Model Solutions – Case Study Competition 2012 ... · PDF fileDepending on load conditions or congestion on the path. ... Compresses the digital speech signal, ... the loops will

Model Solutions – Case Study Competition 2012 – Preliminary Round

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Part 1 – Technical exercises

Exercise 3 (24 points in total)

Exercise 3-1: RF frontend

a) 33 dBm + 1 dB – 0.7 dB = 33.3 dBm33.3 dBm = 2.14 W

b) 3dB attenuation means that half of the input power is dissipated => 30.3 dBm or 1.07 W

Exercise 3-2: RF frequency conversion

fif2 = 125 MHzfif1 = 2885 MHzfLO1 = 3820 .. 3845 MHz (preferred solution: high LO)fLO1 = 1950 .. 1925 MHz (alternative solution: low LO)

0 947 2885 3832 4779 f [MHz]

0 125 2760 2885 f [MHz] Example for fin=947 MHz

Exercise 3-3: Hardware requirements

1 radio frame = 10 ms.Number of samples in a radio frame 30720072.30*ms10Nframe MHz Number of sample needed for measurement 6143991-N*2N frame tmeasuremen

The picture shows the recording window needed to capture one frame completely. Here it is frame n.

Frame n-1 Frame n Frame n+1 Frame n+2

1 sample2 frames -1 sample 2 frames -1 sample


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Exercise 3-4: Measurement bandwidth

The exercises 3-4 to 3-7 are meant to build some awareness for the hardware of a measurement re-ceiver. The developer has the task to find the optimal operating point of the receiver for a certain ex-pected input power (-50 dBm). Optimal here means with the largest dynamic range.

General principle of the receiver:Connector -> Attenuator (“makes level”) -> RF converter (“makes frequency”) -> Analog-to-Digital-converter -> digital baseband.

The RF converter expects the signal always in the same power range. Only then can the maximumdynamic range be used. The RF attenuators are used to put the input signal to the desired power range.The RF attenuator settings lead to an RF reference level.

The diagram has powers on both axes. The students may not be accustomed to such a view. A hintmay be useful here.The diagram shows how the choice of the attenuators (“RF ref level”) affects the dynamic range that thereceiver has in the digital domain.

Each input power (UE transmit power) can be expressed as a line of tuples (RF reference level; levelrelative to digital full scale). A tuple is a point in the diagram.For constant input power there is the following relation:RF reference level + level relative to digital full scale = const [dBm]This is valid at least in the linear operating range.

3-4See picture, Measurement BW = 18 MHz, the curve is moved by dB610*1810log*10 hence

dBmdBdBm 44.5210*1810log*10125 6 dBmdBdBm 44.2410*1810log*1097 6

Exercise 3-5: Power versus Pref

See picture. A signal power of -50 dBm can be expressed as a line crossing the two points (0 dBm Pref ,-50 dBfs) and (-50 dBm Pref , 0 dBfs).


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P [dBfs]

Pref [dBm]

0

-125

-50 400-15

-97

PN @ BW = 1 Hz

-9PDynamic @ BW = 1 Hz

and Pref = 0 dBm

-52.44PN @ BW = 18 MHz

POFF POWER

Nonlinear measurementrange due to PAPR

-41

PDynamic @ BW = 18 MHzand Pref = -41 dBm

-24.44

Exercise 3-6: Power

The light green and dark green lines are not parallel. To find the optimal operating point, move on thelight green line to the left as far as possible without going into the nonlinear operating range.

Pref = -41 dBm gives the largest dynamic range and leaves 9 dB backoff for PAPR.

Exercise 3-7: Power dynamic

The dynamic range can be calculated with the linear equation

dBxY 44.52)15(*

501544.2444.52

With dBmx 41

dBY 64.31

Hence, the dynamic range at dBm41Pref is 31.64 dB

Exercise 3-8: Sampling

Number of bits needed is bitYld Y 6210log

20/10 20/

.

Or simpler bitbitdBdBY 6/6

.


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Exercise 4-1: Modulation symbols

Total number of modulation symbols in a subframe 67212*14*4 modulation symbols

4: because 4 resource blocks are used (see text)A resource block has 12 subcarriers.There are 14 OFDM symbols in a subframe

Modulation symbols for data 5288*312*9*4 modulation symbols

The first two OFDM symbols are used for control data, leaving 12.Of the remaining 12, 3 OFDM symbols have 4 pilot subcarriers each.

Exercise 4-2: Transport blocks

CQI = 3: QPSK has 2 bits per modulation symbol, coding rate = 193/1024

03.1991024193*2*528

The next lower transport block size is 176

CQI = 14: 64QAM has 6 bits per modulation symbol, coding rate = 873/1024

84.27001024873*6*528

The next lower transport block size is 2536

The next lower block size is chosen in the table because the transport block error probability shall notbe exceeded.

Exercise 4-3: Energy per transport block

20 dBm -> 100 mWEnergy per Subframe: 100mW * 1 ms528 out of 672 modulation symbols are used for data

Energy used for data transmission

µJmsmWEdata 57.78672528*1*100

Energy per bit for CQI = 3: bitnJbitsµJEb /446

17657.78

Energy per bit for CQI = 14: bitnJbitsµJEb /31

253657.78


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Exercise 4-4: Error probability

The goal is to use just as much energy per bit as needed. If the block error rate after the first transmis-sion is 0, then the energy per bit may be too high.In this case, either the transport block size can be increased or the link to the UE could use less re-sources in frequency or power.With the HARQ procedure, the 10% of transport blocks not correctly received with the first transmissionare most likely correctly received when combining first and second transmissions in the receiver.


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Exercise 5-1: Mobile network mobility

Idle mode cell reselection LTEWCDMA correct

Exercise 5-2: Message sequence chart

Exercise 5-3: Network handover 1

Message No.: 11677, MobililtyFromEUTRACommand

Exercise 5-4: Network handover 2

This is a handover failure test case. The UE is unable to perform the handover procedure. It comesback to LTE with a RRC Connection Reestablishment procedure (No. 14125).


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Exercise 6-1: Circuit- and packet-switched networks

Circuit Switched:- Data will be sent as a continuous stream- Fixed allocation of resources fort he complete session/connection, even during time where no

data needs to be sent (no speech for example) -> due to that its expensive- All data follows a fixed path from sender to receiver

Packet Switched:- Data will be separated in packets and sent individually from sender to receiver.- Resources in the network are only allocated when data will be sent (if there is no speech then

no data needs to be send)- As every packet will be sent individually every packet can use a different path from sender to

receiver. Depending on load conditions or congestion on the path.- This can result in different delays and jitter (variation of transmit ion time) of the individual

transmission time for the packets.

Exercise 6-2: VoIP components

Mic – Converts the acoustic signal into an electric analog signalADC – Converts the analog signal into a digital oneCodec – Compresses the digital speech signal, optimizes it fort he transfer over the mobile net-work and creates packets (every 20ms as described in the introduction)RTP/UDP/IP – Header which describes the content of the packet, its source and destination.

Buffer – Buffers the received packetsDAC – Converts the digital into an analogue signalSpeaker – Converts analogue electrical signal into an acoustic signal

Exercise 6-3: VoIP transmission path

The IP based transmission path contains for example routers, which decide on a per packet base, whichway the packet should take to reach its destination. As the diagram indicates there is more than onepossible way. This decision is done based on load, time or costs of the network provider.Different processing times of the individual components lead to time difference (jitter) at the receiverside.

Exercise 6-4: Receiving VoIP 1

The codec and the ADC in the receiver require a stable and constant data stream (like in the circuitswitched world). This requires an additional jitter buffer which regulates this, hence a constant datastream will be delivered to the codec/DAC.


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Exercise 6-5: Receiving VoIP 2

The max. Jitter is 38ms, hence the jitter buffer should buffer the packets min. 38ms

Delay and Jitter

020

4060

80100

120140

160180

203222

256280 290

315

360 358

400 410

203 202216 220 210 215

240218

240 230

1 014 18 8 13

3816

38 28

0

50

100

150

200

250

300

350

400

450

1 2 3 4 5 6 7 8 9 10

packet #

time

[ms]

Sending TimeReceiving Time

Delay + Jitter

JitterLinear (Receiving Time)


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Exercise 7-1: Multi – CMW 1

2 CMWs are required.

Exercise 7-2: Multi - CMW 2

3 CMWs are required.

Exercise 7-3: Inter-RAT Scenario

With 2 CMWs it is possible to simulate these possible Scenarios:LTE FDD – LTE TDD – WCDMA – GSMLTE FDD – WCDMA – GSM- CDMA


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Exercise 8-1: Hardware dependency

There was a bug in this exercise. Correct would have been: Given was the output for a Little Endianmachine, and asked was the output for a Big Endian machine.

The expected output of that program is the following:

The bytes of the union’s uint32_t member “mI” can be accessed addressing from low to high through theindexes 0 through 3 of the char-sized union member “mC”.So u.mC[0] gives us the lowest-address byte value of u.CI, u.mC[1] the byte value on next higher ad-dress, and so on.The byte order within a word on a big-endian architecture is defined the way that the most significantbits of the data is located at the lowest address, and for little-endian machines, it’s just the other wayaround: the bits with the highest significance are at the highest address.Therefore, with a little-endian addressing, the output will be as shown above.

Exercise 8-2: Local variables and shadowing

The program terminates properlyThe output of that program is

The inner loops run from 3 to 200 respective from k (= [3, ... ,200]) to 10, with “k” being modified in theinner loop. As there is never anything done to decrement “k”, the loops will eventually terminate.The variable k from the scope of the main() function is just shadowed by the instances of “k” within thenested loops. Therefore, the inner “k”'s do not affect he outer one, which just runs from 0 to 100.

Exercise 8-3: Boolean expression 1

Minimizing:The expression

( !(a&&b) || (b||c) )can be written as

( (!a||!b) || (b||c) )which can be written as

!a||!b||b||cwhich can be written as

!a||true||cwhich is always

trueRegardless of the values of a, b, c.

0xAABBCCDD = {0xAA,0xBB,0xCC,0xDD}

k = 100


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Exercise 8-4: Boolean expression 2

Therefore, the output of that program is

Exercise 8-5: Find the bug

The cause of the bug is the order in which the declaration of the member variables A and B is written inthe declaration of the class CDataObject.The order in which the initializer list of the constructor is executed is defined by the order of membervariables declaration. The declaration order and hence the initialization order is A, then B. As we areinitializing A from B's value, we effectively initialize A from a non-initialized value, which is random. Notethat the order in which the initializer list is written is completely irrelevant. The only key is the order ofdeclaration, not the order of the definition.

There are two ways to fix this problem:

Re-order the declaration of A and B inside the declaration of the class CDataObject.

Move the initialisation of A and B from the initializer list of the constructor inside the body of theconstructor.

f(false,false,false) = truef(false,false,true ) = truef(false,true, false) = truef(false,true, true ) = truef(true, false,false) = truef(true, true, false) = truef(true, false,false) = truef(true, true, true ) = true

class CDataObject{public:

CDataObject( const std::string Name, const int Val ): Name( Name ), B( Val ), A( 22+B )

{}

virtual void print(){std::cout << "{" << Name << ", A=" << A << ", B=" << B << "}" << std::endl;

}

private:std::string Name;int B; // note the orderint A; // of B and A

};


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class CDataObject{public:

CDataObject( const std::string Name, const int Val ): Name( Name )

{B = Val; // note the initialization has moved intoA = 22+B; // the constructor body

}

virtual void print(){std::cout << "{" << Name << ", A=" << A << ", B=" << B << "}" << std::endl;

}

private:std::string Name;int A;int B;};

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Model Solutions – Case Study Competition 2012 ... · PDF fileDepending on load conditions or congestion on the path. ... Compresses the digital speech signal, ... the loops will