Why does dsp.CICInterpolator prepend zeros?

2020 2 27
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更新时间：2020 2 27
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The output of dsp.CICInterpolator is exactly what I expect, except with N zeros padded at the start. What is the idea behind these extra zeros?
In the literature, we usually see that the group delay of a CIC filter is N*(R*M-1)/2. However, with the prepended zeros, dsp.CICInterpolator has a group delay of N*(R*M-1)/2 + N.
The example code provided here uses a lot of magic numbers. In particular, it plots y(1,4:end) (i.e. discarding the first 3 samples). This produces the correct alignment because the group delay in this example is N*(R*M-1)/2 + N = 3. However, if we remove the prepended zeros, then we get N*(R*M-1)/2 as I would expect.
I can't find any specific details on the implementation. Can anyone shed any light on this?
Many thanks.

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Harry
Harry 2020-2-27

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I guess this ultimately comes down to how we view the initial state of the integrator stages. If we view the N integrator stages as starting with "valid" input values of 0, then we will always clock out N zeros at the start.
This makes a lot of practical sense, as it's often convenient to keep output data lengths the same as input (or, for an interpolator, exactly R times greater). Therefore, we have to impose some assumption before the start and/or after the end of the input.
Similar assumptions are made by the filter function and I find they work very well in practice. If and when we care about group delay, then this can be managed separately.

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Harry
Harry
2020-2-27

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Harry
Harry
2020-2-27

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