Bits to keep: output FIR

If any MSB is not used at all then you need to discard it freely. You can easily know the maximum swing of your output by finding out the maximum input value and the gain of your filter(sum of coefficients in hardware).

If one MSB is used scarcely then you may opt to discard it with saturation.

In principle it is better not to saturate. To avoid saturation you can lift up the filter gain to occupy that bit well or lift down to avoid it completely.

Once you know how many MSBs to chop off then you discard the LSBs accordingly. You may round up for LSBs but this is optional to improve the result.

Thanks Kaz!

OUTPUT = 22 (14-22=8)

I have 12 bit (cpl 2) input then => 2^(12-1)-1 = 2.047

The sum coeffiecients (101) = 0,98707...

2.047*0,98707 = 2.020,5420... <- the max value of convolution (right?)

then 11 bit

bits to keep = 14

LSB = 0

MSB = 8 TRUNCATE

Is correct?

thank a lot of!!

Fabio

your input statement is correct:

When using 2's complement(signed), your 12 bit input can swing between 0 + 2047

your coeff statement is incomplete:

dc gain of your filter(strictly speaking your convolution) in hardware is the sum of all coefficients as scaled for hardware. Your figure of 0.98707 suggests a normalised dc gain.

What happens to this gain in hardware depends on how you scale there, your convolution gain goes up by same factor used to scale coeffs.

for example if your scale factor = 1022 (511/.5, assuming peak normalised coeff = .5) then convolution gain becomes .98707 x 1022 = 1009 in hardware.

then max convolution output = 1009 x +2047 = + 2064984

this requires 22 bits (2^21 = 2097152)

Hence you need to keep all Msbs and truncate 8 Lsb

You can fine tune the convolution sum under your control through this scale factor.

A practical scale factor is the one that keeps unity filter dc gain after Lsb truncation i.e. a scale factor that equals your division(Lsb truncation).

In my example I removed 8 Lsb(this means division by 2^8 = 256). Hence the filter which was scaled up by 1022 during convolution but then divided down by 256 after truncation gets a dc gain of 1022/256 = nearly 4 in hardware. In short: scale the normalised coeffs by a value such that their sum = 2^n (512 if 10 bits used) rather than the peak.

It will help if you break above as three stages:

normalisation(at modelling)

scaling then convolution(mult-add in hardware)

truncation(division by lsb truncation) to match signal resolution

Details of gain control described here:

http://www.digital-filters.co.uk/static/filter_gain

!!!! Thanks Kaz very much !!!!

Your example is my case, because i used 10 bit for the filter !!!

Fabio ^_^