> "Jerry Avins" <jya@ieee.org> wrote in message
> news:ca6dnRjSZ4rEYMfbnZ2dnUVZ_o2vnZ2d@rcn.net...
>> mariking wrote:
>>> Hi
>>>
>>> I need to take a standard 8bit ADC which samples at 40Msps increase the
>>> resolution by about 4 or 5 bits. I know that oversampling spreads the
>>> quantization noise over a wider bandwidth thereby decreasing the SQNR. So
>>> i can increase the resolution through decimation? How do I increase the
>>> resolution through decimation? It decimates the signal frequency but i
>>> don't physically know how to increase it because if i send 8 bits in i
>>> should just receive 8 bits out?
>> With a small amount of noise present in or added to the input signal (AND
>> PROVIDED THAT THE LINEARITY OF THE CONVERTER IS 12 BITS ACCURATE) the the
>> average of N readings has sqrt(N) better precision, making it worth
>> keeping log2(n) more bits. If you keep the most significant 12 bits from
>> the sum of 256 dithered readings taken in quick succession you can expect
>> them to be reasonably accurate. The extra 4 bits represent an accuracy
>> improvement of 16:1, for which you pay with a total of 16^2 readings.
>>
>> Jerry
>> --
>
> And, a minor point, you don't get 8 bits out for 8 bits in because of the
> addition of many samples in the decimation filtering.
> That is, in order to properly decimate, you need to first lowpass filter.
> The lowpass filter is implemented with a number of multiplies and adds. The
> simplest lowpass filter is a simple averager. Thus, no multiplies but still
> probably many adds. Thus more bits at the output.
I once had an Analog Devices programmable data acquisition system that
offered 12- or 16-bit conversions. The shorter conversions were done
with a successive-approximation converter at high speed. The 16-bit
conversions were about as fast as the integrating converters of the day.
(16- and 18- bit conversions were typically done with dual-slope
converters.) In fact there was only one A/D on the system. a 12-bit
converter with 16-bit precision on the bit transitions. To get a 12-bit
reading, noise with a triangular distribution was added to the input,
and a burst of 265 readings added to an accumulator. If all the bits had
been kept, that could have resulted in as many as 24 bits, depending on
the reading. The low 8 bits were discarded after addition was complete,
yielding the 16-bit result. Since the signal being measured was made
stationary by the sample-and-hold, an appropriate ramp could have
replaces the "noise".
When a measuring instrument is far more accurate than the uncertainty
introduced by independent measurement errors, the accuracy of a series
of averaged readings increases with the square root of their number. I
don't know how replacing the average with a another kind of low-pass
filter affects the outcome.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by Fred Marshall●May 28, 20072007-05-28
"Jerry Avins" <jya@ieee.org> wrote in message
news:ca6dnRjSZ4rEYMfbnZ2dnUVZ_o2vnZ2d@rcn.net...
> mariking wrote:
>> Hi
>>
>> I need to take a standard 8bit ADC which samples at 40Msps increase the
>> resolution by about 4 or 5 bits. I know that oversampling spreads the
>> quantization noise over a wider bandwidth thereby decreasing the SQNR. So
>> i can increase the resolution through decimation? How do I increase the
>> resolution through decimation? It decimates the signal frequency but i
>> don't physically know how to increase it because if i send 8 bits in i
>> should just receive 8 bits out?
>
> With a small amount of noise present in or added to the input signal (AND
> PROVIDED THAT THE LINEARITY OF THE CONVERTER IS 12 BITS ACCURATE) the the
> average of N readings has sqrt(N) better precision, making it worth
> keeping log2(n) more bits. If you keep the most significant 12 bits from
> the sum of 256 dithered readings taken in quick succession you can expect
> them to be reasonably accurate. The extra 4 bits represent an accuracy
> improvement of 16:1, for which you pay with a total of 16^2 readings.
>
> Jerry
> --
And, a minor point, you don't get 8 bits out for 8 bits in because of the
addition of many samples in the decimation filtering.
That is, in order to properly decimate, you need to first lowpass filter.
The lowpass filter is implemented with a number of multiplies and adds. The
simplest lowpass filter is a simple averager. Thus, no multiplies but still
probably many adds. Thus more bits at the output.
Fred
Reply by Jerry Avins●May 28, 20072007-05-28
mariking wrote:
> Hi
>
> I need to take a standard 8bit ADC which samples at 40Msps increase the
> resolution by about 4 or 5 bits. I know that oversampling spreads the
> quantization noise over a wider bandwidth thereby decreasing the SQNR. So
> i can increase the resolution through decimation? How do I increase the
> resolution through decimation? It decimates the signal frequency but i
> don't physically know how to increase it because if i send 8 bits in i
> should just receive 8 bits out?
With a small amount of noise present in or added to the input signal
(AND PROVIDED THAT THE LINEARITY OF THE CONVERTER IS 12 BITS ACCURATE)
the the average of N readings has sqrt(N) better precision, making it
worth keeping log2(n) more bits. If you keep the most significant 12
bits from the sum of 256 dithered readings taken in quick succession you
can expect them to be reasonably accurate. The extra 4 bits represent an
accuracy improvement of 16:1, for which you pay with a total of 16^2
readings.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by mariking●May 28, 20072007-05-28
Hi
I need to take a standard 8bit ADC which samples at 40Msps increase the
resolution by about 4 or 5 bits. I know that oversampling spreads the
quantization noise over a wider bandwidth thereby decreasing the SQNR. So
i can increase the resolution through decimation? How do I increase the
resolution through decimation? It decimates the signal frequency but i
don't physically know how to increase it because if i send 8 bits in i
should just receive 8 bits out?
Please help.
Thank you very much.
_____________________________________
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