homework4a_solution

# homework4a_solution - Solution : P N = FkT B = 8(1 . 38 ×...

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ECE 5325/6325 Fall 2009: Homework 4(a) Solution (4a) Extend the GSM uplink example from the lecture 7 notes, and consider the GSM downlink, using the same system parameters with one diFerence. The diFerence is that the noise ±gure of a mobile receiver is 8 (linear) instead of the BS receiver, which we said had a noise ±gure of 2. What is EIRP of the base station is required for the identical range of 6 km?
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Unformatted text preview: Solution : P N = FkT B = 8(1 . 38 × 10-23 J/K )(294 K )(200 × 10 3 Hz ) =-141 . 9(dBW), 6.0 dB higher than it was for the uplink. So to achieve the same range as the downlink, we need the transmit power to be 6.0 dB higher as well. Since for the uplink, P t = 0 dBW, now we need P t to be 6 dBW. The EIRP is P t + G t = 6 dBW + 12 dBd = 6 dBW + 14 . 15 dBd = 20 . 15 dBW or about 103 W EIRP....
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## This note was uploaded on 02/25/2011 for the course ECE 5325 taught by Professor Dr.nealpatwari during the Spring '10 term at University of Utah.

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