Chapter 3

# Chapter 3 - Chapter 3 1 d = vt 2 r r = d 2h d Equivalent...

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Chapter 3 1. d = vt r + r 0 = d + 2 h 2 d Equivalent low-pass channel impulse response is given by c ( τ,t ) = α 0 ( t ) e - 0 ( t ) δ ( τ - τ 0 ( t )) + α 1 ( t ) e - 1 ( t ) δ ( τ - τ 1 ( t )) α 0 ( t ) = λ G l 4 πd with d = vt φ 0 ( t ) = 2 πf c τ 0 ( t ) - φ D 0 τ 0 ( t ) = d/c φ D 0 = R t 2 πf 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 4 π ( r + r 0 ) = λR G l 4 π ( d + 2 h 2 d ) with d = vt φ 1 ( t ) = 2 πf c τ 1 ( t ) - φ D 1 τ 1 ( t ) = ( r + r 0 ) /c = ( d + 2 h 2 d ) /c φ D 1 = R t 2 πf 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 0 c delay spread( T m ) = x + x 0 - l c = p ( h t + h r ) 2 + d 2 - p ( h t - h r ) 2 + d 2 c when d ± ( h t + h r ) T m = 1 c 2 h t h r d h t = 10 m, h r = 4 m, d = 100 m 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.

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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 ] = Z Z x 2 + y 2 z 2 1 2 πσ 2 e - ( x 2 + y ) 2 2 σ 2 dxdy = 2 π Z 0 z Z 0 1 2 πσ 2 e - r 2 2 σ 2 rdrdθ = 1 - e - z 2 2 σ 2 ( z 0) df z ( z ) dz = z σ 2 e - z 2 2 σ 2 Rayleigh For Power: F z 2 ( z )= P [ Z z ] = 1 - e - z 2 2 σ 2 f z ( z )= 1 2 σ 2 e - z 2 σ 2 Exponential 6. For Rayleigh fading channel Pr ( P r < P 0 ) = 1 - e - P 0 / 2 σ 2 2 σ 2 = - 80 dBm, P 0 = - 95 dBm, Pr ( P r < P 0 ) = 0 . 0311 2 σ 2 = - 80 dBm, P 0 = - 90 dBm, Pr ( P r < P 0 ) = 0 . 0952 7. For Rayleigh fading channel P outage = 1 - e - P 0 / 2 σ 2 0 . 01 = 1 - e - P 0 /P r P r = - 60 dBm 8. 2 σ 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 z 0 0 z σ 2 e - - ( z 2 + s 2 ) 2 σ 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 z 0 p Ricean Z ( z ) where p Ricean Z ( z ) = 2 z ( K + 1) Pr exp - K - ( K + 1) z 2 Pr I 0 ˆ 2 z r 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 / (2 K + 1). CDF of Nakagami-m distribution is F Nakagami-m Z ( z ) = Z z 0 p Nakagami-m Z ( z ) where p Nakagami-m Z ( z ) = 2 m m z 2 m - 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

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## This note was uploaded on 01/11/2012 for the course EE 359 taught by Professor Goldsmith during the Fall '08 term at Stanford.

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Chapter 3 - Chapter 3 1 d = vt 2 r r = d 2h d Equivalent...

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