Shannon Capacity of a GSM Channel

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: John (YA)

John has over 15 years of Research and Development experience in the field of Wireless Communications. He has worked for a number of companies around the world including Qualcomm Inc. USA. He has an MS in Electrical Engineering from Virginia Tech USA and has published his work in international journals and conferences.

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