Chapter 2

Chapter 2 - Chapter 2 1 P r = P t • √ G l 4 πd ‚ 2 = c/f c = 0 06 10 3 = P t • 4 π 10 ‚ 2 ⇒ P t = 4 39 KW 10 3 = P t • 4 π

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Unformatted text preview: Chapter 2 1. P r = P t • √ G l λ 4 πd ‚ 2 λ = c/f c = 0 . 06 10- 3 = P t • λ 4 π 10 ‚ 2 ⇒ P t = 4 . 39 KW 10- 3 = P t • λ 4 π 100 ‚ 2 ⇒ P t = 438 . 65 KW 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. Δ φ = 2 π ( x + x- l ) λ x + x- l = p ( h t + h r ) 2 + d 2- p ( h t- h r ) 2 + d 2 = d s h t + h r d ¶ 2 + 1- s h t- h r d ¶ 2 + 1 d h t ,h r , we need to keep only first order terms ∼ d 1 2 s h t + h r d ¶ 2 + 1 - 1 2 s h t- h r d ¶ 2 + 1 = 2( h t + h r ) d Δ φ ∼ 2 π λ 2( h t + h r ) d 4. Signal nulls occur when Δ φ = (2 n + 1) π 2 π ( x + x- l ) λ = (2 n + 1) π 2 π λ h p ( h t + h r ) 2 + d 2- p ( h t- h r ) 2 + d 2 i = π (2 n + 1) p ( h t + h r ) 2 + d 2- p ( h t- h r ) 2 + d 2 = λ 2 (2 n + 1) Let m = (2 n + 1) p ( h t + h r ) 2 + d 2 = m λ 2 + p ( 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 p ( h t- h r ) 2 + d 2 x = ( h t + h r ) 2 , y = ( h t- h r ) 2 , x- y = 4 h t h r x = m 2 λ 2 4 + y + mλ p y + d 2 ⇒ d = s • 1 mλ x- m 2 λ 2 4- y ¶‚ 2- y d = s 4 h t h r (2 n + 1) λ- (2 n + 1) λ 4 ¶ 2- ( h t- h r ) 2 ,n ∈ Z 5. h t = 20 m h r = 3 m f c = 2 GHz λ = c f c = 0 . 15 d c = 4 h t h r λ = 1600 m = 1 . 6 Km 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 . 22 μs )) G r = G l = 1 h t = h r = 8 m f c = 900 MHz,λ = c/f c = 1 / 3 R =- 1 delay spread = x + x- l c = 0 . 022 × 10- 6 s ⇒ 2 q 8 2 + ( d 2 ) 2- d c = 0 . 022 × 10- 6 s ⇒ d = 16 . 1 m ∴ τ = d c = 53 . 67 ns α 1 = λ 4 π √ G l l ¶ 2 = 2 . 71 × 10- 6 α 2 = λ 4 π √ 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...
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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 2 - Chapter 2 1 P r = P t • √ G l 4 πd ‚ 2 = c/f c = 0 06 10 3 = P t • 4 π 10 ‚ 2 ⇒ P t = 4 39 KW 10 3 = P t • 4 π

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