# A'ment 2 - 1 reaches P r,min P r,min =-29log 10(d min d min...

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MOHAMAD HILAL BIN ZULKIFLI 0813405 Wireless Communication (CSC 4204) Section 1 Individual Assignment 2 Q3.4 a) The number of duplex channel is: S = b t = 20 x 10 6 = 400 Channels b d 2 x 5 x 10 3 b) 400/4 = 100 channels For N = 4, total number of channels available per cell = 400 = 100 Channels 4

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Q3.5 a) N = 3 , So: S = ( ) 4 = 13.5 = 11.3dB < 15dB I 6 N = 4 , So: S = ( ) 4 = 24 = 13.8dB < 15dB I 6 N = 7 , So: S = ( ) 4 = 73.5 = 18.6dB > 15dB I 6 Q3.6 a) Cluster size The average distance between the mobile on the fringe of the serving cell and the first tier of co-channel cells Cell numbers which are located in the first tier The average distance compare to the value of D = QR, where Q = N 3 N = 1 2R 6 3 R N = 3 3 2 R 18 3R N = 4 4R 24 2 3 R N = 7 2 7 R 42 21 R N = 12 4 3 R 72 6R Q3.7 a) d min : received power at BS 1 reaches P
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Unformatted text preview: 1 reaches P r,min P r,min = -29log 10 (d min ) : d min = 10-Pr,min/29 = 1083m d HO : received power at BS 1 reaches P r,HO P r,HO = -29log 10 (d HO ) : d HO = 10-PrHO/29 So, 1083 - 10-PrHO/29 > 4.5 Seconds 22.22 (m/s) S = 15dB, n = 4; N = ? I Distance travel during handoff = d min – d HO > 4.5 Seconds Mobile speed v P HO > -86.8 dBm Thus, Δ = P r,HO – P r,min Δ > 1.2 dB Q3.10 a) b) 90% or 100 Erlangs = 90 Erlangs 90 = UAu = U(0.1) : U = 900 Users c) Offered: 90E ; C = 100. So, 0.03 or 3% e) 24 MHz = 400 Channels 2.30 kHz 400 Channels = 100 Channels / Cells 4 Cells 2500 km 2 = 500 Cells : 500 x 900 Users/Cells = 450,000 users 5 km 2...
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