Mike,
I’ll have to disagree with your statement that “it’s not the discontinuity…”.
A more careful and correct statement could have been something along these
lines: "instead of explaining the observed phenomenon in terms of an endpoint
discontinuity, one could also note that the phenomenon will be observed
whenever the tone frequency is not an integral multiple of the FFT bin width,
i.e. whenever the tone does not fall in the exact center of a bin.”
I’ll leave it as an exercise for you to convince yourself that these two
explanations are two ends of the same stick.
Sometimes it’s better to think before you respond.
Steve k9an
> On Jan 10, 2017, at 7:21 AM, Michael Black <mdblac...@gmail.com> wrote:
>
> It's not the discontinuity -- it's the bin values.
> If the desired frequency is not an EXACT multiple of the bin value you'll get
> a reduced peak and leakage....one of the downfalls of DFTs. As Joe noted a
> larger FFT would have more resolution but is likely not going to improve
> things.
> A rectangular window preserves minimum main lobe width at the expense of
> higher leakage. Other windows will increase peak value and have reduced
> leakage but also have less accuracy due to spreading of the main peak. The
> leakage should not be much of a factor for this frequency calibration unless
> you are borderline on the peak SNR where a Triangular window would be best
> but picks up less then 1dB over a rectangular window -- see
> http://www.bores.com/courses/advanced/windows/files/windows.pdf
> <http://www.bores.com/courses/advanced/windows/files/windows.pdf>
>
> There's already a fit function in the code which is giving better than the
> bin width in accuracy. Could be improved on slightly perhaps with a cubic
> spline but looks good enough as-is to me and I doubt a spline would really
> improve much at all.
>
> On my freq cal I see numerous RMS entries < 0.1Hz -- Average RMS is 0.27 --
> so the peak fit being done is doing pretty well.
>
> de Mike W9MDB
>
> On Mon, Jan 9, 2017 at 5:23 PM, Steven Franke <s.j.fra...@icloud.com
> <mailto:s.j.fra...@icloud.com>> wrote:
> Since Joe is probably eating dinner at the moment…
>
> With 55296 point FFT, a 1600 Hz tone (7.5 sa/period) has 55296/7.5=7372.8
> cycles in a record - so there is an endpoint discontinuity (i.e. a
> discontinuity between the first and last points of the record).
>
> A 1500 Hz tone (8 samples per period) has 6912 cycles in a record, so there’s
> no endpoint discontinuity.
>
> I think that what you’re seeing is the result of using a non-windowed FFT.
> It’s easily addressed, if necessary.
>
> Steve
>
>
> > On Jan 9, 2017, at 11:07 PM, Bill Somerville <g4...@classdesign.com
> > <mailto:g4...@classdesign.com>> wrote:
> >
> > On 09/01/2017 22:28, Bill Somerville wrote:
> >> ... for sure the three
> >> magnitudes going into peakup are much flatter when the error is greater ...
> > Hi Joe,
> >
> > Here are some examples, these are DFT magnitudes (s vector) for five
> > bins around a 1500Hz tone:
> >
> > 0.107814305 0.175234795 9.38052759E-02 17124122.0
> > 8.95743668E-02 8.78201425E-02 4.62990999E-02
> > 7.97733366E-02 0.128112748 0.307582766 17126434.0
> > 0.167245105 1.00381924E-02 1.45341307E-02
> > 5.32349795E-02 7.02278912E-02 0.130553111 17125166.0
> > 0.135780737 6.56466857E-02 0.104219861
> >
> > and here is the same for a 1600Hz tone:
> >
> > 79235.5156 191998.594 973516.500 15560843.0
> > 432274.188 128688.836 60769.3555
> > 79641.3359 192280.516 972233.375 15562208.0
> > 432276.500 128727.242 60738.6406
> > 79411.3125 192080.391 972825.562 15561087.0
> > 431534.812 128623.734 60596.7227
> >
> > These numbers are from print *, s(ipk-2:ipk+2) with DF set to the tone
> > and FTol at 1000Hz.
> >
> > 73
> > Bill
> > G4WJS.
> >
> >
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