The square root of the inverse Hilbert transform convoluted with the
hypotenuse of the frequency, integrated over +-infinity time and you can
show Why this is correct. :-)
Actually the Fourier transform of a square wave shows odd harmonics that
decrease in frequency, so a reasonable approximation requires summing
a certain number of these harmonics, to recreate a square wave, But probably
more import is the phase shift introduce with lower bandwidth systems.
http://homepages.inf.ed.ac.uk/rbf/CVonline/LOCAL_COPIES/OWENS/LECT4/node2.html
I do not know of any experimental listening tests of this, and I don't think I
can
hear past 12Khz these days. At one time, I could hear 16KHz sine wave
but my wife doesn't think I can hear at all. Maybe she just needs more
bandwidth? Unfortunately bandwidth usually means a lot of work.
I just wish someone would inform the mobile phone developers that
too much compression turns voices into noise.
Wayne - say what?
At 06:20 PM 1/24/2009, you wrote:
>Why?
>
>Chuck Norcutt
>
>Ken Norton wrote:
>
>>
>> It takes 160kHz of bandwidth to pass a a 20kHz square wave. Only 20kHz to
>> pass a 20kHz sine wave.
>--
--
_________________________________________________________________
Options: http://lists.thomasclausen.net/mailman/listinfo/olympus
Archives: http://lists.thomasclausen.net/mailman/private/olympus/
Themed Olympus Photo Exhibition: http://www.tope.nl/
|