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Explorer
Explorer
6,374 Views
Registered: ‎09-28-2012

Question about partial results in multiplications

Hi,

 

I read a tutorial about efficient implement FIR with transposed structure. I don't understand one line:

 

The same data being multiplied by all the coefficients, partial results can be used for many multiplications (regardless of symmetry)

 

I agree that the same data is multiplied by many coefficients. Partial results come from each bit (0 or 1) of every coefficient. Then, it needs the number of FIR taps of MUX to direct the partial results (multiplied by 0 or 1). I don't think the cost of MUX will be obvious lower than the original multiplied by each coef.

 

Maybe my understanding is wrong. Please correct me if you know my question.

Thanks,

 

Here is the tutorial attached. 

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1 Reply
Explorer
Explorer
6,368 Views
Registered: ‎09-28-2012

Re: Question about partial results in multiplications

Excuse me. I realize that I was wrong. It doesn't need MUX. There is only routing between the partial results to each taps. Thanks,

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