UNIVERSITY OF SASKATCHEWAN EE456: Digital Communications FINAL EXAM, 9:00AM 12:00PM, December 9, 2010 (open-book) Examiner: Ha H.

Transcription

1 Name: Page 1 UNIVERSIY OF SASKACHEWAN EE456: Digital Communication FINAL EXAM, 9:00AM 1:00PM, December 9, 010 (open-book) Examiner: Ha H. Nguyen Permitted Material: Only textbook and calculator here are 5 quetion. All quetion are of equal value (with part mark indicated) but not necearily of equal difficulty. Full mark hall only be given to olution that are properly explained and jutified. 1. (NRZI ) Non-return-to-zero-invere (NRZI) i a binary baeband ignalling cheme ued in magnetic recording. he NRZI ignal waveform witche between the two amplitude level of V volt and V volt a follow: he tranition from one amplitude level to another (V to V or V to V ) occur only when the information bit i 1. No tranition occur when the information bit i a 0 (i.e., the amplitude level remain the ame a the previou ignal level). [] (a) Draw the NRZI waveform for the information equence { }. Aume that the amplitude level before the tranmiion of the above equence i V volt. [] (b) Generally, the ignal tranmitted in any bit period depend on what happened previouly. hu there i memory and therefore a tate diagram and a trelli. Draw a tate diagram that decribe NRZI modulation cheme. A a hint, there are two tate. Alo a tate i defined a what do you need to know from the pat, which together with preent input (bit 1 or bit 0) enable you to determine the output (+V or V volt). Label the tranition for the tate with the input bit and the output ignal. [] (c) Now draw the trelli correponding to the above tate diagram. Start at t = 0 and aume that before t = 0 the voltage level at the modulator output i V volt. [] (d) Aume that the ource bit are equally likely and that V b = 1 joule. Draw the ignal pace diagram for the two ignal ued in NRZI. hen ue thi ignal pace diagram and the trelli of (c), equence demodulate the following et of output from a matched filter for the firt 4 bit interval: r (1) = 0.; r () = 0.6; r () = 0.8; r (4) = 0. (volt).

2 Name: Page. (Aymmetric QPSK ) Aymmetric contellation provide a imple olution for unequal error protection, where bit are deemed to be important can be protected more than bit of leer importance (e.g., in multimedia application). Conider an aymmetric QPSK contellation a hown in Figure 1, where the mapping of two bit b 1 b into each ignal point i alo hown. he information bit are equally likely. A uual, the ignal et i ued for communication over an ideal AWGN channel with two-ided power pectral denity N 0 / (watt/hz). φ( t) = in( π fct) b1b b1b ( t ) 1 ( t) d 1 d E 0 π / φ1( t) = co( π fct) ( t) b1b 0 0 ( t) 4 b1b 1 0 Figure 1: Aymmetric QPSK. [] (a) Clearly draw the deciion boundary and the deciion region of the receiver that minimize the ymbol error probability. [4] (b) Determine the ymbol error probability of the receiver in (a) a a function of E /N 0. Hint: Argue that Pr[ymbol error] = Pr[ymbol error 1 (t)]. [] (c) Determine the bit error probabilitie for bit b 1 and b eparately. Which bit i more protected and why? Hint: Determine the deciion boundarie for bit b 1 and b eparately. [] (d) Aume that N 0 = 10 6 watt/hz. How large doe E need to be et to achieve Pr[b 1 i in error] 10? Hint: Q 1 (10 ) =.09.

3 Name: Page. (8-QAM contellation) DOCSIS. (Data Over Cable Service Interface Specification) i the latet tandard in providing high-peed data ervice via cable modem. Conider an 8-QAM contellation in Figure, which i ued in the uptream channel, i.e., from a cable modem (CM) to a cable modem termination ytem (CMS). Figure : 8-QAM contellation. [] (a) Compute the average energy for the contellation. [] (b) Your friend i intereted in comparing the above 8-QAM contellation with the 8-PSK contellation. What i the minimum ditance of the 8-PSK contellation if it ha the ame average energy a that of the 8-QAM contellation in Fig.? Baed on thi reult and the minimum ditance of the 8-QAM contellation, what i a better contellation in term of power efficiency and why? [] (c) Neatly draw the minimum-ditance deciion boundary for the 8-QAM contellation. Which ignal in thi contellation are mot uceptible to error and why? [] (d) It i not poible to pecify a Gray mapping for the 8-QAM contellation of Fig.. However, it i poible to pecify a mapping with only two Gray-code violation. o thi end, tart by mapping the ignal at ( 1, 1), ( 4, 4), (+4, +4) and (+1, +1) with 100, 010, 011 and 101, repectively. Complete mapping for the remaining 4 ignal.

4 Name: Page 4 4. (ISI ) ranmiion from a cable modem termination ytem (CMS) to a cable modem (CM) i done via downtream channel. In DOCSIS.0, the bandwidth of each downtream channel i 6 MHz, while the center carrier frequency range from 111 MHz to 867 MHz. Fig. how the ideal bandlimited model of one uch a channel. H C ( f ) 1 6 MHz -114 MHz -108 MHz 108 MHz 114 MHz f Figure : Model of one downtream channel in DOCSIS.0. [6] (a) One modulation cheme pecified in DOCSIS.0 downtream channel i 64-QAM, which i ued together with quare-root raied coine haping with a roll-off factor β = What are the ymbol rate and bit rate that can be upported by uch a 64-QAM ytem o that ISI i avoided? Explain your anwer by clearly plotting the overall pectrum of the ignal (i.e., after the receive filter). Identify all the relevant frequencie. [4] (b) Another modulation cheme pecified in DOCSIS.0 i 56-QAM. It i ued with quare-root raied coine haping with a roll-off factor β = 0.1. What are the ymbol rate and bit rate that can be achieved with zero ISI by thi ytem? Hint: he frequency-domain Nyquit zero-isi criterion for paband channel can be decribed a in Figure 4, where i the deigned ymbol interval. It i mathematically tated a follow: ( S R f + k ) = contant for < f <. k= β Bandwidth W SR ( f ) 1 SR f SR f SR f f c fc + fc + fc + fc + f Figure 4: Nyquit zero-isi criterion for paband channel.

5 Name: Page 5 5. (Bandwidth-Power Plane) [6] (a) Determine the location of -FSK and 4-FSK modulation cheme on a bandwidthpower plane at the target ymbol error probability of Hint: Figure 8. in the textbook hould be helpful. [4] (b) One of the powerful, tate-of-the-art channel coding technique i the low-denity parity-check (LDPC) code. An LDPC code i baically a well-deigned long block code that can be decoded by an iterative um-product algorithm Symbol error rate Uncoded SNR norm (db) Figure 5. Performance of two LDPC code for tranmiion over the AWGN channel. Solid (dahed) line denote the cae with a block length N of 59 (4489) bit. Curve 1: 16-QAM; curve : 56-QAM, and curve : QAM. For decoding, the um-product algorithm i employed, with the number of iteration limited to 0. Conider Fig. 5 (ee footnote 1 for the ource) and uppoe that you are intereted in the dahed curve that i labeled with number 1 (i.e., marker). A indicated in the figure caption, thi curve i the performance of a ytem that employ 16- QAM and a LDPC code of block length N = 4489 bit. It i alo known that uch an LDPC code map (or encode) every block of K = 40 information bit into a block of N = 4489 bit. Repreent the above ytem on a bandwidthpower plane for the ymbol error probability of What i the gap in E b /N 0 between the above ytem and the Shannon channel capacity? Remark: Fig. 5 plot the ymbol error rate veru the normalized ignal-to-noie ratio, SNR norm. hi SNR norm i defined in the referenced paper a follow: SNR norm = η E b, η 1 N 0 where η i the average number of information bit that each ymbol carrie. 1 E. Eleftheriou, S. Olcer and H. Sadjadpour, Application of Capacity-Approaching Coding echnique to Digital Subcriber Line, IEEE Communication Magazine, pp , Apr. 004.

6 Name: Page 6 You can ue the following figure to provide olution if you want Channel capacity limit, C/W r b /W (bit//hz) Shannon limit Bandwidth limited region: r b /W>1 Power limited region: r b /W< SNR per bit, E /N (db) b 0