[hpsdr] A/D

[Henry Vredegoor]

Hi Leon, All,

Not sure if I understand you correctly.

As far as I know and understand there is no principal difference in sampling
requirements for a signal in an digital oscilloscope for a signal at 30 MHz
or an SSB signal in an SDR at 30 MHz.

For an SDR receiver the HF signal will be first sampled at 30 MHz, the same
as for the scope.
After this the stream of samples in an SDR will usually be digitally down
converted (DDC) to a lower sample rate.
The lower sample rate being such that it will both accommodate for a
practical size for further processing as well as the bandwidth required for
say a 196 KHz band segment. (For e.g. transport to and final processing and
display in a PC)

I don't know how a scope would further digitally process the signal after
the initial A/D conversion.
I could imagine a similar process (DDC) as in an SDR.

Not sure though if a continuous sampling process would always be required or
would be implemented in a (cheaper-) digital scope!

I think it would be correct and practical to say that if you can
CONTINUOUSLY sample a signal of a maximum frequency of 30 MHz (60 Ms/s
sample rate / sample frequency 60 MHz) you have a bandwidth of 30 MHz?

73's,

Henry.






> -----Original Message-----
> From: Leon [mailto:leon355 at btinternet.com] 
> Sent: zaterdag 21 april 2007 12:44
> To: Henry Vredegoor
> Subject: Re: [hpsdr] A/D
> 
> 
> Message----- Original Message ----- 
> From: Henry Vredegoor
> To: hpsdr at hpsdr.org
> Sent: Saturday, April 21, 2007 11:21 AM
> Subject: Re: [hpsdr] A/D
> 
> 
> ***** High Performance Software Defined Radio Discussion List *****
> 
> 
> 
> 
> 
> Hi Frank, All,
> 
> Being basically an analog guy too and not at all an expert on 
> this, I can 
> point you only to the Nyquist Theorem or Sampling Theorem, e.g.
> 
> http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem
> 
> A lot of mathematic's!
> 
> What this all comes down to, as I understand it, is that this 
> guy (and 
> others) proved mathematically that EVEN WITH ONLY TWO samples 
> per period of 
> the signal to be be digitized, it can be reconstructed 
> correctly from these 
> two samples!
> 
> So for 10 meters / 30 MHz you would need a minimum of two 
> samples per one 
> single period of 30 MHz.
> So your minimum sampling frequency would be 2 X 30 MHz = 60 MHz.
> 
> This is why you always see the sampling frequency of e.g. an 
> ADC is divided 
> by 2 to get the highest possible signal frequency that can be 
> digitized 
> correctly.
> 
> Confusing can be the fact that within SDR's we have QSD's 
> with I and Q 
> signals, each coming from their own ADC. (e.g. as is the case 
> in a PC with a 
> stereo sound card)
> In that case you have 2 channels and thus 2 samples each 
> moment you sample, 
> so in fact doubling the sample rate/frequency.
> Then you have a maximum signal frequency of the signal that 
> can be digitized 
> correctly equal to the full sampling frequency of a single 
> channel, so no 
> division by 2.
> 
> Hope this helps; Guru's, please correct me if I am telling this not 
> correctly...... ;-)
> 
> 
> ---------------------------
> 
> As I said previously, the sampling theorem applies to the 
> *bandwidth* of the 
> received signal, so a 130 Ms/s ADC won't have any problems 
> with a 3 KHz 
> bandwidth SSB signal at 30 MHz. Digitising a 30 MHz signal 
> (for a scope, for 
> instance) would require a minimum of 60 Ms/s, of course.
> 
> Leon
> --
> Leon Heller
> Amateur radio call-sign G1HSM
> Yaesu FT-817ND and FT-857D transceivers
> Suzuki SV1000S motorcycle
> leon355 at btinternet.com
> http://www.geocities.com/leon_heller 
> 


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