Tele4653 l2

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<ul><li> 1. TELE4653 Digital Modulation &amp;CodingDigital ModulationWei Zhang w.zhang@unsw.edu.auSchool of Electrical Engineering and TelecommunicationsThe University of New South Wales</li></ul> <p> 2. Outline PAM (ASK) PSK QAM FSK TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.1/2 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 MemoryTELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.2/2 4. Denitions 1 Signaling Interval: Ts . Signaling (Symbol) Rate: Rs =Ts .Ts Bit Interval: Tb = kfor a signal carrying k bits of information. Bit Rate: R = kRs = Rs logM . 2 M Average signal energy: Eavg = m=1 pm Emwith pm being the probability of the mth signal.Eavg Eavg Average energy per bit: Ebavg = k = logM.2 TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.3/2 5. PAMThe signal waveform may be represented assm (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 byAm = 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 &amp; Coding - Lecture 2. March 8, 2010. p.4/2 6. PAMThe energy of signal sm (t) is given by Em = A2 p2 (t)dt = A2 Epm m(3) The average signal energy is MMEpEavg = p m Em = A2 mM m=1m=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 &amp; Coding - Lecture 2. March 8, 2010. p.6/2 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 issm (t) = sml (t)ej2fc t = Am g(t) cos(2fc t)(6)The energy of signal sm (t) is given byA2Em = m Eg . (7) 2Moreover, (M 2 1)Eg (M 2 1)EgEavg =6 , Ebavg =6 logM .(8)2TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.7/2 8. Bandpass PAM - ExpansionFor basedband PAM, the expansion of sm (t) = Am p(t) is sm (t) = AmEp (t) (9)where p(t) (t) =.(10)EpFor bandpass PAM, the expansion of sm (t) =sml (t)ej2fc tis [Tutorial 1]Egsm (t) = Am(t) (11) 2where 2(t) =g(t) cos(2fc t). (12) EgTELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.9/2 9. Bandpass PAM - dminThe bandpass PAM can be represented as the one-dimensionalEvector: 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 Mdmin =2Eg =E 2 1 bavg(14)Mwhere in the last equality Eq. (8) is used. TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.10/2 10. Phase ModulationThe bandpass PM signal waveform may be represented as 2(m1) jsm (t) = g(t)eMej2fc t , 1 m M2(m 1) = g(t) cos + 2fc t (15) M 2(m1)Let m =M, m = 1, 2, , M . Then,sm (t) = g(t) cos m cos(2fc t) g(t) sin m sin(2fc t).(16)Digital phase modulation is usually called phase-shift keying(PSK).The PSK signals have equal energy, Eavg = Em = 1 Eg .2TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.11/2 11. Phase Modulation - Expansion2(m1)jThe expansion of PM signal s(t) = g(t)eMej2fc t isEgEg sm (t) = cos(m )(t) +sin(m )(t) (17)2 2where [see Tutorial 1]2 (t) =g(t) cos(2fc t) (18)Eg 2 (t) = g(t) sin(2fc t) (19)Eg TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.13/2 12. Phase Modulation - dminThe bandpass PM can be represented as the two-dimensionalEg Egvector: sm = 2 cos m , 2 sin m , m = 1, 2, , M . Notem = 2(m1) for m = 1, 2, , M .MThe Euclidean distance between any pair of signal points is 2Egdmn =sm sn = | cos m cos n |2 + | sin m sin n |22 =Eg [1 cos(m n )].(20) TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.14/2 13. Phase Modulation - dminFor adjacent signal points |m n| = 1, it has2 dmin =Eg 1 cos( )(21)M 2 =2Eg sin(22) M2= 2 Ebavg log2 M sin(23) M For large values of M , we have sinM M,and then 2dmin 2 Ebavg log2 M 2 (24) M TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.15/2 14. QAMThe quadrature amplitude modulation (QAM) signal waveformmay be expressed as sm (t) = (Ami + jAmq )g(t)ej2fc t , m = 1, 2, , M = Ami g(t) cos(2fc t) Amq g(t) sin(2fc 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 &amp; Coding - Lecture 2. March 8, 2010. p.16/2 15. QAMAlternatively, the QAM signal may be expressed assm (t) =rm g(t)ejm ej2fc t = rm g(t) cos(m + 2fc t),(26)where rm =A2 + A2 and m = tan1 (Amq /Ami ). mi mqIt is apparent that QAM signal can be viewed as combinedamplitude rm and phase m modulation. TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.17/2 16. QAM - ExpansionSimilar to PSK case, (t) in (18) and (t) in (19) can be used asorthonormal basis for expansion of QAM signals [Tutorial 1] EgEg sm (t) = Ami(t) + Amq(t)(27) 2 2which results in vector representation of the form Eg Eg sm = (sm1 , sm2 ) = Ami, Amq (28) 22and 2 EgEm = s m =A2 + A 2 . mimq (29) 2 TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.19/2 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 of1, 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 &amp; Coding - Lecture 2. March 8, 2010. p.20/2 18. Square QAMIn the case of square QAM (i,e., M = 4, 16, 64, 256, ) withamplitudes of 1, 3, , ( M 1) M M1 EgEavg = (A2 + A2 ) mnM 2 m=1 n=1 Eg 2M (M 1) M 1 ==Eg(32)2M33andM 1Ebavg = Eg(33) 3 log2 M 6 log2 M dmin = Ebavg (34)M 1 TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.22/2 19. PAM, PSK, QAMFor bandpass PAM, PSK, and QAM, the signaling schemes areof the general form sm (t) =Am g(t)ej2fc t , m = 1, 2, , M (35)For PAM, Am is real, equal to 1, 3, , (M 1) j 2 (m1)For PSK, Am is complex and equal to eMFor QAM, Am = Ami + jAmq .Therefore, PAM and PSK can be considered as special cases ofQAM.TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.23/2 20. FSKThe FSK signal waveform is sm (t) = sml (t)ej2fc t , 1 m M, 0 t T 2E =cos (2fc t + 2mf t)(36) T2E j2mf 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 &amp; Coding - Lecture 2. March 8, 2010. p.24/2 21. FSKFSK is an orthogonal signaling if [ sml (t), snl (t) ] = 0, m = n.T2E sml (t), snl (t) = ej2(mn)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 1f = 2T .TELE4653 - Digital Modulation &amp; Coding - Lecture 2. March 8, 2010. p.25/2</p>