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Re: [OM] Peter Lik again

Subject: Re: [OM] Peter Lik again
From: Chuck Norcutt <chucknorcutt@xxxxxxxxxxxxxxxx>
Date: Sat, 24 Jan 2009 19:39:54 -0500
If I had known the answer involved Fourier transforms (and I guess I 
should have) I wouldn't have asked the question.  :-)

Chuck Norcutt

ws wrote:
> 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.
>> -- 
> 
> 
> 
> 
-- 
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