无线通信英文概要.doc

1.Show that the wave function satisfies the differential EM wave equationS：Similarly,Clearly A and B are equivalent provided that：2. Show that the wave functionsatisfies the differential EM wave equationS：Since Magnetic field only varies in z direction (ey for unit vector of y-axis Orthogonal to the variation ex for the x-axis in case of electric field) we haveBut j2= -1NowBut j2= -1ThereforeAs for Question 1 for A and B to be equal we need the following to be true:Which is as before:Tutorial 61.Briefly (1-2 sentences) explain the following terms:(a) The Poynting Vector(b) Forward Error Correction(c) Doppler Spread(d) Mutual orthogonality (of a set of vectors)(e) Inter-symbol interference(f) Polarisation (of an electromagnetic field)(g) Antenna noise temperature(h) Antenna sidelobes(i) Channel equalisationS：(a) Poynting vector, P, is the power flow through area S, that results in the change in stored energy within S At any point in space, P gives the magnitude and direction of power flow Pis the cross product of E and H, it is a vector, and is orthogonal to both E and H ie Poynting vector is defined as P = E x H It has dimensions of watts / square metreb) This is where single or double or multi-bit errors can be accounted for and corrected by using coding /code words, for example even and odd parity can be used to detect single bit errors; some techniques allow for more bit errors to be detected and correctedeg spacing code words (Hamming codes) so far apart that multiple bit errors can be detected but this is at the cost of creating extra overhead bits and reducing available bandwidth for actual data there is an entire area of error coding within wireless which deals with this!(c) Doppler shift is the modulation (i.e. variation) of the nominal frequency of an electromagnetic signal resulting from relative motion of the transmitter, receiver and/or reflectors.(d) This is where the dot product of two vectors is zero, indicating that in their n-dimensional subspace that they are perpendicular to each other (for example if in a 2 or 3 dimensional coordinate system we essentially extend this into an n-dimensional system for orthogonal codes where n3). This is not true of their shifted versions the reason for Multiple Access Interference (MAI)(e) This is where one symbol interferes with another (due in wireless to multipath) in the consecutive signaling of adjacent symbols transmitted in the same symbol stream(f) Polarisation is the geometric orientation or alignment of the E and H vectors of a time-varying electromagnetic field. It can be linear (most often the intention) or circular (E field rotates in a circle over time in direction of travel (if E is x this direction would probably be z) or elliptic (which can be shown to be the general case with linear and circular being special cases of elliptic).(g) Noise temperature of an antenna is the temperature of a resistor which produces an equivalent amount of noise power per unit bandwidth. It is a function of two other functions: antenna gain and environmental temperature each as a function of solid angle.(h) Antennas have a radiation pattern which, ideally, focuses all energy in a specific direction. In practice, as well as a primary direction of maximum gain, there are other directions (in， ) in which there are secondary maxima in the gain function. These are called the antenna sidelobes.(i) Equalisation is the correction of symbol distortion which results from a non-ideal channel impulse / frequency response. Such distortion is an inevitable consequence of finite channel bandwidth2.Suppose that we are attempting to measure the distance between the Earth and the moon. To do this, we will use a large radio-telescope antenna with a maximum gain of 70 dB as both transmitter and receiver. A microwave pulse will be generated with a power level of 1 kW at a frequency of 4 GHz. The effective area of the moon is 6.61011 m2 (not the geometric area - this is the effective area of an isotropic reflector, one which re-radiates all of the incident power equally in all directions). If the receiving antenna has a noise temperature of 25 K, determine the SNR at the receiver if the distance to the moon is 3.8108 m, if the medium loss coefficient Cm is 0.7 and there are no polarisation losses. Assume the antenna is 100% efficient.S：Noise power：3.(a)Calculate the reflection coefficients and VSWR for a junction between a 73 ohm antenna and a 75 ohm coaxial cable.(b)An engineer has designed a four-channel CDMA system with the following chip sequences: Is this a good choice of chip sequences? Briefly explain why/why not.S：Zt=73，Z0=75，HenceIf we reverse Zt and Z0 we will have =0.0135=0.0135,so:The individual chip sequences are not orthogonal - for example, the dot product of+ 1,+ 1,+ 1,+ 1 1,+1,1,1is -2, which is non-zero. Indeed, if any pair of these sequences are not orthogonal, then the set is not mutually orthogonal, and it cannot be used for CDMATutorial 71.Assuming that SIR required for the satisfactory voice performance on the forward channel of a cellular system is 18 dB, what is the frequency reuse factor and the corresponding cluster size for the maximum system capacity if the path loss exponent is:(a) = 4 (b) = 3 (c) = 2Take to account that the system is fully developed, i.e. there are 6 interfering co-channel cells in the first tier.S：Worst Case scenarioOnly considering the interfering cells The path loss exponent is equal to 4, and the number of cochannel interfering cells in a fully developed system is 6,(i0 = 6).I J N Q S/I in dBAverage Worstcase1 1 3 3.00 11.30 8.032 1 7 4.58 18.65 17.263 0 9 5.19 20.84 19.742 2 12 6.00 23.34 22.572. Explain the handoff process in a cellular network. What are the tradeoffs between the handoff initiation threshold and the minimum power required to operate.S: Threshold at which handoffs are made: Optimal signal level at which to initiate the handoff needs to be specified by system designers. The value of this level is derived from the value of minimum usable signal strength, Pmin-usable, for acceptable voice quality (- 90dBm to -100dBm). A signal slightly stronger, Phandoff, is used as a threshold at which handoff is initiated. The margin, = Phandoff - Pmin-usable, cannot be too small or too large. too small - the time for doing handoff might be insufficient, too large - there may be unnecessary handoffs.3.Given the average call holding time for a mobile call is 30 sec. and the average time a user spends in a cell is 8 minutes. Calculate the channel holding (i.e. the effective call holding time) The channel holding time is related to the call holding time and the cell dwell time by the following formula 1/u = 1/(t+d) where u, t and d are the respective holding and dwell rates. Hence t = 1/30, and d =1/60x8 Hence the channel holding time u = 28.23 sec4. A city has a population of two million residents. Three competing mobile networks provide cellular service in the area. System A has 394 cells with 19 channels each, system B has 98 cells with 57 channels each and system C has 49 cells with 100 channels each. Find the number of users that can be supported at 2% blocking if each user averages two calls per hour at an average duration of 3 minutes.S: System A Given: Blocking probability is 0.02 C=19 Traffic intensity is Using the Erlang B chart with C=19 and GoS = 0.02, we get a total load A = 12.33 ErlangsIf each user has an intensity of 0.1 Erlangs, the total number of users = 12.33/0.1 = 123 users.Total number of users is= total number of cells x users per cell = 394 x 123 = 48,462 users System B Given: Blocking probability is 0.02 C=57 Traffic intensity is Using the Erlang B chart with C=57 and GoS = 0.02, we get a total load A = 46.82 ErlangsIf each user has an intensity of 0.1 Erlangs, the total number of users = 46.82/0.1 = 468 users.Total number of users is= total number of cells x users per cell= 98 x 468 = 45,864 users System C Given: Blocking probability is 0.02 C=100 Traffic intensity is Using the Erlang B chart with C=100 and GoS = 0.02, we get a total load A = 87.97 ErlangsIf each user has an intensity of 0.1 Erlangs, the total number of users = 87.97/0.1 = 879 users.Total number of users is= total number of cells x users per cell= 49 x 879 = 43,071 user5.A city has an area of 1300 square miles and is covered by a cellular system using a seven-cell reuse pattern (assume hexagon cells). Each cell has a radius of 4 miles and the city is allocated 40 MHz of spectrum with a full duplex channel bandwidth of 60 kHz. Assume a GoS of 2% is specified. If the offered traffic load per user is 0.03 Erlangs, compute (a) the number of cells in the service area, (b) The number of channels per cell, (c) traffic intensity of each cell, (d) Maximum carried traffic, (e) the total number of users that can be served at 2% GoS, (f) the number of mobiles per channel, where channels are reused and (g) the theoretical maximum of users that can be served at any one time by the systemS: (a) cell radius = 4 miles and total area = 1300 miles square. Area of hexagon is = 2.5981R2= 41.57 square miles per call. Total number of cell = 1300/41.57 = 31 cells. (b) Total number of Channels = allocated spectrum (40 MHz)/channel width (60 kHz) = 666.66 channels since there is a reuse distance of 7 then the total per cell is given by 666 66/7 = 95 channels/cell 666.66/cell. (c) C=95 and GoS = 0.02, Using tables traffic intensity = 83.134 Erlangs (d) Maximum traffic is = cell number x carried load per cell = 31 x 83.134 = 2577.154 Erlangs. (e) Total number of users = Total load/ load per user = 2577.154/0.03 =85,905 users. (f) Number of mobiles/channel = Total users/total channels = 85,905/666.66 = 128 mobiles (g) Maximum number of users at any one time is when all the channels are occupied = C x number of cells = 95 x 31 = 2,945Tutorial 81.In an Omni directional CDMA cellular system, the E/N = 20 dB. If 100 users, each with a baseband data rate of 13 kbps are to be accommodated, determine the minimum bandwidth of the spread spectrum chip sequence.S:Given N, R and E/N, we can determine the channel rate1980=W /13000 W=25.74MHzRepeat the previous problem for the case where voice activity is 40%W=0.4Wold W=0.425.74MHz W=10.3MHzW=990.41300020 W=10.296MHz3.Repeat the previous problem where the CDMA cell employs a tri-sectored antenna system.S: Need to support 100 users, hence each sector supports 33.33 users.Wproblem2=10.3MHz WWproblem2/3 W10.3/33.43MHzW=32220.4 1300020 W=3.36MHz4.For a bit stream of 1011011000, determine the output from a rate 1/2, constraint length 2, convolution encoder. (that uses the same generating polynomial as provided in the lecture).S: Using the state diagram or the generator polynomial. Polynomial 1 = 1 + x + x2, Polynomial 2 = 1+ x2 The first two bits in the register all always 00. The bit stream is shifted one bit at a time and the resulting 3 bits are multiplied by each polynomial. The results for each polynomial are added. For the bit stream 1011011000 The output is 11 10 00 01 01 00 01 01 11 00Tutorial 91.Given that the throughout equation for ALOHA is given by T = Re-2R，Show that the maximum throughput occurs at R=0.5S:Differentiating: T = Re-2Rthus:If we set this equal to zero we find a minima or maxima:To show that it is a maximum throughput we calculate the second derivative:so if R=0.5which indicates a maximum throughput occurs at R=0.5 as required2. Determine the maximum throughput that can be achieved using slotted ALOHA.S：For slotted ALOHA: T = ReRSetting derivative to zero to find maxima (second derivative will be negative):eR (1 R) =0 R =1Setting R=1 into throughput equation:3.Ten thousand airline reservation stations are competing for the use of a singleslotted ALOHA channel. The average station makes 18 requests/hour, and a slottime is 125 microseconds.(a) What is the channel load?(b) What is the expected number of attempts per request to get serviced?S：(a)R=t=(1810000/3600) 125sec R=0.00625 (b)E=eR=e0.00625=1.0063Tutorial 101. Given the following packet reservation MAC protocol.Reservation packets are used to reserve time slots for the actual data in the same time frame. There are 1000 reservation slots and there are 90 data slots. Each data slot can hold a 1 kbyte packet, while the reservation slots hold a 10 byte reservation packet. Determine the average number of successful packets of this scheme if no collisions occur after a data slot is reserved. In the above it is assumed there are 20 users and on average attempt to send 1 data packet every frame time.S: Assuming each user transmits one reservation packet, the traffic intensity or load on the reservation slots is going to be 20/1000 = 0.02. The throughput of reservation packets is then going to beT = R.eR = 0.02e0.02 = 0.0196 Hence, determine the proportion of packets that are successful after being reserved ( Noting 20000/(10000+90000)=0.2;20000 bytes for data packets, 10000 bytes for reservation packets and 90000 bytes in total for data packets gives the 0.2 as first term: )T=0.2 0.0196/0.02=0.196Number of packets 19.6 packets on average2.Compare this result to slotted ALOHA, where the load is the same, but the there are no reservation channels, but the total number of slots is 100. (Hence both schemes have access to the same bandwidth)S: The throughput of slotted ALOHA is going to be T = R.eR= 0.2e0.2 = 0.164 Hence on average 16.4 packets get through, compare to 19.6 packets for PRMA3.Repeat the previous problems when the load is high at 80%, i.e. 80 users with 1 packet per frame to transmit. Compare the performance of the two protocols now.S: Now since there are 80 reservation packets and 1000 opportunities to transmit, then the intensity is going to be 0.08. Hence T is going to beT = 0.08e0.08 = 0.080.923 = 0.0738 Now the number of data packets that are successful is going to