Doubt in Sigma Delta ADC

[hacksgen]

Hi,

I have a question on sigma delta adc for receivers. In one of the papers I have read the following.

**broken link removed**

Imagine the conversion of a signal centered at 1 MHz with 10-kHz bandwidth. With a 10-MHz sampling rate, a traditional converter would provide five times oversampling;
a bandpass ∑Δ converter would achieve 500 times oversampling.

Can anyone explain the meaning of the sentence.

If a ∑Δ converter is converting the signal with center frequency at 1MHz(which I am assuming in modulated signal downconverted from RF to 1MHz) than would not the oversampling still be the same for ∑Δ as it will only be able to sample the 1MHz signal at 10MHz sampling rate.

 

[luben111]

Hi,
Oversampling is not just to measure many times the same value, it's very deep and complex theory which brings you the answer what to do with the multiple samples:
1. To get higher resolution (more bits) from the conversion
2. To increase the Signal to Noise ratio (SNR)

The oversampling theory also tells you how many additional bits you can get from the samples you have (usually 4^n where n is the number of the additional bits you gain) or how much you can increase the SNR.

Actually if you simply measure 16 times the same value it's not yet oversampling and it's not sure that you'll get 2 more bits, you need also to add noise (in many cases the normal ADC noise is not suitable for oversampling) and to conform many othjer requirements.

It's a common mistake to mix oversampling with averaging (many people believe they are the same) - both look quite similar like math but it's only similarity in the calculations