Solutions to the Exam Digital Communications I given on the 11th of June = 111 and g 2. c 2

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1 Soluions o he Exam Digial Communicaions I given on he 11h of June 2007 Quesion 1 (14p) a) (2p) If X and Y are independen Gaussian variables, hen E [ XY ]=0 always. (Answer wih RUE or FALSE) ANSWER: False. (Only if E(X)=0 or E(Y)=0.) b) (3p) Consider a rae 1/2 convoluional code wih generaor sequences g 1 = 111 and g 2 = 110. Assume ha BPSK modulaion is used o ransmi he coded bis. Draw a shif regiser for he encoder. ANSWER: c) (2p) One advanage of non-coheren demodulaion over coheren demodulaion, is ha he demodulaor does no need o implemen phase esimaion. (Answer wih RUE or FALSE) ANSWER: rue. d) (3p) A receiver ha implemens he ML decision rule is always opimal in he sense of minimum symbol error probabiliy. (Answer wih RUE or FALSE) ANSWER: False. e) (2p) he MAP decision rule is a special case of he ML rule. ANSWER: False. ML is a special case of MAP. c 2 (Answer wih RUE or FALSE) f) (2p) PSK is a special case of QAM. (Answer wih RUE or FALSE) ANSWER: rue. Quesion 2 (12p) Here we sudy a binary sysem wih signal wo alernaives: s 1 and s 2. he bis are assumed o be saisically independen and of equal probabiliy. he sysem is an AWGN channel wih noise power specral densiy given by N 0 /2. c 1

2 s 1 A s 2 A /2 -A a) Please express E b (he average energy per bi) as a funcion of A and. (2p) Soluion: he energy for s 1 is E 1 = 0 s1 2 d= A 2 and he energy for s 2 is E 2 = 0 s2 2 d= A 2. Hence, he average energy per bi is E b = A 2. b) Find a basis for he signal space and draw he signal consellaion ha is he vecors corresponding corresponding o s 1 and s 2. Please label he axis in erms of E b. (4p) Soluion: One possible basis is { 1, 2 } where i =s i / E b. Hence, he signal consellaion is: s 1 =[ E b, 0] and s 2 =[0, E b ] c) Wha is he smalles value of E b / N 0 in db required o reach a bi error probabiliy of 10 3 if we assume ha we are using he opimum receiver (which minimizes he bi error probabiliy)? (4p) Soluion: he maximum likelihood receiver will minimize he bi error probabiliy since all signal

3 alernaives are equally likely. he disance beween he signal alernaives is d = 2 E b and he bi error probabiliy is P b =Q d 2 /2N 0 =Q E b / N 0 =10 3. he required E b / N 0 in db is approximaely 9.80dB. d) Please draw he decision regions for he receiver in par c). (2p) Soluion: he ML receiver is a minimum-disance receiver and he decision regions Z 1 and Z 2 are depiced below. Quesion 3 (12p) In his problem, we will consider a convoluional encoder. I is given in he figure below. he informaion bi b gives rise o hree coded bis c 1, c 2 and c 3, which are ransmied over an AWGN channel using binary PAM wih roo-raised cosine pulses wih roll-off facor =0.2. he PAM pulses are ransmied a a rae of 1000 pulses/second. If he coded bi is 1, hen he ransmied ampliude is posiive. Conversely, if he coded bi is 0, hen he ransmied ampliude is negaive. b c1 c2 a) Draw he sae diagram and a rellis secion of he encoder. Please make sure o label he ransiions. (4p) c3

4 Soluion: he sae diagram and rellis are shown in he figure below. c) Now suppose he sampled oupu of he receiver mached filer is {1.51, 0.63, 0.04, 1.14, 0.56, 0.57, 0.07, 1.53, 0.9, 1.68, 0.9, 0.98, 1.99, 0.04, 0.76} his sequence corresponds o he ransmied sequence c 1, c 2, c 3, c 1, c 2, c 3,... which is generaed from a packe of hree informaion bis. he encoder is assumed o be in he all-zero sae a he beginning and ending of he ransmission. he encoder is forced ino he ending sae by appending wo zero bis o he informaion bi sequence. Use hard decision decoding o esimae he ransmied informaion bis. (8p) Soluion: he hard decoded channel bis are {1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1}. he rellis wih he decoded pah (he pah a minimum Hamming disance from he received channel bis) is show below.

5 We see ha he decoded informaion bis are {0, 0, 1}. Quesion 4 (12p) In his problem, we are considering a binary PAM sysem. he pulse shape is given by g and he daa rae is R b =1/ b. We can hus conclude ha he ime beween consecuive pulses equals b seconds. Furhermore, s = k= b k g k b, where b k { 1, 1} is he k h ransmied bi. he sysem is depiced in he block diagram below. Here, sgn{q} denoes he sign of q, ha is if q is posiive hen sgn{q}=1 and if q is negaive hen sgn{q}=-1. Furhermore, he sampling of he kh bi is performed a ime =k b as indicaed in he figure. he channel is an AWGN channel. he addiive noise is denoed n() and has noise power specral densiy N 0 /2. he decision on he k h bi is given by he sign of y k b, where y is he oupu from he mached filer (which has impulse response g R ). he arge bi error rae is P b =10 4, and he energy per bi is denoed by E b. n() =k b b k Binary PAM g s() g R y() sgn{.} b k g R =g A a) Wha is he maximum possible daa rae if he ransmission mus be ISI-free? Answer wih an expression in. (6p) Soluion: I follows from he block diagram ha y = l = b l x l b v where x =g g R and v =n g R, and he sampled oupu from he mached filer is y k b = l = b l x k b l b v k b =b k x, l k b l x k l b v k b here will be no ISI if x n b =0 for all nonzero n. From he figure, i is obvious ha his is equivalen o n b 2 b. Hence, he maximum daa rae for ISI free ransmission is R b =1 /.

6 x A 2 b) Wha is he required E b / N 0 in db for he ISI-free sysem o mee he bi error arge? (6p) Soluion: he kh noise sample is v k b =n g R =k b. he noise sample will be a zeromean Gaussian random variable wih variance E g N 0 /2, where E g is he energy of g R. g 2 R d= A 2. Clearly, E g is also he energy of g and since b k =±1, we E g = have ha E b =E g. Condiioned on ha b k = 1, a bi error occurs wih probabiliy Pr { y k b 0 b k = 1}=Pr { A 2 n k b 0}=Pr { E b n k b 0} Pr {n k b / E b N 0 /2 2 E b / N 0 }=Q 2 E b / N 0 Due o symmery, P b = P b b k = 1 = P b b k = 1 and E b / N 0 =...=8.39dB. 2