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Wat question papers

  1. 1. THM, FRIEDBERG WAT Question papersWAT translated German Question paper with solution UdaySharma 2/8/2013
  2. 2. WAT Question papers WAT Exam 2002/2003Task 1: free-space propagation (20 points)Between sites in the picture is a radio link can be established.Given:2 GHz frequency, antenna gain = 27dBi each gant, cable loss each aKabel = 2dB. Distance 10km,according obstacles picture (not to scale).looking for:a) What is the antennas must be mounted at least? Consider When the radio field viewing onlythe influence of the free-space propagation.b.) = Which transmit power PTX? [dBm] must send the RF transmitter, if the required receivedpower min. PRX =-10dBm should be?c.) Which wattage corresponds to the one found in b.) power?d) What other effects could, in this wireless field computing a role play?Task 2: Antennas (7 points)An omni-directional cellular antenna with a veritkalen Öffungsbreite of 8 ° (vertical HPBW = 8)and a gain of 11dBi = Gant is a feeder cable Fed (3 dB attenuation) with 10 watts.a) What is the equivalent isotropic radiated power (EIRP) in the main beam direction?b.) The antenna has an electrical tilt of 4 °b1.) Sketch the image in the lower right vertical AntennendiagramUday Sharma Page 2
  3. 3. WAT Question papersb2.) Why is used in this antenna is no mechanical tilt?b3.) Which equivalent isotropic radiated power is at an angle of Radiated 8 ° to the horizontal(see picture below)?Task 3: Link Budget (13 points)Consider the following unbalanced link budget. The transmission power of the base station canbe regulated in 2dB steps. It can be assumed that a potential diversity gain of 3dB.a) Please provide at least three solutions to a balanced link budget produce. Note: + / - 1dBalready considered to be balanced.b.) For each solution the maximum possible path loss.c.) For each solution to an advantage and a disadvantage.Task 4: Traffic dimensioning (20 points)Uday Sharma Page 3
  4. 4. WAT Question papersA GSM network operator has 48 frequencies a GSM 900 license. The traffic model is of anaverage call duration of 120 seconds for Main Busy Hour assumed and blocking probability of2% (pBLOCK = 2%) required.Note that not because of the signaling traffic per TRX 8 all Time slots for voice traffic areavailable:Number of TRX 1 2 3 4 5 6 7 8Time slots for signaling 1 2 2 3 3 4 5 5Time slots for voice traffic (TCH) 7 14 22 29 37 44 52 59Voice traffic [Erlang] at block = 2% 2.9 8.2 14.9 21 28.2 34.6 42.1 48.7a) The operator of its network in rural area plans with a Reuse of ARCS = 16th How big can themax. His cell radius r1, with a traffic density of 10 Mobile subscribers per square kilometer?Appreciate this from the cell surface with A = πr ².(Pure traffic analysis, Omni cells).b.) In the downtown area, the traffic density 500 mobile subscribers per km ². What is now themaximum cell radius r2? What do you recommend as a first step to increase the cell radius?Make a suggestion and calculate the new radius r3. What disadvantage is hereby accepted?c.), to which value the traffic density can be introduced by the Frequency hopping / FrequencyHopping increase if no Planned quality losses now with ARCS of 6 and is the cell radius r2 frompart. b) is assumed? Based on what effects this process?Uday Sharma Page 4
  5. 5. WAT Question papers WAT Exam 2004/2005Task 1: Assessment of range (10 points)The maximum range of a new radio system is to be interpreted to estimate. When theselected modulation method was the following relationship between recoverablethroughput and signal to noise ratio at the input of Demodulator determined.The channel bandwidth (equivalent noise bandwidth) of the modulated signal amounts to B= 5MHz. The rate amounts to 2.4 GHz. It provides a relatively cheap receiver amplifier with aNoise figure of 15dB insert. The transmit and receive antenna have a Profit from each 5dBi.The cable losses are negligible and the Operating temperature is T = 290 ° K.Maximum range of which is at a transmission power of 100mW likely to achieve therespective data rates of 1, 2, 5 and 10 Mbit /s?Noise Floor: 10log (kTB/1mW) =-107dBmAt the receiver input:-92dBmMax path loss =-20dBm (-92dBm) +2 * 5dBi SNR = 122dB SNRMAPL @ 1Mbit / s = 122dB-5dB =-117dBmMAPL @ 2Mbit / s = 122dB-15dB =-107dBmMAPL @ 5 Mbit / s = 122dB-20dB =-102dBmMAPL @ 10Mbit / s = 122dB-25dB =-97dBmMaximum range in free space propagation:Task 2: comprehension questions (18 points) What do the abbreviations FDD and TDD. Which of the two techniques is better suited forpacket-switched Internet traffic? Reason! Which constraint is the equation for free-space propagation valid? microcells on mobile networks in areas of high traffic density employed. What is the decisiveadvantage over macro cells? What is the reason for a GSM link budget a "fade margin" introduced? Which unit of the gain of an antenna is given? Explain the meaning of this definition.Uday Sharma Page 5
  6. 6. WAT Question papers What is the qualitative relationship between the income of a Antenna and the horizontal andvertical opening width? Why is the description by means of horizontal and vertical Antenna patterns only anapproximate description of actual antenna pattern. What is the qualitative relationship between ARCS, interference and Capacity in cellular mobileradio networks, such as e.g. GSM? What distinguishes the two GSM Data Services: GPRS and HSCSD? How can a Antennentilt berealized. What is the reason in the mobile cross-polar antennas are used? Give three reasons trigger a handover to a GSM network can. Which "handover cause" indicatesa good network planning? What will be introduced in the GSM network "location areas"? What is the radius of a UMTS cell depends on the number of active Participants? What does the acronym CSMA / CA. Name a radio network that this Method used and brieflydescribe why it is used and the basic functioning. When 802.11g wireless standard, the symbol rate is always 0.25MBaud. Describe how a datarate of 48MBit / s and 54 Mbit / s is realized.Task 3: GSM dimensions (15 points)It shall be a GSM900 network can be dimensioned. The future network operator has aLicense for 48 frequencies.The cities and towns are to be supplied in the following categories divided:City Category 1: 6 Erl / km ²City Category 2: 2 Erl / km ²Location Category 3: 0.1 Erl / km ²When the following issues of the sub-tasks a) to c) by reference. assumed omnidirectional sites! Thehexagonal cell surface can A = 2.6 R ² can be estimated.a) What is the maximum allowable cell radius in the three categories, when the Blocking probability2% and the average Interference probability should be about 2.5%? Remember to leave an therequired transport for signaling.Because demand for interference: ARCS = 12 I.e. max. 4 carriers per site.I.e. at 2% Blockierwarscheinlichkeit and transport: 21 Erlang per cell. MaxCell area = 21 = 3.5km ² ² Erl/6Erl/km / 10.5km ² / 210km ²Cell radius = 1.16km and 2km or 8.98 kmb.) Make a balanced budget on link by following Link budget balance. Proper emphasis on units!How large is the maximum achievable path loss when for an indoor supplyUday Sharma Page 6
  7. 7. WAT Question papersfollowing additional losses should be considered?City Category 1: aGebäude = 20dB 131dBCity Category 2: aGebäude = 10dB 141dBLocation Category 3: aGebäude = 0dB 151dBWith what care you can count on the likelihood of cell edge? 50%c.) Wie groß sind nun die Zellradien für die Städte der Kategorien 1 bis 3 unterInto account the results from part a) and b.)?There are as specified Morphokorrekturfaktoren. Expect a Frequency of 900MHz, HBTS = 30m, MS =1.5m.Category 1: LMorpho 0dB = 1.35km so limited traffic with 1.16kmCategory 2: LMorpho 3dB = 3.15km so traffic limited to 2kmCategory 3: LMorpho = 7dB 7.88km so far-reaching limitsTask 4: Bundling profit (7 points)A mobile subscriber phone during an observation period of 2 Hours on average 12min. a) What is the transport 10 students cause this behavior? 10*0.1Erl=1Erl How many traffic channels would be required for a blocking probability of PBLOCK = 0% necessary? 10 Channels How many traffic channels would be for a blocking probability of pBLOCK = 4% Necessary? 4 Channelsb.) What traffic causing 100 people with this behavior?100*0.1Erl=10ErlHow many traffic channels would be for a blocking probability of pBLOCK = 4% Necessary?15 ChannelsWhy is not the required number of channels also a factor of the number of participants up against asub-task)?Bundling profit:Uday Sharma Page 7
  8. 8. WAT Question papers WAT Exam 2006/2007Task 1: comprehension questions (25 points) 1. Name two situations where a location area update in Wireless network is performed. In what state is it the Mobile station? - Mobile station is in idle mode. If mobile station switched or is the location area changes or cyclical. 2. The TDMA frame contains a 26 long in addition to the payload bit "Training sequence code". Why? - For equalizer for channel equalization. 3. How can a given bandwidth, noise figure F and at a known Modulation method, the sensitivity of a radio system for a desired bit error rate can be estimated? - Can of BER at given Modverfahren required PS / PN (actually first be determined Eb / No then in Ps / Pn) convert. Then there is the sensitivity Prx = 10log (ktB/1mW) + R + Ps / Pn = -174dBm/Hz +10 logB / [Hz] + F [dB] + Ps / Pn [dB] 4. What is the physical effect is responsible for frequency selective fading and so fast fading is caused? - Multipath, and Movement 5. What is the statistical distribution function for fast fading is under Line of sight (LOS) and Non Line of Sight (NLOS) conditions described? - Rice and Rayleigh 6. What are the physical effects are the so-called "log-normal" distributed Fading responsible? - Shadowing, diffraction at larger obstacles 7. What is the abbreviation EIRP (available in English) and how this size is defined? - Effective Isotropic Radiated Power. Power is supplied by the antenna minus cable losses plus antenna gain. 8. What is the three measures for GSM, a link budget balanced are bounded in the uplink? - TMA, AD, reducing the BTS power 9. A UMTS data channel) has a bit rate (binary modulation of 12.2 kbit / s As large, the process gain [dB], and the spreading factor is at a chip rate of - 3.84 Mchip / s 10logRc/Rb = 25dB spreading factor S = Rc / Rb ≈ 3.14 10. Why is a "blank" cell in the UMTS uplink and full UMTS cell Downlink limited (empty and full in the sense of a lot and little traffic). - An empty cell is in an interference limited system limited transmission power of the mobile station is limited. Must at a full cell the base station with its finite transmit power several participants supply, thereby limiting the increase comes from the downlink increasing number of participants concluded. 11. What is the job of a rake receiver for UMTS? - Channel equalization to prevent frequency-selective fading and Intersymbol interference. It becomes even made a profit.Uday Sharma Page 8
  9. 9. WAT Question papers 12. What does the term OFDM, which is why this method is particularly useful in Mobile channel advantageous? - Orthotogonal frequency division multiplexing. Since been extended by the Symbol duration (parallel transmission of data on different frequencies) caused by the multipath problem of intersymbol interference is reduced. 13. What is the access method in WLAN in ad hoc mode. Why is not the same procedure as used in Ethernet? - CSMA / CA with DCF and PCF. Because of the hidden station problem can not CSMA / CD can be used. 14. n) What are the three main application scenarios in WiMAX - Fixed, Nomadic, MobileTask 2: profit by sectorization (14 points)An omnidirectional site with RCS = 12 to be sectored. Estimate in from the following sub-tasks, so if acapacity increase at consistent interference is achieved. Take the case in the lecture derivedanalytical relationship between C / I and RCS with hexagonal Cell structures: a) Due to the sectorization, the number of co-channel interferers of k = 6 on k = 2 is reduced. The propagation exponent amounts to n = 4 which Site Reuse factor RCS can be chosen after sectorization, if the interference situation will remain about the same? Compare the Capacity C [Erl] the locations at B = 36 frequencies and 8 traffic channels per frequency and pBLOCK = 2%. Ans : Omnisite: C / I = sqrt (3 * 12) ^ 4/6 = 216 36/12 * 8 = 24 traffic channels thus 6.16 Erlang Sector Site: 216 = sqrt (3RCS ) ^ 4/2 thus RCS ≈ 7 and almost int *36/7+ * 8 ≈ 40 TCH per location or 5 Frequencies. This can for example be 2/2/1 spread to sectors. So 16/16/8 TCHs and thus 9.8 / 9.8 / 3.6 = 23.2 Erlang Erlang Figure 1: sectorization one omnidirectional siteUday Sharma Page 9
  10. 10. WAT Question papersb.) Loosen the task a part) of a spreading coefficient of n = 3Omnisite:C / I = sqrt (3 * 12) ^ 3/6 = 36 to 36/12 * 8 = 24 traffic channels thus 16.6 ErlangSector Site:36 = sqrt (3RCS ) ^ 3/2 thus RCS ≈ 5.7 and almost int *36/5, 7+ * 8 ≈ 50 TCH per location, or 6.3Frequencies. This can for example be 2/2/2oder3 distributed among the sectors. So TCHs 16/16/16and 3 * 9.8 Erlang = 29.4 Erlang.c.) Why is the Sektorisierungsgewinn in Teilaufgbe b.) higher than in Tasks a part)? May also occurthe case that the through sectorization Capacity is reduced at each location? If so, why?Because the positive result of the lower coefficient of expansion of the influence of lowerInterference is significant, as the negative effect of the reduced profit pooling. Yes when thenegative effect of lower profit pooling predominates. E.G. at n = 5, or if a larger bandwidth.Task 3: GSM dimensions (26 points)It was created for a GSM900 network an offer. Check the expected power quality in terms ofcoverage probability, Blocking probability and probability of interference. The cities and regions tobe supplied are divided into three categories. It for the respective areas following location numbershave been offered:• City Category 1:10 three-sector sites for 100 km ²• suburban Category 2:Omnistandorte for 15 300 km ²• environment Category 3:Omnistandorte for 10 500 km ²Note: The surface of a hexagonal cell Omnistandortes can with A = 2.6 R ² estimated. The area of asector of a three-sector site can A ² = 0.65 D can be estimated according to figure 2. The necessaryoverlap of the Cells should be neglected. Use the following dispersion model to determine the pathloss in Part of a task):L [dB] = 135 +30 log (R / [km]) - LMorphoa) What are the respective cell radii R (category 2 and 3) and the cell diameter D (Cat. 1)?D=sqrt(100/10/3/0,65)=2,26km, R=sqrt(300/15/2,6)=2,77km, R=sqrt(500/10/2,6)=4,38kmDue to the given location and square footage numbers What therefore is the maximum allowablepath loss for Pcov = 50% at the cell edge in the three regions?Always 145 DBCategory 2: LMorpho = 3.3 dBCategory 3: LMorpho = 9.2 dBb.) Ask for Omnistandorte the related link balanced budget on and enter in the correspondingcolumn in the correct units!Uday Sharma Page 10
  11. 11. WAT Question papersHow large is therefore the maximum possible path loss for the omnidirectional Locations. How big itis for the sectorized sites when a sector antenna is used with a gain of 18dBi?148dB and 155 dBWith what probability can supply about the cell edge in the respective categories 1 to 3 expected, ifwe follow on Standard deviations out? Note: What is the latest "Lognormal Margin "?Category 1: σ = 10dB 10dB Lognormal margin: Pcov = 84.1%Category 2: σ = 3dB 3dB Lognormal margin: Pcov = 84.1%Category 3: Lognormal σ = 3dB 3dB Margin: Pcov = 84.1%Estimate the network-wide coverage probability. Pcov (NW)=95%c.) What is the probability of interference pint when a Blocking probability of pBLOCK = 2% isrequired and the network operator Has license for 48 GSM frequencies. Consider the following whencalculating traffic traffic for signaling and use for estimating the pint empirical diagram from thelecture. Go out of oncoming traffic densities and the given cell surfaces:A1 = 3.33 km ², 20km ² = A2, A3 = 50km ²Category 1: 6.3 Erl / km ² traffic per cell 21, Erl, 4 TRX, ARCS = 48/4 = 12, PT = 5%Category 2: 1.05 Erl / km ² traffic per cell 21, Erl, 4 TRX, ARCS = 48/4 = 12, PT = 5%Category 3: 0.42 Erl / km ² traffic per cell 21, Erl, 4 TRX, ARCS = 48/4 = 12, PT = 5%d) How many participants can be found in the cells and in the whole Mobile Network (Cat. 1 to 3)can be supplied, if from an average call duration 1 minute is taken during rush hour? How big is theassumed proportion of mobile phone users, if the population of the entire Region100.000 is?1min Holding Time => 16mErl/Tln => 1260 pts per cell => (3 * 10 +15 +10) * 1260 = 55 * 1260 =69300 pts total. Thus, 69.3% penetration rate.Uday Sharma Page 11
  12. 12. WAT Question papersTask 4: mobile channel (10 points)a) It is in a radio channel Power Delay was gem profiles. Fig.3 determined. What is the value of στ(RMS delay spread)? Is frequency selective Expected fading, when a data rate of 1 Mbit / s for abinary modulation (and BT = 1) is expected? Thus intersymbol interference to and the use of anequalizer expect meaningful?τ = (1 * 1 +0.1 * 2 +1 * 3) / (1 +0.1 +1) = 2μs"Τ ² = (1 * 1 +0.1 * 4 +1 * 9) / (1 +0.1 +1) = 4.952μsστ = sqrt ("τ ²" ² τ) = 0.976μsFrequency selective fading, when Ts <10στ = 9.76μsTs = 1/1MHz = 1usThus: frequency selective fading and intersymbol interference. The use of a EQ is so usefulb.) For the channel estimation of the equalizer to work properly, the need for Duration of the TDMAframe, the channel impulse response be invariant (condition ) for slow fading. How long is at amaximum. speed of 100km / h and f = 900MHz, the maximum frame length to be so? How many bitsper TDMA frame would therefore max. make sense?TTDMA <= λ/2/v=c/f/2/v=5990μs somit 5990 Bits.Uday Sharma Page 12
  13. 13. WAT Question papers WAT Exam 2007/2008Task 1: Assessment of range (11 points)The achievable throughput of a wireless system is assessed. It has the following Connection betweenthe realizable throughput and signal to noise ratio at the input of the demodulator is determined.Fig. 1: Throughput as a function of signal to noise ratio @ input ofDemodulatorFurther the following parameters are given:• Noise bandwidth: 20MHz• Output power PTX=100mW• Noise Figure F=7dB• Antenna gain of transmitter and receiver antenna: 5dBi each.• Cable attenuation: 1,5dB each• T = 290°K.The following propagation model has been calibrated: L[dB]=135+30log(d/km)Which is the max . possible throughput in the following distances:d1=160m, d2=230m and d3=290m?Pn=-174dBm/Hz+10log20*10^6=-101dBmEIRP=PTX-acable+Gant=23,5dBmUday Sharma Page 13
  14. 14. WAT Question papersIRP=PN+F+SNR+acable-GANT=-97,5dBm+SNRMAP=135dB+30logd/km=EIRP-IRP=121dB-SNRAlso SNR=-14dB-30logd/km somitd1=160m, SNR=10dB,DR=4MBit/sd2=230m, SNR=5dB, DR=1,8MBit/sd3=290m, SNR=2dB, DR=1Mbit/sProblem 2: Questions (18 points)a.) Explain the difference between duplex and half duplex.b.) Why does cell sectorisation increase the capacity? Is it also possible, that capacitydecreases if cells are sectorised? Explain why?c.) Explain the meaning of the abbreviations FDD and TDD. Which of these techniquesshould be used for packet switched traffic. Explain!d.) Under which circumstances is the formula for “free space propagation” valid?e.) What is the purpose of a fading margin in a GSM Linkbudget?f.) Which „unit“ is used to define the gain of an antenna? Explain the term antenna gain.g.) The radiation pattern of antennae is usually described by means of a horizontal and avertical pattern. Why is this only an approximation of the real radiation characteristic?h.) Explain the term frequency selective fading in the time and in the frequency domain.i.) Explain the term fast fading in the time and in the frequency domain.j.) Explain the difference between „Large-Scale-Fading“ and „Small-Scale-Fading“ and whatis th the name of the statistical distribution of the power level (PDF) in both cases?Problem 3: Dimensioning Problem (GSM) (21 points)a.) A future mobile operator purchased a GSM900 licence with B=36 frequencies. It isforeseen to cover a city with 25 000 inhabitants. The penetration rate (=percentage of users,which use cell phones) is 70% which are homogeneously spread over an area of 500km².During main busy hour a user uses his phone for approximately 3 minutes. The blockingprobability should not exceed 2%. For the network design only omnidirectional cells will beused. Compute the cell radius based on traffic requirements for the following three casesa1.) ARCS = 18a2.) ARCS = 12a3.) ARCS = 6Hint: Signalling traffic can be neglected. To compute the hexagonal cell area, use thefollowing formula: A = 2,6R².b.) Consider the incomplete link budget in the table below and the following parameters:Pcov=95% at the cell border, standard deviation σ=5dB. PTx,max=45dBm (adjustable in -2dBsteps), diversity gain and max. gain of an optional tower mounted amplifier (TMA): 3dBeach. Complete the link budget!Uday Sharma Page 14
  15. 15. WAT Question papersc.) Compute the cell radius based on the results of the link budget above? Forcomputation use the following propagation model: L[dB]=135+35,3log(R/[km]).d.) Which ARCS should be recommended based on the results of problem a.) andc.)? How many sites are necessary? Which Quality of Service do you expect for thisnetwork: Pint, Pcov and Pblock?At d = 1.68 km = 7.33 km ² A follows, and thus 68.2 LocationsThis 256 Tln / cell or 12.8 Erl / cell to form 20 TCHs so 3 TRX(2 and 4, are too little too much).With ARCS = 12 both conditions are met, pint = 5%, pBLOCK will be better than 2% since 24and 20 not available TCHsis. Pcov at cell edge remains at 95%.Uday Sharma Page 15
  16. 16. WAT Question papersProblem 4: Mobile Communication Channel (6 points)The local spatial average of a power delay profile is measured at 900 MHz as shown in fig. 2a.) Compute the „mean excess delay“ and the „rms delay spread“.Sketch both figures into Figure 2 above.Tau=0,53μsTau²quer=7,64*10^-13sSigmatau=0,7μsb.) Is it possible to transmit over this channel a 256 QAM signal with a bit rate of2 Mbit/s without using an equalizer?ISI = TS if less than 10simatau 7μsHere: So 8Bit/Symbol Symbol rate 2 = Mbit/s/8Bit/symbol 1/4MBaud and thus symbol duration =4μsThus frequency-selective fadingUday Sharma Page 16

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