[hpsdr] General question - sampling rate vs bits on ADC

[Alberto I2PHD]

On 5/17/2011 10:29 PM, David Kirkby wrote:
> On 17 May 2011 20:27, Alberto I2PHD <i2phd at weaksignals.com <mailto:i2phd at weaksignals.com>> wrote:
>
>     On 5/17/2011 5:00 PM, Dr. David Kirkby wrote:
>>     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.
>     /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..
>     the rest is noise...
>
>     Then, there is the processing gain to be taken into account. as you correctly said, the noise is spread up to the
>     Nyquist frequency,
>     and when you decimate, you throw away a lot of it.
>
>     The following image is taken from a presentation I gave last year here in Milano, Italy
>
>     http://sundry.i2phd.com/procgain.gif
>
>     /
>     /--
>     / /*73 Alberto I2PHD*/
>
/
/
>
> /Thank you for that. Do you have a reference for a derivation of that formula? /
Dave,

     ir is all explained in this document from Analog Devices :

http://sundry.i2phd.com/MT-001.pdf
> /The actual card we intend using is a Asus D2X XONAR  which has the specification here:
>
> http://www.asus.com/Multimedia/Audio_Cards/Xonar_D2X/#specifications/ /
>
> That's supposed to have a S/N of 118 dB, which according to my calculations is 19.6 bits.
> (Below is in Mathematica, where Log[10,x] is log base 10 of x. I used NSolve to solve for n.).
>
> In[6]:= NSolve[ 118 == 20 Log[10,2^n],n]/ /
>
> Out[6]= {{n -> 19.5994}}
>
> Is that correct?
> /
According to that document, the ENOB (Effective Number of Bits) can be computed as :

SNR = 6.02N + 1.76dB, over the dc to fs/2 bandwidth.      Solving for N, with SNR = 118 dB :
N = (118 - 1.76)/6.02  which gives N = 19.31 bits

That document explains a lot about ADC resolution and processing gain. Very useful

73  Alberto  i2PHD


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