequency components, but their amplitudes would still be a sinc function.

Mark

To add to all the other replies, if you want to get a "hands-on" feel for this, check out my DaqGen freeware signal generator. You can set up any sort of rectangular pulse you want (or just about any other sort of waveform) and toggle between waveform and spectrum views. (DaqGen uses your sound card, so you can listen to the signal as well... but I'd guess it might be really annoying in this case!)

Best regards,

Bob Masta dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis