Please help me to get good understanding of this statement from book Analog Integrated Circuit Design by ken martin....chapter 14 oversampling converters page536second paragraph last line...


"The reason for the SNR improvement  through use of oversampling is that when quantized samples are added together the signal portion adds linearly, whereas he noise portion adds as the square root of sum of the squares."

thanks in advance...!

Maybe it's easier to understand from straight probability.

The signal is roughly the same in each oversampled data point (if is isn't you aren't oversampling).  So, the signal data are highly correlated and if you take 10 samples you get 10 times each sample.  They just add up.

Noise is different though.  We model quantization noise as a white process, so its amplitude at any time is unknown, but we do know its statistics (or its average amplitude for a long time).  We define the noise power as sigma-squared, where sigma is the standard deviation.  Now, if we take 10 samples of a noise process, elementary statistics indicates we add the noise power, not the amplitudes (since you can't very well add quantities that are unknown).  

Assume you signal is a voltage.  Then, the noise power is in units volts^2, so we have to take the square root of the noise power to get the noise voltage (remember the power of a voltage signal is (V^2)/R and we always normalize R to 1 in noise considerations.

Then, the signal-to-noise improvement you get is just 10/sqrt(10) is our example.  In general it is OSR/sqrt(OSR) (since we took OSR number of samples) which just equals sqrt(OSR).  And there you have it.