ECE-T 512 Winter 2007 : Homework #4Midterm Practice ProblemsMarch 11, 2007125.4.3a)For the Friis’ law to hold, the separation of the two antennas should be at least one Rayleigh distanceapart. Rayleigh distance is calculated as:dR=2L2aλ, whereLais tha largest dimension of the antenna. In our case,La= 15m, andλ= 0.3m(1 GHzcarrier frequency), so Rayleigh distance isdR= 1500m >90m, so for the specified distance, the Friis’ lawcannot be used.b) The effective area of a parabolic antenna isAeff= 0.55A, where A is the physical area.So, in ourcase,Aeff= 0.55πr2= 97.193m2. The antenna gain is then given as:G=4πAeffλ2= 13570.70∼= 41.33dB. If the Friis’ law was used for the link budget, then the received powerwould be calculated as:PRX=PT XGT XGRXλ2(4πd)2⇒PRXdB=PT XdB+GRXdB+GRXdB+ 20 log(lambda4πd⇒PRXdB=PT XdB+ 82.66-71.53⇒PRXdB=PT XdB+ 11.13dB.In other words, the received power is more (!) than the transmitted one.
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