<div><br></div>Hi John,<div><br></div><div>Thanks for the comments and clarifications. My confusion was Equation 8 in this Analog Devices paper:</div><div><br></div><div><a href="http://www.analog.com/static/imported-files/tutorials/MT-006.pdf">http://www.analog.com/static/imported-files/tutorials/MT-006.pdf</a></div>
<div><br></div><div><div>NF = 10log10F = PFS(dBm) + 174 dBm – SNR – 10 log10[fs/2B] – 10 log10 B</div>
<div><br></div><div>But the equation is a little misleading as pointed out by Ulrich, N1UL. It seems that, similarly to an analog receiver, the NF at the first stage will dominate and the total NF will not be changed throughout the receiver. The NF will degrade the signal at each stage. The SNR however can be improved with processing gain. Would this be a correct statement?</div>
<div><br></div><div>73,</div><div><br></div><div>-- Edson</div><div><br></div><div><br></div><br><div class="gmail_quote">On Sun, Mar 27, 2011 at 8:01 PM, John Miles <span dir="ltr"><<a href="mailto:jmiles@pop.net" target="_blank">jmiles@pop.net</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div><div></div><div>***** High Performance Software Defined Radio Discussion List *****<br>
<br>
</div></div><br>
<div>
<div><span><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><font color="#0000ff" size="2" face="Arial"></font></span> </div>
<div><span><font color="#0000ff" size="2" face="Arial">--
john, KE5FX</font></span></div>
<blockquote style="border-left:#0000ff 2px solid;padding-left:5px;margin-left:5px;margin-right:0px" dir="ltr">
<div dir="ltr" align="left"><font size="2" face="Tahoma">-----Original Message-----<br><b>From:</b>
<a href="mailto:hpsdr-bounces@lists.openhpsdr.org" target="_blank">hpsdr-bounces@lists.openhpsdr.org</a>
[mailto:<a href="mailto:hpsdr-bounces@lists.openhpsdr.org" target="_blank">hpsdr-bounces@lists.openhpsdr.org</a>]<b>On Behalf Of </b>John
Miles<br><b>Sent:</b> Sunday, March 27, 2011 3:54 PM<br><b>To:</b>
<a href="mailto:hpsdr@openhpsdr.org" target="_blank">hpsdr@openhpsdr.org</a><br><b>Subject:</b> Re: [hpsdr] DDC Noise
Figure<br><br></font></div>
<div><span><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><font color="#0000ff" size="2" face="Arial"></font></span> </div>
<div><span><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><font color="#0000ff" size="2" face="Arial"></font></span> </div>
<div><span><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><font color="#0000ff" size="2" face="Arial"></font></span> </div>
<div><span><font color="#0000ff" size="2" face="Arial">--
john, KE5FX</font></span></div>
<div><span><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" dir="ltr">
<div dir="ltr" align="left"><font size="2" face="Tahoma">-----Original Message-----<br><b>From:</b>
<a href="mailto:hpsdr-bounces@lists.openhpsdr.org" target="_blank">hpsdr-bounces@lists.openhpsdr.org</a>
[mailto:<a href="mailto:hpsdr-bounces@lists.openhpsdr.org" target="_blank">hpsdr-bounces@lists.openhpsdr.org</a>]<b>On Behalf Of </b>Edson
Pereira<br><b>Sent:</b> Sunday, March 27, 2011 12:30 PM<br><b>To:</b>
<a href="mailto:hpsdr@openhpsdr.org" target="_blank">hpsdr@openhpsdr.org</a><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></div>
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