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Below response shows sinc1,sinc2 and sinc3 filter response. I understand that notches in the frequency response happens at the output data rate of the sigma-delta ADC.

  1. But I am not able to figure out why there is a notch ? All articles I read mentions that notch happens at integral multiples of output data rate but doesn't give explanation.
  2. Also why notch width increases as order of the sinc filter increases ?

Please let me know if I am missing basic concepts that I should read before finding answer to this question.

enter image description here

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  • $\begingroup$ How do you define sync1, sync2 and sync3 ? $\endgroup$ – Hilmar Jun 28 '20 at 11:11
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    $\begingroup$ Mr. Sajjanar's "sinc2" is actually a 3rd-order sinc and his "sinc3" is actually a 5th-order sinc. $\endgroup$ – Richard Lyons Jun 28 '20 at 11:25
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    $\begingroup$ Looks like B-spline rather than sinc. $\endgroup$ – Olli Niemitalo Jun 28 '20 at 12:25
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    $\begingroup$ B-splines in one domain are powers of sinc in the other domain. $\endgroup$ – robert bristow-johnson Jun 29 '20 at 1:48
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Look at the equation for the frequency magnitude response of a sinc filter. It will be of the form: $$\frac{\sin(\alpha)}{ \alpha}$$ Look carefully at the $\alpha$ term and determine for what frequencies the $\sin(\alpha)$ numerator will be equal to zero. The magnitude response notches occur at those frequencies.

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Rick's answer is a good frequency-domain answer: the sinc function has zeros, so the filter's frequency-domain response has zeros.

It may be easier to do this in the time domain, though. A $1^{st}$ order sinc filter in the frequency domain is a "boxcar" filter in the time domain, with an impulse response of $$h(\tau) = \begin{cases} \frac{1}{T} & -\frac{T}{2} < \tau < \frac{T}{2} \\ 0 & \text{otherwise} \end{cases}$$

This is called a "boxcar" filter because if you plot it, it's just a box sitting on the x axis.

This impulse response, in turn, just describes a moving-average filter: $$y(t) = \frac{1}{T}\int_{t - \frac{T}{2}}^{t + \frac{T}{2}} x(\tau) d\tau$$

So why all the hard-seeming math? Because if you take the average of an integer number of cycles of a sine wave, no matter the starting and stopping points, the result is zero. So your filter quite naturally has zeros for any frequency where there are an integer number of cycles over a span of $T$.

The reason the notches get deeper for higher-order filters is that an $n^{th}$-order sinc filter (if you do it the normal way, apparently not an your author has done) can be realized by cascading $n$ moving average filters (i.e., take the output of one and feed it into the input of the next). So when the frequency is close to an integer number of cycles in a span of $T$, each stage attenuates signals at that frequency by the same amount -- the result is that the attenuation is just ever much more so as you increase stages (i.e., filter order).

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