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I just want to ask, whenever you reduce a bandwidth of signal by filtering, I wonder why do you need to improve the number of bits used to represent that signal. The number of bit growth is $$\frac{1}{2}\log_2\left(\frac{BW_{in}}{BW_{out}}\right)$$

Is it to represent to small noise due to filtering. Thanks!

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  • $\begingroup$ Is there a specific paper or book you're using? Where is this formula from? $\endgroup$ – Phonon Oct 24 '13 at 1:50
  • $\begingroup$ Perhaps my knowledge is limited, but I never knew that bandwidth and bits used for representing a signal are related. Further, if noise is a deciding factor, I don't think we need to increase the number of bits, since noise in general may be white or have more high frequency content, which means that filtering only reduces the noise. $\endgroup$ – Vishwanath Oct 24 '13 at 4:09
  • $\begingroup$ It's from my mentor, he said it has something to do with the signal to noise ratio improvement since you filter out noise, hence you need to add a certain number of bits to achieve that Signal to Noise Ration. I really can't understand it quite well. $\endgroup$ – Cordic Oct 25 '13 at 8:04
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Consider a two-sample moving average of one-bit data with values {0,1}. The bandwidth is reduced by a factor of two. The possible outputs are {0,0.5,1}. You have gained half a bit, as predicted by the formula.

John

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  • $\begingroup$ How come you reduce the bandwidth by two, you have those 3 possible outputs --> (0, 0.5, 1)? $\endgroup$ – Cordic Oct 25 '13 at 8:08
  • $\begingroup$ (0 + 0) / 2 = 0, (1 + 0) / 2 = 0.5, (1 + 1) / 2 = 1 $\endgroup$ – John Oct 30 '13 at 0:51

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