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Midterm2007Solutions

# Midterm2007Solutions - SOLUTION to ELEN 4703 Midterm Exam...

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Unformatted text preview: SOLUTION to ELEN 4703 Midterm Exam 3/19/07 Prof. Diament 1. Weakest in-cell power level is P(R), while the co—channel interference level is P(D), where D/R =(3N)“2 and N = 13. So P(D)/P(R) = (D/R)'“ = (3N)‘“’2 ~_~ (39)"“’2 is —28 dB or (n/2)1010g(39) = 28 and n = 5.6/log(39) = 3.52 2. For 1.o.s. 2 Glek2/(4md)2 For 1.o.s. + ground reﬂection: Glczhﬁhzzld“ so we need the ratio 13.0411.Lnd = [ Gage/(41rd)2 1 / [ Gerhlzhf/d‘ ] = 1.de / h12h22(41c)2 expressed in dB or 20 log [ M / 41: hlhz]. At 1.1 GHz, 7. = (3/11) 111, so Pros/Pm in dB = 20 log [ (3/11) (3300) / 41:: (2.2)(13)] = 16.5 dB 3. The GOS spec is met if erlb(12, 7.5) < 0.02 . 4. D = (41t/P)(dP/dQ),,,,.x If the northerly radiation intensity is (dP/dQ)N, the southern one is then 0.5(dP/dQ)N and the total radiated power is P = 0.45 (dP/dQ)N + (0.75)(0.5)(dP/dQ)N = 0.825 (dP/dQ)N. While (dP/dQ)max == (dP/dQ)N, so D = 41d0.825 = 15.23 5. The noise threshold = kTBF + SNR = — 80.0 dBm + SNR = 7 needs to be at most at the level such that z = ('y— u) / G = [ ’y - (~— 45 dBm ) ] / (5.5 dB) is such that Q(z) = 0.98 . From the given data, 2 = — 2.054 so that max SNR = — (2.054)(5.5) + 80 — 45 = 23.7 dB a. ...
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