1
$\begingroup$

I suggested to someone that one can create a sinc filter that is ideal enough to be indistinguishable from the ideal, given certain limitations with the data in the first place.

Is that true?

It was only a suggestion yet it was met with unhelpfulness.

$\endgroup$
3
  • $\begingroup$ i understand that a sinc filter has some sort of mathematical existence - but how can that be so unless it can be applied to some data ? $\endgroup$
    – user15987
    May 31, 2015 at 7:50
  • $\begingroup$ Maybe you should post your comment as a port of the original answer? $\endgroup$
    – jojeck
    May 31, 2015 at 8:53
  • $\begingroup$ @jojek this is how i think of it... suppose we know the value of pi thru math proof etc., but our computers for some reason can only compute it to some finite amount. if i have a picture and want to draw a circle round it, i would only need the computer to approximate pi close enough to cope with the resolution. ?? $\endgroup$
    – user15987
    May 31, 2015 at 13:58

4 Answers 4

3
$\begingroup$

An ideal system (such as a briwckwall filter) is the one which can be described theoretically (mathematically) but cannot be realized practically (physically).

The ideality of the sinc filter, aka the brickwall, results from its frequency-domain definition: an LTI low-pass filter with no ripples in the pass & stop bands and zero transition width.

Having this specification, an ideal lowpass filter, can be described mathematically, but cannot be realized using any practical techniques. One can find the impulse response of an ideal lowpass filter by using any suitable mathematical tools such as the the frequency to time-domain transformation of the given frequency reponse of the filter.

It results that an ideal low-pass filter has an impulse response in the form of a sinc(x) function, which extends from minus to plus infinity in time, a manifestation of its ideality.

Thus, when a sinc filter is to be implemented in practice, it can only be approximated; by truncatation in time for the sinc filter case. The approximation gets better as the length of the truncation increases which, however, is an undesired consequence.

Instead of forcing approximations to ideal sinc filters this way, it is practically more effective to use other techniques, such as weighted windowing, to realize those ideal filters, which follows the conversion of filter characteristics from those of the ideal one to that of the realizable one.

$\endgroup$
2
  • $\begingroup$ hi Fat32 - is any ad hoc implementation always going to be identifiable - so if i block filter something below 200 hertz, can i really do so without leaving any trace of what was there for spying eyes ? $\endgroup$
    – user15987
    May 31, 2015 at 13:53
  • $\begingroup$ possibly not... I'm not sure but. $\endgroup$
    – Fat32
    May 31, 2015 at 17:40
2
$\begingroup$

the "ideal" property of a $\operatorname{sinc}(\cdot)$ filter is in reference to its Fourier Transform

$$ \mathscr{F} \left\{ 2 f_0 \ \operatorname{sinc}(2 f_0 t) \right\} \ = \ \operatorname{rect}\left( \frac{f}{2f_0} \right) $$

where

$$ \operatorname{sinc}(x) \triangleq \begin{cases} \frac{\sin(\pi x)}{\pi x}, & \text{if }x \ne 0 \\ 1, & \text{if }x = 0 \end{cases} $$

and

$$ \operatorname{rect}(x) \triangleq \begin{cases} 1, & \text{if }|x|<\frac{1}{2} \\ 0, & \text{if }|x|>\frac{1}{2} \end{cases} $$

the $\operatorname{sinc}(\cdot)$ filter perfectly and totally excludes all frequency components above $f_0$ and perfectly and totally leaves all frequency components below $f_0$ unmolested. that's why it's ideal. it's an ideal (and impossible to attain) filtering operation.

$\endgroup$
2
  • 1
    $\begingroup$ I would have thought that the OP knows everything that you wrote down. However, you didn't actually answer the OP's question: whether it is possible to "create a sinc filter that is ideal enough to be indistinguishable from the ideal". $\endgroup$
    – Matt L.
    Jun 1, 2015 at 7:01
  • $\begingroup$ @MattL. please never over-estimate my intelligience haha $\endgroup$
    – user15987
    Jun 2, 2015 at 6:08
1
$\begingroup$

A sinc filter is unstable and not causal, so as such it can't be implemented. In discrete-time you can in principle approximate it arbitrarily closely by applying a (very long) window to the ideal filter impulse response, and shifting it such that it becomes causal. The latter will add a delay but it will not affect the magnitude response.

$\endgroup$
14
  • $\begingroup$ i think you mean by "causal" implementable in real time ? or, not.. ? $\endgroup$
    – user15987
    May 31, 2015 at 13:59
  • $\begingroup$ @user3293056: Exactly, it means that the current output sample only depends on the current and on past input samples, but not on future input samples. $\endgroup$
    – Matt L.
    May 31, 2015 at 14:43
  • $\begingroup$ i might agree that a sinc filter is acausal and unrealizable, but i have no idea what is meant by "A sinc filter is unstable ...". i don't think that stability has anything to do with a sinc filter. i've always thunk that stability is a property related to recursion (or feedback). is there some standard recursive theoretical implementation of a sinc filter? i am unaware of it, if there is. $\endgroup$ May 31, 2015 at 22:29
  • 1
    $\begingroup$ Since filter is unstable as mentioned by Matt L. For details please refer to dsp.stackexchange.com/questions/1032/is-ideal-lpf-bibo-unstable $\endgroup$
    – Oliver
    Jun 1, 2015 at 1:44
  • 3
    $\begingroup$ "I don't really believe that this is new to you, is it?" ... no, just a sloppy memory. having $\max \left\{ |y[n]| \right\} < \infty$ is not the same as $\max \left\{ |h[n]| \right\} < \infty$ . i stand (or sit) corrected. $\endgroup$ Jun 1, 2015 at 20:00
0
$\begingroup$

If you mean that the filter impulse response is a sinc function, then you are talking about a function with infinite support, i.e, the length of the filter is infinite. This is impractical to implement, but you can approximate it by truncation. The longer the segment resulting from your truncation, the better the approximation.

Therefore, your statement is rather correct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.