Chapter 4

Chapter 4 - Chapter 4 1. C = B log 2 1 + S N B C = log 2...

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Unformatted text preview: Chapter 4 1. C = B log 2 1 + S N B C = log 2  1+ S N B  1 B As B by LHospitals rule C = S N 1 ln2 2. B = 50 MHz P = 10 mW N = 2 10- 9 W/Hz N = N B C = 6.87 Mbps. P new = 20 mW, C = 13.15 Mbps (for x 1, log(1 + x ) x ) B = 100 MHz, Notice that both the bandwidth and noise power will increase. So C = 7 Mbps. 3. P noise = 0 . 1 mW B = 20 MHz (a) C user 1 base station = 0 . 933 B = 18 . 66 Mbps (b) C user 2 base station = 3 . 46 B = 69 . 2 Mbps 4. (a) Ergodic Capacity (with Rcvr CSI only)= B [ 6 i =1 log 2 (1 + i ) p ( i )] = 2.8831 B = 57.66 Mbps. (b) p out = Pr ( < min ) C o = (1- p out ) B log 2 (1 + min ) For min > 20dB, p out = 1, C o = 0 15dB < min < 20dB, p out = .9, C o = 0.1 B log 2 (1 + min ), max C o at min 20dB. 10dB < min < 15dB, p out = .75, C o = 0.25 B log 2 (1 + min ), max C o at min 15dB. 5dB < min < 10dB, p out = .5, C o = 0.5 B log 2 (1 + min ), max C o at min 10dB. 0dB < min < 5dB, p out = .35, C o = 0.65 B log 2 (1 + min ), max C o at min 5dB.- 5dB < min < 0dB, p out = .1, C o = 0.9 B log 2 (1 + min ), max C o at min 0dB. min <- 5dB, p out = 0, C o = B log 2 (1 + min ), max C o at min -5dB. Plot is shown in Fig. 1. Maximum at min = 10dB, p out =0.5 and C o = 34.59 Mbps. 5. (a) We suppose that all channel states are used 1 = 1 + 4 X i =1 1 i p i = 0 . 8109 1 - 1 4 > true S ( i ) S = 1 - 1 i 0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 2.5 3 3.5 x 10 7 P out Capacity (bps) Figure 1: Capacity vs P out S ( ) S = 1 . 2322 = 1 1 . 2232 = 2 1 . 1332 = 3 . 2332 = 4 C B = 4 X i =1 log 2 i p ( i ) = 5 . 2853 bps/Hz (b) = 1 E [1 / ] = 4 . 2882 S ( i ) S = i S ( ) S = . 0043 = 1 . 0029 = 2 . 4288 = 3 4 . 2882 = 4 C B = log 2 (1 + ) = 2 . 4028 bps/Hz (c) To have p out = 0 . 1 or 0.01 we will have to use all the sub-channels as leaving any of these will result in a p out of at least 0.2 truncated channel power control policy and associated spectral efficiency are the same as the zero-outage case in part b . To have p out that maximizes C with truncated channel inversion, we get max C B = 4 . 1462 bps/Hz p out = 0 . 5 6. (a) SNR recvd = P ( d ) P noise = 10 dB w.p. . 4 5 dB w.p. . 3 dB w.p. . 2- 10 dB w.p. . 1 Assume all channel states are used 1 = 1 + 4 X i =1...
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Chapter 4 - Chapter 4 1. C = B log 2 1 + S N B C = log 2...

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