Goldsmith - Wireless Communications - Solutions Manual

Chapter 11. 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. P b = 10 −12 1 2γ = 10 −12 γ = 10 12 2 = 5 ×10 11 (very high) 4. Geo: 35,786 Km above earth ⇒RTT = 2×35786×10 3 c = 0.2386s Meo: 8,000- 20,000 Km above earth ⇒RTT = 2×8000×10 3 c = 0.0533s Leo: 500- 2,000 Km above earth ⇒RTT = 2×500×10 3 c = 0.0033s Only Leo satellites as delay = 3.3ms < 30ms 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) ⇒ 0.5 2 2 1 ×$1 + 0.5 2 ×$1 + 6 1 0.8 ×0.2 5 ×$0.2 + 6 2 0.8 2 ×0.2 4 ×$0.4+ 1 − 6 1 0.8 ×0.2 5 ×$0.2 − 6 2 0.8 2 ×0.2 4 ×$0.4 ×$0.6 ⇒$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 capacity (data rate per unit bandwidth) increases. 10. (a) 100 cells, 100 users/cell ⇒ 10,000 users (b) 100 users/cell ⇒ 2500 cells required 100km 2 Area/cell = 2500cells ⇒ Area cell = .04km 2 (c) From Rappaport or iteration of formula, we get that 100 channels cell ⇒ 89 channels cell @P b = .02 Each subscriber generates 1 30 of an Erlang of traffic. Thus, each cell can support 30 ×89 = 2670 subscribers Macrocell: 2670 ×100 ⇒267, 000 subscribers Microcell: 6,675,000 subscribers (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 W imax ∼ 40Mbps ∴ number of CDPD lines ∼ 2 ×10 3 Chapter 2 1. P r = P t _√ G l λ 4πd _ 2 λ = c/f c = 0.06 10 −3 = P t _ λ 4π10 _ 2 ⇒P t = 4.39KW 10 −3 = P t _ λ 4π100 _ 2 ⇒P t = 438.65KW Attenuation is very high for high frequencies 2. d= 100m h t = 10m h r = 2m delay spread = τ = x+x −l c = 1.33× 3. ∆φ = 2π(x +x−l) λ x +x −l = _ (h t +h r ) 2 +d 2 − _ (h t −h r ) 2 +d 2 = d _ _ ¸ _ h t +h r d _ 2 + 1 − ¸ _ h t −h r d _ 2 + 1 _ _ d h t , h r , we need to keep only first order terms ∼ d _ _ _ _ _ 1 2 ¸ _ h t +h r d _ 2 + 1 _ _ − _ _ 1 2 ¸ _ h t −h r d _ 2 + 1 _ _ _ _ _ = 2(h t +h r ) d ∆φ ∼ 2π λ 2(h t +h r ) d 4. Signal nulls occur when ∆φ = (2n + 1)π 2π(x +x −l) λ = (2n + 1)π 2π λ _ _ (h t +h r ) 2 +d 2 − _ (h t −h r ) 2 +d 2 _ = π(2n + 1) _ (h t +h r ) 2 +d 2 − _ (h t −h r ) 2 +d 2 = λ 2 (2n + 1) Let m = (2n + 1) _ (h t +h r ) 2 +d 2 = m λ 2 + _ (h t −h r ) 2 +d 2 square both sides (h t +h r ) 2 +d 2 = m 2 λ 2 4 + (h t −h r ) 2 +d 2 +mλ _ (h t −h r ) 2 +d 2 x = (h t +h r ) 2 , y = (h t −h r ) 2 , x −y = 4h t h r x = m 2 λ 2 4 +y +mλ _ y +d 2 ⇒d = ¸ _ 1 mλ _ x −m 2 λ 2 4 −y __ 2 −y d = ¸ _ 4h t h r (2n + 1)λ − (2n + 1)λ 4 _ 2 −(h t −h r ) 2 , n ∈ Z 5. h t = 20m h r = 3m f c = 2GHz λ = c f c = 0.15 d c = 4h t h r λ = 1600m = 1.6Km 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. 6. Think of the building as a plane in R 3 The length of the normal to the building from the top of Tx antenna = h t The length of the normal to the building from the top of Rx antenna = h r In this situation the 2 ray model is same as that analyzed in the book. 7. h(t) = α 1 δ(t −τ) +α 2 δ(t −(τ + 0.22µs)) G r = G l = 1 h t = h r = 8m f c = 900MHz, λ = c/f c = 1/3 R = −1 delay spread = x +x −l c = 0.022 ×10 −6 s ⇒ 2 _ 8 2 + _ d 2 _ 2 −d c = 0.022 ×10 −6 s ⇒d = 16.1m ∴ τ = d c = 53.67ns α 1 = _ λ 4π √ G l l _ 2 = 2.71 ×10 −6 α 2 = _ λ 4π √ RG r x +x _ 2 = 1.37 ×10 −6 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 G r (gain of reflected path) is decreased, the asymptotic behavior of P r 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 G r 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 0 20 40 60 −150 −100 −50 0 50 1 0 * l o g 1 0 ( P r ) 10 * log10(d) G r = 1 0 20 40 60 −150 −100 −50 0 50 1 0 * l o g 1 0 ( P r ) 10 * log10(d) G r = .316 0 20 40 60 −150 −100 −50 0 50 1 0 * l o g 1 0 ( P r ) 10 * log10(d) G r = .1 0 20 40 60 −150 −100 −50 0 50 1 0 * l o g 1 0 ( P r ) 10 * log10(d) G r = .01 0 20 40 60 0 100 200 300 400 P h a s e ( d e g ) 10 * log10(d) Figure 1: Problem 8 9. As indicated in the text, the power fall off with distance for the 10-ray model is d −2 for relatively large distances 10. The delay spread is dictated by the ray reaching last d = _ (500/6) 2 + 10 2 = 83.93m Total distance = 6d = 503.59m τ 0 = 503.59/c = 1.68µs L.O.S ray d = 500m τ 0 = 500/c = 1.67µs ∴ delay spread = 0.01µs 11. f c = 900MHz λ = 1/3m G = 1 radar cross section 20dBm 2 = 10 log 1 0σ ⇒σ = 100 d=1 , s = s = _ (0.5d) 2 + (0.5d) 2 = d √ 0.5 = √ 0.5 Path loss due to scattering P r P t = _ λ √ Gσ (4π) 3/2 ss _ 2 = 0.0224 = −16.498dB Path loss due to reflection (using 2 ray model) P r P t = _ R √ G s +s _ 2 _ λ 4π _ 2 = 3.52 ×10 −4 = −34.54dB d = 10 P scattering = −56.5dB P reflection = −54.54dB d = 100 P scattering = −96.5dB P reflection = −74.54dB d = 1000 P scattering = −136.5dB P reflection = −94.54dB Notice that scattered rays over long distances result in tremendous path loss 12. P r = P t K _ d 0 d _ γ →simplified P r = P t _√ G l 4π _ 2 _ λ d _ 2 →free space ∴ when K = _ √ G l 4π _ 2 and d 0 = λ The two models are equal. 13. P noise = −160dBm f c = 1GHz, d 0 = 1m, K = (λ/4πd 0 ) 2 = 5.7 ×10 −4 , λ = 0.3, γ = 4 We want SNR recd = 20dB = 100 ∵ Noise power is 10 −19 P = P t K _ d 0 d _ γ 10 −17 = 10K _ 0.3 d _ 4 d ≤ 260.7m 14. d = distance between cells with reused freq p = transmit power of all the mobiles _ S I _ uplink ≥ 20dB (a) Min. S/I will result when main user is at A and Interferers are at B d A = distance between A and base station #1 = √ 2km d B = distance between B and base station #1 = √ 2km _ S I _ min = P _ Gλ 4πd A _ 2 2P _ Gλ 4πd B _ 2 = d 2 B 2d 2 A = (d min −1) 2 4 = 100 ⇒ d min − 1 = 20km ⇒ d min = 21km since integer number of cells should be accommodated in distance d ⇒d min = 22km (b) P γ P u = k _ d 0 d _ γ ⇒ _ S I _ min = Pk _ d 0 d A _ γ 2Pk _ d 0 d B _ γ = 1 2 _ d B d A _ γ = 1 2 _ d min −1 √ 2 _ γ = 1 2 _ d min −1 √ 2 _ 3 = 100 ⇒d min = 9.27 ⇒ with the same argument ⇒d min = 10km (c) _ S I _ min = k _ d 0 d A _ γ A 2k _ d 0 d B _ γ B = (d min −1) 4 0.04 = 100 ⇒d min = 2.41km ⇒ with the same argument d min = 4km 15. f c = 900MHz, h t = 20m, h r = 5m, d = 100m Large urban city PL largecity = 353.52dB small urban city PL smallcity = 325.99dB suburb PL suburb = 207.8769dB rural area PL ruralarea/countryside = 70.9278dB As seen , path loss is higher in the presence of multiple reflectors, diffractors and scatterers 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 −60 −50 −40 −30 −20 −10 0 Plot for Gr = 0 dB y = −15x+8 y = −35x+56 Figure 2: Problem 16 16. Piecewise linear model for 2-path model. See Fig 2 17. P r = P t −P L (d) − 3 i FAF i − 2 j PAF j FAF =(5,10,6), PAF =(3.4,3.4) P L (d)K _ d 0 d _ γ 0 = 10 −8 = −8dB −110 = P t −80 −5 −10 −6 −3.4 −3.4 ⇒P t = −2.2dBm 18. (a) P r P t dB = 10 log 10 K −10r log 1 0 d d 0 using least squares we get 10 log 10 K = −29.42dB γ = 4 (b) PL(2Km) = 10 log 10 K −10r log 10 d = −161.76dB (c) Receiver power can be assumed to be Gaussian with variance σ 2 ψdB X ∼ N(0, σ 2 ψdB ) Prob(X < −10) = Prob _ X σ ψdB < −10 σ ψdB _ = 6.512 ×10 −4 19. Assume free space path loss parameters f c = 900MHz →λ = 1/3m σ ψdB = 6 SNR recd = 15dB P t = 1W g = 3dB P noise = −40dBm ⇒P recvd = −55dB Suppose we choose a cell of radius d µ(d) = P recvd (due to path loss alone) = P t _√ G l λ 4πd _ 2 = 1.4 ×1− −3 d 2 µ dB = 10 log 1 0(µ(d)) P(P recd (d) > −55) = 0.9 P _ P recd (d) −µ dB σ ψdB > −55 −µ dB 6 _ = 0.9 ⇒ −55 −µ dB 6 = −1.282 ⇒µ dB = −47.308 ⇒µ(d) = 1.86 ×10 −5 ⇒d = 8.68m 20. MATLAB CODE Xc = 20; ss = .01; y = wgn(1,200*(1/ss)); for i = 1:length(y) x(i) = y(i); for j = 1:i x(i) = x(i)+exp(-(i-j)/Xc)*y(j); end end 21. Outage Prob. = Prob. [received power dB Tp dB ] Tp = 10dB (a) outageprob. = 1 −Q _ Tp −µ ψ σ ψ _ = 1 −Q _ −5 8 _ = Q _ 5 8 _ = 26% (b) σ ψ = 4dB, outage prob < 1% ⇒ Q _ Tp −µ ψ σ ψ _ > 99% ⇒ Tp −µ ψ σ ψ < −2.33 ⇒ µ ψ ≥ 19.32dB (c) σ ψ = 12dB, Tp −µ ψ σ ψ < −6.99 ⇒µ ψ ≥ 37.8dB 0 20 40 60 80 100 120 140 160 180 200 −60 −50 −40 −30 −20 −10 0 10 20 White Noise Process Filtered Shadowing Processing Figure 3: Problem 20 (d) For mitigating the effect of shadowing, we can use macroscopic diversity. The idea in macroscopic diversity is to send the message from different base stations to achieve uncorrelated shadowing. In this way the probability of power outage will be less because both base stations are unlikely to experience an outage at the same time, if they are uncorrelated. 22. C = 2 R 2 R _ r=0 rQ _ a +b ln r R _ dr To perform integration by parts, we let du = rdr and v = Q _ a +b ln r R _ . Then u = 1 2 r 2 and dv = ∂ ∂r Q _ a +b ln r R _ = ∂ ∂x Q(x)| x=a+b ln(r/R) ∂ ∂r _ a +b ln r R _ = −1 √ 2π exp(−k 2 /2) b r dr. (1) where k = a +b ln r R . Then we get C = 2 R 2 _ 1 2 r 2 Q _ a +b ln r R _ _ R r=0 + 2 R 2 R _ r=0 1 2 r 2 1 √ 2π exp(−k 2 /2) b r dr (2) = Q(a) + 1 R 2 R _ r=0 r 2 1 √ 2π exp(−k 2 /2) b r dr (3) = Q(a) + 1 R 2 R _ r=0 1 √ 2π R 2 exp _ 2(k −a) b _ exp(−k 2 /2) b r dr (4) = Q(a) + a _ k=−∞ 1 √ 2π exp _ −k 2 2 + 2k b − 2a b _ dk (5) = Q(a) + exp _ −2a b + 2 b 2 _ a _ k=−∞ 1 √ 2π exp _ − 1 2 (k − 2 b ) 2 _ dk (6) = Q(a) + exp _ 2 −2ab b 2 _ _ 1 −Q _ a − 2 b 1 __ (7) = Q(a) + exp _ 2 −2ab b 2 _ Q _ 2 −ab b _ (8) (9) Since Q(−x) = 1 −Q(x). 23. γ = 3 d 0 = 1 k = 0dB σ = 4dB R = 100m P t = 80mWP min = −100dBm = −130dB P γ (R) = P t K _ d 0 d _ γ = 80 ×10 −9 = −70.97dB a = P min −P γ (R) σ = 14.7575 b = 10γ log 10 e σ ψdB = 3.2572 c = Q(a) +e 2−2ab b 2 Q _ 2 −2ab b _ 1 24. γ = 6 σ = 8 P γ (R) = 20 +P min a = −20/8 = −2.5 b = 10 ×6 ×log 10 e 8 = 20.3871 c = 0.9938 25. γ/σ ψ dB 2 4 6 4 0.7728 0.8587 0.8977 8 0.6786 0.7728 0.8255 12 0.6302 0.7170 0.7728 Since P r (r) ≥ P min for all r ≤ R, the probability of non-outage is proportional to Q _ −1 σ _ , and thus decreases as a function of σ. Therefore, C decreases as a function of σ. Since the average power at the boundary of the cell is fixed, C increases with γ, because it forces higher transmit power, hence more received power at r < R. Due to these forces, we have minimum coverage when γ = 2 and σ = 12. By a similar argument, we have maximum coverage when γ = 6 and σ = 4. The same can also be seen from this figure: 2 3 4 5 6 4 6 8 10 12 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 γ σ ψ dB C o v e r a g e Figure 4: Problem 25 The value of coverage for middle point of typical values i.e. γ = 4 and σ = 8 can be seen from the table or the figure to be 0.7728. Chapter 3 1. d = vt r +r = d + 2h 2 d Equivalent low-pass channel impulse response is given by c(τ, t) = α 0 (t)e −jφ 0 (t) δ(τ −τ 0 (t)) +α 1 (t)e −jφ 1 (t) δ(τ −τ 1 (t)) α 0 (t) = λ √ G l 4πd with d = vt φ 0 (t) = 2πf c τ 0 (t) −φ D 0 τ 0 (t) = d/c φ D 0 = _ t 2πf D 0 (t)dt f D 0 (t) = v λ cos θ 0 (t) θ 0 (t) = 0 ∀t α 1 (t) = λR √ G l 4π(r+r ) = λR √ G l 4π(d+ 2h 2 d ) with d = vt φ 1 (t) = 2πf c τ 1 (t) −φ D 1 τ 1 (t) = (r +r )/c = (d + 2h 2 d )/c φ D 1 = _ t 2πf D 1 (t)dt f D 1 (t) = v λ cos θ 1 (t) θ 1 (t) = π −arctan h d/2 ∀t 2. For the 2 ray model: τ 0 = l c τ 1 = x +x c ∴ delay spread(T m ) = x +x −l c = _ (h t +h r ) 2 +d 2 − _ (h t −h r ) 2 +d 2 c when d (h t +h r ) T m = 1 c 2h t h r d h t = 10m, h r = 4m, d = 100m ∴ T m = 2.67 ×10 −9 s 3. Delay for LOS component = τ 0 = 23 ns Delay for First Multipath component = τ 1 = 48 ns Delay for Second Multipath component = τ 2 = 67 ns τ c = Delay for the multipath component to which the demodulator synchronizes. T m = max m τ m −τ c So, when τ c = τ 0 , T m = 44 ns. When τ c = τ 1 , T m = 19 ns. 4. f c = 10 9 Hz τ n,min = 10 3×10 8 s ∴ min f c τ n = 10 10 3×10 8 = 33 1 5. Use CDF strategy. F z (z)= P[x 2 +y 2 ≤ z 2 ] = _ _ x 2 +y 2 ≤z 2 1 2πσ 2 e −(x 2 +y) 2 2σ 2 dxdy = 2π _ 0 z _ 0 1 2πσ 2 e −r 2 2σ 2 rdrdθ = 1 −e −z 2 2σ 2 (z ≥ 0) df z (z) dz = z σ 2 e − z 2 2σ 2 →Rayleigh For Power: F z 2(z)=P[Z ≤ √ z] = 1 −e −z 2 2σ 2 f z (z)= 1 2σ 2 e −z 2σ 2 →Exponential 6. For Rayleigh fading channel Pr(P r < P 0 ) = 1 −e −P 0 /2σ 2 2σ 2 = −80dBm, P 0 = −95dBm, Pr(P r < P 0 ) = 0.0311 2σ 2 = −80dBm, P 0 = −90dBm, Pr(P r < P 0 ) = 0.0952 7. For Rayleigh fading channel P outage = 1 −e −P 0 /2σ 2 0.01 = 1 −e −P 0 /P r ∴ P r = −60dBm 8. 2σ 2 = -80dBm = 10 −11 Target Power P 0 = -80 dBm = 10 −11 Avg. power in LOS component = s 2 = -80dBm = 10 −11 Pr[z 2 ≤ 10 −11 ] = Pr[z ≤ 10 −5 √ 10 ] Let z 0 = 10 −5 √ 10 = _ z 0 0 z σ 2 e − −(z 2 +s 2 ) 2σ 2 I 0 _ zs σ 2 _ dz, z ≥ 0 = 0.3457 To evaluate this, we use Matlab and I 0 (x) = besseli(0,x). Sample Code is given: clear P0 = 1e-11; s2 = 1e-11; sigma2 = (1e-11)/2; z0 = sqrt(1e-11); ss = z0/1e7; z = [0:ss:z0]; pdf = (z/sigma2).*exp(-(z.^2+s2)/(2*sigma2)).*besseli(0,z.*(sqrt(s2)/sigma2)); int_pr = sum(pdf)*ss; 9. CDF of Ricean distribution is F Ricean Z (z) = _ z 0 p Ricean Z (z) where p Ricean Z (z) = 2z(K + 1) Pr exp _ −K − (K + 1)z 2 Pr _ I 0 _ 2z _ K(K + 1) Pr _ , z ≥ 0 For the Nakagami-m approximation to Ricean distribution, we set the Nakagami m parameter to be (K + 1) 2 /(2K + 1). CDF of Nakagami-m distribution is F Nakagami-m Z (z) = _ z 0 p Nakagami-m Z (z) where p Nakagami-m Z (z) = 2m m z 2m−1 Γ(m)Pr m exp _ −mz 2 Pr _ , z ≥ 0, m ≥ 0.5 We need to plot the two CDF curves for K = 1,5,10 and Pr =1 (we can choose any value for Pr as it is the same for both the distributions and our aim is to compare them). Sample code is given: z = [0:0.01:3]; K = 10; m = (K+1)^2/(2*K+1); Pr = 1; pdfR = ((2*z*(K+1))/Pr).*exp(-K-((K+1).*(z.^2))/Pr).*besseli(0,(2*sqrt((K*(K+1))/Pr))*z); pdfN = ((2*m^m*z.^(2*m-1))/(gamma(m)*Pr^m)).*exp(-(m/Pr)*z.^2); for i = 1:length(z) cdfR(i) = 0.01*sum(pdfR(1:i)); cdfN(i) = 0.01*sum(pdfN(1:i)); end plot(z,cdfR); hold on plot(z,cdfN,’b--’); figure; plot(z,pdfR); hold on plot(z,pdfN,’b--’); As seen from the curves, the Nakagami-m approximation becomes better as K increases. Also, for a fixed value of K and x, prob(γ<x) for x large is always greater for the Ricean distribution. This is seen from the tail behavior of the two distributions in their pdf, where the tail of Nakagami-distribution is always above the Ricean distribution. 10. (a) W = average received power Z i = Shadowing over link i P ri = Received power in dBW, which is Gaussian with mean W, variance σ 2 (b) P outage = P [P r,1 < T ∩ P r,2 < T] = P [P r,1 < T] P[P r,2 < T] since Z 1 , Z 2 independent = _ Q _ W −T σ __ 2 = _ Q _ σ __ 2 (c) P out = ∞ _ −∞ P [P r,1 ≤ T , P r,2 < T|Y = y] f y (y) dy P r,1 |Y = y ~ N _ W +by,a 2 σ 2 _ , and [P r,1 |Y = y] ⊥ [P r,2 |Y = y] P outage= ∞ _ −∞ _ Q _ W +by −T aσ __ 2 1 √ 2πσ e − y 2 2σ 2 dy 0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Ricean Nakagami −m 0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 K = 5 Ricean Nakagami−m 0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 K = 10 Ricean Nakagami−m 0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 K = 1 Ricean Nakagami − m 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 K = 5 Ricean Nakagami−m 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 K = 10 Ricean Nakagami−m 3.5 4 4.5 5 5.5 6 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 10 −31 Tail Behavior K = 10 Ricean Nakagami−m Figure 1: The CDF and PDF for K = 1, 5, 10 and the Tail Behavior let y σ = u = ∞ _ −∞ 1 √ 2π _ Q _ W −T +buσ aσ __ 2 e −u 2 2 du = ∞ _ −∞ 1 √ 2π _ Q _ +byσ aσ __ 2 e −y 2 2 dy (d) Let a = b = 1 √ 2 , σ = 8, = 5. With independent fading we get P out = _ Q _ 5 8 __ 2 = 0.0708. With correlated fading we get P out = 0.1316. Conclusion : Independent shadowing is prefarable for diversity. 11. There are many acceptable techniques for this problem. Sample code for both the stochastic tech- nique(preferred) and the Jake’s technique are included. Jakes: Summing of appropriate sine waves %Jake’s Method close all; clear all; %choose N=30 N=30; M=0.5*(N/2-1); Wn(M)=0; beta(M)=0; %We choose 1000 samples/sec ritemp(M,2001)=0; rqtemp(M,2001)=0; rialpha(1,2001)=0; fm=[1 10 100]; Wm=2*pi*fm; for i=1:3 for n=1:1:M for t=0:0.001:2 %Wn(i)=Wm*cos(2*pi*i/N) Wn(n)=Wm(i)*cos(2*pi*n/N); %beta(i)=pi*i/M beta(n)=pi*n/M; %ritemp(i,20001)=2*cos(beta(i))*cos(Wn(i)*t) %rqtemp(i,20001)=2*sin(beta(i))*cos(Wn(i)*t) ritemp(n,1000*t+1)=2*cos(beta(n))*cos(Wn(n)*t); rqtemp(n,1000*t+1)=2*sin(beta(n))*cos(Wn(n)*t); %Because we choose alpha=0,we get sin(alpha)=0 and cos(alpha)=1 %rialpha=(cos(Wm*t)/sqrt(2))*2*cos(alpha)=2*cos(Wm*t)/sqrt(2) %rqalpha=(cos(Wm*t)/sqrt(2))*2*sin(alpha)=0 rialpha(1,1000*t+1)=2*cos(Wm(i)*t)/sqrt(2); end end %summarize ritemp(i) and rialpha ri=sum(ritemp)+rialpha; %summarize rqtemp(i) rq=sum(rqtemp); %r=sqrt(ri^2+rq^2) r=sqrt(ri.^2+rq.^2); %find the envelope average mean=sum(r)/2001; subplot(3,1,i); time=0:0.001:2; %plot the figure and shift the envelope average to 0dB plot(time,(10*log10(r)-10*log10(mean))); titlename=[’fd = ’ int2str(fm(i)) ’ Hz’]; title(titlename); xlabel(’time(second)’); ylabel(’Envelope(dB)’); end 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −10 −5 0 5 fd = 1 Hz E n v e l o p e ( d B ) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −20 −10 0 10 fd = 10 Hz E n v e l o p e ( d B ) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −20 −10 0 10 fd = 100 Hz E n v e l o p e ( d B ) Figure 2: Problem 11 Stochastic: Usually two guassian R.V.’s are filtered by the Doppler Spectrum and summed. Can also just do a Rayleigh distribution with an adequate LPF, although the above technique is prefered. function [Ts, z_dB] = rayleigh_fading(f_D, t, f_s) % % function [Ts, z_dB] = rayleigh_fading(f_D, t, f_s) % generates a Rayleigh fading signal for given Doppler frequency f_D, % during the time perios [0, t], with sampling frequency f_s >= 1000Hz. % % Input(s) % -- f_D : [Hz] [1x1 double] Doppler frequency % -- t : simulation time interval length, time interval [0,t] % -- f_s : [Hz] sampling frequency, set to 1000 if smaller. % Output(s) % -- Ts : [Sec][1xN double] time instances for the Rayleigh signal % -- z_dB : [dB] [1xN double] Rayleigh fading signal % % Required parameters if f_s < 1000; f_s = 1000; % [Hz] Minumum required sampling rate end; N = ceil( t*f_s ); % Number of samples % Ts contains the time instances at which z_dB is specified Ts = linspace(0,t,N); if mod( N, 2) == 1 N = N+1; % Use even number of samples in calculation end; f = linspace(-f_s,f_s,N); % [Hz] Frequency samples used in calculation % Generate complex Gaussian samples with line spectra in frequency domain % Inphase : Gfi_p = randn(2,N/2); CGfi_p = Gfi_p(1,:)+i*Gfi_p(2,:); CGfi = [ conj(fliplr(CGfi_p)) CGfi_p ]; % Quadrature : Gfq_p = randn(2,N/2); CGfq_p = Gfq_p(1,:)+i*Gfq_p(2,:); CGfq = [ conj(fliplr(CGfq_p)) CGfq_p ]; % Generate fading spectrum, this is used to shape the Gaussian line spectra omega_p = 1; % this makes sure that the average received envelop can be 0dB S_r = omega_p/4/pi./(f_D*sqrt(1-(f/f_D).^2)); % Take care of samples outside the Doppler frequency range, let them be 0 idx1 = find(f>=f_D); idx2 = find(f<=-f_D); S_r(idx1) = 0; S_r(idx2) = 0; S_r(idx1(1)) = S_r(idx1(1)-1); S_r(idx2(length(idx2))) = S_r(idx2(length(idx2))+1); % Generate r_I(t) and r_Q(t) using inverse FFT: r_I = N*ifft(CGfi.*sqrt(S_r)); r_Q = -i*N*ifft(CGfq.*sqrt(S_r)); % Finally, generate the Rayleigh distributed signal envelope z = sqrt(abs(r_I).^2+abs(r_Q).^2); z_dB = 20*log10(z); % Return correct number of points z_dB = z_dB(1:length(Ts)); 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −30 −20 −10 0 1 Hz 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −30 −20 −10 0 10 10Hz 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −60 −40 −20 0 20 l00 Hz Figure 3: Problem 11 12. P r = 30dBm f D = 10Hz P 0 = 0dBm, t z = e ρ 2 −1 ρf D √ 2π = 0.0013s P 0 = 15dBm, t z = 0.0072s P 0 = 25dBm, t z = 0.0264s 13. In the reader, we found the level crossing rate below a level by taking an average of the number of times the level was crossed over a large period of time. It is easy to convince that the level crossing rate above a level will have the same expression as eq. 3.44 in reader because to go below a level again, we first need to go above it. There will be some discrepancy at the end points, but for a large enough T it will not effect the result. So we have L Z (above) = L Z (below) = √ 2πf D ρe −ρ 2 And, t Z (above) = p(z > Z) L Z (above) p(z > Z) = 1 −p(z ≤ Z) = 1 −(1 −e −ρ 2 ) = e −ρ 2 t Z (above) = 1 √ 2πf D ρ The values of t Z (above) for f D = 10,50,80 Hz are 0.0224s, 0.0045s, 0.0028s respectively. Notice that as f D increases, t Z (above) reduces. 14. Note: The problem has been solved for T s = 10µs P r = 10dB f D = 80Hz R 1 : −∞≤ γ ≤ −10dB, π 1 = 9.95 ×10 −3 R 2 : −10dB ≤ γ ≤ 0dB, π 2 = 0.085 R 3 : 0dB ≤ γ ≤ 5dB, π 3 = 0.176 R 4 : 5dB ≤ γ ≤ 10dB, π 4 = 0.361 R 5 : 10dB ≤ γ ≤ 15dB, π 5 = 0.325 R 6 : 15dB ≤ γ ≤ 20dB, π 6 = 0.042 R 7 : 20dB ≤ γ ≤ 30dB, π 7 = 4.54 ×10 −5 R 8 : 30dB ≤ γ ≤ ∞, π 8 = 3.72 ×10 −44 N j → level crossing rate at level A j N 1 = 0, ρ = _ 0 10 N 2 = 19.85, ρ = _ 0.1 10 N 3 = 57.38, ρ = _ 1 10 N 4 = 82.19, ρ = _ 10 0.5 10 N 5 = 73.77, ρ = _ 10 10 N 6 = 15.09, ρ = _ 10 1.5 10 N 7 = 0.03, ρ = _ 10 2 10 N 8 = 0, ρ = _ 10 3 10 MATLAB CODE: N = [0 19.85 57.38 82.19 73.77 15.09 .03 0]; Pi = [9.95e-3 .085 .176 .361 .325 .042 4.54e-5 3.72e-44]; T = 10e-3; for i = 1:8 if i == 1 p(i,1) = 0; p(i,2) = (N(i+1)*T)/Pi(i); p(i,3) = 1-(p(i,1)+p(i,2)); elseif i == 8 p(i,1) = (N(i)*T)/Pi(i); p(i,2) = 0; p(i,3) = 1-(p(i,1)+p(i,2)); else p(i,1) = (N(i)*T)/Pi(i); p(i,2) = (N(i+1)*T)/Pi(i); p(i,3) = 1-(p(i,1)+p(i,2)); end end % p = % % 0 0.0199 0.9801 % 0.0023 0.0068 0.9909 % 0.0033 0.0047 0.9921 % 0.0023 0.0020 0.9957 % 0.0023 0.0005 0.9973 % 0.0036 0.0000 0.9964 % 0.0066 0 0.9934 % 0 0 1.0000 15. (a) S(τ, ρ) = _ _ _ α 1 δ(τ) ρ = 70Hz α 2 δ(τ −0.022µsec) ρ = 49.5Hz 0 else The antenna setup is shown in Fig. 15 From the figure, the distance travelled by the LOS ray is d and the distance travelled by the first multipath component is 2 ¸ _ d 2 _ 2 + 64 Given this setup, we can plot the arrival of the LOS ray and the multipath ray that bounces off the ground on a time axis as shown in Fig. 15 So we have 2 ¸ _ d 2 _ 2 + 8 2 −d = 0.022 ×10 −6 3 ×10 8 ⇒4 _ d 2 4 + 8 2 _ = 6.6 2 +d 2 + 2d(6.6) ⇒d = 16.1m f D = v cos(θ)/λ. v = f D λ/ cos(θ). For the LOS ray, θ = 0 and for the multipath component, θ = 45 o . We can use either of these rays and the corresponding f D value to get v = 23.33m/s. (b) d c = 4h t h r λ d c = 768 m. Since d d c , power fall-off is proportional to d −2 . (c) T m = 0.022µs, B −1 = 0.33µs. Since T m B −1 , we have flat fading. 16. (a) Outdoor, since delay spread ≈ 10 µsec. Consider that 10 µsec ⇒ d = ct = 3km difference between length of first and last path (b) Scattering function S(τ, ρ) = F ∆t [A c (τ, ∆t)] = 1 W rect _ 1 W ρ _ for 0 ≤ τ ≤ 10µsec The Scattering function is plotted in Fig. 16 8m 8m d meters Figure 4: Antenna Setting 0.022 us 0 t t0 = (d/3e8) t1 = 2 sqrt[(d/2)^2+8^2]/3e8 t1 t0 Figure 5: Time Axis for Ray Arrival (c) Avg Delay Spread = ∞ 0 τA c (τ)dτ ∞ 0 A c (τ)dτ = 5µsec RMS Delay Spread = ¸ ¸ ¸ ¸ _ ∞ 0 (τ−µ Tm ) 2 A c (τ)dτ ∞ 0 A c (τ)dτ = 2.89µsec Doppler Spread = W 2 = 50 Hz (d) β u > Coherence BW ⇒ Freq. Selective Fading ≈ 1 T m = 10 5 ⇒ β u > = 10 5 kHz Can also use µ Tm or σ Tm instead of T m (e) Rayleigh fading, since receiver power is evenly distributed relative to delay; no dominant LOS path (f) t R = e ρ 2 −1 ρF D √ 2π with ρ = 1, f D = W 2 → t r = .0137 sec (g) Notice that the fade duration never becomes more than twice the average. So, if we choose our data rate such that a single symbol spans the average fade duration, in the worst case two symbols will span the fade duration. So our code can correct for the lost symbols and we will have error-free transmission. So 1 t R = 72.94 symbols/sec 17. (a) T m ≈ .1msec = 100µsec B d ≈ .1Hz Answers based on µ T m or σ T m are fine too. Notice, that based on the choice of either T m , µ T m or σ T m , the remaining answers will be different too. (b) B c ≈ 1 T m = 10kHz ∆f > 10kHz for u 1 ⊥ u 2 (c) (∆t) c = 10s (d) 3kHz < B c ⇒ Flat 30kHz > B c ⇒ Freq. Selective rho tau W/2 -W/2 10 us S(tau,rho) Figure 6: Scattering Function Chapter 4 1. C = Blog 2 _ 1 + S N 0 B _ C = log 2 1+ S N 0 B 1 B As B →∞ by L’Hospital’s rule C = S N 0 1 ln 2 2. B = 50 MHz P = 10 mW N 0 = 2 ×10 −9 W/Hz N = N 0 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 = 0.1mW B = 20MHz (a) C user1→base station = 0.933B = 18.66Mbps (b) C user2→base station = 3.46B = 69.2Mbps 4. (a) Ergodic Capacity (with Rcvr CSI only)= B[ 6 i=1 log 2 (1 +γ i )p(γ i )] = 2.8831×B = 57.66 Mbps. (b) p out = Pr(γ < γ min ) C o = (1-p out )Blog 2 (1 +γ min ) For γ min > 20dB, p out = 1, C o = 0 15dB < γ min < 20dB, p out = .9, C o = 0.1Blog 2 (1 +γ min ), max C o at γ min ≈ 20dB. 10dB < γ min < 15dB, p out = .75, C o = 0.25Blog 2 (1 +γ min ), max C o at γ min ≈ 15dB. 5dB < γ min < 10dB, p out = .5, C o = 0.5Blog 2 (1 +γ min ), max C o at γ min ≈ 10dB. 0dB < γ min < 5dB, p out = .35, C o = 0.65Blog 2 (1 +γ min ), max C o at γ min ≈ 5dB. −5dB < γ min < 0dB, p out = .1, C o = 0.9Blog 2 (1 +γ min ), max C o at γ min ≈ 0dB. γ min < −5dB, p out = 0, C o = Blog 2 (1 +γ min ), max C o at γ min ≈ -5dB. Plot is shown in Fig. 1. Maximum at γ min = 10dB, p out =0.5 and C o = 34.59 Mbps. 5. (a) We suppose that all channel states are used 1 γ 0 = 1 + 4 i=1 1 γ i p i ⇒ γ 0 = 0.8109 1 γ 0 − 1 γ 4 > 0 ∴ true S(γ i ) S = 1 γ 0 − 1 γ i 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 2.5 3 3.5 x 10 7 P out C a p a c i t y ( b p s ) Figure 1: Capacity vs P out S(γ) S = _ ¸ ¸ _ ¸ ¸ _ 1.2322 γ = γ 1 1.2232 γ = γ 2 1.1332 γ = γ 3 0.2332 γ = γ 4 C B = 4 i=1 log 2 _ γ i γ 0 _ p(γ i ) = 5.2853bps/Hz (b) σ = 1 E[1/γ] = 4.2882 S(γ i ) S = σ γ i S(γ) S = _ ¸ ¸ _ ¸ ¸ _ 0.0043 γ = γ 1 0.0029 γ = γ 2 0.4288 γ = γ 3 4.2882 γ = γ 4 C B = log 2 (1 +σ) = 2.4028bps/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 = 4.1462bps/Hz p out = 0.5 6. (a) SNR recvd = P γ (d) P noise = _ ¸ ¸ _ ¸ ¸ _ 10dB w.p. 0.4 5dB w.p. 0.3 0dB w.p. 0.2 −10dB w.p. 0.1 Assume all channel states are used 1 γ 0 = 1 + 4 i=1 1 γ i p i ⇒ γ 0 = 0.4283 > 0.1 ∴ not possible Now assume only the best 3 channel states are used 0.9 γ 0 = 1 + 3 i=1 1 γ i p i ⇒ γ 0 = 0.6742 < 1 ∴ ok! S(γ) S = _ ¸ ¸ _ ¸ ¸ _ 1.3832 γ = γ 1 = 10 1.1670 γ = γ 2 = 3.1623 0.4832 γ = γ 3 = 1 0 γ = γ 4 = 0.1 C/B = 2.3389bps/Hz (b) σ = 0.7491 C/B = log 2 (1 +σ) = 0.8066bps/Hz (c) _ C B _ max = 2.1510bps/Hz obtained by using the best 2 channel states. With p out = 0.1 + 0.2 = 0.3 7. (a) Maximize capacity given by C = max S(γ): S(γ)p(γ)dγ=S _ γ Blog _ 1 + S(γ)γ S _ p(γ)dγ. Construct the Lagrangian function L = _ γ Blog _ 1 + S(γ)γ S _ p(γ)dγ −λ _ S(γ) S p(γ)dγ Taking derivative with respect to S(γ), (refer to discussion section notes) and setting it to zero, we obtain, S(γ) S = _ 1 γ 0 − 1 γ γ ≥ γ 0 0 γ < γ 0 Now, the threshold value must satisfy _ ∞ γ 0 _ 1 γ 0 − 1 γ _ p(γ)dγ = 1 Evaluating this with p(γ) = 1 10 e −γ/10 , we have 1 = 1 10γ 0 _ ∞ γ 0 e −γ/10 dγ − 1 10 _ ∞ γ 0 e −γ/10 γ dγ (1) = 1 γ 0 e −γ 0 /10 − 1 10 _ ∞ γ 0 10 e −γ γ dγ (2) = 1 γ 0 e −γ 0 /10 − 1 10 EXPINT(γ 0 /10) (3) where EXPINT is as defined in matlab. This gives γ 0 = 0.7676. The power adaptation becomes S(γ) S = _ 1 0.7676 − 1 γ γ ≥ 0.7676 0 γ < 0.7676 (b) Capacity can be computed as C/B = 1 10 _ ∞ 0.7676 log (γ/0.7676) e −γ/10 dγ = 2.0649 nats/sec/Hz. Note that I computed all capacites in nats/sec/Hz. This is because I took the natural log. In order to get the capacity values in bits/sec/Hz, the capacity numbers simply need to be divided by natural log of 2. (c) AWGN capacity C/B = log(1 + 10) = 2.3979 nats/sec/Hz. (d) Capacity when only receiver knows γ C/B = 1 10 _ ∞ 0 log (1 +γ) e −γ/10 dγ = 2.0150 nats/sec/Hz. (e) Capacity using channel inversion is ZERO because the channel can not be inverted with finite average power. Threshold for outage probability 0.05 is computed as 1 10 _ ∞ γ0 e −γ/10 dγ = 0.95 which gives γ 0 = 0.5129. This gives us the capacity with truncated channel inversion as C/B = log _ 1 + 1 1 10 _ ∞ γ 0 1 γ e −γ/10 dγ _ ∗ 0.95 (4) = log _ 1 + 1 1 10 EXPINT(γ 0 /10) _ ∗ 0.95 (5) = 1.5463 nats/s/Hz. (6) (f) Channel Mean=-5 dB = 0.3162. So for perfect channel knowledge at transmitter and receiver we compute γ 0 = 0.22765 which gives capacity C/B = 0.36 nats/sec/Hz. With AWGN, C/B = log(1 + 0.3162) = 0.2748 nats/sec/Hz. With channel known only to the receiver C/B = 0.2510 nats/sec/Hz. Capacity with AWGN is always greater than or equal to the capacity when only the reciever knows the channel. This can be shown using Jensen’s inequality. However the capacity when the transmitter knows the channel as well and can adapt its power, can be higher than AWGN capacity specially at low SNR. At low SNR, the knowledge of fading helps to use the low SNR more efficiently. 8. (a) If neither transmitter nor receiver knows when the interferer is on, they must transmit assuming worst case, i.e. as if the interferer was on all the time, C = Blog _ 1 + S N 0 B +I _ = 10.7Kbps. (b) Suppose we transmit at power S 1 when jammer is off and S 2 when jammer is off, C = Bmax _ log _ 1 + S 1 N o B _ 0.75 + log _ 1 + S 2 N o B +I _ 0.25 _ subject to 0.75S 1 + 0.25S 2 = S. This gives S 1 = 12.25mW, S 2 = 3.25mW and C = 53.21Kbps. (c) The jammer should transmit −x(t) to completely cancel off the signal. S = 10mW N 0 = .001 µW/Hz B = 10 MHz Now we compute the SNR’s as: γ j = |H j | 2 S N 0 B This gives: γ 1 = |1| 2 10 −3 0.001×10 −6 10×10 6 = 1, γ 2 = .25, γ 3 = 4, γ 4 = 0.0625 To compute γ 0 we use the power constraint as: j _ 1 γ 0 − 1 γ j _ + = 1 First assume that γ 0 < 0.0625, then we have 4 γ 0 = 1 + _ 1 1 + 1 .25 + 1 4 + 1 .0625 _ ⇒γ 0 = .1798 > 0.0625 So, our assumption was wrong. Now we assume that 0.0625 < γ 0 < .25, then 3 γ 0 = 1 + _ 1 1 + 1 .25 + 1 4 _ ⇒γ 0 = .48 > 0.25 So, our assumption was wrong again. Next we assume that 0.25 < γ 0 < 1, then 2 γ 0 = 1 + _ 1 1 + 1 4 _ ⇒γ 0 = .8889 < 1 This time our assumption was right. So we get that only two sub-bands each of bandwidth 10 MHz are used for transmission and the remaining two with lesser SNR’s are left unused. Now, we can find capacity as: C = j:γ j ≥γ 0 Blog 2 _ γ j γ 0 _ This gives us, C = 23.4 Mbps. 9. Suppose transmit power is P t , interference power is P int and noise power is P noise . In the first strategy C/B = log 2 _ 1 + P t P int +P noise _ In the second strategy C/B = log 2 _ 1 + P t −P int P noise _ Assuming that the transmitter transmits -x[k] added to its message always, power remaining for actual messages is P t −P int The first or second strategy may be better depending on P t P int +P noise ≷ P t −P int P noise ⇒P t −P int −P noise ≷ 0 P t is generally greater than P int +P noise , and so strategy 2 is usually better. 1 0.5 2 0.25 f (in MHz) fc fc+10 fc+20 fc-10 fc-20 H(f) Figure 2: Problem 11 10. We show this for the case of a discrete fading distribution C = Σlog _ 1 + (1 +j) 2 P j N 0 B _ L = i log _ 1 + (1 +j) 2 P j N 0 B _ −d j _ _ j P j −P _ _ ∂L ∂P j = 0 ⇒ (1 +j) 2 P j N 0 B = 1 λ (1 +j) 2 N 0 B −1 letγ j = (1 +j) 2 P N 0 B ⇒ P j P = 1 λP − 1 γ j denote 1 γ 0 = 1 λP ∴ P j P = 1 γ 0 − 1 γ j subject to the constraint ΣP j P = 1 11. S = 10mW N 0 = .001 µW/Hz B = 10 MHz Now we compute the SNR’s as: γ j = |H j | 2 S N 0 B This gives: γ 1 = |1| 2 10 −3 0.001×10 −6 10×10 6 = 1, γ 2 = .25, γ 3 = 4, γ 4 = 0.0625 To compute γ 0 we use the power constraint as: j _ 1 γ 0 − 1 γ j _ + = 1 First assume that γ 0 < 0.0625, then we have 4 γ 0 = 1 + _ 1 1 + 1 .25 + 1 4 + 1 .0625 _ ⇒γ 0 = .1798 > 0.0625 So, our assumption was wrong. Now we assume that 0.0625 < γ 0 < .25, then 3 γ 0 = 1 + _ 1 1 + 1 .25 + 1 4 _ ⇒γ 0 = .48 > 0.25 So, our assumption was wrong again. Next we assume that 0.25 < γ 0 < 1, then 2 γ 0 = 1 + _ 1 1 + 1 4 _ ⇒γ 0 = .8889 < 1 This time our assumption was right. So we get that only two sub-bands each of bandwidth 10 MHz are used for transmission and the remaining two with lesser SNR’s are left unused. Now, we can find capacity as: C = j:γ j ≥γ 0 Blog 2 _ γ j γ 0 _ This gives us, C = 23.4 Mbps. 12. For the case of a discrete number of frequency bands each with a flat frequency response, the problem can be stated as max s.t. i P(f i )≤P i log 2 _ 1 + |H(f i )| 2 P(f i ) N 0 _ denote γ(f i ) = |H(f i )| 2 P(f i ) N 0 L = i log 2 _ 1 +γ(f i ) P(f i ) P _ +λ( P(f i )) denote x i = P(f i ) P , the problem is similar to problem 10 ⇒x i = 1 γ 0 − 1 γ(f i ) where γ 0 is found from the constraints i _ 1 γ 0 − 1 γ(f i ) _ = 1 and 1 γ 0 − 1 γ(f i ) ≥ 0∀i 13. (a) C=13.98Mbps MATLAB Gammabar = [1 .5 .125]; ss = .001; P = 30e-3; N0 = .001e-6; Bc = 4e6; Pnoise = N0*Bc; hsquare = [ss:ss:10*max(Gammabar)]; gamma = hsquare*(P/Pnoise); for i = 1:length(Gammabar) pgamma(i,:) = (1/Gammabar(i))*exp(-hsquare/Gammabar(i)); end gamma0v = [1:.01:2]; for j = 1:length(gamma0v) gamma0 = gamma0v(j); sumP(j) = 0; for i = 1:length(Gammabar) a = gamma.*(gamma>gamma0); [b,c] = max(a>0); gammac = a(find(a)); pgammac = pgamma(i,c:length(gamma)); Pj_by_P = (1/gamma0)-(1./gammac); sumP(j) = sumP(j) + sum(Pj_by_P.*pgammac)*ss; end end [b,c] = min(abs((sumP-1))); gamma0ch = gamma0v(c); C = 0; for i = 1:length(Gammabar) a = gamma.*(gamma>gamma0ch); [b,c] = max(a>0); gammac = a(find(a)); pgammac = pgamma(i,c:length(gamma)); C = C + Bc*ss*sum(log2(gammac/gamma0ch).*pgammac); end (b) C=13.27Mbps MATLAB Gammabarv = [1 .5 .125]; ss = .001; Pt = 30e-3; N0 = .001e-6; Bc = 4e6; Pnoise = N0*Bc; P = Pt/3; for k = 1:length(Gammabarv) Gammabar = Gammabarv(k); hsquare = [ss:ss:10*Gammabar]; gamma = hsquare*(P/Pnoise); pgamma = (1/Gammabar)*exp(-hsquare/Gammabar); gamma0v = [.01:.01:1]; for j = 1:length(gamma0v) gamma0 = gamma0v(j); a = gamma.*(gamma>gamma0); [b,c] = max(a>0); gammac = a(find(a)); pgammac = pgamma(c:length(gamma)); Pj_by_P = (1/gamma0)-(1./gammac); sumP(j) = sum(Pj_by_P.*pgammac)*ss; end [b,c] = min(abs((sumP-1))); gamma0ch = gamma0v(c); a = gamma.*(gamma>gamma0ch); [b,c] = max(a>0); gammac = a(find(a)); pgammac = pgamma(c:length(gamma)); C(k) = Bc*ss*sum(log2(gammac/gamma0ch).*pgammac); end Ctot = sum(C); Chapter 6 1. (a) For sinc pulse, B = 1 2T s ⇒T s = 1 2B = 5 ×10 −5 s (b) SNR = P b N 0 B = 10 Since 4-QAM is multilevel signalling SNR = P b N 0 B = E s N 0 BT s = 2E s N 0 B _ ∵ BT s = 1 2 _ ∴ SNR per symbol = E s N 0 = 5 SNR per bit = E b N 0 = 2.5 (a symbol has 2 bits in 4QAM) (c) SNR per symbol remains the same as before = E s N 0 = 5 SNR per bit is halved as now there are 4 bits in a symbol E b N 0 = 1.25 2. p 0 = 0.3, p 1 = 0.7 (a) P e = Pr(0 detected, 1 sent — 1 sent)p(1 sent) +Pr(1 detected, 0 sent — 0 sent)p(0 sent) = 0.7Q _ d min √ 2N 0 _ + 0.3Q _ d min √ 2N 0 _ = Q _ d min √ 2N 0 _ d min = 2A = Q _ _ ¸ 2A 2 N 0 _ _ (b) p( ˆ m = 0|m = 1)p(m = 1) = p( ˆ m = 1|m = 0)p(m = 0) 0.7Q _ _ A+a _ N 0 2 _ _ = 0.3Q _ _ A−a _ N 0 2 _ _ , a > 0 Solving gives us ’a’ for a given A and N 0 (c) p( ˆ m = 0|m = 1)p(m = 1) = p( ˆ m = 1|m = 0)p(m = 0) 0.7Q _ _ A _ N 0 2 _ _ = 0.3Q _ _ B _ N 0 2 _ _ , a > 0 Clearly A > B, for a given A we can find B (d) Take E b N 0 = A 2 N 0 = 10 In part a) P e = 3.87 ×10 −6 In part b) a=0.0203 P e = 3.53 ×10 −6 In part c) B=0.9587 P e = 5.42 ×10 −6 Clearly part (b) is the best way to decode. MATLAB CODE: A = 1; N0 = .1; a = [0:.00001:1]; t1 = .7*Q(A/sqrt(N0/2)); t2=.3*Q(a/sqrt(N0/2)); diff = abs(t1-t2); [c,d] = min(diff); a(d) c 3. s(t) = ±g(t) cos 2πf c t r = ˆ r cos ∆φ where ˆ r is the signal after the sampler if there was no phase offset. Once again, the threshold that minimizes P e is 0 as (cos ∆φ) acts as a scaling factor for both +1 and -1 levels. P e however increases as numerator is reduced due to multiplication by cos ∆φ P e = Q _ d min cos ∆φ √ 2N 0 _ 4. A 2 c _ T b 0 cos 2 2πf c tdt = A 2 c _ T b 0 1 + cos 4πf c t 2 = A 2 c _ ¸ ¸ ¸ _ T b 2 + sin(4πf c T b ) 8πf c . ¸¸ . →0 as f c 1 _ ¸ ¸ ¸ _ = A 2 c T b 2 = 1 x(t) = 1 +n(t) Let prob 1 sent =p 1 and prob 0 sent =p 0 P e = 1 6 [1.p 1 + 0.p 0 ] + 2 6 [0.p 1 + 0.p 0 ] + 2 6 [0.p 1 + 0.p 0 ] + 1 6 [0.p 1 + 1.p 0 ] = 1 6 [p 1 +p 0 ] = 1 6 (∵ p 1 +p 0 = 1 always ) 5. We will use the approximation P e ∼ (average number of nearest neighbors).Q _ d min √ 2N 0 _ where number of nearest neighbors = total number of points taht share decision boundary (a) 12 inner points have 5 neighbors 4 outer points have 3 neighbors avg number of neighbors = 4.5 P e = 4.5Q _ 2a √ 2N 0 _ (b) 16QAM, P e = 4 _ 1 − 1 4 _ Q _ 2a √ 2N 0 _ = 3Q _ 2a √ 2N 0 _ (c) P e ∼ 2×3+3×2 5 Q _ 2a √ 2N 0 _ = 2.4Q _ 2a √ 2N 0 _ (d) P e ∼ 1×4+4×3+4×2 9 Q _ 3a √ 2N 0 _ = 2.67Q _ 3a √ 2N 0 _ 6. P s,exact = 1 − _ 1 − 2( √ M −1) √ M Q _ _ 3γ s M −1 __ 2 Figure 1: Problem 5 P s,approx = 4( √ M −1) √ M Q _ _ 3γ s M −1 _ approximation is better for high SNRs as then the multiplication factor is not important and P e is dictated by the coefficient of the Q function which are same. MATLAB CODE: gamma_db = [0:.01:25]; gamma = 10.^(gamma_db/10); M = 16; Ps_exact=1-exp(2*log((1-((2*(sqrt(M)-1))/(sqrt(M)))*Q(sqrt((3*gamma)/(M-1)))))); Ps_approx = ((4*(sqrt(M)-1))/sqrt(M))*Q(sqrt((3*gamma)/(M-1))); semilogy(gamma_db, Ps_exact); hold on semilogy(gamma_db,Ps_approx,’b:’); 7. See figure. The approximation error decreases with SNR because the approximate formula is based on nearest neighbor approximation which becomes more realistic at higher SNR. The nearest neighbor bound over-estimates the error rate because it over-counts the probability that the transmitted signal is mistaken for something other than its nearest neighbors. At high SNR, this is very unlikely and this over-counting becomes negligible. 8. (a) I x (a) = _ ∞ 0 e −at 2 x 2 +t 2 dt since the integral converges we can interchange integral and derivative for a¿0 ∂I x (a) ∂a = _ ∞ 0 −te −at 2 x 2 +t 2 dt x 2 I x (a) − ∂I x (a) ∂a = _ ∞ 0 (x 2 +t 2 )e −at 2 x 2 +t 2 dt = _ ∞ 0 e −at 2 dt = 1 2 _ π a 0 5 10 15 20 25 30 10 −300 10 −200 10 −100 10 0 P s Problem 2 − Symbol Error Probability for QPSK Approximation Exact Formula 0 1 2 3 4 5 6 7 8 9 10 10 −3 10 −2 10 −1 10 0 SNR(dB) P s Problem 2 − Symbol Error Probability for QPSK Approximation Exact Formula Figure 2: Problem 7 (b) Let I x (a) = y, we get y −x 2 y = − 1 2 _ π a comparing with y +P(a)y = Q(a) P(a) = −x 2 , Q(a) = − 1 2 _ π a I.F. = e P(u)u = e −x 2 a ∴ e −x 2 a y = _ − 1 2 _ π a e −x 2 u du solving we get y = π 2x e ax 2 erfc(x √ a) (c) erfc(x √ a) = I x (a) 2x π e −ax 2 = 2x π e −ax 2 _ ∞ 0 e −at 2 x 2 +t 2 dt a = 1 erfc(x) = 2x π e −ax 2 _ ∞ 0 e −at 2 x 2 +t 2 dt = 2 π _ π/2 0 e −x 2 /sin 2 θ dθ Q(x) = 1 2 erfc(x/ √ 2) = 1 π _ π/2 0 e −x 2 /2sin 2 θ dθ 9. P = 100W N 0 = 4W, SNR = 25 P e = Q( √ 2γ) = Q( √ 50) = 7.687 ×10 −13 data requires P e ∼ 10 −6 voice requires P e ∼ 10 −3 so it can be used for both. with fading P e = 1 4γ b = 0.01 So the system can’t be used for data at all. It can be used for very low quality voice. 10. T s = 15µsec at 1mph T c = 1 B d = 1 v/λ = 0.74s T s ∴ outage probability is a good measure. at 10 mph T c = 0.074s T s ∴ outage probability is a good measure. at 100 mph T c = 0.0074s = 7400µs > 15µs outage or outage combined with average prob of error can be a good measure. 11. M γ (s) = _ ∞ 0 e sγ p(γ)dγ = _ ∞ 0 e sγ 1 γ e −γ/γ dγ = 1 1 −γs 12. (a) When there is path loss alone, d = √ 100 2 + 500 2 = 100 √ 6 ×10 3 P e = 1 2 e −γ b ⇒γ b = 13.1224 P γ N 0 B = 13.1224 ⇒P γ = 1.3122 ×10 −14 P γ P t = _ √ Gλ 4πd _ 2 ⇒4.8488W (b) x = 1.3122 ×10 −14 = −138.82dB P γ,dB ∼ N(µP γ , 8), σ dB = 8 P(P γ,dB ≥ x) = 0.9 P _ P γ,dB −µP γ 8 ≥ x −µP γ 8 _ = 0.9 ⇒Q _ x −µP γ 8 _ = 0.9 ⇒ x −µP γ 8 = −1.2816 ⇒µP γ = −128.5672dB = 1.39 ×10 −13 13. (a) Law of Cosines: c 2 = a 2 +b 2 −2ab cos C with a,b = √ E s , c = d min , C = Θ = 22.5 c = d min = _ 2E s (1 −cos 22.5) = .39 √ E s Can also use formula from reader (b) P s = α m Q _√ β m γ s _ = 2Q _ _ d min 2 2N o _ = 2Q( √ .076γ s ) α m = 2, β m = .076 (c) P e = ∞ _ 0 P s (γ s )f(γ s )dγ s = ∞ _ 0 α m Q( √ β m γ s )f(γ s )dγ s Using alternative Q form = α m π π 2 _ 0 _ 1 + gγ s (sinφ) 2 _ −1 dφ with g = β m 2 = α m 2 _ 1 − _ gγ s 1+gγ s _ = 1 − _ .038γ s 1+.038γ s = 1 .076γ s , where we have used an integral table to evaluate the integral (d) P d = P s 4 (e) BPSK: P b = 1 4γ b = 10 −3 , ⇒ γ b = 250, 16PSK: From above get γ s = 3289.5 Penalty = 3289.5 250 = 11.2dB Also will accept γ b (16PSK) = 822 ⇒= 5.2dB 14. P b = _ ∞ 0 P b (γ)p(γ)dγ P b (γ) = 1 2 e −γ P b = 1 2 _ ∞ 0 e −γb p γ (γ)dγ = 1 2 M But from 6.65 M γ (s) = _ 1 − sγ m _ −m ∴ P b = 1 2 _ 1 + γ m _ −m For M = 4, γ = 10 P b = 3.33 ×10 −3 15. %Script used to plot the average probability of bit error for BPSK modulation in %Nakagami fading m = 1, 2, 4. %Initializations avg_SNR = [0:0.1:20]; gamma_b_bar = 10.^(avg_SNR/10); m = [1 2 4]; line = [’-k’, ’-r’, ’-b’] for i = 1:size(m,2) for j = 1:size(gamma_b_bar, 2) Pb_bar(i,j) = (1/pi)*quad8(’nakag_MGF’,0,pi/2,[],[],gamma_b_bar(j),m(i),1); end figure(1); semilogy(avg_SNR, Pb_bar(i,:), line(i)); hold on; end xlabel(’Average SNR ( gamma_b ) in dB’); ylabel(’Average bit error probability ( P_b ) ’); title(’Plots of P_b for BPSK modulation in Nagakami fading for m = 1, 2, 4’); legend(’m = 1’, ’m = 2’, ’m = 4’); function out = nakag_MGF(phi, gamma_b_bar, m, g); %This function calculates the m-Nakagami MGF function for the specified values of phi. %phi can be a vector. Gamma_b_bar is the average SNR per bit, m is the Nakagami parameter %and g is given by Pb(gamma_b) = aQ(sqrt(2*g*gamma_b)). out = (1 + gamma_b_bar./(m*(sin(phi).^2)) ).^(-m); SNR = 10dB M BER 1 2.33×10 −2 2 5.53×10 −3 4 1.03×10 −3 16. For DPSK in Rayleigh fading, P b = 1 2γ b ⇒γ b = 500 N o B = 3 ×10 −12 mW ⇒P target = γ b N 0 B = 1.5 ×10 −9 mW = -88.24 dBm Now, consider shadowing: P out = P[P r < P target ] = P[Ψ < P target −P r ] = Φ _ P target −P r σ _ ⇒Φ −1 (.01) = 2.327 = P target −P r σ P r = −74.28 dBm = 3.73 ×10 −8 mW = P t _ λ 4πd _ 2 ⇒ d = 1372.4 m 17. (a) γ 1 = _ 0 w.p. 1/3 30 w.p. 2/3 γ 2 = _ 5 w.p. 1/2 10 w.p. 1/2 In MRC, γ Σ = γ 1 +γ 2 . So, γ Σ = _ ¸ ¸ _ ¸ ¸ _ 5 w.p. 1/6 10 w.p. 1/6 35 w.p. 1/3 40 w.p. 1/3 (b) Optimal Strategy is water-filling with power adaptation: S(γ) S = _ 1 γ 0 − 1 γ , γ ≥ γ 0 0 γ < γ 0 Notice that we will denote γ Σ by γ only hereon to lighten notation. We first assume γ 0 < 5, 4 i=1 _ 1 γ 0 − 1 γ i _ p i = 1 ⇒ 1 γ 0 = 1 + 4 i=1 p i γ i ⇒γ 0 = 0.9365 < 5 So we found the correct value of γ 0 . C = B 4 i=1 log 2 _ γ i γ 0 _ p i C = 451.91 Kbps (c) Without, receiver knowledge, the capacity of the channel is given by: C = B 4 i=1 log 2 (1 +γ i )p i C = 451.66 Kbps Notice that we have denote γ Σ by γ to lighten notation. 18. (a) s(k) = s(k −1) z(k −1) = g k−1 s(k −1) +n(k −1) z(k) = g k s(k) +n(k) From equation 5.63 , the input to the phase comparator is z(k)z (k −1) = g k g ( k −1) s(k)s (k −1) +g k s(k −1)n k−1 + g ( k −1) s (k −1)n k +n k n k−1 but s(k) = s(k −1) s(k)s (k −1) = |s k | 2 = 1 (normalized) (b) ¯ n k = s k−1 n k ¯ n k−1 = s k−1 n k−1 ¯ z = g k g k−1 +g k ¯ n k−1 +g k−1 ¯ n k φ x (s) = p 1 p 2 (s −p 1 )(s −p 2 ) = A s −p 1 + B s −p 2 A = (s −p 1 )φ x (s)| s=p 1 = p 1 p 2 p 1 −p 2 B = (s −p 2 )φ x (s)| s=p 2 = p 1 p 2 p 2 −p 1 (c) Relevant part of the pdf φ x (s) = p 1 p 2 (p 2 −p 1 )(s −p 2 ) ∴ p x (x) = p 1 p 2 (p 2 −p 1 ) L −1 _ 1 (s −p 2 ) _ = p 1 p 2 (p 2 −p 1 ) e p 2 x , x < 0 (d) P b = prob(x < 0) = p 1 p 2 (p 2 −p 1 ) _ 0 −∞ e p 2 x dx = − p 1 p 2 −p 1 (e) p 2 −p 1 = 1 2N 0 [γ b (1 −ρ c ) + 1] + 1 2N 0 [γ b (1 +ρ c ) + 1] = γ b + 1 N 0 [γ b (1 −ρ c ) + 1][γ b (1 +ρ c ) + 1] ∴ P b = γ b (1 −ρ c ) + 1 2(γ b + 1) (f) ρ c = 1 ∴ P b = 1 2(γ b + 1) 19. γ b 0 to 60dB ρ c = J 0 (2πB D T) with B D T = 0.01, 0.001, 0.0001 where J 0 is 0 order Bessel function of 1 st kind. P b = 1 2 _ 1 +γ b (1 −ρ c ) 1 +γ b _ when B D T = 0.01, floor can be seen about γ b = 40dB when B D T = 0.001, floor can be seen about γ b = 60dB when B D T = 0.0001, floor can be seen between γ b = 0 to 60dB 20. Data rate = 40 Kbps Since DQPSK has 2 bits per symbol. ∴ T s = 2 40×10 3 = 5 ×10 −5 sec DQPSK Gaussian Doppler power spectrum, ρ c = e −(πB D T) 2 B D = 80Hz Rician fading K = 2 ρ c = 0.9998 P floor = 1 2 _ 1 − ¸ (ρ c / √ 2) 2 1 −(ρ c / √ 2) 2 _ exp _ − (2 − √ 2)K/2 1 −ρ c / √ 2 _ = 2.138 ×10 −5 21. ISI: Formula based approach: P floor = _ σT m T s _ 2 Since its Rayleigh fading, we can assume that σ T m ≈ µ T m = 100ns P floor ≤ 10 −4 which gives us _ σT m T s _ 2 ≤ 10 −4 T s ≥ σ T m _ P floor = 10µsec So, T s ≥ 10µs. T b ≥ 5µs. R b ≤ 200 Kbps. Thumb - Rule approach: µ t = 100 nsec will determine ISI. As long as T s µ T , ISI will be negligible. Let T s ≥ 10 µ T . Then R ≤ 2bits symbol 1 T s symbols sec = 2Mbps Doppler: B D = 80 Hz P floor = 10 −4 ≥ 1 2 _ _ 1 − ¸ ¸ ¸ _ _ ρ c / √ 2 _ 2 1 − _ ρ c / √ 2 _ 2 _ _ ⇒ρ c ≥ 0.9999 But ρ c for uniform scattering is J 0 (2πB D T s ), so ρ c = J 0 (2πB D T s ) = 1 −(πf D T s ) 2 ≥ 0.9999 ⇒T s ≤ 39.79µs T b ≤ 19.89µs. R b ≥ 50.26 Kbps. Combining the two, we have 50.26 Kbps ≤ R b ≤ 200 Kbps (or 2 Mbps). 22. From figure 6.5 with P b = 10 −3 , d = θ T m /T s , θ T m = 3µs BPSK d = 5 ×10 −2 T s = 60µsec R = 1/T s = 16.7Kbps QPSK d = 4 ×10 −2 T s = 75µsec R = 2/T s = 26.7Kbps MSK d = 3 ×10 −2 T s = 100µsec R = 2/T s = 20Kbps Chapter 7 1. P s = 10 −3 QPSK, P s = 2Q( √ γ s ) ≤ 10 −3 , γ s ≥ γ 0 = 10.8276. P out (γ 0 ) = M i=1 _ 1 −e − γ 0 γ i _ γ 1 = 10, γ 2 = 31.6228, γ 3 = 100. M = 1 P out = _ 1 −e − γ 0 γ 1 _ = 0.6613 M = 2 P out = _ 1 −e − γ 0 γ 1 _ _ 1 −e − γ 0 γ 2 _ = 0.1917 M = 3 P out = _ 1 −e − γ 0 γ 1 _ _ 1 −e − γ 0 γ 2 _ _ 1 −e − γ 0 γ 3 _ = 0.0197 2. p γ Σ (γ) = M γ _ 1 −e −γ/γ ¸ M−1 e −γ/γ γ = 10dB = 10 as we increase M, the mass in the pdf keeps on shifting to higher values of γ and so we have higher values of γ and hence lower probability of error. MATLAB CODE gamma = [0:.1:60]; gamma_bar = 10; M = [1 2 4 8 10]; fori=1:length(M) pgamma(i,:) = (M(i)/gamma_bar)*(1-exp(-gamma/gamma_bar)).^... (M(i)-1).*(exp(-gamma/gamma_bar)); end 3. P b = _ ∞ 0 1 2 e −γ p γ Σ (γ)dγ = _ ∞ 0 1 2 e −γ M γ _ 1 −e −γ/γ _ M−1 e −γ/γ dγ = M 2γ _ ∞ 0 e −(1+1/γ)γ _ 1 −e −γ/γ _ M−1 dγ = M 2γ M−1 n=0 _ M −1 n _ (−1) n e −(1+1/γ)γ dγ = M 2 M−1 n=0 _ M −1 n _ (−1) n 1 1 + n + γ = desired expression 0 10 20 30 40 50 60 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 γ p γ Σ ( γ ) M = 1 M = 2 M = 4 M = 8 M = 10 Figure 1: Problem 2 4. p γ Σ (γ) = _ Pr{γ 2 < γ τ , γ 1 < γ} γ < γ τ Pr{γ τ ≤ γ 1 ≤ γ} + Pr{γ 2 < γ τ , γ 1 < γ} γ > γ τ If the distribution is iid this reduces to p γ Σ (γ) = _ P γ 1 (γ)P γ 2 (γ τ ) γ < γ τ Pr{γ τ ≤ γ 1 ≤ γ} + P γ 1 (γ)P γ 2 (γ τ ) γ > γ τ 5. P b = _ ∞ 0 1 2 e −γ p γ Σ (γ)dγ p γ Σ (γ) = _ _ 1 −e −γ T /γ _ 1 γ e −γ r /γ γ < γ T _ 2 −e −γ T /γ _ 1 γ e −γ r /γ γ > γ T P b = 1 2γ _ 1 −e −γ T /γ _ _ γ T 0 e −γ/γ e −γ dγ + 1 2γ _ 2 −e −γ r /γ _ _ ∞ γ T e −γ/γ e −γ dγ = 1 2(γ + 1) _ 1 −e −γ T /γ + e −γ T e −γ T /γ _ 6. P b P b (10dB) P b (20dB) no diversity 1 2(γ+1) 0.0455 0.0050 SC(M=2) M 2 M−1 m=0 (−1) m   M −1 m   1+m+γ 0.0076 9.7 ×10 −5 SSC 1 2(γ+1) _ 1 −e −γ T /γ + e −γ T e −γ T /γ _ 0.0129 2.7 ×10 −4 As SNR increases SSC approaches SC 7. See MATLAB CODE: gammab_dB = [0:.1:20]; gammab = 10.^(gammab_dB/10); M= 2; 0 2 4 6 8 10 12 14 16 18 20 10 −7 10 −6 10 −5 10 −4 10 −3 10 −2 10 −1 10 0 γ avg P b ,a v g ( D P S K ) M = 2 M = 3 M = 4 Figure 2: Problem 7 for j = 1:length(gammab) Pbs(j) = 0 for m = 0:M-1 f = factorial(M-1)/(factorial(m)*factorial(M-1-m)); Pbs(j) = Pbs(j) + (M/2)*((-1)^m)*f*(1/(1+m+gammab(j))); end end semilogy(gammab_dB,Pbs,’b--’) hold on M = 3; for j = 1:length(gammab) Pbs(j) = 0 for m = 0:M-1 f = factorial(M-1)/(factorial(m)*factorial(M-1-m)); Pbs(j) = Pbs(j) + (M/2)*((-1)^m)*f*(1/(1+m+gammab(j))); end end semilogy(gammab_dB,Pbs,’b-.’); hold on M = 4; for j = 1:length(gammab) Pbs(j) = 0 for m = 0:M-1 f = factorial(M-1)/(factorial(m)*factorial(M-1-m)); Pbs(j) = Pbs(j) + (M/2)*((-1)^m)*f*(1/(1+m+gammab(j))); end end semilogy(gammab_dB,Pbs,’b:’); hold on 8. γ Σ = 1 N 0 _ M i=1 a i γ i _ 2 M i=1 a 2 i ≤ 1 N 0 a 2 i γ 2 i a 2 i = γ 2 i N 0 Where the inequality above follows from Cauchy-Schwartz condition. Equality holds if a i = cγ i where c is a constant 9. (a) γ i = 10 dB = 10, 1 ≤ i ≤ N N = 1, γ = 10, M = 4 P b = .2e −1.5 γ (M−1) = .2e −15/3 = 0.0013. (b) In MRC, γ Σ = γ 1 + γ 2 + . . . + γ N . So γ Σ = 10N P b = .2e −1.5 γ Σ (M−1) = .2e −5N ≤ 10 −6 ⇒N ≥ 2.4412 So, take N = 3, P b = 6.12 ×10 −8 ≤ 10 −6 . 10. Denote N(x) = 1 √ 2π e −x 2 /2 , Q (x) = −N(x) P b = _ ∞ 0 Q( _ 2γ)dP(γ) Q(∞) = 0, P(0) = 0 d dγ Q( _ 2γ) = −N( _ 2γ) √ 2 2 √ γ = − 1 √ 2π e −γ 1 2 √ γ P b = _ ∞ 0 1 √ 2π e −γ 1 2 √ γ P(γ)dγ P(γ) = 1 −e −γ/γ M k=1 (γ/γ) k−1 (k −1)! 1 _ ∞ 0 1 √ 2π e −γ 1 2 √ γ dγ = 1 2 2 _ ∞ 0 1 √ 2π e −γ 1 2 √ γ e −γ/γ M k=1 (γ/γ) k−1 (k −1)! dγ = M k=1 1 (k −1)! _ 1 2 √ π _ ∞ 0 e −γ 1+ 1 γ γ −1/2 _ γ γ _ k−1 dγ _ Denote A = _ 1 + 1 γ _ −1/2 = M−1 m=0 1 m! 1 2 √ pi _ ∞ 0 e −γ/A 2 γ −1/2 _ γ γ _ m dγ let γ/A 2 = u = M−1 m=0 1 m! 1 2 √ pi _ ∞ 0 e −u u −1/2 A _ uA 2 γ _ m A 2 du = A 2 + M−1 m=1 _ 2m−1 m _ A 2m 2 2m A γ m P b = 1 −A 2 − M−1 m=1 _ 2m−1 m _ A 2m+1 2 2m γ m 11. DenoteN(x) = 1 √ 2π e −x 2 /2 Q (x) = [1 −φ(x)] = −N(x) P b = _ ∞ 0 Q( _ 2γ)dP(γ) = _ ∞ 0 1 √ 2π e −γ 1 √ 2γ P(γ)dγ _ ∞ 0 1 √ 2π e −γ 1 √ 2γ dγ = 1 √ π Γ _ 1 2 _ = 1 2 (1) _ ∞ 0 1 √ 2π e −γ 1 √ 2γ e −2γ/γ dγ = 1 2 _ 1 + 2 γ (2) _ ∞ 0 1 √ 2π e −γ 1 √ 2γ _ πγ γ e −γ/γ _ 1 −2Q __ 2γ γ __ dγ = 1 2 √ γ 1 B √ Aγ (3) where A = 1 + 2 γ , B = 1 + 1 γ overall P b = 1 2 _ 1 − ¸ 1 − 1 (1 + γ) 2 _ 12. P b P b (10dB) P b (20dB) no diversity 1 2 _ 1 − _ γ b 1+γ b _ 0.0233 0.0025 two branch SC _ Q( √ 2γ)p γ Σ dγ 0.0030 3.67 ×10 −5 two branch SSC _ Q( √ 2γ)p γ Σ dγ 0.0057 1.186 ×10 −4 two branch EGC _ Q( √ 2γ)p γ Σ dγ 0.0021 2.45 ×10 −5 two branch MRC _ Q( √ 2γ)p γ Σ dγ 0.0016 0.84 ×10 −5 As the branch SNR increases the performance of all diversity combining schemes approaches the same. MATLAB CODE: gammatv = [.01:.1:10]; gammab = 100; gamma = [0:.01:50*gammab]; for i = 1:length(gammatv) gammat = gammatv(i); gamma1 = [0:.01:gammat]; gamma2 = [gammat+.01:.01:50*gammab]; tointeg1 = Q(sqrt(2*gamma1)).*((1/gammab)*(1-exp(-gammat/gammab)).*exp(-gamma1/gammab)); tointeg2 = Q(sqrt(2*gamma2)).*((1/gammab)*(2-exp(-gammat/gammab)).*exp(-gamma2/gammab)); anssum(i) = sum(tointeg1)*.01+sum(tointeg2)*.01; end 13. gammab_dB = [10]; gammab = 10.^(gammab_dB/10); Gamma=sqrt(gammab./(gammab+1)); pb_mrc =(((1-Gamma)/2).^2).*(((1+Gamma)/2).^0+2*((1+Gamma)/2).^1); pb_egc = .5*(1-sqrt(1-(1./(1+gammab)).^2)); 0 2 4 6 8 10 12 14 16 18 20 10 −5 10 −4 10 −3 10 −2 10 −1 γ P b ( γ ) MRC EGC dB penalty ~ .5 dB Figure 3: Problem 13 14. 10 −3 = P b = Q( √ 2γ b ) ⇒4.75, γ = 10 MRC P out = 1 −e −γ 0 /γ M k=1 (γ 0 /γ) k−1 (k−1)! = 0.0827 ECG P out = 1 −e −2γ R − √ πγ R e −γ R (1 −2Q( √ 2γ R )) = 0.1041 > P out,MRC 15. P b,MRC = 0.0016 < 0.0021P b,EGC 16. If each branch has γ = 10dB Rayleigh γ Σ = overall recvd SNR = γ 1 +γ 2 2 ∼ γe −γ/(γ/2) (γ/2) 2 γ ≥ 0 BPSK P b = _ ∞ 0 Q( _ 2γ)p γ Σ dγ = 0.0055 17. p(γ) where _ ∞ 0 p(γ)e −xγ dγ = 0.01γ √ x we will use MGF approach P b = 1 π _ π/2 0 Π 2 i=1 M γ i _ − 1 sin 2 φ _ dφ = 1 π _ π/2 0 (0.01γ sinφ) 2 dφ = (0.01γ) 2 4 = 0.0025 18. P b = _ 1 −π 2 _ 3 2 m=0 _ l + m m __ 1 + π 2 _ m ; π = _ γ 1 + γ Nakagami-2 fading M γ _ − 1 sin 2 φ _ = _ 1 + γ 2 sin 2 φ _ −2 P b = 1 π _ π/2 0 _ M γ _ − 1 sin 2 φ __ 3 dφ, γ = 10 1.5 = 5.12 ×10 −9 MATLAB CODE: gammab = 10^(1.5); Gamma = sqrt(gammab./(gammab+1)); sumf = 0; for m = 0:2 f = factorial(2+m)/(factorial(2)*factorial(m)); sumf = sumf+f*((1+Gamma)/2)^m; end pb_rayleigh = ((1-Gamma)/2)^3*sumf; phi = [0.001:.001:pi/2]; sumvec = (1+(gammab./(2*(sin(phi).^2)))).^(-6); pb_nakagami = (1/pi)*sum(sumvec)*.001; 19. P b = 1 π _ π/2 0 _ 1 + γ 2 sin 2 φ _ −2 _ 1 + γ sin 2 φ _ −1 dφ gammab_dB = [5:.1:20]; gammabvec = 10.^(gammab_dB/10); for i = 1:length(gammabvec) gammab = gammabvec(i); phi = [0.001:.001:pi/2]; sumvec = ((1+(gammab./(2*(sin(phi).^2)))).^(-2)).*((1+... (gammab./(1*(sin(phi).^2)))).^(-1)); pb_nakagami(i) = (1/pi)*sum(sumvec)*.001; end 5 10 15 20 10 −7 10 −6 10 −5 10 −4 10 −3 10 −2 γ avg (dB) P b a v g Figure 4: Problem 19 20. P b = 2 3 Q _ _ 2γ b (3) sin _ π 8 __ α = 2/3, g = 3 sin 2 _ π 8 _ M γ _ − g sin 2 φ _ = _ 1 + gγ sin 2 φ _ −1 P b = α π _ π/2 0 _ 1 + gγ sin 2 φ _ −M dφ MATLAB CODE: M = [1 2 4 8]; alpha = 2/3; g = 3*sin(pi/8)^2; gammab_dB = [5:.1:20]; gammabvec = 10.^(gammab_dB/10); for k = 1:length(M) for i = 1:length(gammabvec) gammab = gammabvec(i); phi = [0.001:.001:pi/2]; sumvec = ((1+((g*gammab)./(1*(sin(phi).^2)))).^(-M(k))); pb_nakagami(k,i) = (alpha/pi)*sum(sumvec)*.001; end end 5 10 15 20 10 −15 10 −10 10 −5 10 0 γ avg (dB) P b a v g Figure 5: Problem 20 21. Q(z) = 1 π _ π/2 0 exp _ − z 2 sin 2 φ _ dφ , z > 0 Q 2 (z) = 1 π _ π/4 0 exp _ − z 2 2 sin 2 φ _ dφ , z > 0 P s (γ s ) = 4 π _ 1 − 1 √ M __ π/2 0 exp _ − gγ s sin 2 φ _ dφ − 4 π _ 1 − 1 √ M 2 __ π/4 0 exp _ − gγ s sin 2 φ _ dφ P s = _ ∞ 0 P s (γ Σ )p γ Σ (γ Σ )dγ Σ = 4 π _ 1 − 1 √ M __ π/2 0 _ ∞ 0 exp _ gγ Σ sin 2 φ _ p γ Σ (γ)dγ Σ dφ − 4 π _ 1 − 1 √ M _ 2 _ π/4 0 _ ∞ 0 exp _ gγ Σ sin 2 φ _ p γ Σ (γ)dγ Σ dφ But γ Σ = γ 1 + γ 2 + . . . + γ M = Σγ i = 4 π _ 1 − 1 √ M __ π/2 0 Π M i=1 M γ i _ − g sin 2 φ _ dφ − 4 π _ 1 − 1 √ M _ 2 _ π/4 0 Π M i=1 M γ i _ − g sin 2 φ _ dφ 22. Rayleigh: M γ s (s) = (1 −sγ s ) −1 Rician: M γ s (s) = 1+k 1+k−sγ s exp _ k s γ s 1+k−sγ s _ MPSK P s = _ (M−1)π/M 0 Mγ s _ − g sin 2 φ _ dφ →no diversity Three branch diversity P s = 1 π _ (M−1)π/M 0 _ 1 + gγ sin 2 φ _ −1 _ (1 + k) sin 2 φ (1 + k) sin 2 φ + gγ s exp _ − kγ s g (1 + k) sin 2 φ + gγ s __ 2 dφ g = sin 2 _ π 16 _ = 0.1670 MQAM: Formula derived in previous problem with g = 1.5 16−1 = 1.5 15 P s = 0.0553 MATLAB CODE: gammab_dB = 10; gammab = 10.^(gammab_dB/10); K = 2; g = sin(pi/16)^2; phi = [0.001:.001:pi*(15/16)]; sumvec=((1+((g*gammab)./(sin(phi).^2))).^(-1)).*((((... (1+K)*sin(phi).^2)./((1+K)*sin(phi).^2+... g*gammab)).*exp(-(K*gammab*g)./((1+K)*sin(phi).^2+g*gammab))).^2); pb_mrc_psk = (1/pi)*sum(sumvec)*.001; g = 1.5/(16-1); phi1 = [0.001:.001:pi/2]; phi2 = [0.001:.001:pi/4]; sumvec1=((1+((g*gammab)./(sin(phi1).^2))).^... (-1)).*(((((1+K)*sin(phi1).^2)./((1+K)*... sin(phi1).^2+g*gammab)).*exp(-(K*gammab*g)./((... 1+K)*sin(phi1).^2+g*gammab))).^2); sumvec2=((1+((g*gammab)./(sin(phi2).^2))).^(-1)).*((((... (1+K)*sin(phi2).^2)./((1+K)*sin(phi2).^2+... g*gammab)).*exp(-(K*gammab*g)./((1+K)*sin(phi2).^2+g*gammab))).^2); pb_mrc_qam = (4/pi)*(1-(1/sqrt(16)))*sum(sumvec1)*.001 - ... (4/pi)*(1-(1/sqrt(16)))^2*sum(sumvec2)*.001; 5 10 15 20 10 −9 10 −8 10 −7 10 −6 10 −5 10 −4 10 −3 10 −2 10 −1 10 0 P s a v g Figure 6: Problem 22 23. MATLAB CODE: M = [1 2 4 8]; alpha = 2/3; g = 1.5/(16-1); gammab_dB = [5:.1:20]; gammabvec = 10.^(gammab_dB/10); for k = 1:length(M) for i = 1:length(gammabvec) gammab = gammabvec(i); phi1 = [0.001:.001:pi/2]; phi2 = [0.001:.001:pi/4]; sumvec1 = ((1+((g*gammab)./(1*(sin(phi1).^2)))).^(-M(k))); sumvec2 = ((1+((g*gammab)./(1*(sin(phi2).^2)))).^(-M(k))); pb_mrc_qam(k,i) = (4/pi)*(1-(1/sqrt(16)))*sum(sumvec1)*.001 - ... (4/pi)*(1-(1/sqrt(16)))^2*sum(sumvec2)*.001; end end Chapter 10 1. (a) (AA H ) T = (A H ) T .A T = (A T ) T A T = AA H ∴ (AA H ) H = AA H For AA H , λ = λ, i.e. eigen-values are real AA H = QΛQ H (b) X H AA H X = (X H A)(X H A) H = X H A ≥ 0 ∴ AA H is positive semidefinite. (c) I M +AA H = I M +QΛQ H = Q(I + Λ)Q H A H positive semidefinite ⇒λ i ≥ 0∀i ∴ 1 +λ i > 0∀i ∴ I M +AA H positive definite (d) det[I M +AA H ] = det[I M +QΛQ H ] = det[Q(I M + Λ M )Q H ] = det[I M + Λ M ] = Π Rank(A) i=1 (1 +λ i ) det[I N +A H A] = det[I N + ¯ QΛ ¯ Q H ] = det[ ¯ Q(I N + Λ N ) ¯ Q H ] = det[I N + Λ N ] = Π Rank(A) i=1 (1 +λ i ) ∵ AA H and A H A have the same eigen-value ∴ det[I M +AA H ] = det[I N +A H A] 2. H = UΣV T U = _ _ −0.4793 0.8685 −0.1298 −0.5896 −0.4272 −0.6855 −0.6508 −0.2513 0.7164 _ _ Σ = _ _ 1.7034 0 0 0 0.7152 0 0 0 0.1302 _ _ V = _ ¸ ¸ _ −0.3458 0.6849 0.4263 −0.5708 0.2191 0.0708 −0.7116 −0.6109 0.0145 −0.2198 0.3311 −0.9017 _ ¸ ¸ _ 3. H = UΣV T Let U = _ _ 1 0 0 1 0 0 _ _ V = _ _ 1 0 0 1 0 0 _ _ Σ = _ 1 0 0 2 _ ∴ H = _ _ 1 0 0 0 2 0 0 0 0 _ _ 4. Check the rank of each matrix rank(H I ) = 3 ∴ multiplexing gain = 3 rank(H 2 ) = 4 ∴ multiplexing gain = 4 5. C = R H i=1 log 2 _ 1 + λ i ρ M t _ Constraint V i = ρ λ i = constant ∴ ∂C ∂λ i = ρ M t ln 2 1 (1 + λ i ρ M t ) − ρ M t ln 2 1 (1 + λ i ρ M t ) = 0 ⇒λ i = λ j ∴ when all R H singular values are equal, this capacity is maximized. 6. (a) Any method to show H ≈ UΛV is acceptable. For example: D = _ _ .13 .08 .11 .05 .09 .14 .23 .13 .10 _ _ where : d ij = ¸ ¸ ¸ H ij −H H ij ¸ ¸ ¸ ×100 (b) precoding filter M = V −1 shaping filter F = U −1 F = _ _ −.5195 −.3460 −.7813 −.0251 −.9078 .4188 −.8540 .2373 .4629 _ _ M = _ _ −.2407 −.8894 .3887 −.4727 −.2423 −.8473 −.8478 .3876 .3622 _ _ Thus Y = F(H)MX + FN = U ∗ UΛVV ∗ X +U ∗ N = ΛX +U ∗ N (c) P i P = 1 γ o − 1 γ i for 1 γ i > 1 γ o , 0 else γ i = λ i 2 P N o B = 94.5 for i = 1, 6.86 for i = 2, .68 for i = 3 Assume γ 2 > γ 0 > γ 3 since γ 3 = .68 is clearly too small for data transmission P i P = 1 ⇒ 2 γ 0 − 1 γ 1 − 1 γ 2 = 1 ⇒γ 0 = 1.73 P 1 P = .5676 P 2 P = .4324 C = B _ log 2 _ 1 +γ 1 P 1 P _ + log 2 _ 1 +γ 2 P 2 P _¸ = 775.9 kbps (d) With equal weight beamforming, the beamforming vector is given by c = 1 √ (3) [1 1 1]. The SNR is then given by: SNR = c H H H Hc N 0 B = (.78)(100) = 78. (1) This gives a capacity of 630.35 kbps. The SNR achieved with beamforming is smaller than the best channel in part (c). If we had chosen c to equal the eigenvector corresponding to the best eigenvalue, then the SNR with beamforming would be equal to the largest SNR in part(c). The beamforming SNR for the given c is greater than the two smallest eigenvalues in part(c) because the channel matrix has one large eigenvalue and two very small eigenvalues. 7. C = max Blog 2 det[I Mγ +HR X H H ] R X : T γ (R X ) = ρ If the channel is known to the transmitter, it will perform an SVD decomposition of H as H = UΣV HR X H H = (UΣV )R X (UΣV ) H By Hadamard’s inequality we have that for A ∈ n×n det(A) ≤ Π n i=1 A ii with equality iff A is diagonal. We choose R X to be diagonal, say = Ω then det(I MR +HR X H H ) = det(I + ΩΣ 2 ) ∴ C = max i ρ i ≤ρ Bσ i log 2 (1 +λ i ρ i ) where √ λ i are the singular values. 8. The capacity of the channel is found by the decomposition of the channel into R H parallel channels, where R H is the rank of the channel matric H. C = max ρ i : i ρ i ≤ρ i Blog 2 (1 +λ i ρ i ) where √ λ i are the R H non-zero singular values of the channel matrix H and ρ is the SNR constraint. γ i = λ i ρ Then the optimal power allocation is given as P i P = _ 1 γ 0 − 1 γ i γ i ≥ γ 0 0 γ i < γ 0 (2) for some cut-off value γ 0 . The resulting capacity is given as C = i:γ i ≥γ 0 Blog 2 (γ i /γ 0 ) For H = _ ¸ ¸ _ 1 1 −1 1 1 1 −1 −1 1 1 1 1 1 1 1 −1 _ ¸ ¸ _ R H = 3, γ 1 = 80, γ 2 = 40, γ 3 = 40. We first assume that γ 0 is less than the minimum γ i which is 40. γ 0 = 3 1 + 3 i=1 1 γ i which gives γ 0 = 2.8236 < min i γ i , hence the assumption was correct. C B = 12.4732 bits/sec/Hz For H = _ ¸ ¸ _ 1 1 1 −1 1 1 −1 1 1 −1 1 1 1 −1 −1 −1 _ ¸ ¸ _ R H = 4, γ 1 = 40, γ 2 = 40, γ 3 = 40, γ 4 = 40. We first assume that γ 0 is less than the minimum γ i which is 40. γ 0 = 4 1 + 4 i=1 1 γ i which gives γ 0 = 3.6780 < min i γ i , hence the assumption was correct. C B = 13.7720 bits/sec/Hz 9. H = _ ¸ ¸ ¸ ¸ _ h 11 . . . h 1M t . . . . . . . . . . . . . . . h M r 1 . . . h M r M t _ ¸ ¸ ¸ ¸ _ M r ×M t Denote G = HH T lim M t →∞ 1 M t G ii = lim M t →∞ 1 M t [h i1 . . . h iM t ] _ ¸ ¸ ¸ ¸ _ h i1 . . . h iM t _ ¸ ¸ ¸ ¸ _ = lim M t →∞ 1 M t M t j=1 h ij 2 = E j h ij 2 = σ 2 = 1 ∀i lim M t →∞,i=j 1 M t G ij = lim M t →∞ 1 M t [h i1 . . . h iM t ] _ ¸ ¸ ¸ ¸ _ h j1 . . . h jM t _ ¸ ¸ ¸ ¸ _ = lim M t →∞ 1 M t M t k=1 h ik h jk = E k h ik h jk = E k (h ik )E k (h jk ) = 0 ∀i, j, i = j ∴ lim M→∞ 1 M HH T = I M ∴ lim M→∞ Blog 2 det _ I M + ρ M HH T _ = Blog 2 det [I M +ρI M ] = Blog 2 [1 +ρ] det I M = MBlog 2 [1 +ρ] 10. We find the capacity by randomly generating 10 3 channel instantiations and then averaging over it. We assume that distribution is uniform over the instantiations. MATLAB CODE clear; clc; Mt = 1; Mr = 1; rho_dB = [0:25]; rho = 10.^(rho_dB/10); for k = 1:length(rho) for i = 1:100 H = wgn(Mr,Mt,0,’dBW’,’complex’); [F, L, M] = svd(H); for j = 1:min(Mt,Mr) sigma(j) = L(j,j); end sigma_used = sigma(1:rank(H)); gamma = rho(k)*sigma_used; %% Now we do water filling\ gammatemp = gamma; gammatemp1 = gammatemp; gamma0 = 1e3; while gamma0 > gammatemp1(length(gammatemp1)); gammatemp1 = gammatemp; gamma0 = length(gammatemp1)/(1+sum(1./gammatemp1)); gammatemp = gammatemp(1:length(gammatemp)-1); end C(i) = sum(log2(gammatemp1./gamma0)); end Cergodic(k) = mean(C); end 0 5 10 15 20 25 0 5 10 15 20 25 ρ (dB) C e r g o d i c M t = M r = 3 M t = 2 M r =3 M t = M r = 2 M t = 2 M r =1 M t = M r = 1 Figure 1: Problem 10 11. We find the capacity by randomly generating 10 4 channel instantiations and then averaging over it. We assume that distribution is uniform over the instantiations. MATLAB CODE clear; clc; Mt = 1; Mr = 1; rho_dB = [0:30]; rho = 10.^(rho_dB/10); for k = 1:length(rho) for i = 1:1000 H = wgn(Mr,Mt,0,’dBW’,’complex’); [F, L, M] = svd(H); for j = 1:min(Mt,Mr) sigma(j) = L(j,j); end sigma_used = sigma(1:rank(H)); gamma = rho(k)*sigma_used; C(i) = sum(log2(1+gamma/Mt)); end Cout(k) = mean(C); pout = sum(C<Cout(k))/length(C); while pout > .01 Cout(k) = Cout(k)-.1; pout = sum(C<Cout(k))/length(C); end if Cout(k)<0; Cout(k) = 0; end end 0 5 10 15 20 25 30 0 5 10 15 20 25 ρ (dB) C o u t a g e M t = M r = 3 M t = 2 M r =3 M t = M r = 2 M t = 2 M r =1 M t = M r = 1 Figure 2: Problem 11 12. P (u n < X) = P _ M r i=1 u i n i < X _ = M r i=1 u i P (n i < X) = P (n i < X) ∴ the statistics of u n are the same as the statistics of each of these elements 13. Σ x = u Hvx = u Hv 2 x 2 = v H H H u H u Hvx 2 = v H H H Hvx 2 = v H Q H Qvx 2 ≤ λ max x 2 with equality when u, v are the principal left and right singular vectors of the channel matrix H ∴ SNR max = λ max x 2 N = λ max ρ 14. H = _ _ 0.1 0.5 0.9 0.3 0.2 0.6 0.1 0.3 0.7 _ _ When both the transmitter and the receiver know the channel, for beamforming, u and v correspond to the principal singular vectors (or the singular vectors corresponding to the maximum singular value of H). Notice that the singular values of H are the square root of the eigen values of HH H (Wishart Matrix). Using Matlab, we get that the maximum singular value of H is 1.4480 and the singular vectors corre- sponding to this value are u opt = _ _ −0.7101 −0.4641 −0.5294 _ _ and v opt = _ _ −0.1818 −0.4190 −0.8896 _ _ It is easy to check that u T opt u opt = 1 and v T opt v opt = 1 and that u T opt ∗ H ∗ v opt = 1.4480 Since, during beamforming from eq. 10.17 in reader, y = (u T Hv)x +u T n and for a given transmit SNR of ρ, the received SNR is given as SNR rcvd = ρ(u T opt Hv opt ) 2 since, u opt has norm 1, noise power is not increased. For, ρ = 1, SNR is simply (1.4480) 2 = 2.0968. When the channel is not known to the transmitter, it allocates equal power to all the antennas and so the precoding vector (or the optimal weights) at the transmitter is given as v 2 = 1 √ 3 _ _ 1 1 1 _ _ Define h = Hv 2 So, eq. 10.17 in the reader can be written as y = (u T h)x +u T n To maximize SNR we need to find a u 2 of norm 1 such that (u h) is maximized. Using Matlab, we get that the maximum singular value of h is 1.2477 and the singular vector corre- sponding to this value is u 2 = _ _ 0.6941 0.5090 0.5090 _ _ It is easy to check that u T 2 u 2 = 1 and that u T 2 ∗ h = 1.2477 Alternatively, from MRC concept we know that: u 2 = h H ||h|| = _ _ 0.6941 0.5090 0.5090 _ _ where ||h|| is the L 2 norm of h. For a given transmit SNR of ρ, the received SNR is given as SNR rcvd = ρ(u T 2 Hv 2 ) 2 since, u 2 has norm 1, noise power is not increased. For, ρ = 1, SNR is simply (1.2477) 2 = 1.5567. 15. (a) ρ = 10 dB = 10 P e = ρ −d So to have P e ≤ 10 −3 , we should have d ≥ 3, or at least d = 3. Solving the equation that relates diversity gain d to multiplexing gain r at high SNR’s we get d = (M r −r)(M t −r) ⇒3 = (8 −r)(4 −r) Solving for r we get r = 3.35 or 8.64 We have that r ≤ min{M r , M t }, so r ≤ 4 and so r = 3.35. But we know that r has to be an integer. So, we take the nearest integer which is smaller than the calculated value of r, which gives us r=3 . No credits for this part: If we are allowed to assume that equations 10.23 and 10.24 hold at finite SNR’s too and we are given that we can use base 2 for logarithms, we can find the data rate as R = r log 2 (ρ) = 9.96 bits/s/Hz (b) With, r = 3, we can find d as d = (M r −r)(M t −r) = (8 −3)(4 −3) = 5 For this value of d, P e = ρ −d = 10 −5 16. According to SVD of h √ λ = 1.242 ∴ C/B = log 2 (1 +λρ) = log 2 (1 + 1.242 2 .10) = 4.038bps/Hz 17. H = _ .3 .5 .7 .2 _ = _ −.5946 .8041 −.8041 .5946 _ _ .8713 0 0 .3328 _ _ −.8507 .5757 −.5757 −.8507 _ P= 10mW N 0 = 10 −9 W/Hz B = 100 KHz (a) When H is known both at the transmitter and at the receiver, the transmitter will use the optimal precoding filter and the receiver will use the optimal shaping filter to decompose the MIMO chan- nel into 2 parallel channels. We can then do water-filling over the two parallel channels available to get capacity. Finding the γ i ’s γ 1 = λ 2 1 P N 0 B = 75.92 γ 2 = λ 2 2 P N 0 B = 11.08 Finding γ 0 Now, we have to find the cutoff value γ 0 . First assume that γ 0 is less than both γ 1 and γ 2 . Then _ 1 γ 0 − 1 γ 1 _ + _ 1 γ 0 − 1 γ 2 _ = 1 ⇒ 2 γ 0 = 1 + 1 γ 1 + 1 γ 2 ⇒γ 0 = 1 1 + 1 γ 1 + 1 γ 2 = 1.81 which is less than both γ 1 and γ 2 values so our assumption was correct. Finding capacity Now we can use the capacity expression as C = 2 i=1 Blog 2 _ γ i γ 0 _ = 800 Kbps (b) Total is Essentially we have two parallel channels after the precoding filter and the shaping filter are used at the transmitter and receiver respectively. M(γ) = 1 +γK S(γ) S Finding K K = −1.5 ln(5P b ) = .283 γ K = γ 0 /K. Finding γ 0 or γ K We now find the cut-off γ 0 . First assume that γ 0 < {γ 1 , γ 2 }. Notice that γ 1 and γ 2 have already been calculated in part (a) as γ 1 = 75.92 and γ 2 = 11.08. _ 1 γ 0 − 1 γ 1 K _ + _ 1 γ 0 − 1 γ 2 K _ = 1 ⇒ 2 γ 0 = 1 + 1 γ 1 K + 1 γ 2 K ⇒γ 0 = 1 1 + 1 K _ 1 γ 1 + 1 γ 2 _ = 1.4649 which is less than both γ 1 and γ 2 values, so our assumption was correct. γ K = γ 0 /K = 5.1742 Finding Rate R Therefore the total rate, R is given as R = B _ log 2 _ γ 1 γ K _ + log 2 _ γ 2 γ K __ ⇒R = B4.97 This gives that R=497.36 Kbps (Obviously less than ergodic capacity). (c) Since now we use beamforming to get diversity only, the transmitter and the receiver use the principal left and right eigen vectors of the Wishart Matrix HH H . Once this is done the SNR at the combiner output is simply λ max ρ, where λ max is the maximum eigen value of the Wishart Matrix HH H and ρ is P N 0 B Finding γ s As given in the question, λ max is 0.7592 and ρ was calculated to be 100. So we get that γ s = 75.92 . Finding P b When using BPSK, γ s = γ b . Now we can use the expression for P b for BPSK P b = Q _ _ 2γ b _ = Q _√ 2 ×75.92 _ = _ 0 Using the approx. given in the Ques. 3.4 ×10 −35 Using Matlab Credit is given for either value. Finding Rate R Since we are using BPSK and are given that B = 1/T b , we get the rate using BPSK to be R=100 Kbps . Comparing with previous part Comparing with part (b), we can see that the rate R decreases by 397.36 Kbps and the P b im- proves as P b is now 3.4 ×10 −35 ∼ 0 whereas earlier it was 10 −3 . (d) Therefore we see that we can tradeoff rate for robustness of the system. If we are willing to de- crease the rate at which we transmit, we can get more diversity advantage i.e. one strong channel which gives a much less value of P b . 18. (a) clear; clc; Mt = 4; Mr = Mt; rho_dB = [0:20]; rho = 10.^(rho_dB/10); for k = 1:length(rho) for i = 1:1000 H = wgn(Mr,Mt,0,’dBW’,’complex’); [F, L, M] = svd(H); for j = 1:min(Mt,Mr) sigma(j) = L(j,j); end sigma_used = sigma(1:rank(H)); gamma = rho(k)*sigma_used; %% Now we do water filling\ gammatemp = gamma; gammatemp1 = gammatemp; gamma0 = 1e3; while gamma0 > gammatemp1(length(gammatemp1)); gammatemp1 = gammatemp; gamma0 = length(gammatemp1)/(1+sum(1./gammatemp1)); gammatemp = gammatemp(1:length(gammatemp)-1); end C(i) = sum(log2(gammatemp1./gamma0)); end Cergodic(k) = mean(C); end (b) clear; clc; Mt = 4; Mr = Mt; rho_dB = [0:20]; rho = 10.^(rho_dB/10); for k = 1:length(rho) for i = 1:1000 H = wgn(Mr,Mt,0,’dBW’,’complex’); [F, L, M] = svd(H); for j = 1:min(Mt,Mr) sigma(j) = L(j,j); end sigma_used = sigma(1:rank(H)); gamma = rho(k)*sigma_used; C(i) = sum(log2(1+gamma/Mt)); end Cout(k) = mean(C); end 19. using Matlab we get C out = 7.8320 MATLAB CODE clear; clc; Mt = 4; Mr = Mt; rho_dB = 10; rho = 10.^(rho_dB/10); for k = 1:length(rho) for i = 1:1000 0 2 4 6 8 10 12 14 16 18 20 0 5 10 15 20 25 ρ (dB) C C ergodic M t = M r = 1 C out M t = M r = 1 C ergodic M t = M r = 4 C out M t = M r = 4 Figure 3: Problem 18 H = wgn(Mr,Mt,0,’dBW’,’complex’); [F, L, M] = svd(H); for j = 1:min(Mt,Mr) sigma(j) = L(j,j); end sigma_used = sigma(1:rank(H)); gamma = rho(k)*sigma_used; C(i) = sum(log2(1+gamma/Mt)); end Cout(k) = mean(C); pout = sum(C<Cout(k))/length(C); while pout > .1 Cout(k) = Cout(k)-.01; pout = sum(C<Cout(k))/length(C); end if Cout(k)<0; Cout(k) = 0; end end 20. As µ increases, the span of cdf becomes narrower and so capacity starts converging to a single number. MATLAB CODE clear; clc; Mt = 8; Mr = Mt; rho_dB = 10; rho = 10.^(rho_dB/10); for i = 1:1000 H = wgn(Mr,Mt,0,’dBW’,’complex’); [F, L, M] = svd(H); for j = 1:min(Mt,Mr) sigma(j) = L(j,j); end sigma_used = sigma(1:rank(H)); gamma = rho*sigma_used; C(i) = sum(log2(1+gamma/Mt)); end [f,x] = ecdf(C); 6 7 8 9 10 11 12 13 14 15 16 17 18 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Empirical CDF’s of Capacity w/o Tx CSI x F X ( x ) M = 4 M = 6 M = 8 Figure 4: Problem 20 Chapter 11 1. See Fig 1 fc = 100 MHz fc+B fc-B 2B = 100 KHz Figure 1: Band of interest. B = 50 KHz, f c = 100 MHz H eq (f) = 1 H(f) = f Noise PSD = N 0 W/Hz. Using this we get Noise Power = _ f c +B f c −B N 0 |H eq (f)| 2 df (1) = N 0 _ f c +B f c −B f 2 df (2) = N 0 f 3 3 _ (f c +B) (f c −B) (3) = N 0 3 (f c + B) 3 −(f c −B) 3 (4) = 10 21 N 0 W (5) Without the equalizer, the noise power will be 2BN 0 = 10 5 N 0 W. As seen from the noise power values, there is tremendous noise enhancement and so the equalizer will not improve system performance. 2. (a) For the first channel: ISI power over a bit time = A 2 T b /T b = A 2 For the 2nd channel: ISI power over a bit time = A 2 T b ∞ n=1 _ (n+1)T b nT b e −t/T m dt = 2e −1/2 A 2 A S(t) T m /2 1 h T m 1 (t) h 2 (t) t t t Figure 2: Problem 2a (b) No ISI: pulse interval = 11/2µs = 5.5µs ∴ Data rate = 1/5.5µs = 181.8Kbps If baseband signal =100KHz: pulse width = 10µs Data rate = 2/10µs + 10µs = 100Kbps 3. (a) h(t) = _ e − t τ t ≥ 0 0 o.w. (6) τ = 6 µ sec H eq (f) = 1 H(f) H(f) = _ ∞ 0 e − t τ e −j2πft dt (7) = 1 1 τ + j2πf (8) Hence, H eq (f) = 1 τ + j2πf (b) SNR eq SNR ISI = B −B S x (f)|H(f)| 2 |H eq (f)| 2 df B −B N 0 |H eq (f)| 2 df B −B S x (f)|H(f)| 2 df 2BN 0 1 h 10us 1 (t) t 1 X 1us (t) t h 12us 1 (t)*X(t) t 1 Figure 3: Problem 2b Assume S x (f) = S, −B ≤ f ≤ B ⇒ 2BS N 0 2B τ 2 + 8π 2 3 B 3 S B −B |H(f)| 2 df 2BN 0 = 2B 1 τ 2 + 4π 2 3 B 2 1.617 ×10 −6 = 0.9364 = −0.28 dB (c) h[n] = 1 + e −T s τ δ [n −1] + e −2T s τ δ [n −2] + ... H(z) = 1 +e −T s τ z −1 + e −2T s τ z −2 + e −3T s τ z −3 + ... = ∞ n=0 _ e − T s τ z −1 _ n = z z −e − T s τ = 1 1 −e − T s τ z −1 ⇒ H eq (z) = 1 H(z)+N 0 . Now, we need to use some approximation to come up with the filter tap coefficient values. If we assume N 0 0 (the zero-forcing assumption), we get H eq (z) = 1−e − T s τ z −1 . Thus, a two tap filter is sufficient. For T s = 1 30 ms, we have a 0 =1, a 1 = −0.0039 as the tap coefficient values. Any other reasonable way is also accepted. 4. ω i = c i where {c i } is the inverse Z- transform of 1/F(z) Show that this choice of tap weights minimizes ¸ ¸ ¸ ¸ 1 F(z) −(ω −N z N + . . . + ω N z −N ) ¸ ¸ ¸ ¸ 2 . . . (1) at z = e jω If F(z) is of length 2 and monic, say F(z) = 1 −a 1 z then 1 F(z) = 1 + a 1 z −1 + a 2 1 z −2 + . . . where c 1 = a 1 , c 2 = a 2 1 , a 1 < 1 It is easy to see that the coefficients become smaller and smaller. So if we had the opportunity to cancel any (2N+1) coefficients we will cancel the ones that are closest to z 0 . Hence we get that ω i = c i minimizes (1). The result can be similarly proved for length of F(z) greate than 2 or non-monic. 5. (a) H eq (f) = 1 H(f) for ZF equalizer H eq (f) = _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 1 f c −20MHz ≤ f < f c −10MHz 2 f c −10MHz ≤ f < f c 0.5 f c ≤ f < f c + 10MHz 4 f c + 10MHz ≤ f < f c + 20MHz 0 o.w. (9) (b) S=10mW Signal power N = N 0 [1 2 ×10MHz + 2 2 ×10MHz + 0.5 2 ×10MHz + 4 2 ×10MHz] = 0.2125mW ∴ SNR = 47.0588 = 16.73dB (c) T s = 0.0125µsec P b ≤ 0.2e −1.5γ/M−1 or M ≤ 1 + 1.5SNR −ln(5P b ) for P b = 10 −3 M ≤ 14.3228 (M ≥ 4 thus using the formula is reasonable) R = log 2 M T s = 307.2193Mbps (d) We use M=4 non overlapping subchannels, each with B=10MHz bandwidth 1: f c −20MHz ≤ f < f c −10MHz α 1 = 1 2: f c −10MHz ≤ f < f c α 2 = 0.5 3: f c ≤ f < f c + 10MHz α 3 = 2 4: f c + 10MHz ≤ f < f c + 20MHz α 4 = 0.25 Power optimization: γ i = Pα 2 i N 0 B for i = 1, 2, 3, 4 γ 1 = 1000 γ 2 = 250 γ 3 = 4000 γ 4 = 62.5 for P b = 10 −3 K = 0.2831 P i P = _ 1 γ 0 − 1 Kγ i γ i ≥ γ 0 /K 0 γ i ≤ γ 0 /K (10) We can see that all subchannels will be used and P 1 = 2.6523 P 2 = 2.5464 P 3 = 2.6788 P 4 = 2.1225 and γ 0 = 3.7207 thus R = 2B 4 i=1 log(Kγ i /γ 0 ) = 419.9711Mbps 6. (a) F{f(t)} = _ T |f| < 1/T 0 o.w. F Z (f) = 1 T S ∞ n=−∞ F _ f + n T s _ = 1 ∴ folded spectrum of f(t) is flat. (b) y k = y(kT + t 0 ) = ∞ i=−∞ X i f(kT + t 0 −iT) = N i=−N X i f ((k −i)T + t 0 ) = X k sinc(t 0 ) + N+k i=−N+k,i=k X i f ((k −i)T + t 0 ) . ¸¸ . ISI (c) ISI = N+k i=−N+k,i=k X i sin (π(k −i) + t 0 /Tπ) π(k −i) + t 0 /Tπ = sin(πt 0 /T) N i=−N,i=0 1 πt 0 /T −πi = sin(πt 0 /T) N i=1 _ −1 πt 0 /T −πi + 1 πt 0 /T −πi _ = 2 π sin(πt 0 /T) N n=1 n n 2 −t 2 0 /T 2 Thus, ISI →∞ as N →∞ 7. g m (t) = g (−t) = g(t) = sinc(t/T S ), |t| < T s Noise whitening filter : 1 G m (1/z ) 8. J min = 1 − ∞ j=−∞ c j f −j B(z) = C(z)F(z) = F(z)F (z −1 ) F(z)F (z −1 ) + N 0 = X(z) X(z) + N 0 ∴ b 0 = 1 2πj _ B(z) z dz = 1 2πj _ X(z) z[X(z) + N 0 ] dz = T 2π _ π/T −π/T X(e jωT ) X(e jωT ) + N 0 dω ∴ J min = 1 − T 2π _ π/T −π/T X(e jωT ) X(e jωT ) + N 0 dω = T 2π _ π/T −π/T N 0 X(e jωT ) + N 0 dω = T 2π _ π/T −π/T N 0 T −1 ∞ n=−∞ |H(ω + 2πn/T)| 2 + N 0 dω = T s _ −0.5T s −0.5T s N 0 F Σ (f) + N 0 df 9. V W J = _ ∂J ∂w 0 , . . . , ∂J ∂w N _ J = w T M v w −2{V d w } + 1 ∴ ∂J ∂w = 2M v w T −2V d ∂J ∂w = 0 ⇒ 2M v w T = 2V d ⇒ w opt = _ M T v _ −1 V H d 10. J min = T s _ 0.5T s −0.5T s N 0 F Σ (f) + N 0 df ∵ N 0 F Σ (f) + N 0 ≥ 0 ∴ J min ≥ 0 N 0 F Σ (f) + N 0 ≤ N 0 N 0 = 1 ∴ J min ≤ T s _ 0.5T s −0.5T s 1df = 1 ∴ 0 ≤ J min ≤ 1 11. F Σ (f) = 1 T s ∞ n=−∞ F _ f + n T s _ = 1 T s ∞ n=−∞ 1 + 0.5e −j2π f+ n T s + 0.3e −j4π f+ n T s MMSE equalizer : J min = T s _ 0.5/T s −0.5/T s N 0 F Σ (f) + N 0 df DF equalizer : J min = exp _ T s _ 0.5/T s −0.5/T s ln _ N 0 F Σ (f) + N 0 _ _ df 12. (a) G(f) is a sinc(), so theoretically infinite. But 2/T is also acceptable (Null to Null bandwidth) (b) τ T is more likely since T = 10 −9 sec As long as τ > T b , get ISI and so, frequency selective fading (c) Require T b = T m + T ⇒R = 1 T m +T = 49997.5bps (d) H eq (z) = 1 F(z) for ZF equalizer ⇒H eq (z) = 1 d 0 +d 1 z −1 +d 2 z −2 Long division yields the first 2 taps as w 0 = 1/α 0 w 1 = −α 1 /α 2 0 13. (a) H zf (f) = 1 H(f) = _ ¸ ¸ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ ¸ ¸ _ 1 0 ≤ f ≤ 10KHz 2 10KHz ≤ f ≤ 20KHz 3 20KHz ≤ f ≤ 30KHz 4 30KHz ≤ f ≤ 40KHz 5 40KHz ≤ f ≤ 50KHz (b) The noise spectrum at the output of the filter is given by N(f) = N 0 |H eq (f)| 2 , and the noise power is given by the integral of N(f) from -50 kHz to 50 kHz: N = _ 50kHz f=−50kHz N(f)df = 2N 0 _ 50kHz f=0kHz |H eq (f)| 2 df = 2N 0 (1 + 4 + 9 + 16 + 25)(10kHz) = 1.1mW (c) The noise spectrum at the output of the filter is given by N(f) = N 0 (H(f)+α) 2 , and the noise power is given by the integral of N(f) from -50 kHz to 50 kHz. For α = .5 we get N = 2N 0 (.44 + 1 + 1.44 + 1.78 + 2.04))(10kHz) = 0.134 mW For α = 1 we get N = 2N 0 (.25 + .44 + .56 + .64 + .69))(10kHz) = 0.0516 mW (d) As α increases, the frequency response H eq (f) decreases for all f. Thus, the noise power decreases, but the signal power decreases as well. The factor α should be chosen to balance maximizing the SNR and minimizing distortion, which also depends on the spectrum of the input signal (which is not given here). (e) As α → ∞, the noise power goes to 0 because H eq (f) → 0 for all f. However, the signal power also goes to zero. 14. The equalizer must be retrained because the channel de-correlates. In fact it has to be retrained at least every channel correlation time. Benefits of training (a) Use detected data to adjust the equalizer coefficients. Can work without training information (b) eliminate ISI. 15. N = 4 LMS-DFE: 2N+1 operations/iteration ⇒ 9 operations/iteration RLS: 2.5(N) 2 + 4.5N operations/iteration ⇒ 58 operations/iteration Each iteration, one bit sent. The bit time is different for LMS-DFE/RLS, T b (LMS-DFE) <T b (RLS). But time to convergence is faster for RLS. Case 1: f D = 100 Hz ⇒(∆t c ) ≡ 10 msec, must retrain every 5 msec. LMS-DFE: R = 10 7 9 - 1000 bits 5 msecs = 911 Kbps RLS: R = 10 7 58 - 50 bits 5 msec = 162 Kbps Case2: f D = 1000 Hz ⇒ retrain every 0.5 msec R LMS-DFE = 0 bps R RLS = 72.4 Kbps 16. In the adaptive method, we start with some initial value of tap coefficients W 0 and then use the steepest descent method W k+1 = W K −∆G K . . . (1) where ∆ is some small positive number and G K is the gradient of MSE = E| ˆ d k − ˆ ˆ d k | 2 is RW k −p (Notice that 11.37 was a solution of gradient =0 , ∴ RW = p) ∴ G K = RW k −p = −E[ε k Y k ] where Y k = [y k+L . . . y K−L ] T and ε k = ˆ ˆ d k − ˆ d k Approximately (1) can be rewritten as W k+1 = W k +∆ε k Y k Chapter 15 1. City has 10 macro-cells each cell has 100 users ∴ total number of users = 1000 Cells are of size 1 sqkm maximum distance traveled to traverse = √ 2km ∴ time = √ 2 30 = 169.7s In the new setup number of cells = 10 5 microcells total number of users = 1000 ×100 2 users time = √ 2×10 30×10 3 =1.69s ∴ number of users increases by 10000 and handoff time reduces by 1/100 2. See Fig 1 (0,0) (0,1) v u (1,1) (1,0) 60 Figure 1: Problem 2 D 2 = (j20) 2 + (i20) 2 −2(j20)(i20) cos(2π/3) ⇒ D = 2a _ i 2 +j 2 +ij = √ 3R _ i 2 +j 2 +ij 3. diamond shaped cells, R= 100m D min = 600m D = 2KR K = D 2R = 600 2×100 = 3 N = K 2 = 9 (a) number of cells per cluster = N = 9 (b) number of channels per cell = total number/N = 450/9 = 50 4. (a) R=1km D=6km N = A cluster A cell = √ 3D 2 /2 3 √ 3R 2 /2 = 1 3 (D/R) 2 = 1 3 6 2 = 12 number of cells per cluster = N = 12 D SHADED CELLS USE SAME FREQUENCY Figure 2: Problem 3 (b) number of channels in each cell = 1200/12 = 100 (c) _ i 2 +j 2 +ij = 2 √ 3 ⇒ i = 2, j = 2 5. R=10m D=60m γ I = 2 γ 0 = 4 M = 4 for diamond shaped cells SIR a = R −γ I MD −γ 0 = R −2 4D −4 = 32400 SIR b = R −4 4D −4 = 324 SIR c = R −2 4D −2 = 9 SIR a > SIR b > SIR c 6. γ = 2 BPSK P b = 10 −6 → P b = Q( √ 2γ b ) ⇒ γ b = SIR 0 = 4.7534 B = 50MHz each user100KHz = B s SIR = 1 M _ D R _ γ M=6 for hexagonal cells a 1 = 0.167 a 2 = 3 N > 1 a 2 _ SIR 0 a 1 _ 2/γ ⇒ N ≥ 9.4879 ∴ N = 10 C u = 50 7. G = 100 ξ = 1 λ = 1.5 With no sectorization SIR = 1 ξ 3G (N c −1)(1 +λ) = 4.7534 N c = 26.2450 = 26 With sectorization, interference is reduced by a factor of 3 N c = 76.7349 = 76 8. SINR = G N c −1 i=1 X i +N α = p(X i = 1) N ∼ G(0.247N c , 0.078N c ) P o ut = p(SIR < SIR 0 ) (a) P out = p _ G N c −1 i=1 X i +N < SIR 0 _ = p _ N c −1 i=1 X i +N > G SIR 0 _ (b) X = N c −1 i=1 X i then X ∼ Bin(α, N c −1) p(x +N > G/SIR 0 ) = N c −1 N=0 p(n +N > G/SIR 0 |x = n)p(x = n) p(x = n) = _ N c −1 n _ α n (1 −α) N c −1−n p(x +N > G/SIR 0 ) = N c −1 N=0 p(N > G/SIR 0 −n|x = n)p(x = n) = N c −1 N=0 p _ N −0.247N c √ 0.078N c > G SIR 0 −n −0.247N c √ 0.078N c |x = n _ = N c −1 N=0 Q _ G SIR 0 −n −0.247N c √ 0.078N c _ p(x = n) (c) N c = 35 α = 0.5 SIR 0 = 5 G = 150 p = 0.0973 MATLAB for i = 1:length(n) pn(i) = (factorial(Nc-1)./(factorial(n(i)).*factorial(Nc-1-n(i))))... *alpha.^n(i)*(1-alpha).^(Nc-1-n(i)); end sump = 0; for i = 1:length(n) f = ((G/sir0)-n(i)-.247*Nc)/(sqrt(.078*Nc)); sump = sump + .5*erfc((f)/sqrt(2))*pn(i); end (d) If x can be approximated as Gaussian then x ∼ G((N c −1)α, (N c −1)α(1 −α)) x +N ∼ G(0.247N c + (N c −1)α, 0.078N c + (N c −1)α(1 −α)) p(x +N > G/SIR 0 ) = Q _ G SIR 0 −(0.247N c + (N c −1)α) _ 0.078N c + (N c −1)α(1 −α) _ (e) p= 0.0969 (very accurate approximation!) 9. define γ k = g k P k n k +ρ k=j g kj p j k, j ∈ {1, . . . K} where, g k is channel power gain from user k to his base station n k is thermal noise power at user k’s base station ρ is interference reduction factor (ρ ∼ 1/G) g kj is channel power gain from j th interfering transmitter to user k’s base station p k is user k’s Tx power p j is user j’s Tx power define a matrix F such that F kj = _ 0 k = j γ k g kj ρ g k k = j k, j ∈ {1, . . . K} u = _ γ 1 n 1 g 1 , γ 2 n 2 g 2 , . . . , γ K n K g K _ If Perron Ferbinius eigenvalue of F is less than 1 , then a power control policy exists. The optimal power control policy is given to be P = (I −F) −1 u 10. Matlab D = 2:.01:10; R = 1; gamma = 2; Pdes = R^(-gamma); for i = 1:length(D) Pint = 6*(.2*(D(i)-R)^(-gamma)+.2*(D(i)-R/2)^(-gamma)+.2*(D(i))^(-gamma)... +.2*(D(i)+R/2)^(-gamma)+.2*(D(i)+R)^(-gamma)); Pintbest = 6*((D(i)+R)^(-gamma)); Pintworst = 6*((D(i)-R)^(-gamma)); ASE(i) = log(1+Pdes/Pint)/(pi*(.5*D(i))^2); ASEbest(i) = log(1+Pdes/Pintbest)/(pi*(.5*D(i))^2); ASEworst(i) = log(1+Pdes/Pintworst)/(pi*(.5*D(i))^2); end 2 3 4 5 6 7 8 9 10 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 D A S E ( D ) 5 Case ASE Best Case ASE Worst Case ASE Figure 3: Problem 10 11. P t = 5W B = 100KHz N 0 = 10 −16 W/Hz P r = P t K _ d 0 d _ 3 d 0 = 1, K = 100 (a) D=2R 2 users share the band available Each user gets 50KHz BASESTATION USER R=1Km Figure 4: Problem 11a (b) P b = 1 4γ b ⇒ 10 −3 = 1 4γ b ⇒ γ b = 250 If D(n) = 2nR, number of users that share band = 2(2(n-1)+1) ∴ each user gets 100KHz 2(2n−1) = B u (n) interference is only from first tier SIR(n) = P t K _ d 0 R _ 3 N 0 2 B u (n) + 2 _ P t K _ d 0 √ R 2 +D(n) 2 _ 3 _ > 25 using Matlab , n = 4, SIR = 261.9253, D = 8R Matlab Pt = 5; R = 1000; sigma_2 = 1e-16; n = 1; D = 2*n*R; Bu = (100/(2*(2*n-1)))*1e3; K = 100; d0 = 1; Pdes = Pt*K*(d0/R)^3; Pint = 2*(Pt*K*(d0/sqrt(R^2+D^2))^3); Npower = sigma_2*Bu; sir = Pdes/(Npower+Pint); while sir < 250 n = n+1; D = 2*n*R; Bu = (100/(2*(2*n-1)))*1e3; K = 100; d0 = 1; Pdes = Pt*K*(d0/R)^3; Pint = 2*(Pt*K*(d0/sqrt(R^2+D^2))^3); Npower = sigma_2*Bu; sir = Pdes/(Npower+Pint); end (c) ASE = (R 1 +R 2 )/B 2km×2km R 1 = R 2 = B u (1) log(1 +SIR(1)) B u (1) = 50KHz, SIR(1) = 5.5899 ASE = 0.6801bps/Hz/km 2 12. B=100KHz N 0 = 10 −9 W/Hz K = 10 P = 10mW per user (a) 0 ≤ α ≤ 1 α is channel gain between cells. See Matlab If α is large, interference can be decoded and subtracted easily so capacity grows with α as high SNR’s (beyond an α value) . For low SNR values (α less than a value) c decreases with increase in α as interference is increased which cannot be easily decoded due to low SNR. MATLAB CODE: B = 100e3; sigma_2 = 1e-9; P = 10e-3; K = 10; ss = .001; alpha = 0:.01:1; theta = 0:ss:1; for i = 1:length(alpha) capvec = log2(1+(K*P*(1+2*alpha(i)*cos(2*pi*theta)).^2)/(sigma_2*B)); C(i) = (1/K)*sum(capvec)*ss; end 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1 1.02 α C ( α ) / B B = 100 KHz Figure 5: Problem 12a (b) C(K)↓ as K↑ because as the number of mobile per cell increases system resources get shared more and so per user capacity C(K) has to fall. MATLAB CODE: BB = 100e3; sigma_2 = 1e-9; P = 10e-3; K = 1:.1:30; ss = .001; alpha = .5; theta = 0:ss:1; for i = 1:length(K) capvec = log2(1+(K(i)*P*(1+2*alpha*cos(2*pi*theta)).^2)/(sigma_2*B)); C(i) = (1/K(i))*sum(capvec)*ss; end 0 5 10 15 20 25 30 0 1 2 3 4 5 6 K C ( K ) / B B = 100 KHz Figure 6: Problem 12b (c) as transmit power P ↑, capacity C ↑ but gets saturated after a while as the system becomes interference limited. MATLAB CODE: B = 100e3; sigma_2 = 1e-9; P = [0:.1:100]*1e-3; K = 10; ss = .001; alpha = .5; theta = 0:ss:1; for i = 1:length(P) capvec = log2(1+(K*P(i)*(1+2*alpha*cos(2*pi*theta)).^2)/(sigma_2*B)); C(i) = (1/K)*sum(capvec)*ss; end 0 10 20 30 40 50 60 70 80 90 100 0 0.2 0.4 0.6 0.8 1 1.2 1.4 P (in mW) C ( P ) / B B = 100 KHz Figure 7: Problem 12c 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 capacity (data rate per unit bandwidth) increases. 10. (a) 100 cells, 100 users/cell ⇒ 10,000 users (b) 100 users/cell ⇒ 2500 cells required 100km2 Area 2 Area/cell = 2500cells ⇒ cell = .04km (c) From Rappaport or iteration of formula, we get that 100 1 Each subscriber generates 30 of an Erlang of traffic. Thus, each cell can support 30 × 89 = 2670 subscribers Macrocell: 2670 × 100 ⇒ 267, 000 subscribers Microcell: 6,675,000 subscribers (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 channels cell ⇒ 89 channels cell @Pb = .02 Chapter 2 1. Pr = Pt 10−3 = Pt 10−3 = Pt √ Gl λ 4πd λ 4π10 λ 4π100 2 λ = c/fc = 0.06 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 = 3. ∆φ = 2π(x +x−l) λ x+x −l c = 1.33× x +x−l = (ht + hr )2 + d2 − (ht − hr )2 + d2   ht + hr 2 ht − hr 2 = d +1− + 1 d d  ht + hr d 2 d ht , hr , we need to keep only first order terms   1 ∼ d   2 =  ht − hr d 2 1 + 1 −  2   + 1  ∆φ ∼ 2(ht + hr ) d 2π 2(ht + hr ) λ d 4. Signal nulls occur when ∆φ = (2n + 1)π 2π(x + x − l) λ (ht + hr )2 + d2 − (ht + hr )2 + d2 − (ht − hr )2 + d2 = (2n + 1)π = π(2n + 1) λ (2n + 1) 2 λ + 2 (ht − hr )2 + d2 2π λ (ht − hr )2 + d2 = Let m = (2n + 1) (ht + hr )2 + d2 = m square both sides (ht + hr )2 + d2 = m2 x = (ht + hr )2 , y = (ht − hr )2 , x − y = 4ht hr x = m2 ⇒d = d = λ2 + y + mλ y + d2 4 1 mλ x − m2 λ2 −y 4 2 λ2 + (ht − hr )2 + d2 + mλ (ht − hr )2 + d2 4 −y 2 4ht hr (2n + 1)λ − (2n + 1)λ 4 − (ht − hr )2 ,n ∈ Z 5. ht = 20m hr = 3m fc = 2GHz λ = fcc = 0.15 t dc = 4hλhr = 1600m = 1.6Km 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. 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) = α1 δ(t − τ ) + α2 δ(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 c d 2 2 = 0.022 × 10−6 s = 0.022 × 10−6 s 2 82 + −d c ⇒ d = 16.1m d ∴ τ = = 53.67ns c lambda=c/f. counter=1. Gl=1. the asymptotic behavior of Pr tends toward d−2 from d−4 . the lengths of both paths are initially dominated by the difference between the antenna heights (which is 35 meters). dnew=[dc:1:100000]. subplot(2. dc=4*ht*hr/lambda. end hold off . Vec=Gl. d=[1:1:100000]. Also note that the the received power actually increases with distance up to some point. and the dominant factor is the phase difference between the paths. hr=15. This is because for very small distances (i. Thus.5.-20*log10(dnew)). From the plots it can be seen that as Gr (gain of reflected path) is decreased. d = 1).√ λ Gl α1 = 4π l √ λ RGr α2 = 4π x + x 2 = 2.^2+(ht+hr)^2).2.1. ht=50. this causes the two paths to nearly cancel each other out. close all. plot(10*log10(d). plot(10*log10(dnew). f=900e6. r=(d.^2+(ht-hr)^2). the powers of both paths are roughly constant for small values of d.^2. Since R = -1./l+R*Gr... Gr=GR(counter). l=(d. plot(10*log10(dnew). A program to plot the figures is shown below. When the phase difference becomes 180 degrees. hold on. making the phase difference very small.-40*log10(dnew)). R=-1. clear all. The power versus distance curves and a plot of the phase difference between the two paths is shown on the following page. c=3e8. which makes sense since the effect of reflected path is reduced and it is more like having only a LOS path.01].71 × 10−6 2 = 1.e. Pr=(lambda/4/pi)^2*(abs(Vec)).. figure(1).5. GR=[1.10*log10(Pr)-10*log10(Pr(1))).316. Also the variation of power before and around dc is reduced because the strength of the reflected path decreases as Gr decreases. for counter = 1:1:4.^. the first local maxima is achieved. the reflected path is approximately two times the LOS path./r. Additionally.37 × 10−6 8.counter).^.*exp(phd*sqrt(-1)). phd=2*pi/lambda*(r-1). fc = 900M Hz λ = 1/3m G = 1 radar cross section 20dBm2 = 10 log1 0σ ⇒ σ = 100 √ √ d=1 .5d)2 + (0.52 × 10−4 = −34.316 20 40 G = . the power fall off with distance for the 10-ray model is d−2 for relatively large distances 10.54dB distances result in tremendous path loss d0 d 2 γ → simplified λ d 2 → free space .Gr = 1 50 10 * log10(Pr) 0 −50 −100 −150 50 10 * log10(Pr) 0 −50 −100 −150 400 Phase (deg) 300 200 100 0 0 20 40 10 * log10(d) 60 0 20 40 10 * log10(d) 60 10 * log10(Pr) 0 20 40 Gr = .1 10 * log10(d) 60 10 * log10(Pr) 50 0 −50 −100 −150 50 0 −50 −100 −150 0 0 Gr = .5dB Notice that scattered rays over long 12.01µs 11.498dB 2 = 3.5 Path loss due to scattering √ λ Gσ Pr = Pt (4π)3/2 ss Path loss due to reflection (using 2 ray model) √ 2 R G λ Pr = Pt s+s 4π d = 10 Pscattering = −56.93m = 0.54dB Pref lection = −94.54dB Pref lection = −54.59m τ0 = 503.54dB Pref lection = −74. Pr = Pt K Pr = Pt √ Gl 4π 2 (500/6)2 + 102 = 83. As indicated in the text.59/c = 1.68µs L.0224 = −16.01 10 * rlog10(d) 60 20 40 10 * log10(d) 60 Figure 1: Problem 8 9.5 = 0.S ray d = 500m τ0 = 500/c = 1.67µs ∴ delay spread = 0.5d)2 = d 0. s = s = (0.O. The delay spread is dictated by the ray reaching last d = Total distance = 6d = 503.5dB d = 1000 Pscattering = −136.5dB d = 100 Pscattering = −96. 7m 14.G ∴ when K = 4π l and d0 = λ The two models are equal.52dB P Lsmallcity = 325. √ 2 13.41km ⇒ with the same argument dmin = 4km 15. λ = 0. K = (λ/4πd0 )2 = 5. d = distance between cells with reused freq p = transmit power of all the mobiles S I ≥ 20dB uplink (a) Min.3 d 4 d ≤ 260. d = 100m Large urban city small urban city suburb rural area P Llargecity = 353. diffractors and scatterers . S/I will result when main user is at A and √ Interferers are at B dA = distance between A and base station #1 = 2km dB = distance between B and base station √ #1 = 2km S I = min P 2P Gλ 4πdA Gλ 4πdB 2 2 = d2 (dmin − 1)2 B = 100 = 4 2d2 A ⇒ dmin − 1 = 20km ⇒ dmin = 21km since integer number of cells should be accommodated in distance d ⇒ dmin = 22km (b) Pγ d0 =k Pu d 1 dB 2 dA γ γ ⇒ S I = min Pk 2P k d0 dA d0 dB γ γ = ⇒ dmin (c) 1 dmin − 1 γ 1 dmin − 1 √ √ = = 2 2 2 2 = 9.8769dB P Lruralarea/countryside = 70.27 ⇒ with the same argument ⇒ dmin = 10km S I k = min γ d0 dA A γ d0 dB B 3 = 100 = 2k (dmin − 1)4 = 100 0.3. hr = 5m.99dB P Lsuburb = 207. fc = 900M Hz. ht = 20m. path loss is higher in the presence of multiple reflectors. γ = 4 We want SN Rrecd = 20dB = 100 Noise power is 10−19 d0 γ P = Pt K d 10−17 = 10K 0. Pnoise = −160dBm fc = 1GHz.7 × 10−4 .04 ⇒ dmin = 2. d0 = 1m.9278dB As seen . 3.6). Piecewise linear model for 2-path model.Plot for Gr = 0 dB 0 −10 y = −15x+8 −20 −30 −40 y = −35x+56 −50 −60 1 1.6 1.4 − 3.2 2.8 3 Figure 2: Problem 16 16. (a) d = 10 log10 K − 10r log1 0 d0 using least squares we get 10 log10 K = −29.4 1.42dB γ=4 Pr Pt dB (b) PL(2Km) = 10 log10 K − 10r log10 d = −161.4) PL (d)K d0 d γ 0 = 10−8 = −8dB −110 = Pt − 80 − 5 − 10 − 6 − 3.4 ⇒ Pt = −2. PAF =(3. σψdB ) P rob(X < −10) = P rob 19.2dBm 18.4 2. Pr = Pt − PL (d) − 3 F AFi − 2 P AFj i j FAF =(5. Assume free space path loss parameters fc = 900M Hz → λ = 1/3m σψdB = 6 SN Rrecd = 15dB Pt = 1W X σψdB < −10 σψdB = 6.10. See Fig 2 17.2 1.6 2.8 2 2.512 × 10−4 .76dB 2 (c) Receiver power can be assumed to be Gaussian with variance σψdB 2 X ∼ N (0.4. 68m 20.01.4 × 1−−3 = Pt = 4πd d2 = 10 log1 0(µ(d)) µdB P (Precd (d) > −55) = 0. T p − µψ < −6.g = 3dB Pnoise = −40dBm ⇒ Precvd = −55dB Suppose we choose a cell of radius d µ(d) = Precvd (due to path loss alone) √ 2 Gl λ 1.9 P Precd (d) − µdB −55 − µdB > = 0. y = wgn(1.308 ⇒ µ(d) = 1. for i = 1:length(y) x(i) = y(i). [received powerdB Tp = 10dB (a) outageprob. outage prob < 1% ⇒ Q T p − µψ σψ > 99% ⇒ T p − µψ < −2. Outage Prob. for j = 1:i x(i) = x(i)+exp(-(i-j)/Xc)*y(j).86 × 10−5 ⇒ d = 8. MATLAB CODE Xc = 20. ss = .282 6 ⇒ µdB = −47. = Prob. = 1 − Q (b) σψ = 4dB.200*(1/ss)).8dB σψ .99 ⇒ µψ ≥ 37.9 σψdB 6 −55 − µdB ⇒ = −1.33 ⇒ σψ T p − µψ σψ =1−Q −5 8 =Q 5 8 = 26% T pdB ] µψ ≥ 19. end end 21.32dB (c) σψ = 12dB. we can use macroscopic diversity. if they are uncorrelated. 22. In this way the probability of power outage will be less because both base stations are unlikely to experience an outage at the same time. The idea in macroscopic diversity is to send the message from different base stations to achieve uncorrelated shadowing.20 White Noise Process Filtered Shadowing Processing 10 0 −10 −20 −30 −40 −50 −60 0 20 40 60 80 100 120 140 160 180 200 Figure 3: Problem 20 (d) For mitigating the effect of shadowing. 2 C= 2 R R rQ a + b ln r=0 r dr R r To perform integration by parts. we let du = rdr and v = Q a + b ln R . ∂r R ∂x ∂r R r 2π (1) . Then u = 1 r2 and 2 dv = ∂ r ∂ ∂ r −1 b Q a + b ln = Q(x)|x=a+b ln(r/R) a + b ln = √ exp(−k 2 /2) dr. 2572 σψdB 2−2ab b2 c = Q(a) + e 24.7575 σ 10γ log10 e b= = 3.5 10 × 6 × log10 e = 20. γ = 3 d0 = 1 k = 0dB σ = 4dB R = 100m Pt = 80mW Pmin = −100dBm = −130dB Pγ (R) = Pt K a= d0 d γ = 80 × 10−9 = −70.97dB Pmin − Pγ (R) = 14.3871 b= 8 c = 0. γ = 6 σ=8 Q 2 − 2ab b 1 Pγ (R) = 20 + Pmin a = −20/8 = −2. Then we get C = 2 1 2 r r Q a + b ln 2 2 R R 1 R2 1 R2 R R 2 + 2 R r=0 R r=0 1 2 1 b r √ exp(−k 2 /2) dr 2 r 2π (2) = Q(a) + r=0 R 1 b r2 √ exp(−k 2 /2) dr r 2π 1 √ R2 exp 2π 2(k − a) b b exp(−k 2 /2) dr r dk (3) = Q(a) + (4) r=0 a = Q(a) + k=−∞ 1 √ exp 2π −k 2 2k 2a + − 2 b b a (5) = Q(a) + exp −2a 2 + 2 b b 2 − 2ab b2 2 − 2ab b2 k=−∞ 1 1 2 √ exp − (k − )2 dk 2 b 2π a− 1 2 b (6) = Q(a) + exp = Q(a) + exp 1−Q Q (7) (8) (9) 2 − ab b Since Q(−x) = 1 − Q(x).9938 . 23.r where k = a + b ln R . . C decreases as a function of σ. because it forces higher transmit power.7170 6 0. the probability of non-outage is proportional to Q −1 .8 0.e. The same can also be seen from this figure: 0. By a similar argument.25. we have maximum coverage when γ = 6 and σ = 4.85 Coverage 0. we have minimum coverage when γ = 2 and σ = 12. Therefore. γ = 4 and σ = 8 can be seen from the table or the figure to be 0.75 0.6302 4 0.8587 0.7728 0.7 0.6 12 10 8 4 6 σ ψ 6 5 3 4 2 γ dB Figure 4: Problem 25 The value of coverage for middle point of typical values i. hence more received power at r < R. Due to these forces.9 0. and thus σ decreases as a function of σ. Since the average power at the boundary of the cell is fixed.7728 Since Pr (r) ≥ Pmin for all r ≤ R. γ/σψdB 4 8 12 2 0.7728.8977 0.7728 0. C increases with γ.6786 0.8255 0.95 0.65 0. 67 × 10−9 s 3.Chapter 3 1. d = 100m ∴ Tm = 2. hr = 4m. Delay for LOS component = τ0 = 23 ns Delay for First Multipath component = τ1 = 48 ns Delay for Second Multipath component = τ2 = 67 ns τc = Delay for the multipath component to which the demodulator synchronizes. . when τc = τ0 . Tm = 44 ns. When τc = τ1 . Tm = max τm − τc m So. Tm = 19 ns. t) = α0 (t)e−jφ0 (t) δ(τ − τ0 (t)) + α1 (t)e−jφ1 (t) δ(τ − τ1 (t)) G α0 (t) = λ4πd l with d = vt φ0 (t) = 2πfc τ0 (t) − φD0 τ0 (t) = d/c φD0 = t 2πfD0 (t)dt v fD0 (t) = λ cos θ0 (t) θ0 (t) = 0 ∀t√ √ λR G G α1 (t) = 4π(r+r l) = λR 2hl2 with d = vt 4π(d+ √ φ1 (t) = 2πfc τ1 (t) − φD1 2 τ1 (t) = (r + r )/c = (d + 2h )/c d φD1 = t 2πfD1 (t)dt v fD1 (t) = λ cos θ1 (t) h θ1 (t) = π − arctan d/2 ∀t 2. d = vt 2 r + r = d + 2h d Equivalent low-pass channel impulse response is given by c(τ. For the 2 ray model: d ) l c x+x τ1 = c τ0 = x+x −l = c Tm = (ht + hr )2 + d2 − c (ht − hr )2 + d2 ∴ delay spread(Tm ) = when d (ht + hr ) 1 2ht hr c d ht = 10m. For Rayleigh fading channel Poutage = 1 − e−P0 /2σ 0.*exp(-(z.*besseli(0. P r(Pr < P0 ) = 0. s2 = 1e-11. z0 = sqrt(1e-11).0311 2σ 2 = −80dBm.min = 3×108 s ∴ min fc τn = 1010 3×108 = 33 1 5. Fz (z)= P [x2 +y 2 ≤ z 2 ] = x2 +y 2 ≤z 2 −(x2 +y)2 1 e 2σ2 dxdy = 2πσ 2 2π z 0 0 −z 2 −r2 1 e 2 rdrdθ = 1 − e 2σ2 (z ≥ 0) 2πσ 2 2σ df z (z) z − z22 = 2 e 2σ → Rayleigh dz σ For Power: √ −z 2 z] = 1 − e 2 2σ 1 −z fz (z)= 2 e 2 → Exponential 2σ 2σ Fz 2 (z)=P [Z ≤ P r(Pr < P0 ) = 1 − e−P0 /2σ 2 6. pdf = (z/sigma2). 2σ 2 = -80dBm = 10−11 Target Power P0 = -80 dBm = 10−11 Avg. . P0 = −90dBm. σ2 z≥0 = 0. we use Matlab and I0 (x) = besseli(0.0952 7. fc = 109 Hz 10 τn.*(sqrt(s2)/sigma2)). power in LOS component = s2 = -80dBm = 10−11 2 10−5 P r[z 2 ≤ 10−11 ] = P r[z ≤ √ ] 10 Let z0 = 10−5 √ 10 z0 −(z 2 +s2 ) 2σ 2 = 0 z e σ2 − I0 zs dz. z = [0:ss:z0].z.4. sigma2 = (1e-11)/2.01 = 1 − e−P0 /Pr ∴ Pr = −60dBm 8. Sample Code is given: clear P0 = 1e-11. For Rayleigh fading channel 2σ 2 = −80dBm. ss = z0/1e7.3457 To evaluate this. int_pr = sum(pdf)*ss.x).^2+s2)/(2*sigma2)). Use CDF strategy. P r(Pr < P0 ) = 0. P0 = −95dBm. cdfN(i) = 0.01*sum(pdfR(1:i)).’b--’). hold on plot(z.^(2*m-1))/(gamma(m)*Pr^m)). (a) W = average received power Zi = Shadowing over link i Pr i = Received power in dBW. the Nakagami-m approximation becomes better as K increases. CDF of Ricean distribution is Ricean FZ (z) = 0 z pRicean (z) Z K(K + 1) Pr where pRicean (z) = Z 2z(K + 1) (K + 1)z 2 exp −K − I0 Pr Pr 2z .cdfR).*exp(-K-((K+1).2 < T |Y = y] fy (y) dy Pr.*besseli(0.^2))/Pr).1 |Y = y ~ N W + by. and [Pr. Z2 independent = Q (c) Pout = −∞ ∞ W −T σ 2 2 = Q σ P [Pr.2 |Y = y] ∞ Poutage= −∞ Q W + by − T aσ 2 y2 1 √ e− 2σ2 dy 2πσ . we set the Nakagami m parameter to be (K + 1)2 /(2K + 1).pdfR).1 < T ] P [Pr.01*sum(pdfN(1:i)). 10.1 ≤ T .(2*sqrt((K*(K+1))/Pr))*z). m ≥ 0.1 < T ∩ Pr.5 We need to plot the two CDF curves for K = 1.a2 σ 2 . for a fixed value of K and x.2 < T ] = P [Pr.*(z. Pr. which is Gaussian with mean W. CDF of Nakagami-m distribution is Nakagami-m FZ (z) = z 0 pNakagami-m (z) Z where pNakagami-m (z) = Z 2mm z 2m−1 −mz 2 exp .1 |Y = y] ⊥ [Pr. end plot(z.*exp(-(m/Pr)*z. figure. prob(γ<x) for x large is always greater for the Ricean distribution. plot(z. z≥0 For the Nakagami-m approximation to Ricean distribution. As seen from the curves. where the tail of Nakagami-distribution is always above the Ricean distribution.2 < T ] since Z1 . pdfN = ((2*m^m*z. This is seen from the tail behavior of the two distributions in their pdf.cdfN.’b--’).pdfN.9. K = 10. Pr = 1. Γ(m)P rm Pr z ≥ 0. Also.01:3]. hold on plot(z.^2). pdfR = ((2*z*(K+1))/Pr). Sample code is given: z = [0:0.5. m = (K+1)^2/(2*K+1).10 and P r =1 (we can choose any value for Pr as it is the same for both the distributions and our aim is to compare them). variance σ 2 (b) Poutage = P [Pr. for i = 1:length(z) cdfR(i) = 0. 5 2 2.5 0.7 0.2 0.4 1 1.4 0.6 0.5 1 1.2 0.5 4 4.1 0 0 0.4 Ricean Nakagami−m 1.8 0. 10 and the Tail Behavior .5 1 0.6 0.2 0 0 0. 5.6 1.6 0.7 0.5 1 1.5 1 1.5 2 2.6 0.3 0.4 0.5 0.5 Tail Behavior 5 5.3 0.4 0.8 0.5 6 Figure 1: The CDF and PDF for K = 1.7 Ricean Nakagami −m 0.6 0.5 1 1.9 0.5 3 2 x 10 −31 K = 10 1.5 3 0 0 0.6 0.5 2 2.8 1.5 2 2.5 1 1.5 1 1.8 0.1 0.K=5 1 1 Ricean Nakagami−m 0.5 0.8 0.3 0.4 0.1 0.5 3 0 0 0.5 0.2 1 0.9 Ricean Nakagami−m Ricean Nakagami − m 1.5 3 0 0 0.3 0.1 0.5 2 2.6 0.2 0 3.2 0.5 K=5 2 K = 10 Ricean Nakagami−m 0.4 0.2 0.9 0.7 1.8 0.9 1 K = 10 Ricean Nakagami−m 0.8 1.2 0.5 3 0 0 0.5 3 K=1 1 1.5 2 2.4 0.8 0. %choose N=30 N=30.20001)=2*cos(beta(i))*cos(Wn(i)*t) %rqtemp(i. %find the envelope average mean=sum(r)/2001. subplot(3. %ritemp(i. rqtemp(n. Wm=2*pi*fm. Wn(M)=0. With independent fading we get Pout = Q 5 8 2 = 0.let y σ= u ∞ = −∞ 1 √ Q 2π 1 √ 2 W − T + buσ aσ 2 e −u2 2 ∞ du = −∞ 1 √ Q 2π + byσ aσ 2 e −y 2 dy 2 (d) Let a = b = .001:2 %Wn(i)=Wm*cos(2*pi*i/N) Wn(n)=Wm(i)*cos(2*pi*n/N). %summarize rqtemp(i) rq=sum(rqtemp).5*(N/2-1).we get sin(alpha)=0 and cos(alpha)=1 %rialpha=(cos(Wm*t)/sqrt(2))*2*cos(alpha)=2*cos(Wm*t)/sqrt(2) %rqalpha=(cos(Wm*t)/sqrt(2))*2*sin(alpha)=0 rialpha(1. . fm=[1 10 100].2001)=0. %beta(i)=pi*i/M beta(n)=pi*n/M.2001)=0. %r=sqrt(ri^2+rq^2) r=sqrt(ri.2001)=0. There are many acceptable techniques for this problem. rialpha(1.^2). %We choose 1000 samples/sec ritemp(M.^2+rq.1. clear all. 11.i). rqtemp(M.0708.20001)=2*sin(beta(i))*cos(Wn(i)*t) ritemp(n. M=0. With correlated fading we get Pout = 0. Sample code for both the stochastic technique(preferred) and the Jake’s technique are included.1000*t+1)=2*cos(Wm(i)*t)/sqrt(2). = 5. beta(M)=0. Jakes: Summing of appropriate sine waves %Jake’s Method close all. for i=1:3 for n=1:1:M for t=0:0. %Because we choose alpha=0.1000*t+1)=2*sin(beta(n))*cos(Wn(n)*t).1316. σ = 8.1000*t+1)=2*cos(beta(n))*cos(Wn(n)*t). Conclusion : Independent shadowing is prefarable for diversity. end end %summarize ritemp(i) and rialpha ri=sum(ritemp)+rialpha. titlename=[’fd = ’ int2str(fm(i)) ’ Hz’].001:2. %plot the figure and shift the envelope average to 0dB plot(time.time=0:0.(10*log10(r)-10*log10(mean))). title(titlename). xlabel(’time(second)’). end . ylabel(’Envelope(dB)’). time interval [0. f_s = 1000. f_s) % % function [Ts. set to 1000 if smaller.2 0.2 1.8 1 fd = 100 Hz 1. with sampling frequency f_s >= 1000Hz.’s are filtered by the Doppler Spectrum and summed. f_s) % generates a Rayleigh fading signal for given Doppler frequency f_D. % Output(s) % -.4 0. % [Hz] Minumum required sampling rate . % % Input(s) % -. % during the time perios [0.8 2 0 0.8 2 Figure 2: Problem 11 Stochastic: Usually two guassian R.t : simulation time interval length.t] % -.8 1 1.2 0. z_dB] = rayleigh_fading(f_D.f_D : [Hz] [1x1 double] Doppler frequency % -.2 0.2 1. although the above technique is prefered.6 1. z_dB] = rayleigh_fading(f_D. end. t.4 1.8 2 0 0.4 1.6 1. t].f_s : [Hz] sampling frequency.8 1 fd = 10 Hz 1.z_dB : [dB] [1xN double] Rayleigh fading signal % % Required parameters if f_s < 1000.4 1.4 0.fd = 1 Hz 5 Envelope(dB) 0 −5 −10 10 Envelope(dB) 0 −10 −20 10 Envelope(dB) 0 −10 −20 0 0. function [Ts.6 0.Ts : [Sec][1xN double] time instances for the Rayleigh signal % -.V.6 0.4 0.6 0.6 1.2 1. Can also just do a Rayleigh distribution with an adequate LPF. t. CGfq = [ conj(fliplr(CGfq_p)) CGfq_p ]. z_dB = 20*log10(z). % Generate r_I(t) and r_Q(t) using inverse FFT: r_I = N*ifft(CGfi. S_r(idx2(length(idx2))) = S_r(idx2(length(idx2))+1). r_Q = -i*N*ifft(CGfq.f_s. S_r(idx2) = 0. this is used to shape the Gaussian line spectra omega_p = 1. % Return correct number of points z_dB = z_dB(1:length(Ts)). S_r(idx1) = 0. f = linspace(-f_s. % Use even number of samples in calculation % [Hz] Frequency samples used in calculation % Generate complex Gaussian samples with line spectra in frequency domain % Inphase : Gfi_p = randn(2. CGfi = [ conj(fliplr(CGfi_p)) CGfi_p ]. . 2) == 1 N = N+1.:)+i*Gfq_p(2.^2)).:)+i*Gfi_p(2.^2).:). generate the Rayleigh distributed signal envelope z = sqrt(abs(r_I).N).*sqrt(S_r)).N). S_r(idx1(1)) = S_r(idx1(1)-1).N = ceil( t*f_s ). % Finally. % Generate fading spectrum. let them be 0 idx1 = find(f>=f_D). end. CGfq_p = Gfq_p(1. CGfi_p = Gfi_p(1. % Quadrature : Gfq_p = randn(2. % Number of samples % Ts contains the time instances at which z_dB is specified Ts = linspace(0.N/2).N/2).^2+abs(r_Q).:). % Take care of samples outside the Doppler frequency range.*sqrt(S_r))./(f_D*sqrt(1-(f/f_D). if mod( N.t. idx2 = find(f<=-f_D). % this makes sure that the average received envelop can be 0dB S_r = omega_p/4/pi. 2 1.0072s P0 = 25dBm. There will be some discrepancy at the end points.6 0. So we have √ 2 LZ (above) = LZ (below) = 2πfD ρe−ρ And.4 1. tz = 0.1 Hz 0 −10 −20 −30 10 0 −10 −20 −30 20 0 −20 −40 −60 0 0.6 0.0013s ρfD 2π 2 P0 = 15dBm. tz = 0.2 0.8 1 10Hz 1. tz = eρ − 1 √ = 0.2 1.2 0.8 1 l00 Hz 1.4 0. but for a large enough T it will not effect the result.8 1 1.2 0.0264s 13.6 0.4 1. In the reader.8 2 Figure 3: Problem 11 12.8 2 0 0. 3.4 1.44 in reader because to go below a level again.6 1. Pr = 30dBm fD = 10Hz P0 = 0dBm.4 0. tZ (above) = p(z > Z) LZ (above) 2 2 p(z > Z) = 1 − p(z ≤ Z) = 1 − (1 − e−ρ ) = e−ρ 1 tZ (above) = √ 2πfD ρ . we found the level crossing rate below a level by taking an average of the number of times the level was crossed over a large period of time. we first need to go above it.2 1.8 2 0 0.4 0.6 1.6 1. It is easy to convince that the level crossing rate above a level will have the same expression as eq. 085 = 0.2) = 0.The values of tZ (above) for fD = 10. T = 10e-3.2)).50. for i = 1:8 if i == 1 p(i.176 . π1 π2 π3 π4 π5 π6 π7 π8 = 9.19. ρ = N4 = 82. p(i.5 10 102 10 103 10 N2 = 19.3) = 1-(p(i. : 30dB ≤ γ ≤ ∞.03 0].09 .1)+p(i. p(i. p(i.1) = 0.72 × 10−44 Nj → level crossing rate at level Aj N1 = 0. 14.1)+p(i.2)).95e-3 . : −10dB ≤ γ ≤ 0dB. tZ (above) reduces.09.85.2) = (N(i+1)*T)/Pi(i).0028s respectively.2)). ρ = N6 = 15.1 10 1 10 100.3) = 1-(p(i. : 10dB ≤ γ ≤ 15dB. ρ = N5 = 73.85 57. p(i.042 = 4. ρ= 0 10 0.72e-44].38.361 = 0.54 × 10−5 = 3.5 10 10 10 101.95 × 10−3 = 0. : 5dB ≤ γ ≤ 10dB. : 0dB ≤ γ ≤ 5dB.03. N8 = 0. 0.325 = 0.176 = 0.325 . 0.0045s.0224s.361 . : 15dB ≤ γ ≤ 20dB. : 20dB ≤ γ ≤ 30dB.2) = (N(i+1)*T)/Pi(i).54e-5 3.1)+p(i. p(i. Pi = [9. ρ= ρ= MATLAB CODE: N = [0 19. else p(i.38 82.042 4.77 15. ρ = N7 = 0. Note: The problem has been solved for Ts = 10µs Pr = 10dB fD = 80Hz R1 R2 R3 R4 R5 R6 R7 R8 : −∞ ≤ γ ≤ −10dB.77.3) = 1-(p(i. .1) = (N(i)*T)/Pi(i).1) = (N(i)*T)/Pi(i). elseif i == 8 p(i.80 Hz are 0. Notice that as fD increases. p(i.085 . ρ = N3 = 57.19 73. 0023 % 0.5Hz S(τ. (c) Tm = 0.0000  ρ = 70Hz  α1 δ(τ ) α2 δ(τ − 0.0033 % 0.33m/s.0000 0 0 0.end end % p = % % 0 % 0. we can plot the arrival of the LOS ray and the multipath ray that bounces off the ground on a time axis as shown in Fig. ρ) =  0 else The antenna setup is shown in Fig.9973 0.6) ⇒ d = 16. θ = 45o .0023 % 0.33µs.9957 0.9909 0.9801 0. 15 So we have d 2 + 82 − d = 0.9934 1. (b) dc = dc = 768 m.022µs. (a) Outdoor. B −1 . B −1 = 0. θ = 0 and for the multipath component.9921 0. ∆t)] = 1 W rect 1 Wρ for 0 ≤ τ ≤ 10µsec The Scattering function is plotted in Fig.9964 0. we have flat fading.022µsec) ρ = 49.0020 0. Consider that 10 µsec ⇒ d = ct = 3km difference between length of first and last path (b) Scattering function S(τ.62 + d2 + 2d(6. For the LOS ray. power fall-off is proportional to d−2 . Since Tm 16. since delay spread ≈ 10 µsec.0066 % 0 15.022 × 10−6 3 × 108 2 2 ⇒4 d2 + 82 4 = 6.0005 0.1m fD = v cos(θ)/λ. v = fD λ/ cos(θ). the distance travelled by the LOS ray is d and the distance travelled by the first multipath component is 2 d 2 2 + 64 Given this setup. Since d 4ht hr λ dc .0036 % 0.0199 0.0023 % 0. (a) 0. ρ) = F∆t [Ac (τ.0068 0. We can use either of these rays and the corresponding fD value to get v = 23.0047 0. 15 From the figure. 16 . Selective .89µsec Ac (τ )dτ 0 = 50 Hz 1 Tm (d) βu > Coherence BW ⇒ Freq. Notice. (a) Tm ≈ .1Hz Answers based on µTm or σTm are fine too. if we choose our data rate such that a single symbol spans the average fade duration. Selective Fading ≈ Can also use µT m or σT m instead of Tm = 105 ⇒ βu > = 105 kHz (e) Rayleigh fading.1msec = 100µsec Bd ≈ . since receiver power is evenly distributed relative to delay. (b) Bc ≈ T1 = 10kHz m ∆f > 10kHz for u1 ⊥ u2 (c) (∆t)c = 10s (d) 3kHz < Bc ⇒ Flat 30kHz > Bc ⇒ Freq.8m 8m d meters Figure 4: Antenna Setting 0.0137 sec (g) Notice that the fade duration never becomes more than twice the average.94 symbols/sec R 17. So. fD = W 2 → tr = . in the worst case two symbols will span the fade duration. that based on the choice of either Tm . So our code can correct for the lost symbols and we will have error-free transmission. no dominant LOS path (f) tR = eρ √ −1 ρFD 2π 2 with ρ = 1.022 us 0 t0 t0 = (d/3e8) t1 t t1 = 2 sqrt[(d/2)^2+8^2]/3e8 Figure 5: Time Axis for Ray Arrival ∞ τ Ac (τ )dτ (c) Avg Delay Spread = 0 ∞ = 5µsec Ac (τ )dτ ∞ 0 (τ −µT m )2 Ac (τ )dτ ∞ RMS Delay Spread = Doppler Spread = W 2 0 = 2. So t1 = 72. µTm or σTm . the remaining answers will be different too. S(tau,rho) -W/2 W/2 rho 10 us tau Figure 6: Scattering Function Chapter 4 1. C = B log2 1 + C= log2 1+ NSB 1 B 0 S N0 B As B → ∞ by L’Hospital’s rule C= 2. B = 50 MHz P = 10 mW N0 = 2 ×10−9 W/Hz N = N0 B C = 6.87 Mbps. S 1 N0 ln 2 Pnew = 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. Pnoise = 0.1mW B = 20M Hz (a) Cuser1→base station = 0.933B = 18.66M bps (b) Cuser2→base station = 3.46B = 69.2M bps 4. (a) Ergodic Capacity (with Rcvr CSI only)= B[ 6 i=1 log2 (1 + γi )p(γi )] = 2.8831×B = 57.66 Mbps. (b) pout = P r(γ < γmin ) Co = (1-pout )Blog2 (1 + γmin ) For γmin > 20dB, pout = 1, Co = 0 15dB < γmin < 20dB, pout = .9, Co = 0.1Blog2 (1 + γmin ), max Co at γmin ≈ 20dB. 10dB < γmin < 15dB, pout = .75, Co = 0.25Blog2 (1 + γmin ), max Co at γmin ≈ 15dB. 5dB < γmin < 10dB, pout = .5, Co = 0.5Blog2 (1 + γmin ), max Co at γmin ≈ 10dB. 0dB < γmin < 5dB, pout = .35, Co = 0.65Blog2 (1 + γmin ), max Co at γmin ≈ 5dB. −5dB < γmin < 0dB, pout = .1, Co = 0.9Blog2 (1 + γmin ), max Co at γmin ≈ 0dB. γmin < −5dB, pout = 0, Co = Blog2 (1 + γmin ), max Co at γmin ≈ -5dB. Plot is shown in Fig. 1. Maximum at γmin = 10dB, pout =0.5 and Co = 34.59 Mbps. 5. (a) We suppose that all channel states are used 1 =1+ γ0 4 i=1 1 pi ⇒ γ0 = 0.8109 γi 1 1 − > 0 ∴ true γ0 γ4 S(γi ) 1 1 = − γ0 γi S 3.5 x 10 7 3 2.5 Capacity (bps) 2 1.5 1 0.5 0 0 0.2 0.4 Pout 0.6 0.8 1 Figure 1: Capacity vs Pout   1.2322  S(γ)  1.2232 =  1.1332 S   0.2332 C = B (b) 1 σ = E[1/γ] S(γi ) σ = γi S 4 γ γ γ γ = γ1 = γ2 = γ3 = γ4 log2 i=1 γi γ0 p(γi ) = 5.2853bps/Hz = 4.2882   0.0043  S(γ)  0.0029 =  0.4288 S   4.2882 γ γ γ γ = γ1 = γ2 = γ3 = γ4 C = log2 (1 + σ) = 2.4028bps/Hz B (c) To have pout = 0.1 or 0.01 we will have to use all the sub-channels as leaving any of these will result in a pout 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 pout that maximizes C with truncated channel inversion, we get max 6. (a) SN Rrecvd Assume all channel states are used 1 =1+ γ0 4 i=1 C = 4.1462bps/Hz B     pout = 0.5 10dB Pγ (d) 5dB = = 0dB Pnoise    −10dB w.p. w.p. w.p. w.p. 0.4 0.3 0.2 0.1 1 pi ⇒ γ0 = 0.4283 > 0.1 γi ∴ not possible Now assume only the best 3 channel states are used 0.9 =1+ γ0 3 i=1 1 pi ⇒ γ0 = 0.6742 < 1 γi γ γ γ γ = γ1 = γ2 = γ3 = γ4 = 10 = 3.1623 =1 = 0.1 ∴ ok!   1.3832  S(γ)  1.1670 =  0.4832 S   0 C/B = 2.3389bps/Hz (b) σ = 0.7491 C/B = log2 (1 + σ) = 0.8066bps/Hz (c) C B max = 2.1510bps/Hz obtained With pout = 0.1 + 0.2 = 0.3 by using the best 2 channel states. 7. (a) Maximize capacity given by C= max S(γ): S(γ)p(γ)dγ=S γ B log 1 + S(γ)γ S p(γ)dγ. Construct the Lagrangian function L= γ B log 1 + S(γ)γ S p(γ)dγ − λ S(γ) p(γ)dγ S Taking derivative with respect to S(γ), (refer to discussion section notes) and setting it to zero, we obtain, 1 1 S(γ) γ0 − γ γ ≥ γ0 = 0 γ < γ0 S Now, the threshold value must satisfy ∞ γ0 1 1 − γ0 γ p(γ)dγ = 1 Evaluating this with p(γ) = 1 −γ/10 , 10 e we have ∞ γ0 ∞ γ0 10 1 = = = 1 10γ0 e−γ/10 dγ − 1 10 ∞ γ0 e−γ/10 dγ γ (1) (2) (3) 1 −γ0 /10 1 e − γ0 10 e−γ dγ γ 1 −γ0 /10 1 e − EXPINT(γ0 /10) γ0 10 where EXPINT is as defined in matlab. This gives γ0 = 0.7676. The power adaptation becomes S(γ) = S 1 0.7676 − 1 γ 0 γ ≥ 0.7676 γ < 0.7676 7676 log (γ/0.7676) e−γ/10 dγ = 2. (a) If neither transmitter nor receiver knows when the interferer is on.3979 nats/sec/Hz.25S2 = S.25mW and C = 53.22765 which gives capacity C/B = 0.75S1 + 0.95 (4) (5) (6) 1 1 10 EXPINT(γ0 /10) nats/s/Hz. (e) Capacity using channel inversion is ZERO because the channel can not be inverted with finite average power. So for perfect channel knowledge at transmitter and receiver we compute γ0 = 0. In order to get the capacity values in bits/sec/Hz. This is because I took the natural log. Threshold for outage probability 0. (b) Suppose we transmit at power S1 when jammer is off and S2 when jammer is off. C = B max log 1 + subject to 0. This gives us the capacity with truncated channel inversion as C/B = log 1 + = log 1 + = 1. However the capacity when the transmitter knows the channel as well and can adapt its power. (c) AWGN capacity C/B = log(1 + 10) = 2.25mW . Capacity with AWGN is always greater than or equal to the capacity when only the reciever knows the channel. (f) Channel Mean=-5 dB = 0.7Kbps. S2 = 3.05 is computed as 1 10 ∞ γ0 e−γ/10 dγ = 0. can be higher than AWGN capacity specially at low SNR.5129. With AWGN. 8.36 nats/sec/Hz. as if the interferer was on all the time. This gives S1 = 12.5463 1 1 10 ∞ 1 −γ/10 dγ γ0 γ e ∗ 0.3162.21Kbps. C/B = log(1 + 0. i.95 which gives γ0 = 0. C = B log 1 + S N0 B + I = 10. the capacity numbers simply need to be divided by natural log of 2. they must transmit assuming worst case.0649 nats/sec/Hz.3162) = 0.0150 nats/sec/Hz. Note that I computed all capacites in nats/sec/Hz. With channel known only to the receiver C/B = 0. S1 No B 0.e.2748 nats/sec/Hz. At low SNR.2510 nats/sec/Hz.75 + log 1 + S2 No B + I 0. the knowledge of fading helps to use the low SNR more efficiently.25 . (d) Capacity when only receiver knows γ C/B = 1 10 ∞ 0 log (1 + γ) e−γ/10 dγ = 2.(b) Capacity can be computed as C/B = 1 10 ∞ 0.95 ∗ 0. This can be shown using Jensen’s inequality. our assumption was wrong again. then 3 =1+ γ0 1 1 1 + + 1 . our assumption was wrong.25 4 .0625. then we have 4 =1+ γ0 1 1 1 1 + + + 1 . interference power is Pint and noise power is Pnoise . . we can find capacity as: C= j:γj ≥γ0 B log2 γj γ0 This gives us.(c) The jammer should transmit −x(t) to completely cancel off the signal. γ4 = 0.8889 < 1 This time our assumption was right.0625 So. Suppose transmit power is Pt . Next we assume that 0. C = 23.48 > 0. then 2 =1+ γ0 1 1 + 1 4 ⇒ γ0 = .25.0625 < γ0 < . Pt In the first strategy C/B = log2 1 + Pint +Pnoise t In the second strategy C/B = log2 1 + PP−Pint noise Assuming that the transmitter transmits -x[k] added to its message always.4 Mbps.25 < γ0 < 1. Now. power remaining for actual messages is Pt − Pint The first or second strategy may be better depending on Pt Pint + Pnoise Pt − Pint ⇒ Pt − Pint − Pnoise Pnoise 0 Pt is generally greater than Pint + Pnoise .001×10−6 10×106 |Hj |2 S N0 B = 1.25 4 ⇒ γ0 = .0625 To compute γ0 we use the power constraint as: 1 1 − γ0 γj =1 + j First assume that γ0 < 0.001 µW/Hz B = 10 MHz Now we compute the SNR’s as: γj = This gives: γ1 = |1|2 10−3 0.25 So.0625 ⇒ γ0 = . S = 10mW N0 = .1798 > 0. and so strategy 2 is usually better. γ3 = 4.25. Now we assume that 0. So we get that only two sub-bands each of bandwidth 10 MHz are used for transmission and the remaining two with lesser SNR’s are left unused. 9. γ2 = . S = 10mW N0 = . γ2 = .5 0.25. We show this for the case of a discrete fading distribution C = Σ log 1 + (1 + N0 B j)2 P (1 + j)2 Pj N0 B  j  Pj − P  j L= i log 1 + − dj  ∂L ∂Pj (1 + j)2 Pj ⇒ N0 B (1 + j)2 P letγj = N0 B Pj ⇒ P 1 1 denote = γ0 λP Pj ∴ P subject to the constraint ΣPj P 11.001 µW/Hz B = 10 MHz Now we compute the SNR’s as: γj = This gives: γ1 = |1|2 10−3 0.25 fc-20 fc-10 fc fc+10 fc+20 f (in MHz) Figure 2: Problem 11 10. γ4 = 0.0625 To compute γ0 we use the power constraint as: 1 1 − γ0 γj =1 + j .H(f) 2 1 0.001×10−6 10×106 = 0 = 1 (1 + j)2 −1 λ N0 B = 1 1 − λP γj = 1 1 − γ0 γj = 1 |Hj |2 S N0 B = 1. γ3 = 4. the problem is similar to problem 10 ⇒ xi = 1 1 − γ0 γ(fi ) where γ0 is found from the constraints 1 1 − γ0 γ(fi ) = 1 and 1 1 − ≥ 0∀i γ0 γ(fi ) i 13.0625 So. Now we assume that 0.0625 ⇒ γ0 = . our assumption was wrong again.t. For the case of a discrete number of frequency bands each with a flat frequency response. then 2 =1+ γ0 1 1 + 1 4 ⇒ γ0 = . the problem can be stated as |H(fi )|2 P (fi ) max log2 1 + N0 s. So we get that only two sub-bands each of bandwidth 10 MHz are used for transmission and the remaining two with lesser SNR’s are left unused. i P (fi )≤P i denote γ(fi ) = |H(fi N0 )|2 P (f i) L= i log2 1 + γ(fi ) P (fi ) P + λ( P (fi )) denote xi = P (fi ) P .25 4 . we can find capacity as: C= j:γj ≥γ0 B log2 γj γ0 This gives us.0625 < γ0 < . our assumption was wrong.48 > 0. (a) C=13. Now.98Mbps .25 4 ⇒ γ0 = .25 So.1798 > 0.8889 < 1 This time our assumption was right.25.First assume that γ0 < 0. then we have 4 =1+ γ0 1 1 1 1 + + + 1 . Next we assume that 0. C = 23. 12.4 Mbps.0625. then 3 =1+ γ0 1 1 1 + + 1 .25 < γ0 < 1. MATLAB Gammabar = [1 . for j = 1:length(gamma0v) gamma0 = gamma0v(j). P = 30e-3. C = C + Bc*ss*sum(log2(gammac/gamma0ch). [b. Bc = 4e6.01:2]. gamma = hsquare*(P/Pnoise).125]. for i = 1:length(Gammabar) a = gamma.:) = (1/Gammabar(i))*exp(-hsquare/Gammabar(i)). for i = 1:length(Gammabar) pgamma(i. for i = 1:length(Gammabar) a = gamma. Pnoise = N0*Bc. ss = . N0 = .001e-6.*(gamma>gamma0).*pgammac)*ss. end end [b.001e-6./gammac). ss = .001. Pt = 30e-3.5 .5 . [b.001. end gamma0v = [1:.c] = min(abs((sumP-1))). pgammac = pgamma(i. gammac = a(find(a)). Bc = 4e6. gammac = a(find(a)). Pnoise = N0*Bc. N0 = . end (b) C=13.27Mbps MATLAB Gammabarv = [1 .*(gamma>gamma0ch).c] = max(a>0).c] = max(a>0).c:length(gamma)). pgammac = pgamma(i.c:length(gamma)). Pj_by_P = (1/gamma0)-(1. C = 0. sumP(j) = 0. gamma0ch = gamma0v(c).125]. .*pgammac). sumP(j) = sumP(j) + sum(Pj_by_P. hsquare = [ss:ss:10*max(Gammabar)]. sumP(j) = sum(Pj_by_P.*pgammac).01:. C(k) = Bc*ss*sum(log2(gammac/gamma0ch). pgammac = pgamma(c:length(gamma)).*(gamma>gamma0ch). .*(gamma>gamma0). pgamma = (1/Gammabar)*exp(-hsquare/Gammabar). pgammac = pgamma(c:length(gamma)). for k = 1:length(Gammabarv) Gammabar = Gammabarv(k). gamma = hsquare*(P/Pnoise).c] = max(a>0). end Ctot = sum(C). a = gamma.P = Pt/3. a = gamma. gammac = a(find(a)). [b.c] = min(abs((sumP-1)))./gammac). [b. gammac = a(find(a)). hsquare = [ss:ss:10*Gammabar]. gamma0v = [. end [b.*pgammac)*ss. for j = 1:length(gamma0v) gamma0 = gamma0v(j). Pj_by_P = (1/gamma0)-(1.c] = max(a>0).01:1]. gamma0ch = gamma0v(c). 7Q √ 2N0 2N0 2N0 = 2A   2 2A  = Q N0 dmin (b) p(m = 0|m = 1)p(m = 1) = p(m = 1|m = 0)p(m = 0) ˆ ˆ     A + a A − a 0. a = [0:.3Q  .25 2. p1 = 0. p0 = 0.9587 Pe = 5.0203 Pe = 3.a > 0 N0 2 N0 2 Solving gives us ’a’ for a given A and N0 (c) p(m = 0|m = 1)p(m = 1) = p(m = 1|m = 0)p(m = 0) ˆ ˆ     B  A  = 0.5 (a symbol has 2 bits in 4QAM) 0 E (c) SNR per symbol remains the same as before = Ns = 5 0 SNR per bit is halved as now there are 4 bits in a symbol Eb N0 = 1.53 × 10−6 In part c) B=0. 0 sent — 0 sent)p(0 sent) dmin dmin dmin + 0.3. for a given A we can find B E A (d) Take Nb = N0 = 10 0 In part a) Pe = 3. B = 1 2Ts ⇒ Ts = 1 2B = 5 × 10−5 s P (b) SN R = N0bB = 10 Since 4-QAM is multilevel signalling P Es 2Es SN R = N0bB = N0 BTs = N0 B BTs = 1 2 E ∴ SNR per symbol = Ns = 5 0 E SNR per bit = Nb = 2.7Q  = 0.3Q  . (a) For sinc pulse. N0 = .42 × 10−6 Clearly part (b) is the best way to decode. 1 sent — 1 sent)p(1 sent) + P r(1 detected.7*Q(A/sqrt(N0/2)). t1 = .a > 0 0.00001:1].7Q  N0 2 N0 2 Clearly A > B.Chapter 6 1.3Q √ =Q √ = 0.1.87 × 10−6 In part b) a=0. 2 MATLAB CODE: A = 1. .7 (a) Pe = P r(0 detected. s(t) = ±g(t) cos 2πfc t r = r cos ∆φ ˆ where r is the signal after the sampler if there was no phase offset.Q √min 2N0 where number of nearest neighbors = total number of points taht share decision boundary (a) 12 inner points have 5 neighbors 4 outer points have 3 neighbors avg number of neighbors = 4.t2=.p1 + 0.5Q √2a 2N 0 (b) 16QAM.p0 ] + 6 1 [0.3*Q(a/sqrt(N0/2)). [c. diff = abs(t1-t2).5 Pe = 4.4Q 1×4+4×3+4×2 Q 9 √3a 2N0 = 2. the threshold that ˆ minimizes Pe is 0 as (cos ∆φ) acts as a scaling factor for both +1 and -1 levels.exact = 1 − √ 2( M − 1) √ 1− Q M 3γ s M −1 .d] = min(diff). A2 c Tb 0 dmin cos ∆φ √ 2N0 cos2 2πfc tdt = A2 c Tb 0 1 + cos 4πfc t 2   T  b sin(4πfc Tb )  = A2  +  c 8πfc 2  →0 as fc 1 =1 2 x(t) = 1 + n(t) Let prob 1 sent =p1 and prob 0 sent =p0 Pe = 1 [1.p0 ] + [0. We will use the approximation Pe ∼ (average number of nearest neighbors).p1 + 0.p0 ] + 6 6 = A2 Tb c = ( p1 + p0 = 1 always ) d 5. a(d) c 3.p1 + 1. 2×3+3×2 Q 5 1 4 Q √2a 2N0 = 3Q √2a 2N0 √2a 2N0 √2a 2N0 = 2. Pe however increases as numerator is reduced due to multiplication by cos ∆φ Pe = Q 4.67Q √3a 2N0 2 Ps. Pe = 4 1 − (c) Pe ∼ (d) Pe ∼ 6. Once again.p1 + 0.p0 ] 6 1 1 [p1 + p0 ] = 6 6 2 2 [0. (a) ∞ Ix (a) = 0 e−at dt x2 + t2 2 since the integral converges we can interchange integral and derivative for a¿0 ∂Ix (a) ∂a x2 Ix (a) − ∂Ix (a) ∂a ∞ = 0 ∞ −te−at dt x2 + t2 (x2 + t2 )e−at dt = x2 + t2 2 2 ∞ 0 = 0 e−at dt = 2 1 2 π a . gamma = 10.01:25]. Ps_approx = ((4*(sqrt(M)-1))/sqrt(M))*Q(sqrt((3*gamma)/(M-1))). 7. At high SNR.Ps_approx.^(gamma_db/10).Figure 1: Problem 5 √ 4( M − 1) √ = Q M 3γ s M −1 Ps. The approximation error decreases with SNR because the approximate formula is based on nearest neighbor approximation which becomes more realistic at higher SNR. See figure. 8. semilogy(gamma_db.’b:’). Ps_exact=1-exp(2*log((1-((2*(sqrt(M)-1))/(sqrt(M)))*Q(sqrt((3*gamma)/(M-1)))))). hold on semilogy(gamma_db. MATLAB CODE: gamma_db = [0:.approx approximation is better for high SNRs as then the multiplication factor is not important and Pe is dictated by the coefficient of the Q function which are same. The nearest neighbor bound over-estimates the error rate because it over-counts the probability that the transmitted signal is mistaken for something other than its nearest neighbors. Ps_exact). M = 16. this is very unlikely and this over-counting becomes negligible. we get y − x2 y = − comparing with y + P (a)y = Q(a) P (a) = −x2 I. = e P (u)u 1 2 π a .10 0 Problem 2 − Symbol Error Probability for QPSK 10 Ps −100 10 −200 Approximation Exact Formula 10 −300 0 5 10 15 20 25 30 10 0 Problem 2 − Symbol Error Probability for QPSK 10 P s −1 10 −2 Approximation Exact Formula 10 −3 0 1 2 3 4 5 SNR(dB) 6 7 8 9 10 Figure 2: Problem 7 (b) Let Ix (a) = y.F. Q(a) = − 1 2 1 2 π a = e−x 2 2a ∴ e−x a y = solving we get y = (c) − π −x2 u e du a √ π ax2 e erf c(x a) 2x ∞ 0 √ 2x −ax2 2x 2 e erf c(x a) = Ix (a) e−ax = π π a=1 2 2x −ax2 ∞ e−at erf c(x) = e dt π x2 + t2 0 = √ 1 Q(x) = erf c(x/ 2) = 2 2 π 1 π π/2 0 π/2 0 e−at dt x2 + t2 2 e−x e−x 2 /sin2 θ dθ dθ 2 /2sin2 θ . 9 = 0.dB ≥ x) Pγ.dB − µPγ x − µPγ P ≥ 8 8 x − µPγ ⇒Q 8 x − µPγ ⇒ 8 ⇒ µPγ = −128. at 100 mph Tc = 0.076γs ) 2No αm = 2.074s Ts ∴ outage probability is a good measure.3122 × 10 Pγ Pt √ = Gλ 4πd 2 ⇒ 4.5 E c = dmin = 2Es (1 − cos 22. 8).39 Es Can also use formula from reader √ √ dmin 2 (b) Ps = αm Q βm γs = 2Q = 2Q( .9 = −1. 1 with fading Pe = 4γ = 0.dB ∼ N (µPγ .687 × 10−13 data requires Pe ∼ 10−6 voice requires Pe ∼ 10−3 so it can be used for both.b = √ s .3122 × 10−14 = −138. C = Θ = 22. βm = .5672dB = 1. d = 1002 + 5002 = 100 6 × 103 Pe = 1 e−γb ⇒ γb = 13.82dB Pγ. c = dmin . Ts = 15µsec 1 1 at 1mph Tc = Bd = v/λ = 0.5) = .39 × 10−13 = 0.1224 ⇒ Pγ = 1. P = 100W N0 = 4W. at 10 mph Tc = 0.076 .0074s = 7400µs > 15µs outage or outage combined with average prob of error can be a good measure. (a) Law of Cosines: √ c2 = a2 + b2 − 2ab cos C with a.8488W x = 1.2816 (b) 13. 10.01 b So the system can’t be used for data at all.74s Ts ∴ outage probability is a good measure. σdB = 8 P (Pγ.9 = 0. SN R = 25 √ √ Pe = Q( 2γ) = Q( 50) = 7. 11.9.1224 2 Pγ −14 N0 B = 13. (a) When there is path loss alone. It can be used for very low quality voice. ∞ Mγ (s) = = esγ p(γ)dγ 1 esγ e−γ/γ dγ γ 0 ∞ 0 = 1 1 − γs √ √ 12. 5 = 11. 2.2dB 14.gamma_b_bar(j). line = [’-k’. ’m = 4’). m = [1 2 4].038γs = 1 . 4. %Initializations avg_SNR = [0:0. .j) = (1/pi)*quad8(’nakag_MGF’. end figure(1).33 × 10−3 15.^(avg_SNR/10).65 Mγ (s) = ∴ Pb = For M = 4. ∞ Pb = 1 Pb (γ) = e−γ 2 Pb = But from 6.0.:).5 b Penalty = 3289. ’-b’] for i = 1:size(m. where we have used an integral table to evaluate (d) Pd = Ps 4 1 (e) BPSK: P b = 4γ = 10−3 . γ = 10 0 Pb (γ)p(γ)dγ 1 2 ∞ 0 1 e−γb pγ (γ)dγ = M 2 −m −m 1− 1 γ 1+ 2 m sγ m P b = 3. 4’). 2) Pb_bar(i.076γs . gamma_b_bar = 10. ylabel(’Average bit error probability ( P_b ) ’). line(i)).1:20]. ’m = 2’.∞ (c) Pe = ∞ 0 Ps (γs )f (γs )dγs = 0 √ αm Q( βm γs )f (γs )dγs π 2 Using alternative Q form = αm π 1+ 0 gγs (sin φ)2 gγs 1+gγs −1 dφ with g = βm 2 m = α2 1 − the integral = 1− .pi/2. ’-r’.1). 16PSK: From above get γs = 3289.2) for j = 1:size(gamma_b_bar. end xlabel(’Average SNR ( gamma_b ) in dB’). title(’Plots of P_b for BPSK modulation in Nagakami fading for m = 1.[].m(i).2dB 250 Also will accept γb (16P SK) = 822 ⇒= 5.038γs 1+. 2. hold on.[]. legend(’m = 1’. ⇒ γb = 250. semilogy(avg_SNR. Pb_bar(i. %Script used to plot the average probability of bit error for BPSK modulation in %Nakagami fading m = 1. g).p.p. γΣ = γ1 +γ2 .4 m 17.01) = 2. m is the Nakagami parameter %and g is given by Pb(gamma_b) = aQ(sqrt(2*g*gamma_b)). 1/3 30 w.24 dBm Now.28 dBm = ⇒ d = 1372.73 × 10−8 Ptarget −Pr σ mW = Pt λ 2 4πd 0 w. 1/2 10 w. So. gamma_b_bar. 1/2   5   10 γΣ =  35   40 w. w. We first assume γ0 < 5. %This function calculates the m-Nakagami MGF function for the specified values of phi.function out = nakag_MGF(phi. Ptarget −Pr σ 3. Gamma_b_bar is the average SNR per bit. (a) γ1 = γ2 = In MRC.^(-m). %phi can be a vector.^2)) ). w.p.03×10−3 1 16. γ ≥ γ0 0 γ < γ0 1 γ0 Notice that we will denote γΣ by γ only hereon to lighten notation. 1/6 1/6 1/3 1/3 (b) Optimal Strategy is water-filling with power adaptation: S(γ) = S 1 − γ ./(m*(sin(phi). w.p. For DPSK in Rayleigh fading.p. consider shadowing: Pout = P [Pr < Ptarget ] = P [Ψ < Ptarget − Pr ] = Φ ⇒ Φ−1 (. SNR = 10dB M 1 2 4 BER 2.p.p.327 = Pr = −74.53×10−3 1. Pb = 2γb ⇒ γb = 500 −9 −12 mW ⇒ P No B = 3 × 10 target = γ b N0 B = 1. 4 i=1 1 1 − γ0 γi pi = 1 4 i=1 1 ⇒ =1+ γ0 pi γi .p.5 × 10 mW = -88. 2/3 5 w.33×10−2 5. out = (1 + gamma_b_bar. m. 91 Kbps (c) Without. 18. receiver knowledge. the input to the phase comparator is z(k)z (k − 1) = gk g( k − 1) s(k)s (k − 1) + gk s(k − 1)nk−1 + g( k − 1) s (k − 1)nk + nk nk−1 but s(k) = s(k − 1) s(k)s (k − 1) = |sk |2 = 1 (b) nk = sk−1 nk nk−1 = sk−1 nk−1 z = gk gk−1 + gk nk−1 + gk−1 nk p1 p2 A B φx (s) = = + (s − p1 )(s − p2 ) s − p1 s − p2 p1 p2 A = (s − p1 )φx (s)|s=p1 = p1 − p2 p1 p2 B = (s − p2 )φx (s)|s=p2 = p2 − p1 (c) Relevant part of the pdf φx (s) = ∴ px (x) = (d) Pb = prob(x < 0) = p1 p2 (p2 − p1 ) p1 p2 L−1 (p2 − p1 ) (normalized) p1 p2 (p2 − p1 )(s − p2 ) 1 (s − p2 ) = 0 −∞ p1 p2 ep2 x (p2 − p1 ) ep2 x dx = − . 4 C=B i=1 log2 γi γ0 pi C = 451. the capacity of the channel is given by: 4 C=B i=1 log2 (1 + γi )pi C = 451.x < 0 p1 p2 − p1 .⇒ γ0 = 0. (a) s(k) = s(k − 1) z(k − 1) = gk−1 s(k − 1) + n(k − 1) z(k) = gk s(k) + n(k) From equation 5.66 Kbps Notice that we have denote γΣ by γ to lighten notation.9365 < 5 So we found the correct value of γ0 .63 . 0001.001. floor can be seen about γ b = 40dB when BD T = 0. floor can be seen between γ b = 0 to 60dB 20. Rb ≤ 200 Kbps.(e) p2 − p1 = 1 1 γb + 1 + = 2N0 [γ b (1 − ρc ) + 1] 2N0 [γ b (1 + ρc ) + 1] N0 [γ b (1 − ρc ) + 1][γ b (1 + ρc ) + 1] ∴ Pb = (f) ρc = 1 ∴ Pb = 19. Pb = 1 1 + γ b (1 − ρc ) 2 1 + γb γ b (1 − ρc ) + 1 2(γ b + 1) 1 2(γ b + 1) when BD T = 0.01. Ts ≥ 10µs. γ b 0 to 60dB ρc = J0 (2πBD T ) with BD T = 0. ISI: Formula based approach: Pf loor = σTm Ts 2 1 1− = 2 √ √ (ρc / 2)2 (2 − 2)K/2 √ √ exp − = 2.0001 where J0 is 0 order Bessel function of 1st kind. we can assume that σTm ≈ µTm = 100ns Pf loor ≤ 10−4 which gives us σTm 2 ≤ 10−4 Ts σTm Ts ≥ = 10µsec Pf loor So.138 × 10−5 1 − (ρc / 2)2 1 − ρc / 2 Since its Rayleigh fading. 0. 0. Tb ≥ 5µs. ρc = e−(πBD T ) BD = 80Hz Rician fading K = 2 ρc = 0. Data rate = 40 Kbps Since DQPSK has 2 bits per symbol. ∴ Ts = 2 40×103 = 5 × 10−5 sec DQPSK 2 Gaussian Doppler power spectrum. .001.9998 P f loor 21. floor can be seen about γ b = 60dB when BD T = 0.01. 7Kbps MSK d = 3 × 10−2 Ts = 100µsec R = 2/Ts = 20Kbps .89µs. Let Ts ≥ 10 µT . Rb ≥ 50.Thumb . BPSK d = θTm /Ts .9999 ⇒ Ts ≤ 39. Then  ⇒ ρc ≥ 0. ISI will be negligible.7Kbps QPSK d = 4 × 10−2 Ts = 75µsec R = 2/Ts = 26.26 Kbps.5 with Pb = 10−3 .9999 But ρc for uniform scattering is J0 (2πBD Ts ). From figure 6. 22.Rule approach: µt = 100 nsec will determine ISI. so ρc = J0 (2πBD Ts ) = 1 − (πfD Ts )2 ≥ 0. Combining the two. θTm = 3µs d = 5 × 10−2 Ts = 60µsec R = 1/Ts = 16.26 Kbps ≤ Rb ≤ 200 Kbps (or 2 Mbps). we have 50. As long as Ts 2bits 1 R ≤ symbol Ts symbols = 2Mbps sec Doppler: BD = 80 Hz  Pf loor = 10−4 ≥ 1 1− 2 √ 2 ρc / 2 √ 1 − ρc / 2  2 µT .79µs Tb ≤ 19. fori=1:length(M) pgamma(i. γs ≥ γ0 = 10. MATLAB CODE gamma = [0:. the mass in the pdf keeps on shifting to higher values of γ and so we have higher values of γ and hence lower probability of error. (M(i)-1).6228. M =1 γ − 0 Pout = 1 − e γ 1 M =2 γ − 0 Pout = 1 − e γ 1 M =3 γ − 0 Pout = 1 − e γ 1 = 0.*(exp(-gamma/gamma_bar)).:) = (M(i)/gamma_bar)*(1-exp(-gamma/gamma_bar)). γ 3 = 100.^. M Pout (γ0 ) = i=1 1−e − γ0 i γ γ 1 = 10.1917 γ 1−e − γ0 γ 2 1−e − γ0 3 = 0... end 3. pγΣ (γ) = M 1 − e−γ/γ γ γ = 10dB = 10 as we increase M.Chapter 7 1. Ps = 10−3 √ QPSK. ∞ M −1 Pb = = 0 ∞ 0 1 −γ e pγΣ (γ)dγ 2 1 −γ M e 1 − e−γ/γ 2 γ ∞ M −1 e−γ/γ dγ M −1 = = = M 2γ M 2γ M 2 e−(1+1/γ)γ 1 − e−γ/γ M −1 n M −1 n dγ 0 M −1 n=0 M −1 n=0 (−1)n e−(1+1/γ)γ dγ (−1)n 1 = desired expression 1+n+γ . Ps = 2Q( γs ) ≤ 10−3 .8276. M = [1 2 4 8 10].6613 γ 1−e − γ0 2 = 0. gamma_bar = 10.1:60].0197 e−γ/γ 2. γ 2 = 31. ^(gammab_dB/10).08 0. gammab = 10.05 Σ 0. See MATLAB CODE: gammab_dB = [0:.06 pγ (γ) 0. ∞ Pγ1 (γ)Pγ2 (γτ ) γ < γτ P r{γτ ≤ γ1 ≤ γ} + Pγ1 (γ)Pγ2 (γτ ) γ > γτ Pb = 0 1 −γ e pγΣ (γ)dγ 2 1 −γr /γ γe 1 −γr /γ γe pγΣ (γ) = 1 − e−γT /γ 2 − e−γT /γ γ < γT γ > γT ∞ γT Pb = = 6. M= 2.7 × 10−4 As SNR increases SSC approaches SC 7. γ1 < γ} γ < γτ P r{γτ ≤ γ1 ≤ γ} + P r{γ2 < γτ .03 0.0455 0.7 × 10−5 2.1:20].04 0.01 0 0 10 20 30 γ 40 50 60 Figure 1: Problem 2 4. pγΣ (γ) = P r{γ2 < γτ .0129 P b (20dB) 0.0.0076 0. γ1 < γ} γ > γτ If the distribution is iid this reduces to pγΣ (γ) = 5.02 0.1 M=1 M=2 M=4 M=8 M = 10 0. γT 1 1 e−γ/γ e−γ dγ + 1 − e−γT /γ 2 − e−γr /γ 2γ 2γ 0 1 −γT /γ + e−γT e−γT /γ 1−e 2(γ + 1) e−γ/γ e−γ dγ no diversity SC(M=2) SSC M 2 1 2(γ+1) 1 2(γ+1)   Pb M −1 m m=0 (−1) 1+m+γ 1 − e−γT /γ + e−γT e−γT /γ M −1 m   P b (10dB) 0. .0050 9.07 0.09 0. hold on M = 4. Pbs(j) = Pbs(j) + (M/2)*((-1)^m)*f*(1/(1+m+gammab(j))). for j = 1:length(gammab) Pbs(j) = 0 for m = 0:M-1 f = factorial(M-1)/(factorial(m)*factorial(M-1-m)).avg (DPSK) 10 −3 10 −4 10 −5 10 −6 10 −7 0 2 4 6 8 10 γavg 12 14 16 18 20 Figure 2: Problem 7 for j = 1:length(gammab) Pbs(j) = 0 for m = 0:M-1 f = factorial(M-1)/(factorial(m)*factorial(M-1-m)).’b:’).10 0 M=2 M=3 M=4 10 −1 10 −2 Pb.’).Pbs.’b--’) hold on M = 3. Pbs(j) = Pbs(j) + (M/2)*((-1)^m)*f*(1/(1+m+gammab(j))). γΣ = 1 N0 M i=1 ai γi M 2 i=1 ai 2 ≤ 1 N0 a2 i a2 i 2 γi = 2 γi N0 . end end semilogy(gammab_dB. for j = 1:length(gammab) Pbs(j) = 0 for m = 0:M-1 f = factorial(M-1)/(factorial(m)*factorial(M-1-m)). end end semilogy(gammab_dB. hold on 8. end end semilogy(gammab_dB.Pbs.’b-.Pbs. Pbs(j) = Pbs(j) + (M/2)*((-1)^m)*f*(1/(1+m+gammab(j))). 2e−15/3 = 0.12 ×10−8 ≤ 10−6 .5 (Mγ −1) = . Pb = 6. M = 4 Pb = . Equality holds if ai = cγi where c is a constant 9.2e−5N ≤ 10−6 ⇒ N ≥ 2. (a) γi = 10 dB = 10.Where the inequality above follows from Cauchy-Schwartz condition. γΣ = γ1 + γ2 + . So γΣ = 10N Pb = . γ = 10. . 10.4412 So.2e Σ −1.2e −1. + γN . Q (x) = −N (x) ∞ Pb = Q(∞) = 0. Denote N (x) = 2 √1 e−x /2 2π . P (0) = 0 Q( 2γ)dP (γ) 0 √ 1 d 2 1 Q( 2γ) = −N ( 2γ) √ = − √ e−γ √ dγ 2 γ 2 γ 2π ∞ 1 1 √ e−γ √ P (γ)dγ Pb = 2 γ 2π 0 M P (γ) = 1 − e−γ/γ k=1 ∞ (γ/γ)k−1 (k − 1)! 1 0 ∞ 1 1 √ e−γ √ dγ = 2 γ 2π M 1 2 M k=1 2 0 1 (γ/γ)k−1 1 √ e−γ √ e−γ/γ dγ = 2 γ (k − 1)! 2π k=1 Denote A = 1+ 1 γ −1/2 1 1 √ (k − 1)! 2 π ∞ e 0 1 −γ 1+ γ γ −1/2 γ γ k−1 dγ M −1 = m=0 1 1 √ m! 2 pi ∞ 0 e−γ/A γ −1/2 2 γ γ m dγ let γ/A2 = u M −1 = m=0 1 1 √ m! 2 pi M −1 ∞ 0 e−u u−1/2 A uA2 γ m A2 du = Pb = A + 2 1−A − 2 m=1 M −1 m=1 2m − 1 m A2m A 22m γ m A2m+1 22m γ m 2m − 1 m . 1 ≤ i ≤ N N = 1. .5 (M −1) γ = .0013. take N = 3. (b) In MRC. 01.11. gammab = 100. γ 1 1 √ √ 2 γ B Aγ (3) B =1+ 1 (1 + γ)2 overall P b = 12. tointeg1 = Q(sqrt(2*gamma1))./(1+gammab)).^2).01:. no diversity two two two two branch branch branch branch SC SSC EGC MRC 1 2 1 1− 2 1− Pb 1− √ Q( 2γ)pγΣ dγ √ Q( 2γ)pγΣ dγ √ Q( 2γ)pγΣ dγ √ Q( 2γ)pγΣ dγ γb 1+γ b P b (10dB) 0.*((1/gammab)*(2-exp(-gammat/gammab)).186 × 10−4 2.84 × 10−5 As the branch SNR increases the performance of all diversity combining schemes approaches the same. gamma2 = [gammat+.^2)). for i = 1:length(gammatv) gammat = gammatv(i).01:gammat].0057 0.5*(1-sqrt(1-(1. gamma = [0:. pb_mrc =(((1-Gamma)/2). .0016 P b (20dB) 0.*exp(-gamma2/gammab)). gammab_dB = [10]. end 13.0030 0.*exp(-gamma1/gammab)). MATLAB CODE: gammatv = [.01+sum(tointeg2)*. 1 2 DenoteN (x) = √ e−x /2 2π Q (x) = [1 − φ(x)] = −N (x) ∞ ∞ Pb = Q( 2γ)dP (γ) = 0 0 1 1 √ e−γ √ P (γ)dγ 2γ 2π ∞ 0 ∞ 0 1 1 √ e−γ √ dγ = 2γ 2π 0 1 1 √ e−γ √ e−2γ/γ dγ = 2γ 2π 2γ γ dγ = 1 γ ∞ 1 1 √ Γ 2 π 1 2 1+ 2 γ = 1 2 (1) (2) 1 1 √ e−γ √ 2γ 2π πγ −γ/γ e 1 − 2Q γ where A = 1 + 2 .01:. pb_egc = . anssum(i) = sum(tointeg1)*.0025 3.0021 0.67 × 10−5 1.0233 0. gamma1 = [0:.*(((1+Gamma)/2). Gamma=sqrt(gammab.^0+2*((1+Gamma)/2).*((1/gammab)*(1-exp(-gammat/gammab))./(gammab+1)).^(gammab_dB/10). gammab = 10.1:10].45 × 10−5 0. tointeg2 = Q(sqrt(2*gamma2)).01:50*gammab].^1).01:50*gammab]. /(gammab+1)).0021P b.75.5 dB Pb(γ) 10 −3 10 −4 10 −5 0 2 4 6 8 10 γ 12 14 16 18 20 Figure 3: Problem 13 √ 14. Gamma = sqrt(gammab. 10−3 = Pb = Q( 2γb ) ⇒ 4.0016 < 0.0827 k=1 √ √ ECG Pout = 1 − e−2γR − πγR e−γR (1 − 2Q( 2γR )) = 0.EGC 16. π= γ 1+γ Mγ − Pb = MATLAB CODE: gammab = 10^(1. 1 π π/2 0 1 sin2 φ = 1+ 3 γ 2 sin2 φ −2 Mγ − 1 sin2 φ dφ.12 × 10−9 .01γ sin φ)2 dφ π 0 (0.0025 4 m 18.M RC = 0.01γ √ x π/2 0 Pb = Π2 Mγi − i=1 1 sin2 φ dφ = = 1 π/2 (0.0055 0 17.1041 > Pout.5). γ = 10 k−1 0 /γ) MRC Pout = 1 − e−γ0 /γ M (γ(k−1)! = 0. If each branch has γ = 10dB Rayleigh −γ/(γ/2) γΣ = overall recvd SNR = γ1 +γ2 ∼ γe(γ/2)2 γ ≥ 0 2 BPSK ∞ Pb = Q( 2γ)pγΣ dγ = 0.01γ)2 = 0. p(γ) where 0 p(γ)e−xγ dγ = we will use MGF approach 1 π ∞ 0.5 = 5. P b. Pb = Nakagami-2 fading 1−π 2 3 2 m=0 l+m m 1+π 2 .M RC 15.10 −1 MRC EGC 10 −2 dB penalty ~ . γ = 101. for i = 1:length(gammabvec) gammab = gammabvec(i).^(-2)). phi = [0. 2 Pb = Q 3 α = 2/3./(1*(sin(phi).001. sumvec = ((1+(gammab.001. for m = 0:2 f = factorial(2+m)/(factorial(2)*factorial(m)).^2)))). Pb = 1 π π/2 1+ 0 γ 2 sin2 φ −2 1+ γ sin2 φ −1 dφ gammab_dB = [5:.^(gammab_dB/10).001:pi/2]. phi = [0.^(-1)). Mγ g − 2 sin φ π/2 0 2γb (3) sin g = 3 sin2 = π 8 π 8 −1 gγ 1+ sin2 φ −M α Pb = π gγ 1+ sin2 φ dφ . gammabvec = 10. end 10 −2 10 −3 10 Pbavg 10 −4 −5 10 −6 10 −7 5 10 γavg (dB) 15 20 Figure 4: Problem 19 20./(2*(sin(phi).001:pi/2].*((1+. pb_nakagami(i) = (1/pi)*sum(sumvec)*./(2*(sin(phi)..^(-6). sumf = sumf+f*((1+Gamma)/2)^m.001:. end pb_rayleigh = ((1-Gamma)/2)^3*sumf.001:. (gammab. pb_nakagami = (1/pi)*sum(sumvec)*.sumf = 0.^2)))).^2)))). 19..1:20]. sumvec = (1+(gammab. 001:. g = 3*sin(pi/8)^2. phi = [0.001. for k = 1:length(M) for i = 1:length(gammabvec) gammab = gammabvec(i). gammab_dB = [5:. alpha = 2/3.MATLAB CODE: M = [1 2 4 8]. end end 10 0 10 −5 Pb avg 10 −10 10 −15 5 10 γavg (dB) 15 20 Figure 5: Problem 20 .1:20]. sumvec = ((1+((g*gammab).001:pi/2]./(1*(sin(phi).^(-M(k))). pb_nakagami(k. gammabvec = 10.i) = (alpha/pi)*sum(sumvec)*.^2)))).^(gammab_dB/10). Q(z) = Q2 (z) = Ps (γs ) = 1 π 1 π 4 π 4 π Ps = = 0 π/2 exp − 0 π/4 z2 dφ sin2 φ z2 dφ 2 sin2 φ π/2 . .z > 0 gγs dφ − sin2 φ gγs dφ sin2 φ exp − 0 ∞ 1 1− √ M 1 2 1− √ M exp − 0 π/4 exp − 0 Ps (γΣ )pγΣ (γΣ )dγΣ 1 1− √ M 1 1− √ M 1 1− √ M 1 1− √ M π/2 0 2 0 π/2 0 2 0 π/4 0 0 ∞ ∞ 4 π 4 π exp exp gγΣ sin2 φ gγΣ sin2 φ g sin2 φ pγΣ (γ)dγΣ dφ − pγΣ (γ)dγΣ dφ But γΣ = γ1 + γ2 + .5 16−1 MQAM: Formula derived in previous problem with g = P s = 0. + γM = Σγi = 4 π 4 π 22.z > 0 . K = 2.^(gammab_dB/10). . .21. Rayleigh: Mγs (s) = (1 − sγ s )−1 Rician: Mγs (s) = MPSK 1+k 1+k−sγ s ΠM Mγi − i=1 dφ − dφ π/4 ΠM Mγi − i=1 g sin2 φ exp ks γ s 1+k−sγ s (M −1)π/M Ps = Three branch diversity Ps = g = sin2 π 16 1 π 0 M γs − g sin2 φ dφ → no diversity (M −1)π/M 1+ 0 gγ sin2 φ −1 (1 + k) sin2 φ kγ s g exp − (1 + k) sin2 φ + gγ s (1 + k) sin2 φ + gγ s 2 dφ = 0.1670 1.0553 = 1. gammab = 10.5 15 MATLAB CODE: gammab_dB = 10. ^(-1)). g*gammab)). g = 1.^2+.. phi2 = [0..001. 1+K)*sin(phi1).^2). phi1 = [0. alpha = 2/3..001:.001:pi/4]. pb_mrc_psk = (1/pi)*sum(sumvec)*.. (-1))..001. phi1 = [0./((..^2).^.^(gammab_dB/10)./(sin(phi)./((1+K)*.^2). sumvec=((1+((g*gammab)./(sin(phi2)./((1+K)*sin(phi2)..*(((((1+K)*sin(phi1)./(sin(phi1).001:. g = 1.^2+.^2+g*gammab)))..*exp(-(K*gammab*g)..1:20]..^(-1)).. pb_mrc_qam = (4/pi)*(1-(1/sqrt(16)))*sum(sumvec1)*.^2).5/(16-1).^2)))./((1+K)*sin(phi2).001:.*((((. gammabvec = 10..001:.*exp(-(K*gammab*g)..*exp(-(K*gammab*g). .001:pi/2]. for k = 1:length(M) for i = 1:length(gammabvec) gammab = gammabvec(i)./((1+K)*sin(phi).../((1+K)*sin(phi).^2+g*gammab))).^2))). 10 0 10 −1 10 −2 10 −3 10 avg −4 Ps 10 −5 10 −6 10 −7 10 −8 10 −9 5 10 15 20 Figure 6: Problem 22 23. sumvec2=((1+((g*gammab).^2).. g*gammab)).5/(16-1).^2).^2))). phi = [0. gammab_dB = [5:.^2+g*gammab)).001 ..g = sin(pi/16)^2.001:pi/2].^2+g*gammab))). sin(phi1).*((((. (4/pi)*(1-(1/sqrt(16)))^2*sum(sumvec2)*. (1+K)*sin(phi2).001:pi*(15/16)]. (1+K)*sin(phi). sumvec1=((1+((g*gammab). MATLAB CODE: M = [1 2 4 8]. sumvec1 = ((1+((g*gammab). sumvec2 = ((1+((g*gammab).001:.001 .^2)))).phi2 = [0. pb_mrc_qam(k..^(-M(k))).^2))))..i) = (4/pi)*(1-(1/sqrt(16)))*sum(sumvec1)*. end end ./(1*(sin(phi2)./(1*(sin(phi1). (4/pi)*(1-(1/sqrt(16)))^2*sum(sumvec2)*..001.^(-M(k))).001:pi/4]. e.7164   1.Chapter 10 1. (a) (AAH )T = (AH )T .7152 0  Σ= 0 0 0. i.5896 −0.1298 U =  −0.8685 −0.4793 0. λ = λ. eigen-values are real AAH = QΛQH (b) X H AAH X = (X H A)(X H A)H = X H A ≥ 0 ∴ AAH is positive semidefinite.6508 −0.AT = (AT ) AT = AAH ∴ (AAH )H For AAH .4272 −0. (c) IM + AAH = IM + QΛQH = Q(I + Λ)QH AH positive semidefinite ⇒ λi ≥ 0∀i ∴ 1 + λi > 0∀i ∴ IM + AAH positive definite (d) det[IM + AAH ] = det[IM + QΛQH ] = det[Q(IM + ΛM )QH ] = det[IM + ΛM ] = Πi=1 Rank(A) T = AAH (1 + λi ) det[IN + AH A] = det[IN + QΛQH ] = det[Q(IN + ΛN )QH ] = det[IN + ΛN ] = Πi=1 AAH and AH A have the same eigen-value ∴ det[IM + AAH ] = det[IN + AH A] 2.1302  Rank(A) (1 + λi ) .2513 0.6855  −0.7034 0 0 0 0. H = U ΣV T  −0. . Check the rank of each matrix rank(HI ) = 3 ∴ multiplexing gain = 3 rank(H2 ) = 4 ∴ multiplexing gain = 4 5. 6.86 for i = 2.2191 0.7813 F = −.4727 −.0145  −0. (a) Any method to  .23 .3458 0.2198 0.7116 −0.4263  −0. 0 2 P γi = λio B = 94.5 for i = 1.2407 −.8894 .0251 −.8473 −.68 is clearly too small for data transmission .09 D= .05 .68 for i = 3 γ3 = .08 .6849 0.4188  −.8478 .6109 0. C= i=1 RH log2 1 + λi ρ Mt Constraint Vi = ρ λi = constant ∴ ∂C ρ 1 1 ρ = − =0 λi ρ ∂λi Mt ln 2 (1 + ) Mt ln 2 (1 + λi ρ ) Mt Mt ⇒ λi = λj ∴ when all RH singular values are equal.3622 Thus Y = F(H)MX + FN = U ∗ U ΛVV ∗ X + U ∗ N = ΛX + U ∗ N (c) 1 1 1 1 γo − γi for γi > γo . N Assume γ2 > γ0 > γ3 since Pi P = else 6.9017 3. this capacity is maximized.3311 −0.3887 M = −.5708 0. H = U ΣV T Let   1 0 U = 0 1  0 0   1 0 V = 0 1  0 0   Σ= 1 0 0 2  1 0 0 ∴H= 0 2 0  0 0 0 4.8540 .14 where : dij = ij ij × 100 H . −0.11 H −H .3460 −.0708   V =  −0.13 .2423 −.5195 −.13 show H ≈ U ΛV is acceptable.4629   −. For example:  .10 (b) precoding filter M = V −1 shaping filter F = U −1   −.3876 .9078 .2373 . it will perform an SVD decomposition of H as H = U ΣV HRX H H = (U ΣV )RX (U ΣV )H By Hadamard’s inequality we have that for A ∈ n×n det(A) ≤ Πn Aii i=1 with equality iff A is diagonal. C= ρi : max i ρi ≤ρ B log2 (1 + λi ρi ) i √ where λi are the RH non-zero singular values of the channel matrix H and ρ is the SNR constraint. C = max B log2 det[IM γ + HRX H H ] RX : Tγ (RX ) = ρ If the channel is known to the transmitter. The SNR (3) is then given by: SN R = cH H H Hc = (.73 P1 P2 P = . the beamforming vector is given by c = √1 [1 1 1].9 kbps Pi P (d) With equal weight beamforming. N0 B (1) This gives a capacity of 630. The SNR achieved with beamforming is smaller than the best channel in part (c). where RH is the rank of the channel matric H.2 1 1 = 1 ⇒ γ0 − γ1 − γ2 = 1 ⇒ γ0 = 1. i ρi ≤ρ 8. The capacity of the channel is found by the decomposition of the channel into RH parallel channels.35 kbps.78)(100) = 78. say = Ω then det(IM R + HRX H H ) = det(I + ΩΣ2 ) ∴ C = max Bσi log2 (1 + λi ρi ) √ where λi are the singular values. The resulting capacity is given as C= i:γi ≥γ0 B log2 (γi /γ0 ) . then the SNR with beamforming would be equal to the largest SNR in part(c). The beamforming SNR for the given c is greater than the two smallest eigenvalues in part(c) because the channel matrix has one large eigenvalue and two very small eigenvalues.5676 P = . We choose RX to be diagonal. If we had chosen c to equal the eigenvector corresponding to the best eigenvalue. 7. γi = λi ρ Then the optimal power allocation is given as Pi = P 0 1 γ0 − 1 γi γi ≥ γ0 γi < γ 0 (2) for some cut-off value γ0 .4324 C = B log2 1 + γ1 P1 + log2 1 + γ2 P2 P P = 775. . hence the assumption was correct. C = 12. hMr 1   . γ1 = 80. 4 γ0 = 1 1 + 4 γi i=1 which gives γ0 = 3. . hMr Mt M r ×Mt . γ2 = 40.   . . γ1 = 40.4732 bits/sec/Hz B For  1 1 1 −1  1 1 −1 1   H=  1 −1 1 1  1 −1 −1 −1  RH = 4.   ..  9. .For 1  1 H=  1 1   1 −1 1 1 −1 −1   1 1 1  1 1 −1 RH = 3. h1Mt . γ3 = 40.6780 < mini γi . γ2 = 40. γ3 = 40. C = 13.8236 < mini γi .. hence the assumption was correct. We first assume that γ0 is less than the minimum γi which is 40...7720 bits/sec/Hz B h11  . . γ4 = 40.   .. We first assume that γ0 is less than the minimum γi which is 40.  . H =  . . γ0 = 3 1+ 3 1 i=1 γi which gives γ0 = 2.. . ^(rho_dB/10).Mt.’complex’). i = j Mt hj1 . rho = 10. hiMt ]   Mt →∞ Mt  1 = lim Mt →∞ Mt = Ek hik hjk = Ek (hik )Ek (hjk ) = 0 ∀i. L. . . [F. We find the capacity by randomly generating 103 channel instantiations and then averaging over it.0. We assume that distribution is uniform over the instantiations. MATLAB CODE clear. j. rho_dB = [0:25]. . for j = 1:min(Mt. for k = 1:length(rho) for i = 1:100 H = wgn(Mr. hiMt ]   Mt →∞ Mt  1 Mt →∞ Mt lim 2 Mt hi1 . clc. Mt = 1.’dBW’. . .Mr) . Mr = 1. . . .Denote G = HH T  1 Gii = lim Mt →∞ Mt   1 lim [hi1 .i=j lim   1 = lim [hi1 . M] = svd(H). hiMt       = hij j=1 2 = Ej hij = σ2 = 1 ∀i 1 Gij Mt  Mt →∞. hjMt       hik hjk k=1 ∴ lim ∴ lim B log2 det IM M →∞ M →∞ 1 HH T M ρ HH T + M = IM = B log2 det [IM + ρIM ] = B log2 [1 + ρ] det IM = M B log2 [1 + ρ] 10. j). gammatemp1 = gammatemp. end C(i) = sum(log2(gammatemp1. end Cergodic(k) = mean(C).sigma(j) = L(j./gamma0)). gamma0 = 1e3. clc. We assume that distribution is uniform over the instantiations./gammatemp1)). gamma0 = length(gammatemp1)/(1+sum(1. We find the capacity by randomly generating 104 channel instantiations and then averaging over it. . MATLAB CODE clear. gamma = rho(k)*sigma_used. gammatemp1 = gammatemp. end 25 Mt = Mr = 3 Mt = 2 Mr =3 Mt = Mr = 2 Mt = 2 Mr =1 Mt = Mr = 1 20 15 Cergodic 10 5 0 0 5 10 ρ (dB) 15 20 25 Figure 1: Problem 10 11. %% Now we do water filling\ gammatemp = gamma. while gamma0 > gammatemp1(length(gammatemp1)). Mt = 1. end sigma_used = sigma(1:rank(H)). gammatemp = gammatemp(1:length(gammatemp)-1). C(i) = sum(log2(1+gamma/Mt)). end sigma_used = sigma(1:rank(H)). L. pout = sum(C<Cout(k))/length(C). for j = 1:min(Mt.Mt. while pout > . end end 25 Mt = Mr = 3 Mt = 2 Mr =3 Mt = Mr = 2 Mt = 2 Mr =1 Mt = Mr = 1 20 15 Coutage 10 5 0 0 5 10 15 ρ (dB) 20 25 30 Figure 2: Problem 11 .^(rho_dB/10). rho_dB = [0:30]. rho = 10.Mr) sigma(j) = L(j. gamma = rho(k)*sigma_used. end if Cout(k)<0.Mr = 1.j). [F. for k = 1:length(rho) for i = 1:1000 H = wgn(Mr. Cout(k) = 0.’dBW’. pout = sum(C<Cout(k))/length(C).0.01 Cout(k) = Cout(k)-. end Cout(k) = mean(C).1. M] = svd(H).’complex’). Notice that the singular values of H are the square root of the eigen values of HH H (Wishart Matrix).1818 =  −0.6  0. Mr P (u n < X) = P i=1 Mr ui ni < X ui P (ni < X) = i=1 = P (ni < X) ∴ the statistics of u n are the same as the statistics of each of these elements 13.17 in reader.  x 2 = λmax ρ N  0.3 0.5294 and vopt  −0.5 0.4190  −0.9 H =  0.3 0. Σx = = u Hvx u Hv 2 x H 2 2 = vH H H u u Hv x 2 2 = v H H H Hv x = v H QH Qv x ≤ λmax x 2 with equality when u. we get that the maximum singular value of H is 1. during beamforming from eq.7101 uopt =  −0. Using Matlab.2 0.4480 and the singular vectors corresponding to this value are   −0.8896  T It is easy to check that uT uopt = 1 and vopt vopt = 1 and that opt uT ∗ H ∗ vopt = 1.12. v are the principal left and right singular vectors of the channel matrix H ∴ SN Rmax = λmax 14. 10. y = (uT Hv)x + uT n .1 0.4641  −0.4480 opt Since. for beamforming. u and v correspond to the principal singular vectors (or the singular vectors corresponding to the maximum singular value of H).7 When both the transmitter and the receiver know the channel.1 0. noise power is not increased. eq.2477 2 Alternatively. For. SNR is simply (1. When the channel is not known to the transmitter.and for a given transmit SNR of ρ. 10. we get that the maximum singular value of h is 1. noise power is not increased.0968. the received SNR is given as SNRrcvd = ρ(uT Hv2 )2 2 since. (a) ρ = 10 dB = 10 Pe = ρ−d So to have Pe ≤ 10−3 .2477 and the singular vector corresponding to this value is   0. Solving the equation that relates diversity gain d to multiplexing gain r at high SNR’s we get d = (Mr − r)(Mt − r)  .5090  0. we should have d ≥ 3.5090  u2 = ||h|| 0. 15. For a given transmit SNR of ρ. SNR is simply (1. or at least d = 3.2477)2 = 1.6941 u2 =  0. from MRC concept we know that:  0.17 in the reader can be written as y = (uT h)x + uT n To maximize SNR we need to find a u2 of norm 1 such that (u h) is maximized. the received SNR is given as SNRrcvd = ρ(uT Hvopt )2 opt since.5567. For. it allocates equal power to all the antennas and so the precoding vector (or the optimal weights) at the transmitter is given as   1 1   1 v2 = √ 3 1 Define h = Hv2 So. ρ = 1. ρ = 1. Using Matlab. u2 has norm 1.5090 It is easy to check that uT u2 = 1 and that 2 uT ∗ h = 1. uopt has norm 1.5090 hH where ||h|| is the L2 norm of h.6941 =  0.4480)2 = 2. First assume that γ0 is less than both γ1 and γ2 . we can find the data rate as R = r log2 (ρ) = 9.038bps/Hz 17.64 We have that r ≤ min{Mr . the transmitter will use the optimal precoding filter and the receiver will use the optimal shaping filter to decompose the MIMO channel into 2 parallel channels.2422 . so r ≤ 4 and so r = 3.5946 . But we know that r has to be an integer.5757 −.7 .5946 . No credits for this part: If we are allowed to assume that equations 10. H= P = 10mW N0 = 10−9 W/Hz B = 100 KHz (a) When H is known both at the transmitter and at the receiver.35 or 8. Finding the γi ’s λ2 P 1 = 75. r = 3. we take the nearest integer which is smaller than the calculated value of r.08 N0 B . we can find d as d = (Mr − r)(Mt − r) = (8 − 3)(4 − 3) = 5 For this value of d. which gives us r=3 .96 bits/s/Hz (b) With.92 N0 B λ2 P 2 = 11.23 and 10. Then 1 1 − γ0 γ1 + 1 1 − γ0 γ2 =1 .8041 −.8507 γ1 = γ2 = Finding γ0 Now. we have to find the cutoff value γ0 .5757 −.24 hold at finite SNR’s too and we are given that we can use base 2 for logarithms.10) = 4. Pe = ρ−d = 10−5 16. √ According to SVD of h λ = 1.5 .3328 −. We can then do water-filling over the two parallel channels available to get capacity.8507 .242 ∴ C/B = log2 (1 + λρ) = log2 (1 + 1. So.3 .2 = −.⇒ 3 = (8 − r)(4 − r) Solving for r we get r = 3.8713 0 0 . Mt }.8041 .35. Notice that γ1 and γ2 have already been calculated in part (a) as γ1 = 75. γ2 }. First assume that γ0 < {γ1 .81 1 + γ1 + γ2 ⇒ which is less than both γ1 and γ2 values so our assumption was correct. Finding γ0 or γK We now find the cut-off γ0 . Finding capacity Now we can use the capacity expression as 2 C= i=1 B log2 γi γ0 = 800 Kbps (b) Total is Essentially we have two parallel channels after the precoding filter and the shaping filter are used at the transmitter and receiver respectively.2 1 1 =1+ + γ0 γ1 γ2 1 ⇒ γ0 = 1 1 = 1.08. γK = γ0 /K = 5.5 = . 1 1 − γ0 γ1 K ⇒ + 1 1 − γ0 γ2 K =1 2 1 1 =1+ + γ0 γ1 K γ2 K 1 ⇒ γ0 = = 1.4649 1 1 1 1 + K γ1 + γ2 which is less than both γ1 and γ2 values. R is given as R = B log2 γ1 γK + log2 γ2 γK .92 and γ2 = 11. M (γ) = 1 + γK Finding K −1. so our assumption was correct.1742 Finding Rate R Therefore the total rate.283 ln(5Pb ) S(γ) S K= γK = γ0 /K. 92 = 0 Using the approx.Mr) . 18. So we get that γs = 75.e. the transmitter and the receiver use the principal left and right eigen vectors of the Wishart Matrix HHH . Comparing with previous part Comparing with part (b).Mt. γs = γb . one strong channel which gives a much less value of Pb . Once this is done the SNR at the combiner output is simply λmax ρ.7592 and ρ was calculated to be 100. we can get more diversity advantage i. we can see that the rate R decreases by 397.^(rho_dB/10). If we are willing to decrease the rate at which we transmit.0.92 . where λmax is the maximum eigen value of the Wishart Matrix HHH and ρ is NPB 0 Finding γs As given in the question. Finding Pb When using BPSK. (c) Since now we use beamforming to get diversity only. L. rho = 10.4 × 10−35 ∼ 0 whereas earlier it was 10−3 .4 × 10−35 Using Matlab Credit is given for either value.’complex’). Now we can use the expression for Pb for BPSK Pb = Q 2γb = Q √ 2 × 75. [F.97 This gives that R=497. given in the Ques. we get the rate using BPSK to be R=100 Kbps . for j = 1:min(Mt.⇒ R = B4. for k = 1:length(rho) for i = 1:1000 H = wgn(Mr. Finding Rate R Since we are using BPSK and are given that B = 1/Tb . 3.36 Kbps and the Pb improves as Pb is now 3.36 Kbps (Obviously less than ergodic capacity). Mt = 4.’dBW’. M] = svd(H). (d) Therefore we see that we can tradeoff rate for robustness of the system. λmax is 0. Mr = Mt. rho_dB = [0:20]. clc. (a) clear. for j = 1:min(Mt. L. M] = svd(H). end Cout(k) = mean(C).’complex’).sigma(j) = L(j.Mt. end C(i) = sum(log2(gammatemp1. while gamma0 > gammatemp1(length(gammatemp1)). gamma0 = 1e3.j). end sigma_used = sigma(1:rank(H)). Mr = Mt. gamma = rho(k)*sigma_used. Mr = Mt. rho = 10.j). gamma0 = length(gammatemp1)/(1+sum(1. using Matlab we get Cout = 7. end Cergodic(k) = mean(C). Mt = 4. gammatemp1 = gammatemp.8320 MATLAB CODE clear. rho = 10. gammatemp1 = gammatemp./gamma0)).’dBW’. [F./gammatemp1)). end sigma_used = sigma(1:rank(H)). end (b) clear. %% Now we do water filling\ gammatemp = gamma. clc.^(rho_dB/10). gamma = rho(k)*sigma_used. end 19. rho_dB = 10.0. for k = 1:length(rho) for i = 1:1000 H = wgn(Mr. rho_dB = [0:20]. for k = 1:length(rho) for i = 1:1000 . clc.Mr) sigma(j) = L(j. gammatemp = gammatemp(1:length(gammatemp)-1). Mt = 4. C(i) = sum(log2(1+gamma/Mt)).^(rho_dB/10). Cout(k) = 0. MATLAB CODE clear. while pout > . clc. end if Cout(k)<0.Mr) sigma(j) = L(j. As µ increases.j).25 Cergodic Mt = Mr = 1 C M =M =1 out t r Cergodic Mt = Mr = 4 Cout Mt = Mr = 4 20 15 C 10 5 0 0 2 4 6 8 10 ρ (dB) 12 14 16 18 20 Figure 3: Problem 18 H = wgn(Mr. end Cout(k) = mean(C).’dBW’. end end 20. . pout = sum(C<Cout(k))/length(C). pout = sum(C<Cout(k))/length(C).Mt. C(i) = sum(log2(1+gamma/Mt)). L. M] = svd(H). for j = 1:min(Mt. [F. end sigma_used = sigma(1:rank(H)). the span of cdf becomes narrower and so capacity starts converging to a single number.1 Cout(k) = Cout(k)-.0.’complex’).01. gamma = rho(k)*sigma_used. 2 0.6 FX(x) 0. Empirical CDF’s of Capacity w/o Tx CSI 1 M=4 M=6 M=8 0. end [f. M] = svd(H).x] = ecdf(C).j).1 0 6 7 8 9 10 11 12 x 13 14 15 16 17 18 Figure 4: Problem 20 .9 0.0. L.8 0.7 0.4 0. rho_dB = 10. end sigma_used = sigma(1:rank(H)).Mt = 8. rho = 10. C(i) = sum(log2(1+gamma/Mt)). j = 1:min(Mt.5 0.^(rho_dB/10).’dBW’.Mt.Mr) sigma(j) = L(j. for 1:1000 wgn(Mr. Mr = Mt. gamma = rho*sigma_used.3 0. for i = H = [F.’complex’). (a) For the first channel: ISI power over a bit time = A2 Tb /Tb = A2 For the 2nd channel: 2 (n+1)T ISI power over a bit time = Ab ∞ nTb b e−t/Tm dt = 2e−1/2 A2 n=1 T . there is tremendous noise enhancement and so the equalizer will not improve system performance.Chapter 11 1. 2. As seen from the noise power values. the noise power will be 2BN0 = 105 N0 W. B = 50 KHz. See Fig 1 2B = 100 KHz fc-B fc = 100 MHz Figure 1: Band of interest. fc = 100 MHz 1 =f H(f ) fc+B Heq (f ) = Noise PSD = N0 W/Hz. Using this we get Noise Power = = N0 |Heq (f )|2 df fc −B fc +B N0 f 2 df fc −B 3 (fc +B) f 3 (fc −B) fc +B (1) (2) (3) (4) (5) = N0 = N0 (fc + B)3 − (fc − B)3 3 = 1021 N0 W Without the equalizer. 1 τ 1 + j2πf 1 + j2πf τ Heq (f ) = (b) B −B Sx (f )|H(f )|2 |Heq (f )|2 df B 2 −B N0 |Heq (f )| df B 2 −B Sx (f )|H(f )| df SNReq = SNRISI 2BN0 . (6) Heq (f ) = ∞ H(f ) = 0 e− τ e−j2πf t dt (7) (8) = Hence.5µs = 181.8Kbps If baseband signal =100KHz: pulse width = 10µs Data rate = 2/10µs + 10µs = 100Kbps 3.S(t) A T /2 m t h (t) 1 1 T m t h (t) 2 t Figure 2: Problem 2a (b) No ISI: pulse interval = 11/2µs = 5.w.5µs ∴ Data rate = 1/5. (a) h(t) = τ = 6 µ sec 1 H(f ) t e− τ 0 t t≥0 o. .transform of 1/F(z) Show that this choice of tap weights minimizes 1 − (ω−N z N + . (1) . . we need to use some approximation to come up with the filter tap Ts coefficient values. 1 Thus. Now. we have a0 =1. . 1 F (z) where c1 = a1 . + ωN z −N ) F (z) at z = ejω If F(z) is of length 2 and monic. .. c2 = a2 . . 4.9364 = −0. a1 < 1 1 2 .. ωi = ci where {ci } is the inverse Z. Any other reasonable way is also accepted. −B ≤ f ≤ B ⇒ N0 S 2 2B + 8π B 3 3 τ2 B −B 2BS |H(f )|2 df 2BN0 = 2 1 + 4π B 2 3 τ2 2B 1.28 dB (c) h [n] = 1 + e H(z) = 1 + e ∞ −Ts τ δ [n − 1] + e −2Ts τ −2T s τ δ [n − 2] + .617 × 10−6 = 0.h (t) 1 1 10us t X (t) 1 1us t h (t)*X(t) 1 1 12us t Figure 3: Problem 2b Assume Sx (f ) = S. we get Heq (z) = 1−e− τ z −1 . . −3Ts τ −Ts τ z −1 + e z −1 n z −2 + e z − Ts τ z −3 + . a two tap filter is sufficient. 1 1− Ts e− τ z −1 = n=0 e− Ts τ = z−e = ⇒ Heq (z) = 1 H(z)+N0 . a1 = −0..0039 as the tap coefficient values. . say F (z) = 1 − a1 z then 1 = 1 + a1 z −1 + a2 z −2 + . If we assume N0 0 (the zero-forcing assumption). For Ts = 30 ms. It is easy to see that the coefficients become smaller and smaller.2193M bps T (d) We use M=4 non overlapping subchannels.5 for Pb = 10−3 K = 0.5 3: fc ≤ f < fc + 10M Hz α3 = 2 4: fc + 10M Hz ≤ f < fc + 20M Hz α4 = 0. 2. . n Ts FZ (f ) = 1 TS F n=−∞ f+ = 1 ∴ folded spectrum of f(t) is flat.2125mW ∴ SN R = 47.52 × 10M Hz + 42 × 10M Hz] = 0.5γ/M −1 or M ≤ 1 + −ln(5PR) b for Pb = 10−3 M ≤ 14.w. each with B=10MHz bandwidth 1: fc − 20M Hz ≤ f < fc − 10M Hz α1 = 1 2: fc − 10M Hz ≤ f < fc α2 = 0.w. (a) F{f (t)} = T 0 ∞ 4 i=1 log(Kγi /γ0 ) = 419.6523 P2 = 2.5SN Pb ≤ 0. 3. Hence we get that ωi = ci minimizes (1).0588 = 16.9711M bps |f | < 1/T o.25 P α2 i Power optimization: γi = N0 B for i = 1.7207 thus R = 2B 6.1225 and γ0 = 3. So if we had the opportunity to cancel any (2N+1) coefficients we will cancel the ones that are closest to z 0 . (a) Heq (f ) = 1 H(f ) for ZF equalizer   1    2  0. The result can be similarly proved for length of F(z) greate than 2 or non-monic.2831 Pi = P 1 γ0 − 0 1 Kγi γi ≥ γ0 /K γi ≤ γ0 /K (10) We can see that all subchannels will be used and P1 = 2.3228 (M ≥ 4 thus using the formula is reasonable) R = log2sM = 307.5464 P3 = 2. 4 γ1 = 1000 γ2 = 250 γ3 = 4000 γ4 = 62.6788 P4 = 2.73dB (c) Ts = 0. (9) (b) S=10mW Signal power N = N0 [12 × 10M Hz + 22 × 10M Hz + 0.5 Heq (f ) =   4    0 fc − 20M Hz ≤ f < fc − 10M Hz fc − 10M Hz ≤ f < fc fc ≤ f < fc + 10M Hz fc + 10M Hz ≤ f < fc + 20M Hz o.0125µsec 1.2e−1. 5. |t| < Ts 1 Noise whitening filter : G (1/z ) m . ISI → ∞ as N → ∞ 7.i=0 N 1 πt0 /T − πi = sin(πt0 /T ) = 2 sin(πt0 /T ) π i=1 N −1 1 + πt0 /T − πi πt0 /T − πi n2 n − t2 /T 2 0 n=1 Thus.i=k Xi f ((k − i)T + t0 ) ISI (c) N +k ISI = i=−N +k.(b) yk = y(kT + t0 ) ∞ = i=−∞ N Xi f (kT + t0 − iT ) Xi f ((k − i)T + t0 ) i=−N N +k = = Xk sinc(t0 ) + i=−N +k.i=k Xi sin (π(k − i) + t0 /T π) π(k − i) + t0 /T π N = sin(πt0 /T ) i=−N. gm (t) = g (−t) = g(t) = sinc(t/TS ). 0. Jmin = 1 − ∞ j=−∞ cj f−j B(z) = C(z)F (z) F (z)F (z −1 ) = F (z)F (z −1 ) + N0 X(z) = X(z) + N0 1 B(z) ∴ b0 = dz 2πj z 1 X(z) = dz 2πj z[X(z) + N0 ] = ∴ Jmin = 1 − T 2π π/T −π/T T 2π π/T −π/T X(ejωT ) dω X(ejωT ) + N0 X(ejωT ) X(ejωT ) + N0 dω = = T 2π T 2π π/T −π/T π/T N0 jωT ) X(e T −1 + N0 dω N0 dω + 2πn/T )|2 + N0 −π/T −0...5Ts Ts 1df = 1 −0.5Ts ∴ 0 ≤ Jmin ≤ 1 .5Ts −0.8..5Ts ∞ n=−∞ |H(ω = Ts N0 df FΣ (f ) + N0 9. VW J J ∴ = = ∂J ∂J .5Ts Jmin N0 FΣ (f ) + N0 N0 FΣ (f ) + N0 ∴ Jmin = ≥ ≤ ≤ Ts 0 −0.. ∂w0 ∂wN wT Mv w − 2 {Vd w } + 1 ∂J = 2Mv wT − 2Vd ∂w ∂J = 0 ⇒ 2Mv wT = 2Vd ∂w T −1 H ⇒ wopt = Mv Vd 10.5Ts N0 df FΣ (f ) + N0 ∴ Jmin ≥ 0 N0 =1 N0 0. 69))(10kHz) = 0.78 + 2.1mW N0 (c) The noise spectrum at the output of the filter is given by N (f ) = (H(f )+α)2 .11.134 mW For α = 1 we get N = 2N0 (. For α = .3e n −j4π f + T s MMSE equalizer : 0. (a)  1    2   1 Hzf (f ) = = 3 H(f )   4     5 0 ≤ f ≤ 10KHz 10KHz ≤ f ≤ 20KHz 20KHz ≤ f ≤ 30KHz 30KHz ≤ f ≤ 40KHz 40KHz ≤ f ≤ 50KHz (b) The noise spectrum at the output of the filter is given by N (f ) = N0 |Heq (f )|2 .44 + 1.0516 mW . frequency selective fading (c) Require Tb = Tm + T ⇒ R = 1 Tm +T = 49997.04))(10kHz) = 0.5/Ts Jmin = Ts DF equalizer : −0. so theoretically infinite. FΣ (f ) = = 1 Ts 1 Ts ∞ F n=−∞ ∞ f+ n Ts n −j2π f + T s 1 + 0.5/Ts 0. and the noise power is given by the integral of N (f ) from -50 kHz to 50 kHz.44 + 1 + 1.44 + .56 + . (a) G(f) is a sinc().64 + .5bps 1 (d) Heq (z) = F (z) for ZF equalizer 1 ⇒ Heq (z) = d0 +d1 z −1 +d2 z −2 Long division yields the first 2 taps as w0 = 1/α0 2 w1 = −α1 /α0 13. But 2/T is also acceptable (Null to Null bandwidth) (b) τ T is more likely since T = 10−9 sec As long as τ > Tb .5 we get N = 2N0 (.5/Ts N0 df FΣ (f ) + N0 N0 FΣ (f ) + N0 Jmin = exp Ts ln −0. and the noise power is given by the integral of N (f ) from -50 kHz to 50 kHz: 50kHz 50kHz N= f =−50kHz N (f )df = 2N0 f =0kHz |Heq (f )|2 df = 2N0 (1 + 4 + 9 + 16 + 25)(10kHz) = 1.5e n=−∞ + 0.5/Ts df 12. get ISI and so.25 + . Thus. LMS-DFE: R = RLS: R = 107 9 - 1000 bits 5 msecs = 911 Kbps 107 50 bits 58 . . one bit sent. 14. the noise power goes to 0 because Heq (f ) → 0 for all f . the signal power also goes to zero. ∴ RW = p) ∴ GK = RWk − p = −E[εk Yk ] ˆ ˆ ˆ where Yk = [yk+L . The bit time is different for LMS-DFE/RLS. But time to convergence is faster for RLS.4 Kbps 16. (e) As α → ∞. N = 4 LMS-DFE: 2N +1 operations/iteration ⇒ 9 operations/iteration RLS: 2. .5 msec = 162 Kbps Case2: fD = 1000 Hz ⇒ retrain every 0. must retrain every 5 msec. The factor α should be chosen to balance maximizing the SNR and minimizing distortion. (1) ˆ ˆ ˆ where ∆ is some small positive number and GK is the gradient of MSE = E|dk − dk |2 is RWk − p (Notice that 11. Benefits of training (a) Use detected data to adjust the equalizer coefficients.(d) As α increases. In the adaptive method. Case 1: fD = 100 Hz ⇒ (∆tc ) ≡ 10 msec. Can work without training information (b) eliminate ISI.5 msec RLMS-DFE = 0 bps RRLS = 72. . 15. However.5N operations/iteration ⇒ 58 operations/iteration Each iteration. we start with some initial value of tap coefficients W0 and then use the steepest descent method Wk+1 = WK − ∆GK .37 was a solution of gradient =0 . In fact it has to be retrained at least every channel correlation time.5(N )2 + 4. yK−L ]T and εk = dk − dk Approximately (1) can be rewritten as Wk+1 = Wk + ∆εk Yk . the frequency response Heq (f ) decreases for all f . but the signal power decreases as well. which also depends on the spectrum of the input signal (which is not given here). Tb (LMS-DFE) <Tb (RLS). The equalizer must be retrained because the channel de-correlates. the noise power decreases. . 1) 60 (1.0) (0.0) Figure 1: Problem 2 D2 = (j20)2 + (i20)2 − 2(j20)(i20) cos(2π/3) √ ⇒ D = 2a i2 + j 2 + ij = 3R i2 + j 2 + ij 3. R= 100m Dmin = 600m D = 2KR D 600 K = 2R = 2×100 = 3 N = K2 = 9 (a) number of cells per cluster = N = 9 (b) number of channels per cell = total number/N = 450/9 = 50 4. See Fig 1 u (1. diamond shaped cells.Chapter 15 1.1) v (0. City has 10 macro-cells each cell has 100 users ∴ total number of users = 1000 Cells are of size 1 sqkm √ maximum√distance traveled to traverse = 2km 2 ∴ time = 30 = 169.7s In the new setup number of cells = 105 microcells total number of users = 1000 × 1002 users √ 2×10 time = 30×103 =1.69s ∴ number of users increases by 10000 and handoff time reduces by 1/100 2. (a) R=1km D=6km √ 2 3D 1 cluster N = AAcell = 3√3R2/2 = 3 (D/R)2 = 1 62 = 12 3 /2 number of cells per cluster = N = 12 . 7349 = 76 . j = 2 5. G = 100 ξ=1 λ = 1. γ = 2 BP SK √ Pb = 10−6 → Pb = Q( 2γb ) ⇒ γb = SIR0 = 4. interference is reduced by a factor of 3 Nc = 76.7534 B = 50M Hz each user100KHz = Bs γ 1 SIR = M D M=6 for hexagonal cells R a1 = 0.7534 Nc = 26.SHADED CELLS USE SAME FREQUENCY D Figure 2: Problem 3 (b) number of channels in each cell = 1200/12 = 100 √ (c) i2 + j 2 + ij = 2 3 ⇒ i = 2.4879 ∴ N = 10 7.167 a2 = 3 1 N > a2 Cu = 50 SIR0 a1 2/γ ⇒ N ≥ 9. R=10m D=60m γI = 2 γ0 = 4 M = 4 for diamond shaped cells SIRa = R−γI R−2 = = 32400 −γ0 MD 4D−4 R−4 = 324 4D−4 R−2 =9 SIRc = 4D−2 SIRa > SIRb > SIRc SIRb = 6.5 With no sectorization SIR = ξ 3G (Nc 1 − 1)(1 + λ) = 4.2450 = 26 With sectorization. j ∈ {1.078Nc p(x = n) Q − n − 0.078Nc (c) Nc = 35 α = 0.247Nc √ 0.. define γk = nk + ρ gk Pk k=j G SIR0 − (0.078Nc G SIR0 G SIR0 − n − 0. for i = 1:length(n) f = ((G/sir0)-n(i)-./(factorial(n(i)).247*Nc)/(sqrt(. K} . 0. .^(Nc-1-n(i)). end (d) If x can be approximated as Gaussian then x ∼ G((Nc − 1)α.078Nc ) Po ut = p(SIR < SIR0 ) (a) Pout = p Nc −1 i=1 Nc −1 i=1 G Xi +N G Xi +N < SIR0 =p Nc −1 i=1 Xi +N > G SIR0 Nc −1 (b) X = i=1 Xi then X ∼ Bin(α.0969 (very accurate approximation!) 9.247Nc √ > 0. .247Nc .078Nc + (Nc − 1)α(1 − α)) p(x + N > G/SIR0 ) = Q (e) p= 0.*factorial(Nc-1-n(i)))).0973 MATLAB for i = 1:length(n) pn(i) = (factorial(Nc-1).247Nc √ |x = n 0. Nc − 1) p(x + N > G/SIR0 ) = Nc −1 p(n + N > G/SIR0 |x = n)p(x = n) N =0 p(x = n) = Nc − 1 αn (1 − α)Nc −1−n n Nc −1 p(x + N > G/SIR0 ) = N =0 Nc −1 p(N > G/SIR0 − n|x = n)p(x = n) p N =0 Nc −1 = = N =0 N − 0.078*Nc)).5*erfc((f)/sqrt(2))*pn(i). SINR = α = p(Xi = 1) N ∼ G(0. 0.^n(i)*(1-alpha). (Nc − 1)α(1 − α)) x + N ∼ G(0. .078Nc + (Nc − 1)α(1 − α) gkj pj k..247Nc + (Nc − 1)α.8. end sump = 0.5 SIR0 = 5 G = 150 p = 0. sump = sump + . *alpha.247Nc + (Nc − 1)α) 0. then a power control policy exists..2*(D(i)-R/2)^(-gamma)+.5*D(i))^2). ASE(i) = log(1+Pdes/Pint)/(pi*(.5*D(i))^2). K} u= γ nK γ1 n1 γ2 n2 . The optimal power control policy is given to be P = (I − F )−1 u 10. gk is channel power gain from user k to his base station nk is thermal noise power at user k’s base station ρ is interference reduction factor (ρ ∼ 1/G) gkj is channel power gain from j th interfering transmitter to user k’s base station pk is user k’s Tx power pj is user j’s Tx power define a matrix F such that Fkj = k.2*(D(i))^(-gamma).5*D(i))^2).2*(D(i)+R)^(-gamma)). . Pdes = R^(-gamma). +.where. j ∈ {1. ASEworst(i) = log(1+Pdes/Pintworst)/(pi*(.. . gamma = 2. end . Matlab D = 2:.. Pintworst = 6*((D(i)-R)^(-gamma)).. for i = 1:length(D) Pint = 6*(. K g1 g2 gK 0 γk gkj ρ gk k=j k=j If Perron Ferbinius eigenvalue of F is less than 1 .2*(D(i)+R/2)^(-gamma)+. Pintbest = 6*((D(i)+R)^(-gamma)). ..2*(D(i)-R)^(-gamma)+.01:10. . R = 1.. ASEbest(i) = log(1+Pdes/Pintbest)/(pi*(. number of users that share band = 2(2(n-1)+1) ∴ each user gets 100KHz = Bu (n) 2(2n−1) interference is only from first tier SIR(n) = N0 2 Bu (n) Pt K + 2 Pt K d0 R 3 3 > 25 √ d0 R2 +D(n)2 using Matlab .1 0.9253. Pt = 5W B = 100KHz N0 = 10−16 W/Hz Pr = Pt K d0 d 3 d0 = 1.2 ASE(D) 0. Matlab Pt = 5. n = 1.15 0.3 0. D = 8R .35 5 Case ASE Best Case ASE Worst Case ASE 0. R = 1000.0. K = 100 (a) D=2R 2 users share the band available Each user gets 50KHz R=1Km BASESTATION USER Figure 4: Problem 11a 1 1 (b) P b = 4γ ⇒ 10−3 = 4γ ⇒ γb = 250 b b If D(n) = 2nR.25 0. n = 4. sigma_2 = 1e-16.05 0 2 3 4 5 6 D 7 8 9 10 Figure 3: Problem 10 11. SIR = 261. 01:1. For low SNR values (α less than a value) c decreases with increase in α as interference is increased which cannot be easily decoded due to low SNR. Pint = 2*(Pt*K*(d0/sqrt(R^2+D^2))^3). K = 100. sigma_2 = 1e-9. Pdes = Pt*K*(d0/R)^3. See Matlab If α is large. B=100KHz N0 = 10−9 W/Hz K = 10 P = 10mW per user (a) 0 ≤ α ≤ 1 α is channel gain between cells. C(i) = (1/K)*sum(capvec)*ss. P = 10e-3. Bu = (100/(2*(2*n-1)))*1e3. . d0 = 1. Pint = 2*(Pt*K*(d0/sqrt(R^2+D^2))^3). SIR(1) = 5. ss = . K = 100. for i = 1:length(alpha) capvec = log2(1+(K*P*(1+2*alpha(i)*cos(2*pi*theta)). end (c) ASE = (R1 +R2 )/B 2km×2km R1 = R2 = Bu (1) log(1 + SIR(1)) Bu (1) = 50KHz. sir = Pdes/(Npower+Pint). while sir < 250 n = n+1.^2)/(sigma_2*B)). MATLAB CODE: B = 100e3. D = 2*n*R. alpha = 0:.001. interference can be decoded and subtracted easily so capacity grows with α as high SNR’s (beyond an α value) . Npower = sigma_2*Bu. d0 = 1.D = 2*n*R. sir = Pdes/(Npower+Pint).5899 ASE = 0. theta = 0:ss:1. Npower = sigma_2*Bu. Pdes = Pt*K*(d0/R)^3. K = 10.6801bps/Hz/km2 12. Bu = (100/(2*(2*n-1)))*1e3. end B = 100 KHz 1.86 0.02 1 0. P = 10e-3.8 0.5. MATLAB CODE: . ss = .001. end B = 100 KHz 6 5 4 C(K)/B 3 2 1 0 0 5 10 15 K 20 25 30 Figure 6: Problem 12b (c) as transmit power P ↑.^2)/(sigma_2*B)).1 0.9 0.94 C(α)/B 0.82 0 0.98 0. C(i) = (1/K(i))*sum(capvec)*ss.9 1 Figure 5: Problem 12a (b) C(K)↓ as K↑ because as the number of mobile per cell increases system resources get shared more and so per user capacity C(K) has to fall.2 0. capacity C ↑ but gets saturated after a while as the system becomes interference limited.92 0.84 0. for i = 1:length(K) capvec = log2(1+(K(i)*P*(1+2*alpha*cos(2*pi*theta)).7 0.5 α 0. alpha = .1:30.6 0.4 0. MATLAB CODE: BB = 100e3. sigma_2 = 1e-9.3 0. K = 1:.96 0.88 0. theta = 0:ss:1. 2 0 0 10 20 30 40 50 P (in mW) 60 70 80 90 100 Figure 7: Problem 12c . P = [0:.001. alpha = .8 C(P)/B 0.1:100]*1e-3.2 1 0. sigma_2 = 1e-9.B = 100e3. theta = 0:ss:1.5.^2)/(sigma_2*B)). C(i) = (1/K)*sum(capvec)*ss. for i = 1:length(P) capvec = log2(1+(K*P(i)*(1+2*alpha*cos(2*pi*theta)). end B = 100 KHz 1. ss = .4 1. 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