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<DIV><SPAN class=766505922-27032011><FONT color=#0000ff size=2 face=Arial>And
yes, process gain will improve the SNR by 10*log(BW ratio), but that's not
something you'd ordinarily associate with the term 'noise figure'.
</FONT></SPAN></DIV>
<DIV><SPAN class=766505922-27032011><FONT color=#0000ff size=2
face=Arial></FONT></SPAN> </DIV>
<DIV><SPAN class=766505922-27032011><FONT color=#0000ff size=2 face=Arial>--
john, KE5FX</FONT></SPAN></DIV>
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<DIV dir=ltr class=OutlookMessageHeader align=left><FONT size=2
face=Tahoma>-----Original Message-----<BR><B>From:</B>
hpsdr-bounces@lists.openhpsdr.org
[mailto:hpsdr-bounces@lists.openhpsdr.org]<B>On Behalf Of </B>John
Miles<BR><B>Sent:</B> Sunday, March 27, 2011 3:54 PM<BR><B>To:</B>
hpsdr@openhpsdr.org<BR><B>Subject:</B> Re: [hpsdr] DDC Noise
Figure<BR><BR></FONT></DIV>
<DIV><SPAN class=250492422-27032011><FONT color=#0000ff size=2 face=Arial>A
good question. Multiplication by saturated diodes in a conventional
DBM yields a best-case conversion loss of 3.9 dB -- 3 dB for the opposite
sideband which is unused, and 0.9 more dB lost in various distortion
products arising from square-wave switching. I'd expect the
"noise figure" of an ideal DDC to be simply 3 dB, because
sine-wave multiplication wouldn't be accompanied by all the
distortion. </FONT></SPAN></DIV>
<DIV><SPAN class=250492422-27032011><FONT color=#0000ff size=2
face=Arial></FONT></SPAN> </DIV>
<DIV><SPAN class=250492422-27032011><FONT color=#0000ff size=2 face=Arial>In
real life, the multiplier is presumably followed by one or more CIC filters,
and the picture gets murkier there due to bit allocation. You can
lose up to ~6 dB (almost a full bit) if the MSB's precision isn't
fully utilized. In other words, if the MSB is toggled by
only the very strongest signals, it will contribute less than the usual 6.02
dB of dynamic range, and that deficit is effectively part of the conversion
loss.</FONT></SPAN></DIV>
<DIV><SPAN class=250492422-27032011><FONT color=#0000ff size=2
face=Arial></FONT></SPAN> </DIV>
<DIV><SPAN class=250492422-27032011><FONT color=#0000ff size=2
face=Arial>Another concern might be DC offset from the ADCs. If the full
dynamic range isn't available going into the DDC it won't be available at
baseband, either. </FONT></SPAN></DIV>
<DIV><SPAN class=250492422-27032011><FONT color=#0000ff size=2
face=Arial></FONT></SPAN> </DIV>
<DIV><SPAN class=250492422-27032011><FONT color=#0000ff size=2 face=Arial>--
john, KE5FX</FONT></SPAN></DIV>
<DIV><SPAN class=250492422-27032011><FONT color=#0000ff size=2
face=Arial></FONT></SPAN> </DIV>
<BLOCKQUOTE
style="BORDER-LEFT: #0000ff 2px solid; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px"
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<DIV dir=ltr class=OutlookMessageHeader align=left><FONT size=2
face=Tahoma>-----Original Message-----<BR><B>From:</B>
hpsdr-bounces@lists.openhpsdr.org
[mailto:hpsdr-bounces@lists.openhpsdr.org]<B>On Behalf Of </B>Edson
Pereira<BR><B>Sent:</B> Sunday, March 27, 2011 12:30 PM<BR><B>To:</B>
hpsdr@openhpsdr.org<BR><B>Subject:</B> [hpsdr] DDC Noise
Figure<BR><BR></FONT></DIV><BR>
<DIV>Hello,</DIV>
<DIV><BR></DIV>
<DIV>I have been trying to educate myself about SDR using direct sampling
and am having some difficulty in understanding how to calculate the noise
figure of a DDC based SDR system. Since the noise figure takes into account
the bandwidth, would it be correct to say that even though the noise figure
of the ADC is high, the total noise figure of the receiver system can be
improved (by the various decimation stages)? Putting a pre-amp in front of
the ADC would seems to help considerably, at the cost of lower dynamic
range, but if my understanding is correct, the total noise figure will be
much better than that of the ADC alone. Could anyone shed some light? Some
math will be ok.</DIV>
<DIV><BR></DIV>
<DIV>73,</DIV>
<DIV><BR></DIV>
<DIV>-- Edson, pu1jte, n1vtn, jf1afn,
pu2mwd�</DIV></BLOCKQUOTE></BLOCKQUOTE></BODY></HTML>