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, [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 > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm -- regards, ==================================================== GPG key 1024D/4434BAB3 2008-08-24 gpg --keyserver subkeys.pgp.net --recv-keys 4434BAB3 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
