1. THM, FRIEDBERG
WAT Question papers
WAT translated German Question paper with
solution
UdaySharma
2/8/2013

2. WAT Question papers
WAT Exam 2002/2003
Task 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 only
the influence of the free-space propagation.
b.) = Which transmit power PTX? [dBm] must send the RF transmitter, if the required received
power 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 Antennendiagram
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3. WAT Question papers
b2.) 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 can
be 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: + / - 1dB
already 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)
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4. WAT Question papers
A GSM network operator has 48 frequencies a GSM 900 license. The traffic model is of an
average call duration of 120 seconds for Main Busy Hour assumed and blocking probability of
2% (pBLOCK = 2%) required.
Note that not because of the signaling traffic per TRX 8 all Time slots for voice traffic are
available:
Number of TRX 1 2 3 4 5 6 7 8
Time slots for signaling 1 2 2 3 3 4 5 5
Time slots for voice traffic (TCH) 7 14 22 29 37 44 52 59
Voice traffic [Erlang] at block = 2% 2.9 8.2 14.9 21 28.2 34.6 42.1 48.7
a) The operator of its network in rural area plans with a Reuse of ARCS = 16th How big can the
max. 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 the
maximum 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 / Frequency
Hopping increase if no Planned quality losses now with ARCS of 6 and is the cell radius r2 from
part. b) is assumed? Based on what effects this process?
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5. WAT Question papers
WAT Exam 2004/2005
Task 1: Assessment of range (10 points)
The maximum range of a new radio system is to be interpreted to estimate. When the
selected modulation method was the following relationship between recoverable
throughput 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 a
Noise 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 the
respective data rates of 1, 2, 5 and 10 Mbit /s?
Noise Floor: 10log (kTB/1mW) =-107dBm
At the receiver input:-92dBm
Max path loss =-20dBm (-92dBm) +2 * 5dBi SNR = 122dB SNR
MAPL @ 1Mbit / s = 122dB-5dB =-117dBm
MAPL @ 2Mbit / s = 122dB-15dB =-107dBm
MAPL @ 5 Mbit / s = 122dB-20dB =-102dBm
MAPL @ 10Mbit / s = 122dB-25dB =-97dBm
Maximum 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 for
packet-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 decisive
advantage 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.
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6. WAT Question papers
 What is the qualitative relationship between the income of a Antenna and the horizontal and
vertical opening width?
 Why is the description by means of horizontal and vertical Antenna patterns only an
approximate description of actual antenna pattern.
 What is the qualitative relationship between ARCS, interference and Capacity in cellular mobile
radio networks, such as e.g. GSM?
 What distinguishes the two GSM Data Services: GPRS and HSCSD? How can a Antennentilt be
realized.
 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" indicates
a 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 briefly
describe why it is used and the basic functioning.
 When 802.11g wireless standard, the symbol rate is always 0.25MBaud. Describe how a data
rate 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 a
License for 48 frequencies.
The cities and towns are to be supplied in the following categories divided:
City Category 1: 6 Erl / km ²
City Category 2: 2 Erl / km ²
Location Category 3: 0.1 Erl / km ²
When the following issues of the sub-tasks a) to c) by reference. assumed omnidirectional sites! The
hexagonal 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 probability
2% and the average Interference probability should be about 2.5%? Remember to leave an the
required 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. Max
Cell area = 21 = 3.5km ² ² Erl/6Erl/km / 10.5km ² / 210km ²
Cell radius = 1.16km and 2km or 8.98 km
b.) 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 supply
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7. WAT Question papers
following additional losses should be considered?
City Category 1: aGebäude = 20dB 131dB
City Category 2: aGebäude = 10dB 141dB
Location Category 3: aGebäude = 0dB 151dB
With 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 unter
Into 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 1.16km
Category 2: LMorpho 3dB = 3.15km so traffic limited to 2km
Category 3: LMorpho = 7dB 7.88km so far-reaching limits
Task 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 Channels
b.) What traffic causing 100 people with this behavior?
100*0.1Erl=10Erl
How many traffic channels would be for a blocking probability of pBLOCK = 4% Necessary?
15 Channels
Why is not the required number of channels also a factor of the number of participants up against a
sub-task)?
Bundling profit:
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8. WAT Question papers
WAT Exam 2006/2007
Task 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.
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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, Mobile
Task 2: profit by sectorization (14 points)
An omnidirectional site with RCS = 12 to be sectored. Estimate in from the following sub-tasks, so if a
capacity increase at consistent interference is achieved. Take the case in the lecture derived
analytical 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 site
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10. WAT Question papers
b.) Loosen the task a part) of a spreading coefficient of n = 3
Omnisite:
C / I = sqrt (3 * 12) ^ 3/6 = 36 to 36/12 * 8 = 24 traffic channels thus 16.6 Erlang
Sector Site:
36 = sqrt (3RCS ') ^ 3/2 thus RCS' ≈ 5.7 and almost int *36/5, 7+ * 8 ≈ 50 TCH per location, or 6.3
Frequencies. This can for example be 2/2/2oder3 distributed among the sectors. So TCHs 16/16/16
and 3 * 9.8 Erlang = 29.4 Erlang.
c.) Why is the Sektorisierungsgewinn in Teilaufgbe b.) higher than in Tasks a part)? May also occur
the 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 lower
Interference is significant, as the negative effect of the reduced profit pooling. Yes when the
negative 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 of
coverage probability, Blocking probability and probability of interference. The cities and regions to
be supplied are divided into three categories. It for the respective areas following location numbers
have 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 ²
Note: The surface of a hexagonal cell Omnistandortes can with A = 2.6 R ² estimated. The area of a
sector of a three-sector site can A ² = 0.65 D can be estimated according to figure 2. The necessary
overlap of the Cells should be neglected. Use the following dispersion model to determine the path
loss in Part of a task):
L [dB] = 135 +30 log (R / [km]) - LMorpho
a) 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)=4,38km
Due to the given location and square footage numbers What therefore is the maximum allowable
path loss for Pcov = 50% at the cell edge in the three regions?
Always 145 DB
Category 2: LMorpho = 3.3 dB
Category 3: LMorpho = 9.2 dB
b.) Ask for Omnistandorte the related link balanced budget on and enter in the corresponding
column in the correct units!
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How large is therefore the maximum possible path loss for the omnidirectional Locations. How big it
is for the sectorized sites when a sector antenna is used with a gain of 18dBi?
148dB and 155 dB
With what probability can supply about the cell edge in the respective categories 1 to 3 expected, if
we 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% is
required and the network operator Has license for 48 GSM frequencies. Consider the following when
calculating traffic traffic for signaling and use for estimating the pint empirical diagram from the
lecture. Go out of oncoming traffic densities and the given cell surfaces:
A1 = 3.33 km ², 20km ² = A2, A3 = 50km ²
Category 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 the
assumed 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.
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12. WAT Question papers
Task 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 a
binary modulation (and BT = 1) is expected? Thus intersymbol interference to and the use of an
equalizer 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) = 4.952μs
στ = sqrt ("τ ²" ² τ) = 0.976μs
Frequency selective fading, when Ts <10στ = 9.76μs
Ts = 1/1MHz = 1us
Thus: frequency selective fading and intersymbol interference. The use of a EQ is so useful
b.) For the channel estimation of the equalizer to work properly, the need for Duration of the TDMA
frame, the channel impulse response be invariant (condition ) for slow fading. How long is at a
maximum. speed of 100km / h and f = 900MHz, the maximum frame length to be so? How many bits
per TDMA frame would therefore max. make sense?
TTDMA <= λ/2/v=c/f/2/v=5990μs somit 5990 Bits.
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13. WAT Question papers
WAT Exam 2007/2008
Task 1: Assessment of range (11 points)
The achievable throughput of a wireless system is assessed. It has the following Connection between
the 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 of
Demodulator
Further 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=-101dBm
EIRP=PTX-acable+Gant=23,5dBm
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14. WAT Question papers
IRP=PN+F+SNR+acable-GANT=-97,5dBm+SNR
MAP=135dB+30logd/km=EIRP-IRP=121dB-SNR
Also SNR=-14dB-30logd/km somit
d1=160m, SNR=10dB,DR=4MBit/s
d2=230m, SNR=5dB, DR=1,8MBit/s
d3=290m, SNR=2dB, DR=1Mbit/s
Problem 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 capacity
decreases if cells are sectorised? Explain why?
c.) Explain the meaning of the abbreviations FDD and TDD. Which of these techniques
should 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 a
vertical 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 what
is 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 is
foreseen 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 blocking
probability should not exceed 2%. For the network design only omnidirectional cells will be
used. Compute the cell radius based on traffic requirements for the following three cases
a1.) ARCS = 18
a2.) ARCS = 12
a3.) ARCS = 6
Hint: Signalling traffic can be neglected. To compute the hexagonal cell area, use the
following 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 -2dB
steps), diversity gain and max. gain of an optional tower mounted amplifier (TMA): 3dB
each. Complete the link budget!
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15. WAT Question papers
c.) Compute the cell radius based on the results of the link budget above? For
computation 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.) and
c.)? How many sites are necessary? Which Quality of Service do you expect for this
network: Pint, Pcov and Pblock?
At d = 1.68 km = 7.33 km ² A follows, and thus 68.2 Locations
This 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 24
and 20 not available TCHs
is. Pcov at cell edge remains at 95%.
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16. WAT Question papers
Problem 4: Mobile Communication Channel (6 points)
The local spatial average of a power delay profile is measured at 900 MHz as shown in fig. 2
a.) Compute the „mean excess delay“ and the „rms delay spread“.
Sketch both figures into Figure 2 above.
Tau=0,53μs
Tau²quer=7,64*10^-13s
Sigmatau=0,7μs
b.) Is it possible to transmit over this channel a 256 QAM signal with a bit rate of
2 Mbit/s without using an equalizer?
ISI = TS if less than 10simatau 7μs
Here: So 8Bit/Symbol Symbol rate 2 = Mbit/s/8Bit/symbol 1/4MBaud and thus symbol duration =
4μs
Thus frequency-selective fading
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