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about bit overflow in the integrator portion of cic interpolation filter.

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kannan2590

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i am attaching one pdf file based on cic filter.In the bitgrowth section the author has mentioned about the increase in the bits at the output stage .Suppose if i am writing a vhdl code for cic interpolation stage for example let differential delay M=1, Interpolation factor R=8,Input bit size be 23 and number of comb sections and integrator sections be N=9.Then how the output bit size change for each comb section.Can anybody explain me with the above mentioned parameters?
 

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Then how the output bit size change for each comb section.
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A detailed calculation can be found in Hogenauer's classical paper. It says, that the interpolator comb regsisters grow one bit per stage,
 

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