Chapter 2 - Kỹ thuật thông tin vô tuyến

... + 1)π 2 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 ... −1 √ 2 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 √ 2 exp(−k 2 /2) b r dr (2) = Q(a) + 1 R 2 R  ... r 2 1 √ 2 exp(−k 2 /2) b r dr (3) = Q(a) + 1 R 2 R  r=0 1 √ 2 R 2 exp  2( k − a) b  exp(−k 2 /2) b r dr (4) = Q(a) + a  k=−∞ 1 √ 2 exp  −k 2 2 + 2k b − 2a b  dk (5) = Q(a) + exp  −2a...

Chapter 1 - Kỹ thuật thông tin vô tuyến

... ⇒ 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 ... high. 3. P b = 10 − 12 1 2 = 10 − 12 γ = 10 12 2 = 5 × 10 11 (very high) 4. Geo: 35,786 Km above earth ⇒ RT T = 2 35786×10 3 c = 0 .23 86s Meo: 8,00 0- 20 ,000 Km above earth ⇒ RT T = 2 8000×10 3 c = ... 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...

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Chapter 3 - Kỹ thuật thông tin vô tuyến

... F z (z)= P [x 2 +y 2 ≤ z 2 ] =   x 2 +y 2 ≤z 2 1 2 σ 2 e −(x 2 +y) 2 2σ 2 dxdy = 2  0 z  0 1 2 σ 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 ... 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; ... Rayleigh For Power: F z 2 (z)=P [Z ≤ √ z] = 1 − e −z 2 2σ 2 f z (z)= 1 2 2 e −z 2 2 → Exponential 6. For Rayleigh fading channel P r(P r < P 0 ) = 1 − e −P 0 /2 2 2σ 2 = −80dBm, P 0 = −95dBm,...

Chapter 4 - Kỹ thuật thông tin vô tuyến

... 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 Capacity (bps) Figure 1: Capacity vs P out S(γ) S =        1 .23 22 γ = γ 1 1 .22 32 γ = γ 2 1.13 32 γ = γ 3 0 .23 32 γ = γ 4 C ... 2  γ i γ 0  p(γ i ) = 5 .28 53bps/Hz (b) σ = 1 E[1/γ] = 4 .28 82 S(γ i ) S = σ γ i S(γ) S =        0.0043 γ = γ 1 0.0 029 γ = γ 2 0. 428 8 γ = γ 3 4 .28 82 γ = γ 4 C B = log 2 (1 + σ) = 2. 4 028 bps/Hz ... and S 2 when jammer is oﬀ, C = B max  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 .25 S 2 = S. This gives S 1 = 12. 25mW , S 2 = 3 .25 mW and C = 53 .21 Kbps....

Chapter 6 - Kỹ thuật thông tin vô tuyến

... (a) 2x π e −ax 2 = 2x π e −ax 2  ∞ 0 e −at 2 x 2 + t 2 dt a = 1 erf c(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 erf c(x/ √ 2) = 1 π  π /2 0 e −x 2 ... −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 20 0 10 −100 10 0 P s Problem 2 − Symbol Error ... (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:...

Chapter 7 - Kỹ thuật thông tin vô tuyến

... 2 dγ = 1 √ π Γ  1 2  = 1 2 (1)  ∞ 0 1 √ 2 e −γ 1 √ 2 e 2 /γ dγ = 1 2  1 + 2 γ (2)  ∞ 0 1 √ 2 e −γ 1 √ 2  πγ γ e −γ/γ  1 − 2Q   2 γ  dγ = 1 2 √ γ 1 B √ Aγ (3) where A = 1 + 2 ...  −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 ... 1 − A 2 − M−1  m=1  2m − 1 m  A 2m+1 2 2m γ m 11. DenoteN(x) = 1 √ 2 e −x 2 /2 Q  (x) = [1 − φ(x)]  = −N(x) P b =  ∞ 0 Q(  2 )dP (γ) =  ∞ 0 1 √ 2 e −γ 1 √ 2 P (γ)dγ  ∞ 0 1 √ 2 e...

Chapter 10 - Kỹ thuật thông tin vô tuyến

... 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 ergodic M t = M r = 3 M t = 2 M r =3 M t = M r = 2 M t = 2 M r =1 ... Cout(k) = 0; end end 0 5 10 15 20 25 30 0 5 10 15 20 25 ρ (dB) C outage 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 ... According to SVD of h √ λ = 1 .24 2 ∴ C/B = log 2 (1 + λρ) = log 2 (1 + 1 .24 2 2 .10) = 4.038bps/Hz 17. H =  .3 .5 .7 .2  =  −.5946 .8041 −.8041 .5946  .8713 0 0 .3 328  −.8507 .5757 −.5757 −.8507...

Chapter 11 - Kỹ thuật thông tin vô tuyến

... 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 ... < f c + 20 MHz 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 .21 25mW ∴ SNR = 47.0588 = 16.73dB (c) T s = 0.0 125 µsec P b ≤ 0.2e −1.5γ/M−1 ... 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...

Chapter 15 - Kỹ thuật thông tin vô tuyến

... 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)); ... (j20) 2 + (i20) 2 − 2( j20)(i20) cos (2 /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 ...  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 = 324 00 SIR b = R −4 4D −4 = 324 SIR c = R 2 4D 2 =...

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