Wireless Communications Prof. Mostofi Homework #1 Solutions 1. Consider a wireless communications system with the following parameters: • transmitted power: +20 dBm • minimum usable received power: −100 dBm • transmitting antenna gain: 10 dB • receiving antenna gain: 0 dB • carrier frequency: 1 GHz (a) What is the range (the distance over which the wireless system will work) in free space? (b) Assume the signal propagates over an ideal flat surface with reflection coefficient of −1. Compute the range for a transmitting antenna height of ht = 25 ft and a receiving antenna height of hr = 6 ft, and compare with part (a). Solution: 1. (a) From the Friis formula, Pr = gt gr Pt λ 4πd 2 Pr,dBm − Pt,dBm = −100 − 20 = −120 dB, so Pr /Pt = 10−12 . f = 1 GHz gives λ = 0.3 m. gt = 10, gr = 1. r λ gt gr d= = 75.49 km 4π Pr /Pt (b) For the ground reflection case, Pr (ht hr )2 ≈ gt gr Pt d4 Using 0.3048 m = 1 ft, d= (ht hr )2 gt gr Pr /Pt 1/4 = 6.64 km which is 8.8% of the range of part (a). 2. Consider the situation in the figure below where a mobile station is moving in the vicinity of two local scatters, each with a reflection coefficient of 0.5. The reflection coefficient is the ratio of the amplitude of the reflected wave to the direct wave. Let Vr represent the amplitude of the received direct path. Let the transmitted signal be s(t) = cos(wt). 1 path B path A v path C (a) Derive an expression for the resultant received signal from paths A, B, and C. For the next two parts, you may assume the following: • • • • τA = τB = τC , i.e., the time delays of all paths are the same θ = 30◦ (remains constant over our region of interest) v = 60km/hr fc = 900MHz (b) Plot the amplitude variation due to multipath fading and calculate the maximum Doppler frequency (take the delays to be zero). (c) How far apart (in distance) do the nulls in the signal occur? Solution: 2. (a) The received signal is S(t) = Vr cos [(ω − ωd )(t − τA )] + 0.5Vr cos [(ω + ωd cos θ)(t − τB )] +0.5Vr cos [(ω + ωd cos θ)(t − τC )] v where ωd = ω is the maximum Doppler frequency. c (b) Let τA = τB = τC = 0. We have S(t) = Vr cos [(ω − ωd )t] + Vr cos [(ω + ωd cos θ)t] 1 + cos θ 1 − cos θ = 2Vr cos ωd t · cos ω − ωd t 2 2 α−β α+β · cos . So the envelope 2 2 1 + cos θ r(t) = 2Vr cos ωd t , 2 using cos α + cos β = 2 cos and there is a Doppler frequency shift in the carrier of ω∆ = − 1 − cos θ ωd 2 2 or f∆ = − 1 − cos θ fd . 2 2 r(t)/Vr 1.5 1 0.5 0 0 0.01 0.02 0.03 Time [s] 0.04 0.05 v = 60 km/hour and fc = 900 MHz gives cos θ = cos 30◦ = So √ fd = fc · 3 . 2 6 × 104 /3600 v = 9 × 108 · = 50 Hz. c 3 × 108 " 50 r(t) = 2Vr cos 2π · 2 and f∆ = − 1− √ ! # 3 1+ ·t , 2 √ 3 2 · 50 = −3.35 Hz. 2 Then the envelope variation |r(t)| is as shown. (c) The nulls occur when the argument of the cosine term above is equal to (2n+1) π2 for integer n, or, equivalently, the difference in phase should be π. Let d = vz t bet the distance the mobile travels in time t. We can then solve for the distance traveled as follows: (using vz = (60 km/hr)/(3600 sec/hr) = 16.67 m/s) √ ! 50 3 π = 2π · 1+ ∆t 2 2 ∆d = vz ∆t = (16.67) 1 √ = 17.9 cm (50)(1 + 3/2) 3 3. Consider a mobile user moving slowly in a residential area, where there are many random paths arriving at many random angles. Assume there is no large scale signal variation over the distance interval of motion. Assume that from subjective listening tests, the quality of the radio link is acceptable if the signal level falls below the minimum usable signal threshold only 1% of the time. How many dB must the mean of the signal level exceed the minimum usable signal level to ensure acceptable signal quality? Solution: 3. The cumulative distribution function for the Rayleigh random variable is Pr (r ≤ R) = Z R −∞ 2 /2P f (r)dr = 1 − e−R Let the threshold be Rt , then: 2 1 − e−Rt /2P ≤ 0.01 Rt2 ≤ − ln 0.99 2P √ √ P ≥ Rt / −2 ln 0.99 = 7.053Rt √ Since the mean M = 1.25 P , we need M ≥ 1.25 · 7.053 · Rt = 8.8163Rt or M ≥ 8.8163 Rt so the mean must exceed the threshold by 20 log10 8.8163 = 18.9 dB. 4. The power sum of N uncorrelated log-normal interference signals is I= N X Ik = N X 10Xk /10 , k=1 k=1 where Xk is Gaussian distributed with mean µk and variance σk2 . Unfortunately there is no simple analytical formula for the distribution of I. A general approach is to approximate I by another log-normal variable, Iˆ = 10X/10 , where X is Gaussian distributed with some mean µ and variance σ 2 . For Xk ’s with small variances, a good choice of µ and σ can be obtained by equating the mean and variance of I and ˆ I. (a) Assume µk and σk are the same for all k and are equal to µ1 and σ1 , find µ and σ in terms of µ1 and σ1 using the above criterion. (Hint: I = 10X/10 = eαX , E[I n ] = E[enαX ], where α = ln 10/10.) 4 (b) Compute µ and σ for N = 6, µ1 = 1 and σ1 = 4 using the result of (a). (c) Test this approximation with computer simulation (MATLAB is recommended). Generate many samples of I using the data given in (b), then plot the pdf of 10 log I. Compare with N (µ, σ). Repeat for σ1 = 10. Comment on the results. Solution: 4. (a) Let α = ln 10/10, then Y = 10Z/10 = eln(10 Z/10 ) = eZ/10·ln 10 = eαZ . The expectation of Y n for Z ∼ (µ, σ) can be computed as follows: E[Y n ] = E[enαZ ] Z 1 2 2 e−(z−µ) /2σ dz = enαZ · √ 2πσ Z 1 2 2 2 2 nαµ+(nασ) /2 √ e−(z−(µ+nασ )) /2σ dz = e 2πσ nαµ+(nασ)2 /2 = e So ˆ = eαµ+(ασ) /2 E[I] ˆ = E[Iˆ2 ] − E[I] ˆ2 V ar[I] 2 (1) 2 2 = e2αµ+2(ασ) − e2αµ+(ασ) 2 = e2αµ+(ασ) and E[I] = V ar[I] = N X k=1 N X k=1 2 e(ασ) − 1 2 /2 E[Ik ] = N · eαµk +(ασk ) (2) (3) 2αµk +(ασk )2 V ar[Ik ] = N · e (ασk )2 e −1 (4) By equating (1)(3) and (2)(4) respectively, we have the following equations: 2 /2 2 /2 eαµ+(ασ) 2αµ+(ασ)2 e = N · eαµk +(ασk ) 2 2 2 e(ασ) − 1 = N · e2αµk +(ασk ) e(ασk ) − 1 (5) (6) which yields (divide (6) by the square of (5) to get σ): 2 (ασ)2 e(ασk ) − 1 = ln 1 + N αµ = αµk + 5 ! α2 2 (σk − σ 2 ) + ln N 2 (7) (8) (b) Substitute N = 6, µk = 1 and σk = 4 into equations (7)&(8), we get σ = 1.947 and µ = 10.187. (c) The pdf curves agree very well for small standard deviations, e.g. σk = 4, but differ greatly for large standard deviations, e.g. σk = 10. It suggests that this approximation method only applies to sum of log-normals with small variances. 5. Large scale signal variation can be approximated by a log-normal distribution with a standard deviation, σs . Interference also can be approximated by a log-normal distribution with a different standard deviation, σi . Let the signal power be represented by S and the interference power by I. The quantities S and I are independent. Define s = 10logS and i = 10logI then s and i are normally distributed with means s̄ and ī and standard deviations σs and σi . The S/I ratio in dB thus becomes s − i and the a quantity of interest often is p = Pr {s − i < γ} where γ is a signal-to-interference threshold for a system in dB. Derive an expression for p in terms of a function of s̄, ī, σs , σi and γ. Solution: 5. Assume s and i are uncorrelated, then s−i will be Gaussian with mean s̄− ī and variance σs2 +σi2 . So Z γ (x−(s̄−ī))2 − 1 2 2 p p= e 2(σs +σi ) dx 2 2 2π(σs + σi ) −∞ x−(s̄−ī) Substituting y = √ , we have 2 2 σs +σi γ−(s̄−ī) Z √ 2 2 1 −y2 /2 e dy 2π −∞ Z +∞ 1 −y2 /2 e dy = 1 − γ−(s̄−ī) √ 2 2 2π σs +σi ! γ − (s̄ − ī) = 1−Q p σs2 + σi2 p = σs +σi 6. Consider a mobile receiver operating at a frequency of 860 MHz and moving at 100 km/hr in a Rayleigh fading environment with angles of arrival uniformly distributed over 0 − 2π in a horizontal plane. Sketch the Doppler spectrum if a CW signal is transmitted and indicate the maximum frequencies in cycles/second, cycles/meter, and cycles/wavelength. Solution: 6 6. f = 860 MHz, v = 100 km/hr = 105 m/hr = 27.78 m/s 3600 sec/hr 27.78 v (860 × 106 ) = 79.63 Hz fm = f = c 3 × 108 c 3 × 108 λ= = = 0.349 m f 860 × 106 Maximum frequency in cycles/m: 79.63 cycles/sec = 2.87 cycles/m 27.78 m/s Maximum frequency in cycles/wavelenth: (2.87 cycles/m)(0.349 m/wavelength) = 1 cycle/wavelength 860 MHz 79 63 Hz 2 87 1 : : 860 MHz 0 0 860 MHz + 79 63 Hz (cycles/sec) (cycles/m) 2 87 (cycles/wavelength) 1 : : 7. Assume someone in a car is using a receiver operating at 900 MHz. The antenna is a dipole and is mounted vertically on top of the car. There are many paths arriving at the antenna uniformly distributed in angles in the horizontal plane. Only consider small-scale variations. (a) The receiver records samples of the envelope of the received signal every t seconds, where 5 ms< t <15 ms. If the car travels at 36 km/hr, what are the values for t such that the adjacent samples are uncorrelated? 7 (b) What is the correlation coefficient between the adjacent samples if we use the same t found in part (a) but the speed changes to 18 km/hr? Solution: 7. (a) fm = fc v 36 × 103 /3600 = 9 × 108 · = 30 Hz. The nulls of J0 (2πfm t) are c 3 × 108 fm t = 0.383, 0.879, 1.377, . . . which correspond to t = 12.8 ms, 29.3 ms, 45.9 ms, . . . For the given range, 5 ms < t < 15 ms, we choose t = 12.8 ms. (b) fm = 15 Hz. So 2πfm t = 2π · 15 × 0.0128 = 2π · 0.192 = 1.203 for t = 12.8 ms. Therefore the correlation coefficient is J02 (2πfm t) = J02 (1.203) = 0.4487 8. A mobile user is moving in an environment in which signal energy arrives uniformly distributed in angles in the horizontal plane, and the antenna has unity gain. The following receiver is used to measure the power received from a sinusoidal transmission with fc = 900 MHz. Power Meter BPF v Let the power received at the antenna output before the BPF be P . Define W = fc . Consider c two BPFs having the following frequency responses for f > 0: ( 1 |f − fc | < W • BPF1 : |H1 (f )| = 0 otherwise ( 1 fb < |f − fc | < 2W • BPF2 : |H2 (f )| = 0 otherwise (a) If BPF1 is used and the user moves at a speed of v, what is the power at the output of the BPF in terms of P ? (b) Repeat for a user speed of 2v. (c) Now use BPF2 with the speed still 2v. Find the value of fb in terms of W such that the received power at the output of the BPF is the same as in part (b). 8 Solution: 8. (a) When the user moves at a speed of v, the maximum Doppler frequency is W . Therefore all of the energy will pass through BPF1 . The output power is just P . (b) When a user moves at a speed of 2v, the maximum Doppler frequency is 2W . Only the energy in the range 0 < |f − fc | < W will pass through BPF1 . That corresponds to the arrival angle α, where −π < α < π, in the range of 0 < | cos α| < 1/2, or π/3 < |α| < 2π/3(see figure). Since the power is uniformly distributed in angle, we will have P/3 at the output of the BPF1 . (c) The maximum Doppler frequency is again 2W , but now only the energy in the range fb < |f − fc | < 2W will pass. That corresponds to fb /2W < | cos α| < 1. Therefore 0 < |α| < cos−1 (fb /2W ) or π − cos−1 (fb /2W ) < |α| < π(see figure). Obviously we need √ −1 ◦ cos (fb /2W ) = 30 in order to get P/3 at the output of BPF2 , so fb = 3W . (b) (c) 60Æ 30Æ v v Energy Passed Through BPF 9. Assume that a receiving antenna is moving at a uniform speed through a multipath propagation medium that has a very large number of waves arriving from angles uniformly distributed in a plane. Assume that all the arriving angles are confined to the plane and that the received signal amplitude is Rayleigh distributed. The transmitted signal is a single frequency sinusoid at frequency f. The receiving antenna has a beam pattern G(α) as a function of angle α. α is the angle with respect to the direction of motion. Find and Sketch the power spectrum of the received signal for the following two cases: (a) G(α) has a constant gain, G, for 0 ≤ α ≤ π and zero for the rest of α. (b) G(α) has a constant gain, G, for −π/2 ≤ α ≤ π/2 and zero for the rest of α. Solution: 9 9. we know that S(f ) = p p 2 − (f − f )2 fm c × (p(α)G(α) + p(−α)G(−α)) |f − fc | ≤ fm (9) c Where α = cos−1 ( ff−f ). m (a) G(α) = G for 0 ≤ α ≤ π and zero for the rest of α. We assume constant power intensity per unit angle in the plane, i.e. p(α) = will be half of what was derived on page 4-57. S(f ) = q Gp/wm c 2 ) 1 − ( ff−f m |f − fc | ≤ fm 1 . 2π Then S(f ) (10) (b) G(α) = G for −π/2 ≤ α ≤ π/2 and zero for the rest of α. −π/2 ≤ α ≤ π/2 will result in the frequency range fc ≤ f ≤ fc + fm Hence we get the same spectrum derived on page 4-57 only for the frequency range fc ≤ f ≤ fc + fm . S(f ) = q 2Gp/wm 1− c 2 ) ( ff−f m fc ≤ f ≤ fc + fm (11) 10. Consider a wireless system with the following parameters: • transmitted power: +20 dBm • minimum usable received power: −100 dBm • gain of transmitting antenna: 10 dB • gain of receiving antenna: 0 dB • frequency: 2 GHz Assume that this wireless system is being used in a residential environment as described by the data in the figure. For this problem, assume that the standard deviation of the log-normal process with the trend removed is σ = 10 dB. Assume for this problem that there is only one base station and one user. (a) Let Pr (d) represent the mean received power (the solid line) in dBm at distance d, and Pr0 (d) be the free-space received power (the dotted line) in dBm at distance d. Suppose the reference power is Pref in dBm. Then the x-axis of the figure is Pr (d)−Pref or Pr0 (d)−Pref . Find Pref using the dotted line and the Friis formula. Show that we can write Pr (d) = K − 45 log d, then find the value of K for d in feet. 10 Part a 4 3 S(f) /(2pG/w ) m 2 1 0 f −f c f m f +f c c m Part b 8 6 S(f) /(2pG/w ) m 4 2 0 f −f c f m f +f c c m (b) If we ignore the small scale multipath fading, and consider a user traveling on a circle of radius d centered at the base station. Find the value for d such that the system will work (i.e. the received signal power is above the minimum usable level) for half of the time. (c) Repeat (b) if we want the system to work for 90% of the time. Solution: 10. (a) For f = 815 MHz, λ = c = 0.368 m. In free space, f Pr0 (d) = Pt + Gt + Gr + 20 log λ 4πd For d = 500 ft (152.37 m), Pr0 (d) = 20 + 10 + 0 + 20 log 0.368 = −44.32 dBm 4π × 152.37 From the figure, the free-space received power at 500 ft relative to reference is about Pr0 (d) − Pref = −31 dB So Pref = Pr0(d) + 31 = −44.32 + 31 = −13.3 dBm. Since the mean received power has a distance dependence form of 1/dn , in log form, we will have, Pr (d) = K − n log(d) 11 where K is some constant. From figure, we find for d =500 ft, Pr (d) − Pref = −51 dB and for d =1000 ft, Pr (d) − Pref = −64.5 dB. So n = (64.5 − 51)/3 = 44.84 ≈ 45 To find K, note that for d =500 ft, Pr (d) = Pref − 51 = −13.3 − 51 = −64.3 dBm Therefore, K = Pr (d) + 45 log d = −64.3 + 45 log 500 = −64.3 + 121.5 ≈ 57. Then, Pr (d) = 57 − 45 log d dBm (d in ft). (b) Since the signal received at distance d in dBm ∼ N (Pr (d), σ), for the system to work 50% of the time, the mean of the large scale variation should equal to the minimum usable level, i.e. Pr (d) = 42 − 45 log d = −100 dBm =⇒ d = 1430 ft (c) For the system to work 90% of the time, Pr (d) − (−100 dBm) = 0.1 Q σ Since Q(1.28) = 0.1, we have Pr (d) = −100 + 1.28σ = −100 + 1.28 × 10 = −87.2 dBm Substitute into Pr (d) = 42 − 45 log d, we get d = 743 ft. 12