My RX path clock speeds are as below for LTE 10 MHz configuration:
983040000 (BBPLL), 245760000 (ADC_CLK, RHB3), 122880000(RHB2 CLK), 61440000 (RHB1 CLK), 30720000 (RX FIR CLK), 15360000 (RX Sampling rate)
1. Can you please confirm is the below formula correct for RX FIR number of taps calculation?
#of taps = 32 x (ADC_CLK) / (2 x RX FIR CLK)) = 32 x 245.76 / (2 x 30.72) = 128
2. Does this mean 128 taps FIR for each RX channel is available in case of a MIMO / 2RX scenario or has to be divided by 2, as 64 taps for each channel?