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# Tele4653 l2

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### Tele4653 l2

1. 1. TELE4653 Digital Modulation & Coding Digital Modulation Wei Zhang w.zhang@unsw.edu.au School of Electrical Engineering and Telecommunications The University of New South Wales
2. 2. Outline PAM (ASK) PSK QAM FSK TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.1/2
3. 3. Modulation Source information to be transmitted is usually in the form of a binary data stream. The transmission medium, i.e., communication channel suffers from noise, attenuation, distortion, fading, and interference. Digital Modulation - To generate a signal that represents the binary data stream and matches the characteristics of the channel Modulation with Memoryless or with Memory TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.2/2
4. 4. Deﬁnitions 1 Signaling Interval: Ts . Signaling (Symbol) Rate: Rs = Ts . Ts Bit Interval: Tb = k for a signal carrying k bits of information. Bit Rate: R = kRs = Rs logM . 2 M Average signal energy: Eavg = m=1 pm Em with pm being the probability of the mth signal. Eavg Eavg Average energy per bit: Ebavg = k = logM . 2 TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.3/2
5. 5. PAMThe signal waveform may be represented as sm (t) = Am p(t), 1 ≤ m ≤ M (1)where p(t) is a pulse of duration T and Am denotes theamplitude with the mth value, given by Am = 2m − 1 − M, 1 ≤ m ≤ M (2)i.e., the amplitudes are ±1, ±3, ±5, · · · , ±(M − 1).Digital amplitude modulation is usually called amplitude-shiftkeying (ASK). TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.4/2
6. 6. PAMThe energy of signal sm (t) is given by ∞ Em = A2 p2 (t)dt = A2 Ep m m (3) −∞The average signal energy is M M Ep Eavg = p m Em = A2 m M m=1 m=1 (M 2 − 1)Ep = . (4) 3and the average energy per bit is (M 2 − 1)Ep Ebavg = M . (5) 3 log2 TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.6/2
7. 7. Bandpass PAMThe bandpass PAM signals are carrier-modulated bandpasssignals with lowpass equivalents of the form s ml (t) = Am g(t),where Am and g(t) are real. The signal waveform is sm (t) = sml (t)ej2πfc t = Am g(t) cos(2πfc t) (6)The energy of signal sm (t) is given by A2 Em = m Eg . (7) 2Moreover, (M 2 −1)Eg (M 2 −1)Eg Eavg = 6 , Ebavg = 6 logM . (8) 2 TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.7/2
8. 8. Bandpass PAM - ExpansionFor basedband PAM, the expansion of sm (t) = Am p(t) is sm (t) = Am Ep φ(t) (9)where p(t) φ(t) = . (10) EpFor bandpass PAM, the expansion of sm (t) = sml (t)ej2πfc tis [Tutorial 1] Eg sm (t) = Am φ(t) (11) 2where 2 φ(t) = g(t) cos(2πfc t). (12) Eg TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.9/2
9. 9. Bandpass PAM - dminThe bandpass PAM can be represented as the one-dimensional Evector: sm = Am 2g , where Am = ±1, ±3, · · · , ±(M − 1).The Euclidean distance between any pair of signal points is 2 Eg dmn = s m − sn = |Am − An | (13) 2For adjacent signal points |Am − An | = 2, it has 12 log2 M dmin = 2Eg = E 2 − 1 bavg (14) Mwhere in the last equality Eq. (8) is used. TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.10/2
10. 10. Phase ModulationThe bandpass PM signal waveform may be represented as 2π(m−1) j sm (t) = g(t)e M ej2πfc t , 1 ≤ m ≤ M 2π(m − 1) = g(t) cos + 2πfc t (15) M 2π(m−1)Let θm = M , m = 1, 2, · · · , M . Then, sm (t) = g(t) cos θm cos(2πfc t) − g(t) sin θm sin(2πfc t). (16)Digital phase modulation is usually called phase-shift keying(PSK).The PSK signals have equal energy, Eavg = Em = 1 Eg . 2 TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.11/2
11. 11. Phase Modulation - Expansion 2π(m−1) jThe expansion of PM signal s(t) = g(t)e M ej2πfc t is Eg Eg ˜ sm (t) = cos(θm )φ(t) + sin(θm )φ(t) (17) 2 2where [see Tutorial 1] 2 φ(t) = g(t) cos(2πfc t) (18) Eg ˜ 2 φ(t) = − g(t) sin(2πfc t) (19) Eg TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.13/2
12. 12. Phase Modulation - dminThe bandpass PM can be represented as the two-dimensional Eg Egvector: sm = 2 cos θm , 2 sin θm , m = 1, 2, · · · , M . Noteθm = 2π(m−1) for m = 1, 2, · · · , M . MThe Euclidean distance between any pair of signal points is 2 Egdmn = sm − sn = | cos θm − cos θn |2 + | sin θm − sin θn |2 2 = Eg [1 − cos(θm − θn )]. (20) TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.14/2
13. 13. Phase Modulation - dminFor adjacent signal points |m − n| = 1, it has 2π dmin = Eg 1 − cos( ) (21) M 2 π = 2Eg sin (22) M 2 π = 2 Ebavg log2 M × sin (23) M π πFor large values of M , we have sin M ≈ M, and then π2 dmin ≈ 2 Ebavg log2 M × 2 (24) M TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.15/2
14. 14. QAMThe quadrature amplitude modulation (QAM) signal waveformmay be expressed as sm (t) = (Ami + jAmq )g(t)ej2πfc t , m = 1, 2, · · · , M = Ami g(t) cos(2πfc t) − Amq g(t) sin(2πfc t), (25)where Ami and Amq are the information-bearing signalamplitudes of the quadrature carriers and g(t) is the signal pulse. TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.16/2
15. 15. QAMAlternatively, the QAM signal may be expressed as sm (t) = rm g(t)ejθm ej2πfc t = rm g(t) cos(θm + 2πfc t), (26)where rm = A2 + A2 and θm = tan−1 (Amq /Ami ). mi mqIt is apparent that QAM signal can be viewed as combinedamplitude rm and phase θm modulation. TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.17/2
16. 16. QAM - Expansion ˜Similar to PSK case, φ(t) in (18) and φ(t) in (19) can be used asorthonormal basis for expansion of QAM signals [Tutorial 1] Eg Eg ˜ sm (t) = Ami φ(t) + Amq φ(t) (27) 2 2which results in vector representation of the form Eg Eg sm = (sm1 , sm2 ) = Ami , Amq (28) 2 2and 2 Eg Em = s m = A2 + A 2 . mi mq (29) 2 TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.19/2
17. 17. QAM - dminThe Euclidean distance between any pair of signal vectors inQAM is dmn = s m − sn 2 Eg = [(Ami − Ani )2 + (Amq − Anq )2 ]. (30) 2In the case when the signal amplitude take values of √±1, ±3, · · · , ±( M − 1) on both Ami and Amq , the signal spacediagram is rectangular, as shown in Fig. on next page. In thiscase, dmin = 2Eg . (31) TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.20/2
18. 18. Square QAMIn the case of square QAM (i,e., M = 4, 16, 64, 256, · · · ) with √amplitudes of ±1, ±3, · · · , ±( M − 1) √ √ M M 1 Eg Eavg = (A2 + A2 ) m n M 2 m=1 n=1 Eg 2M (M − 1) M −1 = × = Eg (32) 2M 3 3and M −1 Ebavg = Eg (33) 3 log2 M 6 log2 M dmin = Ebavg (34) M −1 TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.22/2
19. 19. PAM, PSK, QAMFor bandpass PAM, PSK, and QAM, the signaling schemes areof the general form sm (t) = Am g(t)ej2πfc t , m = 1, 2, · · · , M (35) For PAM, Am is real, equal to ±1, ±3, · · · , ±(M − 1) j 2π (m−1) For PSK, Am is complex and equal to e M For QAM, Am = Ami + jAmq .Therefore, PAM and PSK can be considered as special cases ofQAM. TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.23/2
20. 20. FSKThe FSK signal waveform is sm (t) = sml (t)ej2πfc t , 1 ≤ m ≤ M, 0 ≤ t ≤ T 2E = cos (2πfc t + 2πm∆f t) (36) T 2E j2πm∆f twhere sml (t) = T e , 1 ≤ m ≤ M and 0 ≤ t ≤ T .FSK signaling is a nonlinear modulation scheme, whereas ASK,PSK, and QAM are linear modulation schemes. TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.24/2
21. 21. FSKFSK is an orthogonal signaling if [ sml (t), snl (t) ] = 0, m = n. T 2E sml (t), snl (t) = ej2π(m−n)∆f t dt (37) T 0and [ sml (t), snl (t) ] = 2Esinc(2T (m − n)∆f ) (38)FSK is an orthogonal signaling when ∆f = k/2T . The minimumfrequency separation ∆f that guarantees orthogonality is 1∆f = 2T . TELE4653 - Digital Modulation & Coding - Lecture 2. March 8, 2010. – p.25/2