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On 5/17/2011 10:29 PM, David Kirkby wrote:<br>
<blockquote
cite="mid:BANLkTi=r=zSDBuQkb6UaN=ExwWu3N5+mSw@mail.gmail.com"
type="cite">
<div class="gmail_quote">On 17 May 2011 20:27, Alberto I2PHD <span
dir="ltr"><<a moz-do-not-send="true"
href="mailto:i2phd@weaksignals.com">i2phd@weaksignals.com</a>></span>
wrote:<br>
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<div class="im"> On 5/17/2011 5:00 PM, Dr. David Kirkby
wrote:
<blockquote type="cite">
<pre>What trade-offs does one get for an SDR? The audio card route is cheaper and
gives more bits, so should I assume give better dynamic range. In contrast, I
can see that with a faster sampling rate, the noise is spread over a wider
range, so its lower in a specific bandwidth. But I would have thought the
quantisation noise would be higher with the faster cards, as they have less bits.
</pre>
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<i><font face="Tahoma">First of all, the nominal 24 bits of
a sound card actually rarely are more than 20... more
often they are in the range 18-19..<br>
the rest is noise...<br>
<br>
Then, there is the processing gain to be taken into
account. as you correctly said, the noise is spread up
to the Nyquist frequency,<br>
and when you decimate, you throw away a lot of it.<br>
<br>
The following image is taken from a presentation I gave
last year here in Milano, Italy<br>
<br>
<a class="moz-txt-link-freetext" href="http://sundry.i2phd.com/procgain.gif">http://sundry.i2phd.com/procgain.gif</a><br>
</font><br>
</i>
<div><i>-- <br>
</i> <i><b>73 Alberto I2PHD</b></i></div>
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<i><br>
</i><br>
<blockquote
cite="mid:BANLkTi=r=zSDBuQkb6UaN=ExwWu3N5+mSw@mail.gmail.com"
type="cite">
<div class="gmail_quote">
<div><br>
<i>Thank you for that. Do you have a reference for a
derivation of that formula? </i><br>
</div>
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</blockquote>
Dave,<br>
<br>
ir is all explained in this document from Analog Devices :<br>
<br>
<a class="moz-txt-link-freetext" href="http://sundry.i2phd.com/MT-001.pdf">http://sundry.i2phd.com/MT-001.pdf</a><br>
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cite="mid:BANLkTi=r=zSDBuQkb6UaN=ExwWu3N5+mSw@mail.gmail.com"
type="cite">
<div class="gmail_quote">
<div><i>The actual card we intend using is a Asus D2X XONAR
which has the specification here:<br>
<br>
<a moz-do-not-send="true"
href="http://www.asus.com/Multimedia/Audio_Cards/Xonar_D2X/#specifications">http://www.asus.com/Multimedia/Audio_Cards/Xonar_D2X/#specifications</a></i>
<i><br>
<br>
That's supposed to have a S/N of 118 dB, which according to
my calculations is 19.6 bits. <br>
(Below is in Mathematica, where Log[10,x] is log base 10 of
x. I used NSolve to solve for n.). <br>
<br>
In[6]:= NSolve[ 118 == 20 Log[10,2^n],n]</i>
<i><br>
<br>
Out[6]= {{n -> 19.5994}}<br>
<br>
Is that correct? <br>
</i></div>
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</blockquote>
According to that document, the ENOB (Effective Number of Bits) can
be computed as :<br>
<br>
SNR = 6.02N + 1.76dB, over the dc to fs/2 bandwidth. Solving
for N, with SNR = 118 dB :<br>
N = (118 - 1.76)/6.02 which gives N = 19.31 bits<br>
<br>
That document explains a lot about ADC resolution and processing
gain. Very useful<br>
<br>
73 Alberto i2PHD<br>
<br>
<br>
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