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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( ). Then, E r ( f,t,r ( t ) ,, )) = < [ ( ,,f )exp { j 2 f (1- r ( t ) /c ) } ] r ( t ) . Moreover, if we assume that r vt , then we get that r ( t ) r + vt cos( ). Thus, the doppler shift is fv cos( ) /c . 2. Let ( x,y,z ) be the position of the mobile in Cartesian coordinates, and ( r,, ) the position in polar coordinates. Then ( x,y,z ) = ( r sin cos ,r sin sin ,r cos ) ( r,, ) = p x 2 + y 2 + z 2 , arctan( y/x ) , arccos( z/ p x 2 + y 2 + z 2 ) = x y- xy x 2 + y 2 =- zr- z r r 2 p 1- ( z/r ) 2 We see that is small for large x 2 + y 2 . Also is small for | z/r | < 1 and r large. If | r/z | = 1 then = 0 or = and v < = r | | so v/r large assures that is small. If r is not very large then the variation of and 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 ) = cos[2 f ( t- r ( t ) /c )] 2 d- r ( t ) + 2 [ d- r ( t )]cos[2 f ( t- r ( t ) /c )] r ( t )[2 d- r ( t )]- cos[2 f ( t + ( r ( t )- 2 d ) /c )] 2 d- r ( t ) 2 Tse and Viswanath: Fundamentals of Wireless Communication 3 = 2 sin[2 f ( t- d/c )]sin[2 f ( r ( t )- d ) /c ] 2 d- r ( t ) + 2 [ d- r ( t )]cos[2 f ( 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 ) = < [ exp { j 2 [ f (1- v/c ) t- fr /c ] } ] r + vt- < [ exp { j 2 [ 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.... View Full Document

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