In the previous post we calculated the Shannon Capacity of a 200kHz GSM channel in AWGN (Additive White Gaussian Noise). However, in a practical scenario the capacity is limited by time varying fading and interference. Let us consider a fading channel with four possible states corresponding to SNRs of 15dB, 10dB, 5dB and 0dB. The probability of these states is 0.50, 0.25, 0.15 and 0.10 respectively. The Shannon Capacity of such a channel is given as (assuming that the channel state information is known at the receiver):

C=Σ B*log2(1+SNRi)* p(SNRi)

C=B*(Σ log2(1+SNRi)* p(SNRi))

C=(200e3)*(log2(1+31.62)*0.50+log2(1+10.00)*0.25+log2(1+3.16)*0.15+log2(1+1)*0.10)

C=757.43kbps

Assuming that only one out of eight time slots is allocated to any user the Shannon Capacity of a GSM channel is reduced to 94.68kbps.

Note: The contribution of the high SNR states dominates the capacity of the channel. The contribution of the four states in terms of percentage capacity is given as 66.38%, 22.84%, 8.14% and 2.64%.

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.