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Unformatted text preview: Chapter 4 1. C = B log 2 â€¡ 1 + S N B Â· C = log 2  1+ S N B  1 B As B â†’ âˆž by Lâ€™Hospitalâ€™s rule C = S N 1 ln2 2. B = 50 MHz P = 10 mW N = 2 Ã— 10 9 W/Hz N = N B C = 6.87 Mbps. P new = 20 mW, C = 13.15 Mbps (for x Â¿ 1, log(1 + x ) â‰ˆ 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 â†’ base station = 0 . 933 B = 18 . 66 Mbps (b) C user 2 â†’ base station = 3 . 46 B = 69 . 2 Mbps 4. (a) Ergodic Capacity (with Rcvr CSI only)= B [ âˆ‘ 6 i =1 log 2 (1 + Î³ i ) p ( Î³ i )] = 2.8831 Ã— B = 57.66 Mbps. (b) p out = Pr ( Î³ < Î³ min ) C o = (1 p out ) B log 2 (1 + Î³ min ) For Î³ min > 20dB, p out = 1, C o = 0 15dB < Î³ min < 20dB, p out = .9, C o = 0.1 B log 2 (1 + Î³ min ), max C o at Î³ min â‰ˆ 20dB. 10dB < Î³ min < 15dB, p out = .75, C o = 0.25 B log 2 (1 + Î³ min ), max C o at Î³ min â‰ˆ 15dB. 5dB < Î³ min < 10dB, p out = .5, C o = 0.5 B log 2 (1 + Î³ min ), max C o at Î³ min â‰ˆ 10dB. 0dB < Î³ min < 5dB, p out = .35, C o = 0.65 B log 2 (1 + Î³ min ), max C o at Î³ min â‰ˆ 5dB. 5dB < Î³ min < 0dB, p out = .1, C o = 0.9 B log 2 (1 + Î³ min ), max C o at Î³ min â‰ˆ 0dB. Î³ min < 5dB, p out = 0, C o = B log 2 (1 + Î³ min ), max C o at Î³ min â‰ˆ5dB. Plot is shown in Fig. 1. Maximum at Î³ min = 10dB, p out =0.5 and C o = 34.59 Mbps. 5. (a) We suppose that all channel states are used 1 Î³ = 1 + 4 X i =1 1 Î³ i p i â‡’ Î³ = 0 . 8109 1 Î³ 1 Î³ 4 > âˆ´ true S ( Î³ i ) S = 1 Î³ 1 Î³ 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 ( Î³ ) S = 1 . 2322 Î³ = Î³ 1 1 . 2232 Î³ = Î³ 2 1 . 1332 Î³ = Î³ 3 . 2332 Î³ = Î³ 4 C B = 4 X i =1 log 2 Î³ i Î³ Â¶ p ( Î³ i ) = 5 . 2853 bps/Hz (b) Ïƒ = 1 E [1 /Î³ ] = 4 . 2882 S ( Î³ i ) S = Ïƒ Î³ i S ( Î³ ) S = . 0043 Î³ = Î³ 1 . 0029 Î³ = Î³ 2 . 4288 Î³ = Î³ 3 4 . 2882 Î³ = Î³ 4 C B = log 2 (1 + Ïƒ ) = 2 . 4028 bps/Hz (c) To have p out = 0 . 1 or 0.01 we will have to use all the subchannels as leaving any of these will result in a p out of at least 0.2 âˆ´ truncated channel power control policy and associated spectral efficiency are the same as the zerooutage 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 Î³ ( 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 Î³ = 1 + 4 X i =1...
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 Fall '08
 Goldsmith
 Power, Natural logarithm, Logarithm, English Channel, max Co, S1 No. 1 Style

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