[Jprogramming] Tau =: 2p1

[bill lam]

That depends on how you want to interpret the problem. From my
understanding, a sampling frequency of 8khz should correspond to 
sampling at intervals of 1/8000 sec (not 1/7999).

Птн, 15 Июл 2011, mijj писал(а):
> "0 and 2p1 are the same point on the unit circle."
> 
> which explains why the signal at those angles have the same value.
> ie. shows that the value at 2p1 is the same as the value at 0.
> 
> in this case we're going round the unit cirlce more than once
> (ie. 194.18Hz signal for 1 second = we've gone round the unit
> circle 194.18 times for the sampling period)
> But, just because we've been at particular angles of the signal
> before, it doesn't mean that there's no sample to be had.
> 
> ...
> 
> oh .. sampling frequency comes to mind .. sampling frequency
> should be greater than 2* the highest frequency in the signal
> (to avoid aliasing).
> 
> 8000 sample points over one second =>
> rate = 7999 [samples / second] > 2 * 194.18 [Hz]
> .. so that's ok
> (note: 3 samples taken over one second =>
> 1 intial sample + rate of 2 sample / second)
> 
> 
> 
> On 2011-07-15 05:11, bill lam wrote:
> > 0 and 2p1 are the same point on the unit circle.  This is related to Nyquist
> > sampling frequency if my memory still serves me.
> >
> > Птн, 15 Июл 2011, mijj писал(а):
> >> in the original:
> >> NB.  We'll sample a sine wave at 194.18 Hz over 1 second.
> >> NB.  Or in other words: our x axis covers 194.18 radians,
> >> NB.  and we'll choose our 8000 sample points from that range:
> >>
> >> that's 8000 sample *points*
> >>
> >> 8000 sample points means 7999 sample *intervals*
> >>
> >> ie. the range of (194.18 * 2p1 ) has
> >> 7999 sample intervals and 8000 sample points
> >>
> >> so, in terms of proportion of a segment:
> >> 7999 intervals and 8000 points:
> >> (i. %<:) 8000   NB. ranges from 0 .. 1
> >>
> >> so ...
> >>
> >> SamplePoints =. 1 o. (194.18 * 2p1 ) * (i. %<:) 8000
> >>
> >>
> >>
> >> On 2011-07-15 03:09, neitzel at gaertner.de wrote:
> >>> mijj>    yeh .. i always confuse samples with sample intervals too ..
> >>> mijj>    should be ...
> >>> mijj>    x =. 1 o. (194.18 * 2p1 ) * (i. %<:) 8000
> >>>
> >>> Should be not.  So, once more:
> >>>
> >>> The sample interval here is 1r8000 sec (=8kHz
> >>> sampling).  For Ian, the one-second signal is "194.14 sine waves",
> >>> and for direct treatment of "sin(x)" function with 1&o., the "x"
> >>> values ("points where the samples are taken") for the periods have
> >>> to be measured in the radians:
> >>>
> >>> 	x =. (194.18 * 2p1 ) * (i. %<:) 8000
> >>>
> >>> At every sample point, the sample value ("sample") will be the
> >>> sine value or some scaling thereof:
> >>>
> >>> 	samples =.  ...   1 o. x
> >>>
> >>> Less confused now?
> >>>
> >>> 						Martin Neitzel
> >>>
> >>> (Before submitting my code last night, I checked a 440 Hz version
> >>> against my tuning fork.  You might get one before posting, too.)
> >>> ----------------------------------------------------------------------
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> >>>
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> >
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-- 
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