# Effect of bandwidth reduction in bit growth

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!

• Is there a specific paper or book you're using? Where is this formula from? – Phonon Oct 24 '13 at 1:50
• 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. – Vishwanath Oct 24 '13 at 4:09
• 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. – Cordic Oct 25 '13 at 8:04

## 1 Answer

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

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