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> 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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