fixed point CIC filter

Thanks a lot
Jing


>sunflowerhj wrote:
>> Hi:
>> Can anyone help me to understand why CIC filter can not be implemented
in
>> floating point?
>> 
>> Mr. Dirk A Bell mentioned:
>> 
>>>>A CIC filter cannot be implemented in floating point. To get it to
>>>>work it must be done in non-saturating integer math, such as is done
>>>>using non-saturating two's complement representation. The integer
>>>>math must wrap around.
>>>>
>>>>Dirk A. Bell
>> 
>> in http://www.dsprelated.com/showmessage/4544/1.php
>> 
>> Could anyone explain it in more details or refer some articles to
read?
>> Thanks for the help in advance.
>> 
>> Jing
>> 
>> 
>> 
>> 
>
>
>It is because the CIC depends on exact integer math in order to exactly 
>cancel samples in the comb part of the filter.  Floating point has some 
>nasty rounding that will prevent exact cancellation.  A CIC also depends

>on the modularity of the number system: we allow the integrator to 
>continuously overflow, but since the number system is modular and the 
>comb section just takes the difference between the current and a delayed

>integrator output, the overflows are of no consequence.  If it were to 
>be done with floating point, a mod function would have to be used to 
>preserve the modularity, otherwise the floating point will keep bumping 
>up the exponent and shifting lsbs off the integrator as it sums without 
>bound
>
>This becomes perhaps a bit more obvious if you consider a first order 
>CIC.  The CIC is a recursive implementation of a boxcar filter, which is

>to say it outputs the sum of the last N samples.  Rather than summing 
>all those samples on every sample interval, it accumulates the samples 
>in an integrator (without regard to overflow).  The comb section 
>subtracts the integrator output of N samples ago from the current 
>integrator output. Since the comb section is computing the difference 
>between integrator outputs, we don't care about integrator overflows as 
>long as the number system is modular (e.g. two's complement or RNS 
>systems), meaning  A+N=A where N is the modulus of the number system. 
>For 2's complement, N=2^k where k is the number of bits.
>
>