Shannon Capacity – Page 2

June 23, 2011August 8, 2011Capacity

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 […]

June 21, 2011November 15, 2011Capacity, WCDMA

CDMA vs FDMA Capacity (Mbps)

We saw previously that the channel capacity of a WCDMA system is severely limited by the Multiple Access Interference (MAI). Now let us consider the case where the 5MHz channel is divided equally among 20 users such that each user has a bandwidth of 250kHz. Keeping the transmit power for the narrow band signal the same as the wideband signal the signal to noise ratio would improve tremendously (since there is no MAI and the noise power is reduced by a factor of 20). Thus each narrowband channel would have a capacity of 2.99Mbps giving a combined capacity of 59.83Mbps […]