Electronics Forum / Basics / February 2006

see http://cnx.org/content/m0068/latest/ 

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

                    y

                    ^
                    |
         ,..........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

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

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Data AcQuisition And Real-Time Analysis
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