chapter04 problems.pdf
9
0
0
Full text
(2) (b)Generally, when d>10(ht+hr), we can say that d>>(ht+hr), and thus may apply two-ray model. For ht=35m, hr=3m, d=250m, d<10(ht+hr)=380m, Hence, the two-ray model could not be applied. For ht=30m, hr=1.5m, d=450m, d>10(ht+hr)=315m Hence, the two-ray model could not be applied. (c) Using the two-ray model, we can see that at large distances, the received power falls off with distance raised to the fourth power or at a rate of 40dB/decade, and the received power and path loss are independent of frequency. 4.9, Prove that in the two-ray ground reflected model, ∆=d”-d’≈2hthr/d. Show when this holds as a good approximation. Solution 1. ∆ = d "− d ' =. (ht + hr ). 2. +d2 −. (ht − hr ). 2. 2. 1. h + hr 2 ⎤ 2 h − hr 2 ⎤ 2 ⎡ ⎡ + d 2 = d ⎢1 + ( t ) ⎥ − d ⎢1 + ( t ) ⎥ d d ⎣ ⎦ ⎣ ⎦ 2. ⎛ h + hr ⎞ ⎛ h + hr ⎞ For d >> ht + hr , ⎜ t ⎟ << 1 , ⎜ t ⎟ << 1 . ⎝ d ⎠ ⎝ d ⎠ Using Taylor series approximation, we have. h − hr 2 ⎤ d 4ht hr 2ht hr ⎡ 1 h + hr 2 ⎤ ⎡ 1 h − hr 2 ⎤ d ⎡ ht + hr 2 ∆ ≈ d ⎢1 + ( t ) ⎥ − d ⎢1 + ( t ) ⎥ = ⎢( ) −( t ) ⎥= • 2 = d d d d d ⎣ 2 ⎦ ⎣ 2 ⎦ 2⎣ d ⎦ 2 4.14, Assuning a receiver is located 10km from a 10W transmitter. The carrier frequency is 1900MHz, free space is assumed, Gt=1, Gr=2, find: (a) the power at the receiver; (b) the magnitude of the E-field at the receiver antenna; (c) the open circuit rms voltage applied to the receiver input assuming that the receiver antenna has a purely real impedance of 50Ω and matched to the receiver; (d) find the received power at the mobile using the Two-ray ground reflection model assuming the height of the transmitting antenna is 50m, receiving antenna is 1.5m above the ground, and the ground reflection is -1. Solution (a). c 3 × 10 8 λ= = = 0.1579m f 1900 × 10 6 PtGtGrλ2 50 × 1 × 2 × 0.1579 2 Pr(d ) = 10 lg( ) = 10 lg = 10 lg(1.58 × 10 −10 ) = −98dBW = −68dBm 2 4 2 (4πd ) (4π × 10 ) (b). λ2 Ae = Gr , 4π (c) V =. Pr(d ) =. E. 2. 120π. Ae ,. E =. Pr(d )120π = 3.67 mV / m Grλ2 / 4π. Pr(d ) × 4 Rant = 1.58 × 1010 × 4 × 50 = 0.178mV. (d) In order to use the two-ray model approximation, the following condition must be held. d>. 20hthr = 9500m λ 2.
(3) Since d=10000m, we can use the following equations for the two-ray ground reflection model,. ht2 hr2 50 2 × 1.5 2 = 50 × 1 × 2 × = 5.625 × 10 −11W 4 2 d 10000 Pr(dBm) = −72.5dBm Pr(W ) = PtGtGr. 4.16, For the knife-edge geometry in the Figure P4.16, show that. 2π∆ 2π ⎡ h 2 (a) φ = = ⎢ λ λ ⎣2 (b). v =α. ⎛ d 1 + d 2 ⎞⎤ ⎜⎜ ⎟⎟⎥ and d d ⎝ 1 2 ⎠⎦. 2d 1 d 2 λ (d 1 + d 2 ). ,. where. v 2π = φ , d 1 , d 2 >> h , h >> λ 2. ,. and. ∆ = p1 + p 2 − ( d 1 + d 2 ) . Solution 2. (a) P1 =. 2 1. 2. d + h = d1. ⎛ h ⎞ ⎛ h ⎞ 1 + ⎜⎜ ⎟⎟ , P2 = d 22 + h 2 = d 2 1 + ⎜⎜ ⎟⎟ ⎝ d1 ⎠ ⎝ d2 ⎠. 2. Since d1, d2>>h>>λ, h/d1, h/d2<<1, using Taylor series approximation, we have. ⎡ 1 ⎛ h ⎞2 ⎤ ⎡ 1 ⎛ h ⎞2 ⎤ 1 h2 1 h2 P1 ≈ d 1 ⎢1 + ⎜⎜ ⎟⎟ ⎥ = d 1 + , P2 ≈ d 2 ⎢1 + ⎜⎜ ⎟⎟ ⎥ = d 2 + 2 d2 2 d1 ⎢⎣ 2 ⎝ d 1 ⎠ ⎥⎦ ⎢⎣ 2 ⎝ d 2 ⎠ ⎥⎦ Therefore, ∆ = P1 + P2 − ( d 1 + d 2 ) ≈ d 1 +. 2π∆ 2π ⎡ h 2 And, φ = = ⎢ λ λ ⎣2. 1 h2 1 h2 h 2 ⎛ d1 + d 2 ⎞ ⎜ ⎟ + d2 + − (d 1 + d 2 ) = 2 d2 2 d1 2 ⎜⎝ d 1 d 2 ⎟⎠. ⎛ d 1 + d 2 ⎞⎤ ⎜⎜ ⎟⎟⎥ . ⎝ d 1 d 2 ⎠⎦. (b) From the definition of υ ,. υ 2π =φ, 2 υ=. Since, tan β =. 2π λ. υ= φ ⎡h2 ⎢ ⎣2. 2 π. ⎛ d1 + d 2 ⎞ 2 ⎤ 2 d1 + d 2 ⎜⎜ ⎟⎟ • ⎥ = h • λ d1d 2 ⎝ d1 d 2 ⎠ π ⎦. .. h h << 1 , tan γ = << 1 , d1 d2. we have, β ≈ tan β =. h h , γ ≈ tan γ = d1 d2. 4.17 A general design rule for microwave links is 55% clearance of the first Fresnel zone. For a 1km link at 2.5GHz, what is the maximum first Fresnel zone radius? What clearance is required for this system? Solution. 3.
(4) λ=. c 30 × 10 8 = = 0.12m f 2.5 × 10 9. For the first Fresnel zone, n=1. The maximum first Fresnel zone radius occurs for d1=d2=500m. Using equation (4.56), the Fresnel zone radius is found to be. rn =. nλd1 d 2 1 × 0.12 × 500 × 500 = = 5.48m d1 + d 2 500 + 500. Thus 55% first Fresnel zone clearance would require at least 5.48×55%=3.01m above the obstruction to the LOS path as shown in the figure below.. Chapter 5 5.1, Determining the maximum and the minimum spectral frequencies received from a stationary GSM transmitter that has a center frequency of exactly 1950.000000MHz, assuming that the receiver is traveling at speed of : (a) 1km/hr; (b) 5km/hr; (c) 100km/hr; and (d) 1000km/hr. Solution. 4.
(5) c 3 × 10 8 m / s = = 0.154m f 1.95 × 10 9 Hz v v f d = cos θ , f d max = λ λ v = 1km / hr , f d = 1.8 Hz v = 5km / hr , f d = 9.03 Hz v = 100km / hr , f d = 180.5 Hz v = 1000km / hr , f d = 1805 Hz c = λf , λ =. ∴. at 1km/hr, spectral edges are 1949.999998MHz and 1950.0000018MHz at 5km/hr, spectral edges are 1949.99999097MHz and 1950.00000903MHz at 100km/hr, spectral edges are 1949.999998MHz and 1950.0001805MHz at 1000km/hr, spectral edges are 1949.998195MHz and 1950.001805MHz. 5.7, If a baseband binary message with a bit rate Rb=100kbps is modulated by an RF carrier using BPSK, answer the following: (a) Find the range of values required for the rms delay spread of the channel such that the received signal is a flat-fading signal. (b) If the modulation carrier frequency is 5.8GHz, what is the coherence time of the channel, assuming a vehicle speed of 30 miles/hr? (c) For your answer in (b), is the channel “fast” or “slow” fading? (d) Given your answer in (b), how many bits are sent while the channel appears “static”? (e) A CDMA Rake receiver is able to exploit multipath when the channel is (circle all that apply) a) flat; b) slow; c) fast; d) frequency selective Solution (a) Ts=1/100kbps=10-5s. 0 ≤ σ τ ≤ 10. If σ τ <. 1 Ts ⇒ flat fading. 10. −4. Ts ≥ 10σ τ , σ τ ≤. Ts 10. v , c = λf , λ = c / f = 3 × 10 8 /(5.8 × 10 9 ) ≈ 0.05m λ 30miles 13 v= • = 13m / s , f d = = 260 Hz 0.05 hr. (b) f d =. Coherence time definition [50%--90%]. Tc ≈. 1 9 1 = 0.004S , Tc ≈ = = 0.00125S . 16πf m 5 f m fm. (c) Here we have, fd=260Hz, Ts≈10-5S, Tc≈10-3S. Slow fading ⇒. Ts << Tc , here , 10 −5 << 10 −3. (d) Pick your Tc, then, the bits sent = Rb Ts =. 10 5 b − 3 10 S ≈ 100 S. 5.
(6) 5.8, For the power delay profiles in the Fig. P5.6, estimate the 90% correlation and 50% correlation coherence bandwidth. Solution For (a), the 90% correlation coherence bandwidth is: Bc0.9≈1/(50στ)=1/(50*27ns)≈740kHz. The 50% correlation coherence bandwidth is: Bc0.5=1/(5στ)=7.4MHz. For (b), Bc0.9=1/(50στ)=1/(50*1.688μs)≈11.85kHz. Bc0.5=1/(5στ)=118.5kHz. 5.13, A vehicle receives a 900MHz transmission while traveling at a constant velocity for 10s. The average fade duration for a signal level 10 dB below the rms level is 1 ms. How far does the vehicle travel during the 10s interval? How many fades does the signal undergo at the rms threshold level during a 10s interval? Assume that the local mean remains constant during travel. Solution For ρ=-10dB≈0.316, 2. 2. 2. eρ − 1 eρ − 1 e 0.316 − 1 τ = ⇒ fm = = ≈ 132.8( Hz ) ρf m 2π ρτ 2π 0.316 × 10 − 3 × 2π v c 1 fm = ⇒ v = fmλ = fm = 132.8 × ≈ 44.3(m / s ) λ fc 3 d = vt = 44.3 × 10 = 443m For ρ = 1 ,. N R = 2π f m ρe − ρ. 2. 2π × 132.8 × 1 × e −1 ≈ 122.4(cros sin gs / s ). So, total number of fade the signal undergo at the rms threshold level during a 10 second interval=NRt=408. Chapter 7 N. 7.1, Use identical notation described in Section 7.3, except now let d k =. ∑w. nk. y nk , and verify that the. n =0. MSE is identical for the multiple input linear filter shown in Fig.P7.1. Solution N. In this case X k =. ∑W. nk. y nk , let Yk = [ y 0 k. y1k. ⋯ y Nk ] , T. n =0. Wk = [W0 k T. T. W01k. ⋯ W Nk ] . T. T. T. We have: X k = Yk • Wk = Wk • Yk , and ek = d k − X k = d k − Yk Wk = d k − Wk Yk We can see that the expression for Xk and ek are the same as equation (7.11) and (7.12), thus the MSE are identical. Using the same method described in section 7.3, we have the optimum might vector Wˆ for MMSE,. Wˆ = R −11 • P. 6.
(7) Where,. [. R = E Yk YkT and. ]. ⎡ y 02k y 0 k y1k ⋯ y 0 k y Nk ⎤ ⎢ ⎥ = E ⎢ y1k y 0 k y12k ⋯ y1k y Nk ⎥ ⎢y y 2 ⎥ ⎣ Nk 0 k y N y1k ⋯ y Nk ⎦. P = E [d k Yk ] = E [d k y 0 k. d k y1k. ⋯ d k y Nk ]. T. 7.2, Consider the two-tap adaptive equalizer shown in Fig.P7.2. (a) Find an expression for MSE in terms of w0, w1, and N. (b) If N>2, find the minimum MSE. (c) If w0=0, w1=-2, and N=4 samples/cycle, what is the MSE? (d) For parameters in (c), what is the MRE if dk=2sin(2 π k/N)? 7.3,For the equalizer in Fig.P7.2, what weight values will produce a rms value of ε k = 2 ? Assume N=5, and express your answer in terms of w0 and w1. Solution. 1⎡ 2π 2π ⎤ MSE = 2 + ⎢W02 + 2 cos( )W0W1 + W12 ⎥ + 2W sin =4 2⎣ N N ⎦ 1⎡ 2π 2π ⎤ 2 + ⎢W02 + 2 cos( )W0W1 + W12 ⎥ + 2W sin =4 2⎣ 5 5 ⎦ 0.5W02 + 0.5W12 + 0.309W0 W1 + 1.902W1 − 2 = 0 Any pair of W0 and W1 that can satisfy the above equation can have the rms value of ε k =2.. Chapter 9 9.1, The GSM TDMA system uses a 270.833kbps data rate to support 8 users per frame. (a) What is the raw data rate provided for each user? (b) If guard time, ramp-up time, and synchronization bits occupy 10.1kbps, determine the traffic efficiency for each user. Solution. 9.5, Assume that a nonlinear amplifier is used to broadcast FDMA transmissions for the US AMPS standard. If control channel 352 and voice channel 360 are simultaneously transmitted by a BS, determine all cellular channels on the forward link that might carry interference due to intermodulation. Solution. 7.
(8) 9.12, In a omnidirectional CDMA cellular system, Eb/N0=20dB is required for each user. If 100 users, each with a baseband data rate of 13kbps, are to be accommodated, determine the minimum channel bit rate of the spread spectrum chip sequence. Ignore voice activity consideration. Solution. 8.
(9) Chapter 11 11.1, Of the following air interface standards, identify all that are digital, analog, TDMA, and CDMA: GSM, AMPS, ETACS, IS-95, IS-136, DECT. Solution. 11.2, Which of the following is NOT true of GSM? Check all that apply. (a) The uplink and downlink channels are separated by 45MHz. (b) There are 8 half-rate users in one timeslot. (c) The peak frequency deviation of the GSM modulator is an integer multiple of the GSM data rate. (d) GSM uses a constant envelope modulation. Solution B, C 11.3, List all the cellular systems that do not support Mobile Assisted Handoff. Solution AMPS, ETACS, N-AMPS 11.7, Which of the following is NOT true of the IS-95 system. (a) No hard limits on capacity. (b) Soft handoff possible. (c) Uses slow frequency hopping. (d) The number of channels that can be accommodated in the forward and reverse links are different. Solution C. 9.
(10)
Related documents
Speci…cally, rational expectations models with standard (time-separable) preference assump- tions are largely unable to explain: why the price dividend ratio of stocks is so
Though prior studies have documented high rates of substance involvement among emergency psychiatry patients, this study extends those findings by further
In this study the α″ stress-induced martensitic transformation and damping behaviour of the superelastic β-Ti–25Ta–25Nb alloy are investigated by tensile tests at room
Using a time series data of market rate, internal rate and non-maturity deposits, we run these models and we estimate the optimal parameters.. Based on results and
Subjecting solutions (4% w/w) of α-lac or β-lg to the starch gelatinisation temperature profile (Section 3.2.4) demonstrated that β-lg resulted in a higher final viscosity
Estimates and 95% confidence intervals (corrected for overdispersion)for the Ψ(stage)p(trophic) community model. A) Effect of vegetation succession stage over occupancy estimate
Manufacturing Industry (CNAE 10-33) Logistics, Highways, Railroads, Ports, Waterways, Airports, Urban Mobility, Transportation, Storage, Mail Services and Others (NACE 49- 53) ICT
On top of the described test of universality and the extraction of TMD distributions, in this work we perform many additional studies of the TMD approach, some of which should be
