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, [email protected] 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.) > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
