The principle underlying the maximum channel capacity is that the maximum channel capacity is equal to the symbol rate times the number of bits per symbol. Or equivalently, it is given by the bandwidth efficiency in bits/Hz times the bandwidth. The bandwidth efficiency depends on the modulation order used among the QAM modulation M-QAM. It is n=log2 M. So, for 64 QAM, n=6. If the channel is coded with a coding rate say 5/6 as hinted by Emil, then the useful data bits will be only five out of sixth which reduces the bandwidth efficiency to 5 instead of 6 because of the redundancy added to the original data bits. Also you can take the reduction of the bandwidth efficiency because of the signaling. pilot and cyclic redundancy bits. This will render the resultant efficiency smaller than 5.
Then given the bandwidth one can calculate the maximum channel capacity. The key of the calculating the input bit rate is the the bandwidth efficiency.
On the other side, to realize this bandwidth efficiency on has to make the Eb/No greater equal to certain minimum value such that the bit error rate remains lower than that required for quality of service.
To achieve the maximum bit rate one has to make the Eb/No greater than certain minimum value among the communication channel.
For whit ad dative noise channels there is performance curve for the different modulation techniques and order. These curves depict the bit error rate as a function of Eb/No and the modulation type and order. From these curves one can determine the minimum required Eb/No for the required modulation and the bit error rate.
I believe that the LTE uplink supports 64-QAM modulation with 5/6 coding rate. With a sufficiently high SNR you will hence be able to communicate at 5 bit/s/Hz. You can multiply that with the bandwidth (e.g., 20 MHz) to get the peak rate.
The principle underlying the maximum channel capacity is that the maximum channel capacity is equal to the symbol rate times the number of bits per symbol. Or equivalently, it is given by the bandwidth efficiency in bits/Hz times the bandwidth. The bandwidth efficiency depends on the modulation order used among the QAM modulation M-QAM. It is n=log2 M. So, for 64 QAM, n=6. If the channel is coded with a coding rate say 5/6 as hinted by Emil, then the useful data bits will be only five out of sixth which reduces the bandwidth efficiency to 5 instead of 6 because of the redundancy added to the original data bits. Also you can take the reduction of the bandwidth efficiency because of the signaling. pilot and cyclic redundancy bits. This will render the resultant efficiency smaller than 5.
Then given the bandwidth one can calculate the maximum channel capacity. The key of the calculating the input bit rate is the the bandwidth efficiency.
On the other side, to realize this bandwidth efficiency on has to make the Eb/No greater equal to certain minimum value such that the bit error rate remains lower than that required for quality of service.
To achieve the maximum bit rate one has to make the Eb/No greater than certain minimum value among the communication channel.
For whit ad dative noise channels there is performance curve for the different modulation techniques and order. These curves depict the bit error rate as a function of Eb/No and the modulation type and order. From these curves one can determine the minimum required Eb/No for the required modulation and the bit error rate.
You can use the Shannon capacity formula but you should be aware that what do you get is the highest possible theological maximum capacity of the communication channel. There is only few communication systems in which one can approach the Shannon limit of the channel capacity such as in voice band modems for transmitting data in the voice band of the telephone line. One used advanced modulation and channel coding methods to achieve that. Namely one used turbo coding. Practical systems still have real channel capacity which may be much lower than the Shannon limit.