Model Solutions Case Study Competition 2012 Preliminary Round Part 1 Technical exercises

Transcription

1 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 dbm 33.3 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 MHz fif1 = 2885 MHz flo1 = MHz flo1 = MHz (preferred solution: high LO) (alternative solution: low LO) f [MHz] f [MHz] Example for fin=947 MHz Exercise 3-3: Hardware requirements 1 radio frame = 10 ms. Number of samples in a radio frame 10 ms *30.72 MHz N frame Number of sample needed for measurement N 2* N frame measurement The picture shows the recording window needed to capture one frame completely. Here it is frame n. 2 frames -1 sample 2 frames -1 sample 1 sample Frame n-1 Frame n Frame n+1 Frame n+2 1

2 Exercise 3-4: Measurement bandwidth The exercises 3-4 to 3-7 are meant to build some awareness for the hardware of a measurement receiver. The developer has the task to find the optimal operating point of the receiver for a certain expected 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-Digitalconverter -> digital baseband. The RF converter expects the signal always in the same power range. Only then can the maximum dynamic 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 hint may be useful here. The diagram shows how the choice of the attenuators ( RF ref level ) affects the dynamic range that the receiver has in the digital domain. Each input power (UE transmit power) can be expressed as a line of tuples (RF reference level; level relative 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 See picture, Measurement BW = 18 MHz, the curve is moved by *log1018*10 db dbm 10 *log1018*10 db dbm 6 97 dbm 10 *log1018*10 db dbm Exercise 3-5: Power versus Pref 10 hence 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). 2

3 P [dbfs] Nonlinear measurement range due to PAPR 0-9 P BW = 1 Hz and P ref = 0 dbm P OFF POWER P BW = 18 MHz and P ref = -41 dbm P BW = 18 MHz P BW = 1 Hz P ref [dbm] Exercise 3-6: Power The light green and dark green lines are not parallel. To find the optimal operating point, move on the light 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 Y *( x 15) With x 41 dbm Y db Hence, the dynamic range at Exercise 3-8: Sampling Y / 20 Number of bits needed is ld10 db P ref 41 dbm is db Y / 20 6 bit log102. Or simpler Y db 6 bit 6 db / bit. 3

4 Exercise 4 (20 points in total) Exercise 4-1: Modulation symbols Total number of modulation symbols in a subframe 4 * 14 * 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 4 * 9 *12 3*8 528 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/ * 2 * The next lower transport block size is 176 CQI = 14: 64QAM has 6 bits per modulation symbol, coding rate = 873/ *6* 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 not be exceeded. Exercise 4-3: Energy per transport block 20 dbm -> 100 mw Energy per Subframe: 100mW * 1 ms 528 out of 672 modulation symbols are used for data Energy used for data transmission 528 E data 100 mw *1 ms * µj 672 Energy per bit for CQI = 3: µj E b 446 nj / bit 176 bits Energy per bit for CQI = 14: µj E b 31 nj / bit 2536 bits 4

5 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 transmission 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 resources in frequency or power. With the HARQ procedure, the 10% of transport blocks not correctly received with the first transmission are most likely correctly received when combining first and second transmissions in the receiver. 5

6 Exercise 5 (14 points in total) Exercise 5-1: Mobile network mobility Idle mode cell reselection LTE WCDMA 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 comes back to LTE with a RRC Connection Reestablishment procedure (No ). 6

7 Exercise 6 (18 points in total) 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 signal ADC Converts the analog signal into a digital one Codec Compresses the digital speech signal, optimizes it fort he transfer over the mobile network 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 packets DAC Converts the digital into an analogue signal Speaker 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, which way the packet should take to reach its destination. As the diagram indicates there is more than one possible 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 receiver side. Exercise 6-4: Receiving VoIP 1 The codec and the ADC in the receiver require a stable and constant data stream (like in the circuit switched world). This requires an additional jitter buffer which regulates this, hence a constant data stream will be delivered to the codec/dac. 7

8 Exercise 6-5: Receiving VoIP 2 The max. Jitter is 38ms, hence the jitter buffer should buffer the packets min. 38ms Delay and Jitter time [ms] packet # Sending Time Receiving Time Delay + Jitter Jitter Linear (Receiving Time) 8

9 Exercise 7 (7 points in total) 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 GSM LTE FDD WCDMA GSM- CDMA 9

10 Exercise 8 (13 points in total) Exercise 8-1: Hardware dependency There was a bug in this exercise. Correct would have been: Given was the output for a Little Endian machine, and asked was the output for a Big Endian machine. The expected output of that program is the following: 0xAABBCCDD = {0xAA,0xBB,0xCC,0xDD} The bytes of the union s uint32_t member mi can be accessed addressing from low to high through the indexes 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 address, and so on. The byte order within a word on a big-endian architecture is defined the way that the most significant bits of the data is located at the lowest address, and for little-endian machines, it s just the other way around: 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 properly The output of that program is k = 100 The inner loops run from 3 to 200 respective from k (= [3,...,200]) to 10, with k being modified in the inner 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 the nested 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 c which can be written as!a true c which is always true Regardless of the values of a, b, c. 10

11 Exercise 8-4: Boolean expression 2 Therefore, the output of that program is f(false,false,false) = true f(false,false,true ) = true f(false,true, false) = true f(false,true, true ) = true f(true, false,false) = true f(true, true, false) = true f(true, false,false) = true f(true, true, true ) = true 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 in the declaration of the class CDataObject. The order in which the initializer list of the constructor is executed is defined by the order of member variables declaration. The declaration order and hence the initialization order is A, then B. As we are initializing A from B's value, we effectively initialize A from a non-initialized value, which is random. Note that the order in which the initializer list is written is completely irrelevant. The only key is the order of declaration, 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. 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; int A; }; // note the order // of B and A Move the initialisation of A and B from the initializer list of the constructor inside the body of the constructor. 11

12 class CDataObject { public: CDataObject( const std::string Name, const int Val ) : Name( Name ) { B = Val; // note the initialization has moved into A = 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; }; 12