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<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff
size=2>Reading back over that, I don't think I phrased it very well. An
example would probably make more sense.</FONT></SPAN></DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff size=2>Say
your ADC is clocked at 10 MHz, and your Nyquist bandwidth is 4 MHz (to allow for
antialiasing filter rolloff in the absence of an oversampling scheme). You
feed the output of the ADC to a 1,000-point FFT. Each bin of the FFT
output window, then, represents 4 kHz of spectrum. The noise level in that
bin can be normalized to 1 Hz by subtracting 10*log(4000), or 36
dB.</FONT></SPAN></DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff size=2>Now
you have to consider what your spectrum display actually <EM>does</EM>. If
you tell it to look at +/-200 kHz of spectrum, mapping that to a 100-pixel-wide
display, then it can just display the contents of the FFT bins on a 1:1
basis (remember, 4 kHz/bin * 100 pixels = 400 kHz, centered on whatever bin
represents the 'tuned' frequency of interest.)</FONT></SPAN></DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff size=2>If you
tell it to zoom in to +/- 20 kHz, you may think of that as a "bandwidth change",
but I'll bet that all the software does in that case is replicate every FFT bin
10 times, maybe with a basic smoothing algorithm (low-pass function) to hide the
stairsteps. Your noise level wouldn't change in that case, because you
didn't actually change the <EM>sampling </EM>bandwidth, just the size of the
chunk of it that you wanted to see.</FONT></SPAN></DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff size=2>Does
that help?</FONT></SPAN></DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff size=2>--
john, KE5FX</FONT></SPAN></DIV>
<DIV><SPAN class=123162503-26102007><FONT face=Arial color=#0000ff size=2>
</FONT></SPAN></DIV>
<BLOCKQUOTE dir=ltr
style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #0000ff 2px solid; MARGIN-RIGHT: 0px">
<DIV class=OutlookMessageHeader dir=ltr align=left><FONT face=Tahoma
size=2>-----Original Message-----<BR><B>From:</B>
hpsdr-bounces@lists.hpsdr.org [mailto:hpsdr-bounces@lists.hpsdr.org]<B>On
Behalf Of </B>John Miles<BR><B>Sent:</B> Thursday, October 25, 2007 6:26
PM<BR><B>To:</B> hpsdr@lists.hpsdr.org<BR><B>Subject:</B> Re: [hpsdr] phase
noise<BR><BR></FONT></DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff
size=2>Frank,</FONT></SPAN></DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff size=2>I
don't have one of the receivers in question, but generally the apparent noise
level is tied to the sampling bandwidth. If your sampling
clock is 10 MHz, your noise bandwidth is 5 MHz, and you can normalize it to 1
Hz by subtracting 67 dB*. </FONT></SPAN></DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff size=2>But
you also have to realize that the noise energy is distributed evenly
between the FFT bins. Digitally speaking, what does this rig do when you
"change bandwidths", as you say? Does that mean it just uses a
different number of bins to draw the spectrum graph? This is
important: if the ADC's sampling clock and front-end bandwidth didn't change,
and the FFT kernel size (# of bins) didn't change, then the noise level in
each bin won't change, either.</FONT></SPAN></DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff size=2>I'm
not personally crazy about the idea of using an 8640B cavity as an LO for
a narrowband receiver LO. They drift, and they are nothing special
when it comes to close-in phase noise (< 10 kHz). Instead, consider
adding a PLL to clean up any DDS spurs at inter-channel offsets, or upgrading
to a DDS with better SFDR. </FONT></SPAN></DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff size=2>--
john, KE5FX</FONT></SPAN></DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff size=2>
<DIV><SPAN class=561270301-26102007><FONT face=Arial color=#0000ff size=2>*:
In reality, FFTs have "equivalent noise bandwidths" just like
conventional filters do, so the noise-normalization function is not
strictly 10*log(BW). This is a property of the window. There's a
decent tech note on that phenomenon here: <A
href="http://www.bores.com/courses/advanced/windows/files/windows.pdf">http://www.bores.com/courses/advanced/windows/files/windows.pdf</A> </FONT></SPAN></FONT></SPAN></DIV></DIV></BLOCKQUOTE></BODY></HTML>