Fundamentals of Wireless Communication 习题答案

【实例简介】
Fundamentals of Wireless Communication 是无线通信教材中的经典。上传的文档即此教材的习题解答。希望对自学者有帮助哦！
Tse and viswanath: Fundamentals of wireless communication 2asin2x/(-d/)sin{2n∫(r()-d)/d,2a-r()cos[2∫(t=r(1)/c) 2d-r(t r(t)2d-r(t) 2 where we applied the identit CoS -cos y SIN (=D)m(2) We observe that the first term of (2. 1)is similar in form to equation(2. 13) in the notes. The second tern 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 Er(, t)= R(a exp,2r[f(1-y/ct-fro/c r[aexpj2T[f (1-u/c)t-f(ro+ 2d)/c3 ro+ut ro+2d+ut 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 (ro +ut) and (ro +2d+ut)and also on the phase difference f 2d/c Over frequency, we observe that when f2d / c is an integer both waves interfere destructively resulting in a small received signal. When f2d/c=(2k+1)/ 2,kE Z these waves interfere constructively resulting in a larger received signal. So when f is varied by c/4d the amplitude of the received signal varies from a minimum to a maximum The variation over frequency is similar in nature to that of section 2.1.3, but since the delay spread is different the coherence bandwidth is also different However there is no variation over time because the Doppler spread is zero EXERCISE 2.4. 1. i)With the given information we can compute the Doppler spread 1-/2|==|cos61-cosb from which we can compute the coherence time 4Ds 4fu cos 41-cos 821 There is not enough information to compute the coherence bandwidth, as it depends on the delay spread which is not given. We would need to know the difference in path length to compute the delay spread I'd and use it to compute Tse and viswanath: Fundamentals of wireless communication 2. From part 1 we see that a larger angular range results in larger delay spread and smaller coherence time. Then, in the richly scattered environment the channe would show a snaller coherence time than in the environment where the reflectors are clustered in a small angular range FX(|SE2.5.1. +(hs-hr )2 h)2/r2≈7(1+ hnr r2+(hs hr)2 (hs+hn)2/r2≈7(1+ (hs+hr (h+h)2-(hs-hx)2h2+12+2h。khr-h2-h2+2h。h 2h,h herefore b= 2hshr Er(, t) Red( 2r(ft-fr1/cJ-expj2(ft-fr2olI Re[aexp 2r(ft- fr1/clll-exp(2Tf(r1-r2)/ Regexp 2r(ft-fri/cll p(727J/c b Regexp{2(ft-f1/C)[1-(1-)2丌f/c米b/r) rlexp(jza)exp[i2(ft-fr1/c)JI f ab sin2(ft-fr/c)+∠a Therefore B= 2T fa b/c 71+(2-71)n11+(72-1)/r] T 71 Therefore if we don't make the approximation of b) we get another term in the expansion that decays as r 3. This term is negligible for large enough r as compared to B/r2 Tse and viswanath: Fundamentals of wireless communication EXERCISE 2.6. 1. Let f2 be the probability density of the distance from the originl at which the photon is absorbed by exactly the 2nd obstacle that it hits. Let a be the location of the first obstacle. then f2(r)=Pphoton absorbed by 2nd obstacle at r] P absorbed by 2nd obstacle at r not absorbed by 1st obstacle at I x not absorbed by 1st, obstacle at, m d r Since the obstacle are distributed according to poisson process which has mem oryless distances between consecutive points. the first term inside the integral is fi(r-c). The second term is the probability that the first obstacle is at c and the photon is not absorbed by it. Thus it is given by (1-nq(a).Thus )= (1-)q(x)f1(-x)da 2. Similarly, we observe that fa+1(r) is given by k+1() P absorbed by(k+ 1)th obst at r not absorbed by lst obst at y IP Inot absorbed by 1st obstacle at m I (1-ma(x)f(r-nd.c 22) 3. Summing up(2. 2) for k=1 to oo, we get =C=0( k= Tse and viswanath: Fundamentals of wireless communication us f(r)-f1(r) q( or equivalently. f( 7(r)+ (1-~)9(x)f(r-m)dm 2.3 4. Using(2.3), we get that (u)=(1-)Q(u)+F(u)Q(), 2.4 where F and Q denote the Fourier transform of f and q respectively. Since the q(a) is known explicitly, its Fourier transform can be directly calculated and it turns out to be: Substituting thin in(2.4), we get Thus, F is of the same form as Q, except for a different parameter n. Thus f(r)=yc-√m 5. Without any loss of generality we can assume that r is positive, then power density at r is given f(d V77 d r A similar calculation for a negative r gives power density at distance r to be r Eⅹ ERCISE2.7. Tse and viswanath: Fundamentals of wireless communication EXERCISE 2.8. The block diagram for the(unmodified )system is k -t)h t=kT 12丌ft 2c 2丌fct 1. Which filter should one redesign? One should redesign the filter at the transmitter. Modifying the filter at the receiver may cause 6(t-kiT)k no longer to be an orthonormal set, resulting in noise on the samples not to be i i d. By leaving e(t-kiT)k at the receiver as an orthonornal set we are assured thethe noise on the sainples is ii d Let the modified filter be g(t). The block diagran for the modified system is t=kT k 9() 2丌fat 2 02n fct (Solution to Part 3: Figure of the various filters at passband) We want to find g(t)such that there is no ISI between samples. Before we continue to find g(t). we depict the desired simplified block diagram for the system with no tSt B k For ease of manipulation, we transform the passband representation of the system to a baseband representation u(t kT A g Bk Where Hb(f) H(f+f)∈ otherwise H(f)is assumed bandlimited between [fc-y, fe+ I We let g(t)=2k gk6(t-kr), and redraw the block diagran Tse and viswanath: Fundamentals of wireless communication 0}-d0(-0) t=kT We now convert the signals and filters from the continuous to discrete time domain k 9k B where hk=b米hb*b-t=kT We justify interchanging the order of w(t) and 6(t), since we know the noise OIl the samples is i.i.d G(a)=H-(2) gives the desired result In sunnary, g(t)=>k 9r 0(t-kT)where gk is given by G(a)=H-(2), and H(a is given by the Z-Transform of hk=0*k hb*8-t=kl EXERCISE 2. 9. Part 1) 10 h[k, 1 l for W=10KHz Figure 2.1: Magnitude of taps, W=10k Hz, time= 1 sec. Two pat hs are completely lumped together Tse and viswanath: Fundamentals of wireless communication Ih[k 1]l for W=100KHz Figure 2.2: Magnitude of taps, W= 100kHz, time= l sec. Two paths are starting to become resolved h[k, 1]l fo 1 0 Figure 2.3: Magnitude of taps, W=IMHz, time= I sec. Two paths are more resolved Tse and viswanath: Fundamentals of wireless communication h[k, 1] for W=3MHZ Figure 2.4: Magnitude of taps, W=3MHz, time 1 sec. Two paths are clearly resolved 【实例截图】
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