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.