tailieunhanh - Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 11

Tham khảo tài liệu 'proakis j. (2002) communication systems engineering - solutions manual (299s) episode 11', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | whereas the average transmitted power of the QAM constellation PQAM 8 4A2 4 A PQAM 2 1 73 2 A The relative power advantage of the PSK constellation over the QAM constellation is 8 gain pPSK . . Zj . 7- ------------T - dB PQAM 2 1 V3 2 2 -72 Problem 1 The number of bits per symbol is 4800 4800 k R 24iS 2 Thus a 4-QAM constellation is used for transmission. The probability of error for an M-ary QAM system with M 2k is 3k8b M - 1 N0 With PM 10 5 and k 2 we obtain qL M No_ 5 X 10 6 ệb N0 2 If the bit rate of transmission is 9600 bps then 9600 2400 4 In this case a 16-QAM constellation is used and the probability of error is PM 1 1 2 Ọ 4 Q 3 X 4 X Eb 15 X N0 2 Thus Q 3 X Eb 15 X N0 1 X 10 5 E 3 N0 3 If the bit rate of transmission is 19200 bps then 19200 2400 8 In this case a 256-QAM constellation is used and the probability of error is Pm 1 1 2 1 16 Q 3 X 8 X Eb 255 X N0 2 With PM 10 5 we obtain ịb N0 198 4 The following table gives the SNR per bit and the corresponding number of bits per symbol for the constellations used in parts a -c . k 2 4 8 SNR db As it is observed there is an increase in transmitted power of approximately 3 dB per additional bit per symbol. Problem 1 Although it is possible to assign three bits to each point of the 8-PSK signal constellation so that adjacent points differ in only one bit this is not the case for the 8-QAM constellation of Figure . This is because there are fully connected graphs consisted of three points. To see this consider an equilateral triangle with vertices A B and C. If without loss of generality we assign the all zero sequence 0 0 . 0 to point A then point B and C should have the form B 0 . 0 1 0 . 0 C 0 . 0 1 0 . 0 where the position of the 1 in the sequences is not the same otherwise B C. Thus the sequences of B and C differ in two bits. 2 Since each symbol conveys 3 bits of information the resulted symbol rate is Rs 90 X 106 3 30 X 106 symbols sec 3 The .