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**Unformatted text preview: **Chapter 1 Solutions to Exercises 1 Chapter 2 Solutions to Exercises Exercise 2.1 . 1. Let r ( t ) = p r 2 + ( vt ) 2 + 2 r vt cos( φ ). Then, E r ( f,t,r ( t ) ,θ,ψ )) = < [ α ( θ,ψ,f )exp { j 2 πf (1- r ( t ) /c ) } ] r ( t ) . Moreover, if we assume that r vt , then we get that r ( t ) ≈ r + vt cos( φ ). Thus, the doppler shift is fv cos( φ ) /c . 2. Let ( x,y,z ) be the position of the mobile in Cartesian coordinates, and ( r,ψ,θ ) the position in polar coordinates. Then ( x,y,z ) = ( r sin θ cos ψ,r sin θ sin ψ,r cos θ ) ( r,ψ,θ ) = ‡ p x 2 + y 2 + z 2 , arctan( y/x ) , arccos( z/ p x 2 + y 2 + z 2 ) · ˙ ψ = x ˙ y- ˙ xy x 2 + y 2 ˙ θ =- ˙ zr- z ˙ r r 2 p 1- ( z/r ) 2 We see that ˙ ψ is small for large x 2 + y 2 . Also ˙ θ is small for | z/r | < 1 and r large. If | r/z | = 1 then θ = 0 or θ = π and v < = r | ˙ θ | so v/r large assures that ˙ θ is small. If r is not very large then the variation of θ and ψ may not be negligible within the time scale of interest even for moderate speeds v. Here large depends on the time scale of interest. Exercise 2.2 . E r ( f,t ) = α cos[2 πf ( t- r ( t ) /c )] 2 d- r ( t ) + 2 α [ d- r ( t )]cos[2 πf ( t- r ( t ) /c )] r ( t )[2 d- r ( t )]- α cos[2 πf ( t + ( r ( t )- 2 d ) /c )] 2 d- r ( t ) 2 Tse and Viswanath: Fundamentals of Wireless Communication 3 = 2 α sin[2 πf ( t- d/c )]sin[2 πf ( r ( t )- d ) /c ] 2 d- r ( t ) + 2 α [ d- r ( t )]cos[2 πf ( t- r ( t ) /c )] r ( t )[2 d- r ( t )] (2.1) where we applied the identity cos x- cos y = 2sin x + y 2 ¶ sin y- x 2 ¶ We observe that the first term of (2.1) is similar in form to equation (2.13) in the notes. The second term of (2.1) goes to 0 as r ( t ) → d and is due to the difference in propagation losses in the 2 paths. Exercise 2.3 . If the wall is on the other side, both components arrive at the mobile from the left and experience the same Doppler shift. E r ( f,t ) = < [ α exp { j 2 π [ f (1- v/c ) t- fr /c ] } ] r + vt- < [ α exp { j 2 π [ f (1- v/c ) t- f ( r + 2 d ) /c ] } ] r + 2 d + vt We have the interaction of 2 sinusoidal waves of the same frequency and different amplitude. Over time, we observe the composition of these 2 waves into a single sinusoidal signal of frequency f (1- v/c ) and constant amplitude that depends on the attenuations ( r + vt ) and ( r + 2 d + vt ) and also on the phase difference f 2 d/c . Over frequency, we observe that when f 2 d/c is an integer both waves interfere destructively resulting in a small received signal. When f 2 d/c = (2 k + 1) / 2 ,k ∈ Z these waves interfere constructively resulting in a larger received signal. So when f is varied by c/ 4 d the amplitude of the received signal varies from a minimum to a maximum....

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