Solution:
1 RB = 12 sub-carriers
1 Sub-carrier = 7 OFDM symbols in
time domain = 7 Resource elements
Thus total OFDM symbols in 1 RB = 12
X 7
With 64 QAM 1 symbol = 6 bits
Thus total bits in 1 RB = 12 X 7 X 6
transmitted in 0.5 msec
With 20 MHZ BW, 100 RB are
transmitted
Thus total bits in 100 RB = 12 X 7 X
6 X 100 transmitted in 0.5 msec
Thus in 1 msec, total bits transmitted = 12 X 7
X 6 X 2 X100
In 1 sec, total transmitted bits = 12
X 7 X 6 X 2 X 100 X 1000 = 100.8 Mbps
This is possible with Single antenna
configuration
But with 4 x 4 MIMO
Maximum
throughout = 4 X 100.8 = 403.2 Mbps