《wireless communications》习题解答.pdf-资料库

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Chapter 1 1. In case of an accident, there is a high chance of getting lost. The transportation cost is very high each time. However, if the infrastructure is set once, it will be very easy to use it repeatedly. Time for wireless transmission is negligible as signals travel at the speed of light. 2. Advantages of bursty data communication (a) Pulses are made very narrow, so multipaths are resolvable (b) The transmission device needs to be switched on for less time. Disadvantages (a) Bandwidth required is very high (b) Peak transmit power can be very high. 3. Pb = 10−12 2γ = 10−12 γ = 1012 1 2 = 5 × 1011 (very high) 4. Geo: 35,786 Km above earth ⇒ RT T = 2×35786×103 = 0.2386s Meo: 8,000- 20,000 Km above earth ⇒ RT T = 2×8000×103 Leo: 500- 2,000 Km above earth ⇒ RT T = 2×500×103 Only Leo satellites as delay = 3.3ms < 30ms c c c = 0.0033s = 0.0533s 5. 6. optimum no. of data user = d optimum no. of voice user = v Three diﬀerent cases: Case 1: d=0, v=6 ⇒ revenue = 60.80.2 = 0.96 Case 2: d=1, v=3 revenue = [prob. of having one data user]×(revenue of having one data user) + [prob. of having two data user]×(revenue of having two data user) + [prob. of having one voice user]×(revenue of having one voice user) + [prob. of having two voice user]×(revenue of having two voice user) + [prob. of having three or more voice user]×(revenue in this case) ⇒ 0.52 × $1 + 0.52 × $1 + 0.8 × 0.25 × $0.2 + 0.82 × 0.24 × $0.4+ 0.8 × 0.25 × $0.2 − 0.82 × 0.24 × $0.4 × $0.6 2 1 1 − 6 1 6 2 6 1 6 2 ⇒ $1.35 Case 3: d=2, v=0 revenue =2 × 0.5 = $1 So the best case is case 2, which is to allocate 60kHz to data and 60kHz to voice.

7. 8. 1. Hand-oﬀ becomes a big problem. 2. Inter-cell interference is very high and should be mitigated to get reasonable SINR. 3. Infrastructure cost is another problem. 9. Smaller the reuse distance, larger the number of users who can use the same system resource and so 10. capacity (data rate per unit bandwidth) increases. (a) 100 cells, 100 users/cell ⇒ 10,000 users (b) 100 users/cell ⇒ 2500 cells required 100km2 Area/cell = 2500cells ⇒ Area cell = .04km2 (c) From Rappaport or iteration of formula, we get that 100 channels cell ⇒ 89 channels cell @Pb = .02 Each subscriber generates 1 Thus, each cell can support 30 × 89 = 2670 subscribers Macrocell: 2670 × 100 ⇒ 267, 000 subscribers Microcell: 6,675,000 subscribers 30 of an Erlang of traﬃc. (d) Macrocell: $50 M Microcell: $1.25 B (e) Macrocell: $13.35 M/month ⇒ 3.75 months approx 4 months to recoop Microcell: $333.75 M/month ⇒ 3.75 months approx 4 months to recoop 11. One CDPD line : 19.2Kbps average Wimax ∼ 40M bps ∴ number of CDPD lines ∼ 2 × 103

Chapter 2 1. Pr = Pt 10−3 = Pt 10−3 = Pt √ Glλ 4πd λ 4π10 λ 4π100 λ = c/fc = 0.06 2 2 ⇒ Pt = 4.39KW 2 ⇒ Pt = 438.65KW Attenuation is very high for high frequencies 2. d= 100m ht = 10m hr = 2m delay spread = τ = x+x−l c = 1.33× 3. ∆φ = 2π(x+x−l) λ x + x − l = d ht, hr, we need to keep only ﬁrst order terms (ht + hr)2 + d2 − 2 = d ht + hr d   1 2 2(ht + hr) ∼ d = ht + hr d + 1  + 1 + 1 − 2 (ht − hr)2 + d2 ht − hr 2  − d 1 2 2  + 1 ht − hr d ∆φ ∼ 2π λ d 2(ht + hr) d

4. Signal nulls occur when ∆φ = (2n + 1)π 2π λ (ht + hr)2 + d2 − (ht + hr)2 + d2 − = (2n + 1)π = π(2n + 1) 2π(x + x − l) λ (ht − hr)2 + d2 (ht − hr)2 + d2 = λ 2 Let m = (2n + 1) (2n + 1) λ 2 + (ht + hr)2 + d2 = m square both sides (ht + hr)2 + d2 = m2 λ2 4 x = (ht + hr)2, y = (ht − hr)2, x − y = 4hthr (ht − hr)2 + d2 + (ht − hr)2 + d2 + mλ (ht − hr)2 + d2 x = m2 λ2 4 1 mλ ⇒ d = d = + y + mλ y + d2 x − m2 λ2 4 − y 4hthr (2n + 1)λ − (2n + 1)λ 4 2 − y 2 − (ht − hr)2 , n ∈ Z 5. ht = 20m hr = 3m fc = 2GHz λ = c fc dc = 4hthr This is a good radius for suburban cell radius as user density is low so cells can be kept fairly large. Also, shadowing is less due to fewer obstacles. λ = 1600m = 1.6Km = 0.15 6. Think of the building as a plane in R3 The length of the normal to the building from the top of Tx antenna = ht The length of the normal to the building from the top of Rx antenna = hr In this situation the 2 ray model is same as that analyzed in the book. 7. h(t) = α1δ(t − τ) + α2δ(t − (τ + 0.22µs)) Gr = Gl = 1 ht = hr = 8m fc = 900M Hz, λ = c/fc = 1/3 R = −1 delay spread = x + x − l 2 − d ⇒ 2 d 82 + c 2 c ⇒ d = 16.1m = 53.67ns ∴ τ = d c = 0.022 × 10−6s = 0.022 × 10−6s

λ 4π √ Gl l λ 4π √ RGr x + x 2 2 = 2.71 × 10−6 = 1.37 × 10−6 α1 = α2 = 8. A program to plot the ﬁgures is shown below. The power versus distance curves and a plot of the phase diﬀerence between the two paths is shown on the following page. From the plots it can be seen that as Gr (gain of reﬂected path) is decreased, the asymptotic behavior of Pr tends toward d−2 from d−4, which makes sense since the eﬀect of reﬂected path is reduced and it is more like having only a LOS path. Also the variation of power before and around dc is reduced because the strength of the reﬂected path decreases as Gr decreases. Also note that the the received power actually increases with distance up to some point. This is because for very small distances (i.e. d = 1), the reﬂected path is approximately two times the LOS path, making the phase diﬀerence very small. Since R = -1, this causes the two paths to nearly cancel each other out. When the phase diﬀerence becomes 180 degrees, the ﬁrst local maxima is achieved. Additionally, the lengths of both paths are initially dominated by the diﬀerence between the antenna heights (which is 35 meters). Thus, the powers of both paths are roughly constant for small values of d, and the dominant factor is the phase diﬀerence between the paths. clear all; close all; ht=50; hr=15; f=900e6; c=3e8; lambda=c/f; GR=[1,.316,.1,.01]; Gl=1; R=-1; counter=1; figure(1); d=[1:1:100000]; l=(d.^2+(ht-hr)^2).^.5; r=(d.^2+(ht+hr)^2).^.5; phd=2*pi/lambda*(r-1); dc=4*ht*hr/lambda; dnew=[dc:1:100000]; for counter = 1:1:4, Gr=GR(counter); Vec=Gl./l+R*Gr./r.*exp(phd*sqrt(-1)); Pr=(lambda/4/pi)^2*(abs(Vec)).^2; subplot(2,2,counter); plot(10*log10(d),10*log10(Pr)-10*log10(Pr(1))); hold on; plot(10*log10(dnew),-20*log10(dnew)); plot(10*log10(dnew),-40*log10(dnew)); end hold off

9. As indicated in the text, the power fall oﬀ with distance for the 10-ray model is d−2 for relatively large Figure 1: Problem 8 (500/6)2 + 102 = 83.93m distances 10. The delay spread is dictated by the ray reaching last d = Total distance = 6d = 503.59m τ0 = 503.59/c = 1.68µs L.O.S ray d = 500m τ0 = 500/c = 1.67µs ∴ delay spread = 0.01µs 11. fc = 900M Hz λ = 1/3m G = 1 radar cross section 20dBm2 = 10 log1 0σ ⇒ σ = 100 d=1 , s = s = Path loss due to scattering (0.5d)2 + (0.5d)2 = d √ 0.5 = √ 0.5 √ λ Gσ (4π)3/2ss Path loss due to reﬂection (using 2 ray model) Pr Pt = 2 √ R G s + s Pr Pt = 2 2 λ 4π = 0.0224 = −16.498dB = 3.52 × 10−4 = −34.54dB Pscattering = −56.5dB Pref lection = −54.54dB d = 10 Pscattering = −96.5dB Pref lection = −74.54dB d = 100 d = 1000 Pscattering = −136.5dB Pref lection = −94.54dB Notice that scattered rays over long distances result in tremendous path loss γ → simpliﬁed 2 2 → free space d0 d Pr = PtK √ Pr = Pt Gl 4π λ d 12. 0204060−150−100−5005010 * log10(Pr)10 * log10(d)Gr = 10204060−150−100−5005010 * log10(Pr)10 * log10(d)Gr = .3160204060−150−100−5005010 * log10(Pr)10 * log10(d)Gr = .10204060−150−100−5005010 * log10(Pr)10 * log10(d)Gr = .0102040600100200300400Phase (deg)10 * log10(d)

√ 2 ∴ when K = The two models are equal. Gl 4π and d0 = λ 13. Pnoise = −160dBm fc = 1GHz, d0 = 1m, K = (λ/4πd0)2 = 5.7 × 10−4, λ = 0.3, γ = 4 We want SN Rrecd = 20dB = 100 ∵ Noise power is 10−19 d0 d γ 4 P = PtK 10−17 = 10K 0.3 d d ≤ 260.7m 14. d = distance between cells with reused freq p = transmit power of all the mobiles uplink S I ≥ 20dB (a) Min. S/I will result when main user is at A and Interferers are at B √ 2km dB = distance between B and base station dA = distance between A and base station #1 = #1 = √ 2km S I 2 2 = d2 B 2d2 A = P 2P Gλ 4πdA Gλ 4πdB min ⇒ dmin − 1 = 20km ⇒ dmin = 21km since integer number of cells should be accommodated in distance d ⇒ dmin = 22km (dmin − 1)2 4 = = 100 γ S I = min = 1 2 γ γ = 3 P k d0 dA 2P k dmin − 1√ 2 d0 dB = 100 (b) (c) γ ⇒ dmin − 1√ 2 γ γ d0 dA k A = = k Pγ Pu γ d0 d = 1 2 1 2 dB dA S I ⇒ dmin = 9.27 ⇒ with the same argument ⇒ dmin = 10km 0.04 ⇒ dmin = 2.41km ⇒ with the same argument dmin = 4km d0 dB 2k min B (dmin − 1)4 = = 100 15. fc = 900M Hz, ht = 20m, hr = 5m, d = 100m Large urban city P Llargecity = 353.52dB small urban city P Lsmallcity = 325.99dB P Lsuburb = 207.8769dB suburb rural area P Lruralarea/countryside = 70.9278dB As seen , path loss is higher in the presence of multiple reﬂectors, diﬀractors and scatterers

Figure 2: Problem 16 16. Piecewise linear model for 2-path model. See Fig 2 17. Pr = Pt − PL(d) −3 i F AFi −2 FAF =(5,10,6), PAF =(3.4,3.4) j P AFj γ = 10−8 = −8dB PL(d)K d0 d 0 −110 = Pt − 80 − 5 − 10 − 6 − 3.4 − 3.4 ⇒ Pt = −2.2dBm 18. dB = 10 log10 K − 10r log1 0 d (a) Pr Pt using least squares we get 10 log10 K = −29.42dB γ = 4 d0 (b) PL(2Km) = 10 log10 K − 10r log10 d = −161.76dB (c) Receiver power can be assumed to be Gaussian with variance σ2 X ∼ N(0, σ2 ψdB) ψdB P rob(X < −10) = P rob X σψdB −10 σψdB < = 6.512 × 10−4 19. Assume free space path loss parameters fc = 900M Hz → λ = 1/3m σψdB = 6 SN Rrecd = 15dB Pt = 1W 11.21.41.61.822.22.42.62.83−60−50−40−30−20−100Plot for Gr = 0 dBy = −15x+8 y = −35x+56

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《wireless communications》习题解答.pdf

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