Chapter 4

# Chapter 4 - Chapter 4 1 C = B log 2 â€ 1 S N B Â C = log...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 4 1. C = B log 2 â€¡ 1 + S N B Â· C = log 2 &#16; 1+ S N B &#17; 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‰ˆ-5dB. 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 sub-channels 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 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 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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

Chapter 4 - Chapter 4 1 C = B log 2 â€ 1 S N B Â C = log...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online