Regarding accuracy and speed. You trade off resolution, speed.
Some questions -
1) What are you using digitized signal for ? Simple harmonic content determination
or magnitude scale is important. FFT for former, wavelet transform for latter.
2) How much T do you want to store signal, eg. how many samples do you need to accumulate.
3) If you want to preserve 3 Khz harmonic content, then meeting Nyquist will take care of
that. @ 6 Khz DelSig can generate 17 bit conversion.
Nyquist doesn't quite cut it.
An SAR converter has a frequency response of sin(x)/x. The Delsig converter in the PSoC3 has a frequency response of (sin(x)/x)^4, where x = PI*f/fsample. For the SAR case, the -3dB bandwidth is 0.44* fsample. For the Delsig case, the -3 dB bandwidth is 0.22*fsample. If you want accuracy = 10% at 3.0 kHz, the minimum sample rate is 24 kHz. This is a simple calculation. This is not a circuit design or implementation or software problem. It is a mathematical certainty of the topology of nth order delta sigma converters. It can be compensated with specifically designed digital filters.
Dennis its so much easier on calculations of we ignore the H(s) response of
other components in the signal path. Its a good thing I did not contribute to the
design on navigation systems to Mars, would have wound up crashing into alpha
centauri. I am glad you caught my error.
Question, in SAR is H(s) a simple sampler or a zero order hold response ?
Which leads me to a suggestion, encorporate either in config tool, or in a table,
1/2 lsb frequency response for all basic resolutions. Just a thought, maybe even
a graphic to remind us (me especially) response properties of the converter.
The 1/2 LSB frequency response is an interesting idea, an interesting way to look at the problem.
We're looking at log(base2)(1/(1-abs((sin(x)/x))^n) where x = pi*f/fsample.The guaranteed accuracy is not as good as you might hope. See attached .xls, where series# = sinx/x order(n).
This is mathematical, entirely different from the settling problem, which is handled nicely by the settling time as specified in the component datasheet.
The spreadsheet attach missing.
I can assure you it is quite more difficult to hit Alpha Centauri than to hit Mars, which tonight will be visible just right to and above the full-moon from your location. Alpha Centauri (proxima) is high up in the north while Mars and Moon traveling in the ecliptic. This means that any object started from Earth gets a huge V0-kick vectored in the ecliptic and it will be VERY hard (costing MUCH energy) to leave that trajectory
Dennis, chrome has a problem with attachments, use IE or Firefox.....