Debugging an FFT... If an FFT time-domain input comprises say, 1.5 cycles of sine starting positive going followed by another 1.5 cycles of sine starting positive going - in other words has 3 cycles with an inversion half way along, then the real and imaginary FFT outputs at the 3-cycle frequency should be zero. That's right, isn't it? Cheers -- Syd
FFT sanity check.
Started by ●July 19, 2012
Reply by ●July 19, 20122012-07-19
On Thu, 19 Jul 2012 09:46:53 +0100, Syd Rumpo <usenet@neonica.co.uk> wrote:>Debugging an FFT... > >If an FFT time-domain input comprises say, 1.5 cycles of sine starting >positive going followed by another 1.5 cycles of sine starting positive >going - in other words has 3 cycles with an inversion half way along, >then the real and imaginary FFT outputs at the 3-cycle frequency should >be zero. > >That's right, isn't it? > >Cheers >-- >Syd >No. http://www.dsprelated.com/showarticle/174.php Eric Jacobsen Anchor Hill Communications www.anchorhill.com
Reply by ●July 19, 20122012-07-19
On Thu, 19 Jul 2012 09:46:53 +0100, Syd Rumpo <usenet@neonica.co.uk> wrote:>Debugging an FFT... > >If an FFT time-domain input comprises say, 1.5 cycles of sine starting >positive going followed by another 1.5 cycles of sine starting positive >going - in other words has 3 cycles with an inversion half way along, >then the real and imaginary FFT outputs at the 3-cycle frequency should >be zero. > >That's right, isn't it? > >Cheers >-- >SydI shouldn't have been so terse with my previous response, and I actually misinterpreted what you'd said, anyway. It does look like bin 3 should be zero. The computation for bin 3 is essentially a correlator, and since the correlations for the first half and the second half will have opposite sign they cancel. The discontinuity adds a lot of other frequency content, though, that gets splattered around among the other bins. Eric Jacobsen Anchor Hill Communications www.anchorhill.com
Reply by ●July 19, 20122012-07-19
On 19/07/2012 17:06, Eric Jacobsen wrote:> On Thu, 19 Jul 2012 09:46:53 +0100, Syd Rumpo <usenet@neonica.co.uk> > wrote: > >> Debugging an FFT... >> >> If an FFT time-domain input comprises say, 1.5 cycles of sine starting >> positive going followed by another 1.5 cycles of sine starting positive >> going - in other words has 3 cycles with an inversion half way along, >> then the real and imaginary FFT outputs at the 3-cycle frequency should >> be zero. >> >> That's right, isn't it? >> >> Cheers >> -- >> Syd > > I shouldn't have been so terse with my previous response, and I > actually misinterpreted what you'd said, anyway. > > It does look like bin 3 should be zero. > > The computation for bin 3 is essentially a correlator, and since the > correlations for the first half and the second half will have opposite > sign they cancel. The discontinuity adds a lot of other frequency > content, though, that gets splattered around among the other bins. > > > Eric Jacobsen > Anchor Hill Communications > www.anchorhill.comThanks, Eric. -- Syd