To make it simple, all waveform other than sine wave is made up of the
fundamental frequency + harmonics, for example a 20KHz square wave will
contain 20KHz sine wave + 40KHz+60KHz+80KHz+100KHz...... sine wave in
different ampitudes, so you need an amplifier of much higher bandwidth to
pass a 20KHz square wave without smoothing out the corners.
All amplifier quoted the bandwidth in sine wave and you also need to note
the power they can handle in such a high frequency, that is called "power
bandwidth", some just quote their bandwidth at 1W power output.
C.H.Ling
----- Original Message -----
From: "Chuck Norcutt"
> 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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