We know that GSM bit rates can vary from a few kbps to a theoretical maximum of 171.2kbps (GPRS). But what is the actual capacity of a 200kHz GSM channel. We can use the Shannon Capacity Theorem to find this capacity.
C=B*log2(1+SNR)
or
C=B*log2(1+P/N)
The noise power can be found by using the following formula:
N=B*No=k*T*B=(1.38e-23)*(293)*(200e3)=8.08e-16W=-121dBm
Let us now assume a signal power 0f -90dBm. This gives us an SNR of 31dB or 1258.9 on linear scale. The capacity can thus be calculated as:
C=200e3*log2(1+1258.9)=2.06Mbps
This is the capacity if all time slots are allocated to a single user. If only one time slot is allocated to a user the capacity would be reduced to 257.48kbps.
Author: Yasir
More than 20 years of experience in various organizations in Pakistan, the USA, and Europe. Worked with the Mobile and Portable Radio Group (MPRG) of Virginia Tech and Qualcomm USA and was one of the first researchers to propose Space Time Block Codes for eight transmit antennas. Have publsihed a book “Recipes for Communication and Signal Processing” through Springer Nature.