Shannon Capacity of LTE in noise and fading can be calculated by using the Shannon Capacity formula:
Where B is the bandwidth and SNR is the signal to noise ratio. We know that LTE can have a maximum channel bandwidth of 20 MHz. We now assume that the signal to noise ratio varies between four states with SNRs of 3 dB, 6 dB, 9 dB and 12 dB with each state having equal probability (25%). Plugging in these values in the above formula we get a theoretical capacity for the four states as:
Since the probability of the four states is equal we simply take the average of the four capacities to find the resultant capacity as 55.67 Mbps. This is the theoretical maximum with one transmit and one receive antenna. However if we have four transmit antennas and four receive antennas and employ spatial multiplexing (a MIMO technique) the capacity would increase four fold to 222.68 Mbps.
Peak LTE data rate can also be calculated using LTE packet structure.
1 Time-slot=0.5 ms (i.e 1 Sub-frame = 1 ms)
1 Time-slot=7 Modulation Symbols (when normal CP length is used)
1 Modulation Symbol=6 bits; if 64 QAM is used as modulation scheme
Data rate for a single carrier
=Number of symbols per time slot*Bits per symbol/Duration of a time slot
If 1200 carriers (100 RBs) are used then the aggregated throughput would be=1200*84kbps=100.8Mbps
If 4×4 MIMO is used then the capacity would increase four fold to=403.2Mbps
With 3/4 channel coding the data rate would be reduced to=302.4Mbps
Theoretically it is possible to achieve peak data rates in excess of 200 or even 300 Mbps but practically this would not happen as all the resources (bandwidth, resource blocks) are not allocated to a single user. Secondly in dynamic channel conditions it will not always be possible to use a high order modulation scheme like 64-QAM and a higher code rate of 3/4. In adverse channel conditions you may have to fall back on QPSK modulation scheme and code rate of 1/3.