程序代写代做代考 CSE 516 (Fall 2016) Practice Set #1

CSE 516 (Fall 2016) Practice Set #1
1. ForachannelwithoutfadingorISIwhatisthemaximumpossibledatarategiven a bandwidth, B = 10MHz, and SNR of 23dB?
Shannon-Hartley theorem for AWGN channel:
Using the formula:
𝑆𝑆 = 𝑅𝑅−𝑛𝑛 = (𝐷𝐷�𝑅𝑅)𝑛𝑛 = (√3𝐾𝐾)𝑛𝑛
𝐶𝐶 = 𝐵𝐵 ∙ 𝑙𝑙𝑙𝑙𝑙𝑙 (1 + 𝑆𝑆�𝑁𝑁) 2
23dB  199.53
𝐶𝐶 = 10𝑒𝑒6 ∙ 𝑙𝑙𝑙𝑙𝑙𝑙 (1 + 199.53) = 10𝑒𝑒6 ∙ 7.65 = 76.5𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
2
2. Log-distancepathlossisindicativeofwhattypeoffading?
Average Large scale fading
3. Diffraction,reflection,andscatteringinduceswhattypeofsmall-scalefading?
Multipath fading
4. Rapidmovementorchangesintheenvironment(channelvariations),transmitter, or receiver is indicative of what type of small-scale fading?
High Doppler spreadFast fading
If BWsig > BWchanFrequency selective, fast fading If BWsig < BWchanFlat, fast fading 5. ConsiderasystemwithadesignconstraintofanS/Iof23dBorhigher.Iftheco- channel interference is limited to the first ring of nearest cells and a path-loss exponent of 3.5 is assumed, what is the frequency reuse factor given hexagonal cells? 𝐼𝐼 ∑𝑖𝑖0 𝐷𝐷−𝑛𝑛 𝑖𝑖0 𝑖𝑖0 𝑖𝑖=1 We know that �√3𝐾𝐾�3.4 ≥ 23 = 102.3 = 199.53 =� 200 6 �√3𝐾𝐾�3.5 ≥ 1200 𝐾𝐾 ≥ 19.2 Thus, K = 20  (This would be satisfied, for example, by i=3, j=3, K = 27) 6. Considerasystemwith10MHzofavailablebandwidth.Eachsimplexcarrieris 30kHz and we have a frequency reuse factor of 12. If there are 275 users in a cell, each generating 3 calls/hour, and each call is held average of 2 minutes, what is the blocking probability in the network? Assume 1 channel/carrier. • Total available bandwidth, Bsys = 10,000,000 Hz • Simplex carrier bandwidth, Bcarr = 30,000 hz • Frequency reuse factor, Freuse = 12 • The number of users in a cell = 275 users • Average call holding time, 1/μ = 2 min • The number of call per user = 3 voice calls/hour ♦ The number of carriers in the system, 𝑁𝑁 = 𝐵𝐵 = 10𝑒𝑒6 = 333.3�332 = 166 pairs for full duplex 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐵𝐵 𝑠𝑠𝑠𝑠𝑠𝑠 30𝑒𝑒3 𝐹𝐹 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 12 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑁𝑁 = 𝑁𝑁 = 166 = 13.8 = 13 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑖𝑖𝑒𝑒𝑐𝑐𝑀𝑀/𝑐𝑐𝑒𝑒𝑙𝑙𝑙𝑙 ♦ The number of carriers per cell, 𝑐𝑐𝑒𝑒𝑐𝑐𝑐𝑐 𝑐𝑐𝑒𝑒𝑟𝑟𝑟𝑟𝑒𝑒 ♦ The number of channels per cell: N = 10 channels/cell ♦ 1/μ = average call holding time = 2min ♦ λ = the number of calls/user * the number of users/cell = 3 calls/hr/cell/user * 1 hr/60 min * 275 users/cell = 13.75 calls/min ♦ Thus, the load, A = λ/μ = 27.5 ♦ The blocking probability: 𝑀𝑀 = 𝐴𝐴𝑁𝑁 = 27.513 = .5540 𝐵𝐵 𝑁𝑁!𝐴𝐴𝑛𝑛 27.5𝑛𝑛 ∑𝑁𝑁 ( ) ∑13 ( ) 𝑛𝑛=0 𝑛𝑛! 𝑛𝑛=0 𝑛𝑛! 7. Considerthesysteminproblem5and6.WhatistheS/Irequiredtogeta blocking probability of 1%? ♦ Find the number of channels, N, such that the blocking probability: 𝐴𝐴𝑁𝑁 27.5 𝑀𝑀=𝑁𝑁!= 13≤.01 𝐵𝐵 ∑ (𝐴𝐴𝑛𝑛) ∑ (27.5𝑛𝑛) 𝑛𝑛=0 𝑛𝑛=0 𝑁𝑁 𝑛𝑛! 13 𝑛𝑛! N >= 39
♦ Find the frequency reuse factor such that
𝐹𝐹166 ≥39 𝑐𝑐𝑒𝑒𝑟𝑟𝑟𝑟𝑒𝑒
Thus, Freuse <= 4 ♦ Thus the required S/I is: 𝑆𝑆 = (�3𝐾𝐾)3.5 = (√3 ∙ 4)3.5 = 12.89 = 11.10𝑑𝑑𝐵𝐵 𝐼𝐼66 for N = 35:45 Pb_den = 0.0; Pb_num = (27.5^N)/factorial(N); for n = 0:N Pb_den = Pb_den + (27.5^n)/factorial(n); end Pb_ans =Pb_num/Pb_den; Pb_ans if Pb_ans < 0.01 N end end