How to realize a sinc function using CIC filter for decimation and interpolation ??? Can I combine interpolation and decimaton methods inorder to get the complete response of a sinc function???

  • $\begingroup$ do you want the $\text{sinc}$ to be in frequency or time domain? $\endgroup$ Jan 25 '17 at 11:27
  • $\begingroup$ The real question, and you should probably edit your question to include that, or ask a new one, is why you want that? $\endgroup$ Jan 25 '17 at 12:30

The frequency/z domain response of a $N$-Stage CIC filter is, invariably [1]:

$$\begin{align} H(z) &= \frac{(1-z^{-RM})^N}{(1-z^{-1})^N}\\ &= \left(\sum\limits_{k=0}^{RM-1}{z^{-k}}\right)^N \end{align}$$

with $R$ being the rate change, and $M$ being the delay length. You can assume $M=1$ (you don't get a special shape if you assume larger $M$, just sharper response).

For $N=1$, you get the "boring" observation that you've built a moving average filter (which, in fact, has sinc-shape in frequency domain). But that's really just due to the fact that that you really did just that – build a recursive MA.

The magnitude response is usually (as in [1]) represented as

$$ \lvert H(f)\rvert = \left\lvert \frac{\sin(\pi M f)}{\pi M f} \right\rvert^N $$

which goes to say you only get the sinc shape for $N=1$. In other words, a moving average. Which really shouldn't surprise you – the moving average is the time-domain rectangle, and the Fourier transform of that is the sinc.

[1] CIC Filter Introduction, Matthew P. Donadio, July 2000. Available Online

  • 2
    $\begingroup$ To clarify, the CIC with N=1 only approximates a Sinc, and only reasonably well for large M. For small M it is what I would call an "aliased Sinc". I mention this minor point not to detract from this good answer but in case anyone wants a precise Sinc function response at least over the limited frequency range applied. That said I believe the final magnitude response is incorrect for general M where M is any positive integer. @MarcusMüller would you agree? $\endgroup$ Apr 25 '17 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.