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. Bellin 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
fixed point CIC filter
Started by ●November 16, 2006
Reply by ●November 16, 20062006-11-16
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.
Reply by ●November 16, 20062006-11-16
Try http://users.snip.net/~donadio/cic.pdf In article <x6ednZLBb-9DLcHYnZ2dnUVZ_v6dnZ2d@giganews.com>, "sunflowerhj" <sunflowerhj@hotmail.com> 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 > > > >
Reply by ●November 16, 20062006-11-16
Ray Andraka wrote:> ... The comb section > subtracts the integrator output of N samples ago from the current > integrator output. ...Not the integrator input from N samples ago from the current integrator output? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 16, 20062006-11-16
Jerry Avins wrote:> Ray Andraka wrote: > >> ... The comb section subtracts the integrator output of N samples >> ago from the current integrator output. ... > > > Not the integrator input from N samples ago from the current integrator > output? > > JerryNope, the boxcar is H(z)= 1 + z^-1 + z^-2 + .... +z^-n. This is a geometric progression of n terms, so H(z) = (1-z^-n)/(1-z^-1) the integrator is, u(z) = x(z) + u(z)*z^-1 = x(z)/(1-z^-1) so the CIC output is y(z) = (1-z^-n)*u(z) = u(z) - u(z)*z^-n
Reply by ●November 17, 20062006-11-17
Ray Andraka wrote:> Jerry Avins wrote: > >> Ray Andraka wrote: >> >>> ... The comb section subtracts the integrator output of N samples >>> ago from the current integrator output. ... >> >> >> Not the integrator input from N samples ago from the current >> integrator output? >> >> Jerry > Nope, the boxcar is H(z)= 1 + z^-1 + z^-2 + .... +z^-n. This is a > geometric progression of n terms, so > > > H(z) = (1-z^-n)/(1-z^-1) > > the integrator is, u(z) = x(z) + u(z)*z^-1 = x(z)/(1-z^-1) > > so the CIC output is y(z) = (1-z^-n)*u(z) = u(z) - u(z)*z^-nIsn't that subtracting the *input* N samples ago from the current output? I don't see a subtraction involving y(z^-n). Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 17, 20062006-11-17
Jerry Avins wrote:> Ray Andraka wrote: > >> Jerry Avins wrote: >> >>> Ray Andraka wrote: >>> >>>> ... The comb section subtracts the integrator output of N >>>> samples ago from the current integrator output. ... >>> >>> >>> >>> Not the integrator input from N samples ago from the current >>> integrator output? >>> >>> Jerry >> >> Nope, the boxcar is H(z)= 1 + z^-1 + z^-2 + .... +z^-n. This is a >> geometric progression of n terms, so >> >> >> H(z) = (1-z^-n)/(1-z^-1) >> >> the integrator is, u(z) = x(z) + u(z)*z^-1 = x(z)/(1-z^-1) >> >> so the CIC output is y(z) = (1-z^-n)*u(z) = u(z) - u(z)*z^-n > > > Isn't that subtracting the *input* N samples ago from the current > output? I don't see a subtraction involving y(z^-n). > > JerryJerry, Subtracting the output of the integrator, not the output of the filter. The CIC is an integrator followed by a comb filter. The comb subtracts the output of the integrator (u) N samples ago from the current output of the integrator. The output of the filter (y) is that difference. integrator comb +---+ +---+ --->| + |--+--+--->| + |------> +---+ | | +---+ + ^ | | ^ - |____| |______|
Reply by ●November 17, 20062006-11-17
Ray Andraka wrote:> Jerry Avins wrote: > >> Ray Andraka wrote: >> >>> Jerry Avins wrote: >>> >>>> Ray Andraka wrote: >>>> >>>>> ... The comb section subtracts the integrator output of N >>>>> samples ago from the current integrator output. ... >>>> >>>> >>>> >>>> >>>> Not the integrator input from N samples ago from the current >>>> integrator output? >>>> >>>> Jerry >>> >>> >>> Nope, the boxcar is H(z)= 1 + z^-1 + z^-2 + .... +z^-n. This is a >>> geometric progression of n terms, so >>> >>> >>> H(z) = (1-z^-n)/(1-z^-1) >>> >>> the integrator is, u(z) = x(z) + u(z)*z^-1 = x(z)/(1-z^-1) >>> >>> so the CIC output is y(z) = (1-z^-n)*u(z) = u(z) - u(z)*z^-n >> >> >> >> Isn't that subtracting the *input* N samples ago from the current >> output? I don't see a subtraction involving y(z^-n). >> >> Jerry > > > > Jerry, > > Subtracting the output of the integrator, not the output of the filter. > > The CIC is an integrator followed by a comb filter. The comb subtracts > the output of the integrator (u) N samples ago from the current output > of the integrator. The output of the filter (y) is that difference. > > integrator comb > +---+ +---+ > --->| + |--+--+--->| + |------> > +---+ | | +---+ > + ^ | | ^ - > |____| |_z^-N_|Oops, I forgot the delay on the - branch into the comb
Reply by ●November 17, 20062006-11-17
Ray Andraka wrote: ...> Oops, I forgot the delay on the - branch into the combNever mind. You finally screwed my head on straight anyway. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 17, 20062006-11-17
Jerry Avins wrote:> Ray Andraka wrote: > > ... > >> Oops, I forgot the delay on the - branch into the comb > > > Never mind. You finally screwed my head on straight anyway. > > JerryMy pleasure? ;-) CIC's are one of the tricks I learned about early in my DSP career, and at the time practically nobody was using them. Nowadays I see them mentioned all over the place. That and CORDIC.
