[Elecraft] Re: Displaying the keying waveshape on Spectrogram.

[Masleid, Michael A.]

Hello George,

>Michael, with all respect, I think you need to review the mechanics of =
FFT
>processing. What you propose as a mechanism for Spectrogram operation =
would
>result in a single data point for the energy integrated over the entire =
FFT
>bin width which you would set equal to the receiver bandwidth..

Exactly so.  There are only two numbers of interest.  First, what is the
signal level of the signal - how much is the K2 putting out?  Second, =
what
is the signal level of the interference.  How much does the K2
interfere with a receiver in the next "channel".

>I don't know your background in this area, but evidently either there =
is
>some considerable confusion or I am not understanding what it is that =
you
>are trying to convey.

I think the confusion is over practical matters.  Lets say I'm using a =
K2.
I set the bandwidth to 80 Hz (AF2 on).  I want to know if I'll hear a =
click if I
tune 80 Hz down from another K2.  I'll use Spectrogram, I'll claim that =
I need
an 80Hz bin.

If I sample at 5.5kHz, the highest frequency I can see analyze is about =
2 kHz.  Not
enough bandwidth to read the audio from the SSB receiver So, I'll sample =
at 11 kHz.
Then the highest frequency I can analyze is 5 kHz.  Good.  If I run a =
1024 FFT size,
I'll see 512 bins (please forgive factor of 2 errors - I do that often). =
 So, each bin
will be about 10 Hz.  That's too big.  I need 8 times fewer bins.  I =
want the FFT width
to be 64!.  I'm not going to get that out of Spectrogram.  Needless to =
say, I won't
get a 16 point or 8 point FFT either.

>As I posted before, there is no meaningful relationship between FFT bin
>width or resolution  and the bandwidth of the receiver in examining the
>spectrum of a keyed waveform.  The resolution or bin width used in
>Spectrogram is not " a narrow filter" as you state. I should not be =
compared
>with the receiver bandwidth in determining the validity and accuracy of =
the
>spectral estimate.

Well, there's the receiver bandwidth of the instrument feeding =
Spectrogram.  That
is very wide, enough so that it doesn't matter.  Then there's the bin =
width of
the FFT, which I still claim as a narrow filter.  Then there's the =
bandwidth of
the receiver used by someone who thinks he hears clicks.  It is this =
test
bandwidth that I'm worried about.  Do contesters use 500 Hz filters?

>You are correct in that what the spectral estimate looks like depends
>heavily on the receiver bandwidth involved in the conversion to audio =
for
>the spectral estimate to be made. Guy is correct in that the bandwidth =
needs
>to be large in order to capture the high-frequency components that
>contribute to keying artifacts.

Perhaps the confusion is over which receiver?  There's the one feeding
Spectrogram.  We agree on the need for wide bandwidth there.

>In most cases, the conclusions derived from spectra of keyed waveforms =
are
>as much a matter of receiver bandwidth as it they are of the nature of =
the
>waveform itself.

So, QRM is in the ear of the beholder? :)  This is the missing step.  =
Once we
have the power in all the little bins from Spectrogram, we need to =
predict what
will happen when that spectrum hits a real receiver.

Normally, you'd take the transfer function of the filter, times the =
voltage per
bin from the FFT, and then do an inverse FFT to get the output of the =
filter.
The trouble is, you need to know the phase information to do this right.

>Please accept these remarks in the spirit in which they are intended =
and
>offered.

How else am I going to learn?  Thank you for your time.


73, Michael, AB9GV