Polyphase Filter Length

[pq7]

In the example for ResamplePolyphaseGetFixedFilter

int inRate=16000; // input frequency
int outRate=8000; // output frequency
int history; // half of filter length
char fname[]="filter.flt\0";
// coefficient file name
{
int size,len,height;
FILE *file; short *pFilter;
IppsresamplingPolyphaseFixed_16s *state;
history=(int)(64.0f*0.5*IPP_MAX(1.0,1.0/(double)outRate/(double)inRate))+1;
...

What is the point of this expression on line 10? How does it ever result in anything other than 1.0?

IPP_MAX(1.0,1.0/(double)outRate/(double)inRate)

Since in C, x/y/z == x/(y*z)

[ShivaniK_Intel]

Hi,

Thanks for reaching out to us.

Thanks for bringing up this to our notice. We will get back to you soon.

Regards

Shivani

[Ruqiu_C_Intel]

Hi,

I am afraid not fully understood your question.

For the line 10 in the example, it calculates the filter length for fixed factor resampling. IPP developer reference guide mentioned that for general description of the polyphase resampling, please see "Multirate Digital Signal Processing" by R. Crochine and L. Rabier

Thanks

[pq7]

Thanks Ruqiu. Let me try to simplify the question for you:

Can you provide sample values for inRate and outRate such that the following expression does not give 1.0?

IPP_MAX(1.0,1.0/(double)outRate/(double)inRate)

[Andrey_B_Intel]

Hi

I agree that result is 1.0 and this equation contains ambiguity. 

The "outRate/inRate" ration  in denominator more appropriate.

IPP_MAX(1.0,   1.0/( (double)outRate/(double)inRate)  )

to increase filter length for decimation only.

[Ruqiu_C_Intel · Accepted solution]

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