Question about Dynamic Range

All I know is: The more points used, the more it looks like the original.

D from BC myrealaddress(at)comic(dot)com BC, Canada Posted to usenet sci.electronics.design

You seem mightily confused.

The dynamic range measures the ratio of the largest to the smallest signal your device can deal with.

To prevent aliasing, you must restrict the bandwidth of your signal to less than half the sampling rate.

These two concepts are completely independent.

Jeroen Belleman

Not completely:

-- Joe

Indeed, you are right. Somehow I think the OP isn't up to that yet.

Jeroen Belleman

This values isn't correct in all cases.

DR is a bit of a fuzzy term because you have to know more information to know what is really being considered.

If the noise is large the DR is determined by the noise not the LSB of the ADC. To prove this lets consider two very extreme cases with audio signals being important.

(1) Imagine that you have 10mV of signal and 100mV of noise. This noise is all up in the 100MHz to 1GHz band. Further imagine that your ADC is only a one bit converter (a comparator) but that it converts at a

(2) Imagine that you have a 24 bit ADC but the bottom 8 bits don't always change as one would expect. In the span near from -1/4096 to +1/4096 it reports the same value. Obviously for this converter it is only really as good as a 12 bit converter in some applications.

There is another version of the DR that is as you are dealing with here. This version is based on the question "what is the biggest signal and smallest signal I can have at the same time?" and leads to quite a different answer than the number of bits in the ADC.

You can expect to see values like 1/sqrt(12) appear in the expression from time to time.

Imagine a slow ramp that takes you up the codes of the ADC. If you subtract the ADC number from the ideal over a span between steps, you get:

Y =3D X - 0.5 (X =3D 0 to 1)

If we work out the RMS

Y^2 =3D X^2 - X + 0.25

(int)Y^2 =3D (1/3)X^2 - (1/2)X^2 + 0.25X

(int)Y^2)(0 to 1) =3D (1/3) - (1/2) + 0.25 - 0

=3D 1/3 - 1/4 =3D (4-3)/(3*4) =3D 1/12

sqrt() =3D 1/sqrt(12)

J.A. Legris wrote:

Dontcha just love guys who entitle Wikipedia pages using em dashes? Is *anyone* seeing that as an unbroken link?

what was posted was not a url.

I can't wikipedia working well if the page titles were constrained to a subset of printable ascii,

Thanks for all answers, i'll try to understand with an example....Considering this adc:

Vref=3V resolution=8bit =>LSB=3/256=11,7mV; DR=6.02*8=48.16dB

and considering as input signal a rect function with maximum amplitude equal to 3V. I want to set the correct sampling frequency in ADC to avoid aliasing (it is only an example, so i don't consider anti aliasing filter). Primarily i shoud obtain fourier transform of this signal, but

If fourier transform is obtained, i should set (sampling frequency)/2. From (sampling frequency)/2 all harmonics should be below quantization error, so i should consider DR*6.02 (because quantization error is - lsb/2

Oh, I think I see where this is going: You say that if the input signal is smaller, high harmonics drop below the quantization noise floor, and you could get away with lower sampling rates. Is that your point?

Anyway, to work this out, the input signal amplitude is supposed to exactly fill the full range of the ADC.

Jeroen Belleman

You're on the right track, but you should consider using an anti- aliasing filter, and also the benefits of oversampling.

The following application note deals with them in some detail. WARNING: PDF file >2MBytes. Also, the file name really is as long as it appears:

-- Joe

This comes right up for me. Agent hand-off to Firefox 3 is smooth.

This is only an example: i would only undestand how to choose the right ADC....so i consider a gven signal without anti-aliasing finlter and oversampling

I know this great application note: my questions are based on what i've read from it... I don't understand how to use DR in frequency domain..

If i have a rect(t/T) signal with amplitude Vref, its transform is Vref*T*sinc(f*T), is it right? If so, at f=0 an harmonic with amplitude Vref*T is present. From this harmonic, i shoud consider DR or DR*6.02 ? I would obtain harmonics lower than quantization noise, so i think DR*6.02, but i'm not really sure...