page 7.1 CHAPTER 7 AMPLITUDE MODULATION Transmit information-b earing message or baseband signal voice-music through a Communications Channel Baseband = band of frequencies representing the original signal for music 20 Hz - 20,000 Hz, for voice 300 - 3,400 Hz write the baseband message signal mt $ M f Communications Channel Typical radio frequencies 10 KHz ! 300 GHz write ct= A cos2f ct c ct = Radio Frequency Carrier Wave A = Carrier Amplitude c fc = Carrier Frequency Amplitude Mo dulation AM ! Amplitude of carrier wavevaries a mean value in step with the baseband signal mt st= A [1 + k mt] cos 2f t c a c Mean value A . c 31 page 7.2 Recall a general signal st= at cos[2f t + t] c For AM at = A [1 + k mt] c a t = 0 or constant k = Amplitude Sensitivity a Note 1 jk mtj < 1or [1 + k mt] > 0 a a 2 f w = bandwidth of mt c 32 page 7.3 AM Signal In Time and Frequency Domain st = A [1 + k mt] cos 2f t c a c j 2f t j 2f t c c e + e st = A [1 + k mt] c a 2 A A c c j 2f t j 2f t c c e + e st = 2 2 A k c a j 2f t c + mte 2 A k c a j 2f t c + mte 2 To nd S f use: mt $ M f j 2f t c e $ f f c j 2f t c e $ f + f c expj 2f tmt $ M f f c c expj 2f tmt $ M f + f c c A c S f = [f f +f +f ] c c 2 A k c a + [M f fc+Mf +f ] c 2 33 page 7.4 st = A [1 + k mt] cos 2f t c a c A c = [1 + k mt][expj 2f t+ expj 2f t] a c c 2 If k mt > 1, then a ! Overmo dulation ! Envelop e Distortion see Text p. 262 - Fig. 7.1 max value of k mt= Percent Mo dulation a Lo ok at AM signal in frequency domain A c S f = [ f f +f +f ] c c 2 k A a c + [M f f +Mf +f ] c c 2 where M f =FT of mt 34 page 7.5 Example 1 p. 265 - Single Tone Mo dulation mt = A cos 2f t f = 1 KHz message frequency m m m st = A 1 + k mt] cos 2f t c a c st = A 1 + cos 2f t] cos 2f t c m c = k A < 1 a m = Mo dulation Factor Percentage Draw st for <1 A 1 + A c max = A A 1 min c A A max min = A + A max min Rewrite st using trig identity 1 1 cos cos = cos + cos + with = f ; = f c m 2 2 st = A [cos 2f t + cos 2 f + f t c c c m 2 + cos 2 f f t] c m 2 A c S f = [ f f +f +f ] c c 2 A c + [ f f f +f +f +f ] c m c m 4 A c + f f + f +f +f f ] c m c m 4 35 page 7.6 2 2 [st] =power into 1 jSfj 2 2 A A c c = Carrier Power 2 2 2 Positive and NegativeFrequencies 2 2 2 A A c c Upp er Sideband 2 = 4 8 Power 2 2 2 A A c c Lower Sideband 2 = 4 8 Power 2 2 2 2 A A 2 c c + USB + LSB Power 8 8 = = 2 2 2 2 2 A A A c c c Total Power 2+ + + 2 8 8 Example Mo dulation Audio Carrier Power Total Audio Power Power Mo dulation USB + LSB Power Total 2 2 2 2 A A A c c c 2 2 100 2 + 2 2 4 4 2+ 100 100 50 150 0.33 100 50 12.5 112.5 0.11 100 20 4 104 0.04 100 0 0 100 0 36 page 7.8 Generation of AM Waves - P. 267 Square Law Mo dulator - 3 stages: Adder, Non-Linearity NL, Bandpass Filter BPF 2 For this example, we use non-linearity v t= a v t+a v t 2 1 1 2 1 Here v t = A cos 2f t + mt 1 c c z }| { 2a 2 ! v t = a A 1+ mt cos 2f t 2 1 c c a 1 2 2 2 +a mt+ a m t+a A cos 2f t 1 2 2 c c The rst bracketed term is desired AM wave at carrier f with c 2a 2 k = a a 1 Remaining terms are at baseband or 2f and are ltered out by BPF. BPF has center c 2 frequency f bandwidth 2w . The cos term has comp onents at baseband and at 2f b ecause c c 1 2 = cos 1 + 2 cos 2 with =2f t c 2 37 page 7.10 Given v t nd V f Curve 2 2 v t = a A cos 2f t a 2 1 c c 2a A mt cos 2f t b 2 c c +a mt c 1 2 +a m t d 2 1 2 +a A 1 + cos 2 2f t e 2 c c 2 | {z } 2 cos 2f t c A c V f = a [ f f +f +f ] 2 1 c c 2 +2a A [M f f +Mf +f ] 2 c c c +a M f 1 2 +a FT[m t] 2 2 a A 2 c [ f + f 2f +f +2f ] + c c 2 Assignment Text: p. 384, p. 7.1, p. 7.3, p. 269 Ex. 2 38 page 7.11 Generation of AM Waves Continued-P. 269-270 Switching Mo dulator Same as square law except non linearityisnow: ! v t v t>0 1 1 v t= 2 0 v t<0 1 v t = A cos 2f t + mt 1 c c = ct+ mt Assume jmtj A c v t ct> 0 1 Thus v t= 2 0 ct< 0 Thus switch dio de on and o at rate f c dio de on only when ct > 0 v t=[A cos2f t + mt]g t 2 c c p where g t is a unip olar square waveat f . p c 39 page 7.12 Recall Fourier Series ELEC 260 p. 106 f t with p erio d T 1 X 2nt 2nt f t = a + a cos + b sin o n n T T n=1 Z T=2 1 f tdt a = o T T=2 Z T=2 2 2nt a = dt f t cos n T T T=2 Z T=2 2 2nt b = f t sin dt n T T T=2 Consider g t with p erio d T and frequency f =1=T p 0 c 0 Z T =4 o 1 1 1 T o = a = 1 dt = o T T 2 2 T =4 o o o T o Z T =4 o 4 2 2nt 2 T 2nt o a = cos dt = sin n T o T T T 2n T T =4 o o o o o 4 " ! ! 2nT =4 1 2nT =4 o o = sin sin n T T o o 8 9 2 > > n =1; 5; 9 ::: n1 < = n 1 n 2 1 = 2 sin = = > > n 2 2n 1 : ; 2 n =3; 7; ::: n n =2m1; m=1;2;3;::: 40 page 7.13 Relab el m ! n " 1 n1 X 1 2 1 2 2n 1f t c with f =1=T g t= + cos c o p 2 2n 1 T o n=1 write g tasFourier cosine series see ELEC 260 p 1 n1 X 1 2 1 g t = + cos[2f t2n 1] p c 2 2n 1 n=1 v t = [A cos 2f t + mt]g t 2 c c p 1 2 = [A cos 2f t + mt] + cos 2f t + ::: c c c 2 A 2 mt c = cos 2f t + mt cos 2f t + + ::: c c 2 2 4 A c 1+ + ::: = mt A cos 2f t c c |{z} 2 A c {z } | 4 = AM wave with k = unwanted comp onents a A c at frequencies removed from f c 41 page 7.14 Demo dulation of AM Waves Square Law Demo dulator No go o d unless mt A i.e. mo dulation very low see P. 271 - Skip this c Envelop e Detector -P. 272 42 page 7.15 7.2 P. 274 - Double Sideband Suppressed Carrier Mo dulation DSBSC Carrier ct= A cos 2f t do es not carry information ! waste of p ower c c Power in mt 1 Recall = max at 100 mo dulation Power in ct 3 ! Suppress the Carrier AM: s t = [1 + mt]ct AM = ct+mt ct k =1 a DSB: st = mtct = A cos 2f tmt c c 1 Sf = A [Mf f +Mf f ] c c c 2 Phase reversal when mt < 0. 43 page 7.16 DSBSC st = ctmt = A cos 2f t mt c c A c = [expj 2f t + expj 2f t]mt c c 2 Recall shifting prop erty mt $ M f expj 2f tmt $ M f f c c A c [Mf +f +Mf f ] S f = c c 2 Single tone mo dulation mt = A cos 2f t m m A m M f = f f +f +f ] m m 2 A m M f f = [ f f f +f +f f ] c m c m c 2 A m M f + f = [ f f + f +f +f +f ] c m c m c 2 44 page 7.18 3.5 Generation of DSBSC Waves -P.