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Re: FFT1 Resolution Bandwidth - back to basics

Hi Leif,

> If you ask for 200 Hz you should get a bandwidth
> around 200 Hz. Usually it will be a bit smaller, but sometimes
> it could be a bit larger.

It would be good if this were so, but it does not seem to be the case.
I can only give you an example with a sample rate of 111.111kHz.
At this rate and a sine^3 window, the possible bandwidths are:
260.6Hz for 1024 points,
130.3Hz for 2048 points,
65.2Hz for 4096 points.

Ask for 261Hz, you get 260.6Hz (and 1024 points),
ask for 260Hz, you get 130.3Hz (and 2048 points),
ask for 200Hz, you get 130.3Hz (and 2048 points),
ask for 131Hz, you get 130.3Hz (and 2048 points),
ask for 130Hz, you get 65.2Hz (and 4096 points).
You NEVER get more than you ask!

I have not looked at the code,
but this makes me think of a calculation of the power of 2 for
the FFT length from the asked bandwidth (and window power),
which is then type converted from floating point to integer
to give an exact power of 2.
As you know, such type conversions simply truncate to the
integer below the floating point value.
If you want correct rounding to the nearest integer above or below,
then you have to add 0.5 to the floating point value before the
I think it would be best if you made it happen this way,
but it does not currently do this, in v2.39 at least.

> Oooh! Look here:http://www.sm5bsz.com/linuxdsp/linrad.htm
> the line: "To compile linrad.exe from source code under Windows,"
> The compilers can be downloaded from my site (15 megabytes).

Thanks, I missed that.
I have some spare download capacity this month, so I will download it
Broadband in rural Australia is hardly broadband at all,
with a limited download (500MB) and slow speed (132kB/s) costing the

73  Guy  VK2KU
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