> 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.com
Thanks, Eric.
--
Syd
Reply by Eric Jacobsen●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
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.com
Reply by Eric Jacobsen●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
>
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