[sdiy] Digital filtering question

[cheater cheater]

On Tue, Aug 24, 2010 at 12:01, Thomas Strathmann <thomas at pdp7.org> wrote:
> On 8/24/10 00:37 , cheater cheater wrote:
>>>
>>> Firstly, it is more helpful to think of the the sinc function as:
>>> sin(pi*x*n) / (pi*x*n)
>>>
>>> Where n is the sample number and x is the "design parameter" that is used
>>> to
>>> dilate the sinc function in the time domain and control the cutoff
>>> frequency
>>> in the frequency domain.
>>>
>>>> My question is: What's the cutoff frequency?
>>>
>>> The cutoff frequency is equal to the Nyquist frequency of the system
>>> (Fs/2)
>>> divided by the number of samples between zero crossings in the sinc
>>> function.   The number of samples between the zero crossings in the sinc
>>> function is equal to 1/x in the above equation.
>>
>> Between which zero crossings? The sin(x)/x function has infinitely many.
>
> It should be the first (two) zero crossing(s) because for a rectangle which
> corresponds to a lowpass filter with cutoff frequency f_c in the frequency
> domain you get a sinc function whose width (distance from zero crossing in
> the fourth quadrant to z.c. in the first quadrant)

OK, I guess what you mean is: the two zero crossings x_1, x_2 such
that |x_1| =< |x_2| < x_n for any other zero crossings x_n.

> is inversely proportional
> to f_c. That's one of the interesting relationships between sinc and
> rectangle mediated by the Fourier transform (the other having to do with the
> area of the sinc between the first zero crossings and the area of the
> rectangle).

That's pretty cool to know! Thanks!

> I hope I got this right (avoiding any derivations on purpose
> because I tend to get the factors wrong anyway when typing into a computer),
> just drinking my morning coffee.
>
>        Thomas
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Cheers,
D.