Principles of Communications

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1 Priniples of Communiations Meixia Tao Dept. of Eletroni Engineering Shanghai Jiao Tong University Chapter 3: Analog Modulation Seleted from Ch 3, Ch , Ch of of Fundamentals of Communiations Systems, Pearson Prentie Hall 005, by Proakis & Salehi

2 Topis to be Covered AM/FM radio FM radio TV broadast Soure Modulator Channel Demodulator Output Amplitude modulation Angle modulation (phase/frequeny) Effet of noise on amplitude modulation Effet of noise on frequeny modulation

3 Modulation What is modulation? Transform a message into another signal to failitate transmission over a ommuniation hannel Generate a arrier signal at the transmitter Modify some harateristis of the arrier with the information to be transmitted Detet the modifiations at the reeiver Why modulation? Frequeny translation Frequeny-division multiplexing Noise performane improvement 3

4 Analog Modulation Charateristis that be modified in sin arrier Amplitude Amplitude modulation Frequeny Phase Angle modulation 4

5 Amplitude Modulation Double-sideband suppressed-arrier AM (DSB-SC) Baseband signal (modulating wave): mt () Carrier wave Modulated wave ( ω θ ) st () = tmt () () = Amt ()os t+ 0 5

6 Spetrum of DSB-SC Signals Spetrum of message M(f) M(0) -W 0 W f Spetrum of DSB-SC USB LSB S(f) (/)A M(0) LSB USB -f -W -f -f +W 0 f -W S f ) = A + [ M ( f f ) + M ( f f )] ( f f +W f Translation of the original message spetrum to 6

7 Bandwidth and Power Effiieny S(f) (/)A M(0) -f -W -f -f +W 0 f -W f f +W f Required hannel bandwidth Required transmit power B = W T/ / T Ps = lim s ( t) dt lim A ( ) os ( / / m t ωt θ0) dt T T = T T T + T A T / A = lim m ( t) [ + os(ω 0) ] T / t + θ dt = Pm T T 7

8 Demodulation of DSB-SC Signals Phase-oherent demodulation s(t) Produt v(t) Low-pass v o (t) modulator filter Loal osillator os( πf t) PLL (phase-loked loop) If there is a phase error φ, then Saled version of message signal Unwanted 8

9 Demodulation of DSB-SC Signals Pilot-tone assisted demodulation Add a pilot-tone into the transmitted signal Filter out the pilot using a narrowband filter 9

10 Carrier wave: Baseband signal (normalized): Conventional AM Modulating wave Modulation index: a Modulated wave a Modulated wave a > overmodulated 0

11 Spetrum of Conventional AM Spetrum of message signal M(f) M(0) -W 0 W f (A /)δ(f+f ) S(f) (/)aa M(0) (A /)δ(f-f ) -f -W -f -f +W 0 f -W f f +W f A Aa S( f) = δ ( f f) + δ ( f + f) + M ( f f) + M ( f + f)

12 Bandwidth and Power Effiieny Required hannel bandwidth Required transmit power B = W arrier power message power Modulation effiieny a AP m apm m + AP m power in sideband E = = = total power A a + ap

13 Example Message signal: Carrier: mt ( ) = 3os(00 πt) + sin(600 πt) 5 t ( ) = os( 0 t) Modulation index: a=0.85 Determine the power in the arrier omponent and in the sideband omponents of the modulated signal 3

14 Demodulation of AM signals Envelop Detetor The simpliity of envelop detetor has made Conventional AM a pratial hoie for AM-radio broadasting 4

15 Single Sideband (SSB) AM Common problem in AM and DSBSC: Bandwidth wastage SSB is very bandwidth effiient H(ω) ω 5

16 Expression of SSB signals The baseband signal an be written as the sum of finite sinusoid signals n mt ( ) = x os( π ft+ θ ), f f i= i i i i Then its USB omponent is A m ( t) x os ( f f ) t ) n = i + i + i i= [ π θ ] After manipulation n n A m( t) = ios( i i) os i sin( i i) sin x π ft+ θ π ft x π ft+ θ π ft i= i= A A = mt ()os π ft mt ()sinπ ft Hilbert transform of m(t) 6

17 About Hilbert Transform x( τ ) xt () xt () xt () = dτ = xt () π t τ πt j, f > 0 H( f) = j, f < 0 0, f = 0 H (ω) 相移 ω 90 ω -90 7

18 Generation of SSB-AM Signal The spetral effiieny of SSB makes it suitable for voie ommuniation over telephone lines (0.3~3.4 khz) Not suitable for signals with signifiant low frequeny omponents 8

19 Vestigial Sideband: VSB VSB is a ompromise between SSB and DSBSC M(f) -W 0 W f VSB signal bandwidth is B = W+f v VSB is used in TV broadasting and similar signals where low frequeny omponents are signifiant S(f) -f 0 f f W f v f v W 9

20 Comparison of AM Tehniques DSB-SC: more power effiient. Seldom used Conventional AM: simple envelop detetor. AM radio broadast SSB: requires minimum transmitter power and bandwidth. Suitable for point-to-point and over long distanes VSB: bandwidth requirement between SSB and DSBSC. TV transmission 0

21 Signal Multiplexing Multiplexing is a tehnique where a number of independent signals are ombined and transmitted in a ommon hannel These signal are de-multiplexed at the reeiver Two ommon methods for signal multiplexing TDM (time-division multiplexing) FDM (frequeny-division multiplexing)

22 FDM LPF: ensure signal bandwidth limited to W MOD (modulator): shift message frequeny range to mutually exlusive high frequeny bands BPF: restrit the band of eah modulated wave to its presribed range

23 FDM appliation in telephone omm. Voie signal: 300~3400Hz Message is SSB modulated. In st -level FDM, signal are staked in frequeny, with a freq. separation of 4 khz between adjaent arriers A omposite 48 khz hannel, alled a group hannel, transmits voie-band signals simultaneously In the next level of FDM, a number of group hannel (typially 5 or 6) are staked to form a supergroup hannel Higher-order FDM is obtained by ombining several supergroup hannels => FDM hierarhy in telephone omm. systems 3

24 Quadrature-Carrier Multiplexing Transmit two messages on the same arrier as ( π ) ( π ) s() t = Am()os t ft + Am ()sin t ft os() and sin() are two quadrature arriers Eah message signal is modulated by DSB-SC Bandwidth-effiieny omparable to SSB-AM Synhronous demodulation of m(t): ( π ) = ( π ) + ( π ) ( π ) st ( )os ft Am( t)os ft Am( t)sin ft os ft A A A = m() t + m()os t 4 ft + m()sin t 4 ft ( π ) ( π ) LPF 4

25 Appliation: AM Radio Broadasting Commerial AM radio uses onventional AM Superheterodyne reeiver: from variable arrier freq of the inoming RF to fixed IF 5

26 Topis to be Covered Soure Modulator Channel Demodulator Output Amplitude modulation Angle modulation (phase/frequeny) Effet of noise on amplitude modulation Effet of noise on frequeny modulation 6

27 Angle Modulation Either phase or frequeny of the arrier is varied aording to the message signal The general form: θ(t): the time-varying phase instantaneous frequeny of s(t): f () t i = f + dθ () t π dt 7

28 Representation of FM and PM signals Phase modulation (PM) θ () t = kmt p () where k p = phase deviation onstant Frequeny modulation (FM) d fi() t f = kfmt () = () t π dt θ where k f = frequeny deviation onstant/frequeny sensitivity The phase of FM is t θ() t = πkf m( τ) dτ 0 8

29 Distinguishing Features of PM and FM No perfet regularity in spaing of zero rossing Constant envelop, i.e. amplitude of s(t) is onstant Relationship between PM and FM m(t) integrator m ( t) dt Phase modulator A C os( πf t) FM wave A [ f t + k p m( t dt] C os π ) m(t) differentiator d dt m(t) Frequeny modulator PM wave A C os [ π f t + πk m( t) ] f os( πf t) Will disuss the properties of FM only A C 9

30 Example: Sinusoidal Modulation Sinusoid modulating wave m(t) FM wave d dt m(t) PM wave 30

31 Example: Square Modulation Square modulating wave m(t) FM wave PM wave 3

32 FM by a Sinusoidal Signal Message Instantaneous frequeny of resulting FM wave Frequeny deviation: Carrier phase t f θ( t) = π ( f( ) ) sin( ) 0 i τ f dτ = π ft m f = βsin( π ft) Modulation index: m m 3

33 Example Problem: a sinusoidal modulating wave of amplitude 5V and frequeny khz is applied to a frequeny modulator. The frequeny sensitivity is 40Hz/V. The arrier frequeny is 00kHz. Calulate (a) the frequeny deviation (b) the modulation index Solution: Frequeny deviation f = k f Am = 40 5 = 00Hz Modulation index β = f f m = 00 =

34 Spetrum Analysis of Sinusoidal FM Wave Rewrite the FM wave as In-phase omponent Quadrature-phase omponent Define the omplex envelop of FM wave ~ jβ sin(πf m ) s ( t) = s ( ) ( ) t I t + jsq t = Ae ~ ( t ) retains omplete information about s(t) s j [ f t + β sin(π f t )] s { A e } [ s t e j f t ] π π Re ~ ( ) m ( ) = Re = t 34

35 ~ jβ sin(πf m ) s ( t) = A t e where is periodi, expanded in Fourier series as ~ j nfmt s ( π t) = e n= n n-th order Bessel funtion of the first kind Hene, π Jn( β) = exp j ( βsin x nx) dx π = n π A (β ) J n 35

36 Substituting into ~ s ( t ) ~ s ( t) = A n= FM wave in time domain J n ( β )exp ( j πnf t) m FM wave in frequeny-domain A S ( f ) = J n( β ) m + m = n [ δ ( f f nf ) + δ ( f + f nf )] 36

37 Properties of Bessel Funtion β β as 0 ) J n ( = odd ), ( even ), ( ) ( n J n J J n n n β β β Property : for small β 0.3 (Narrowband FM) Approximations Substituting above into 0, ) ( ) ( ) ( 0 > n J J J n β β β β [ ] [ ] t f f A t f f A t f A t s m m ) ( os ) ( os ) os( ) ( + + π β π β π Approximate bandwidth = Disuss the similarity between the onventional AM wave and a narrow band FM wave 37

38 General Case Goal: to investigate how and affet the spetrum Fix and vary and are varied.0.0 β = β = 5 f f f f 38

39 General Case Fix and vary is fixed, but is varied.0.0 β = 5 β = f f f f 39

40 Effetive Bandwidth of FW Waves For large B is only slightly greater than For small The spetrum is limited to Carson s Rule: B f + f = ( +β ) f m m 40

41 Effetive Bandwidth of FW Waves 99% bandwidth approximation The separation between the two frequenies beyond whih none of the side-frequenies is greater than % of the unmodulated arrier amplitude i.e where is the max that satisfies J n ( β ) > 0.0 β n max

42 Effetive Bandwidth of FW Waves A universal urve for evaluating the 99% bandwidth As inreases, the bandwidth oupied by the signifiant sidefrequenies drops toward that over whih the arrier frequeny atually deviates, i.e. B beome less affeted by

43 FM by an Arbitrary Message Consider an arbitrary with highest freq. omponent W Frequeny deviation: Modulation index: f= k max mt ( ) f Carson s rule applies as Carson s rule underestimates the FM bandwidth requirement Universal urve yields a onservative result 43

44 Example In north Ameria, the maximum value of frequeny deviation is fixed at for ommerial FM broadasting by radio. Take, typially the maximum audio frequeny of interest in FM transmission, the modulation index is Using Carson s rule, Using universal urve, 44

45 Exerise ( ) 4 Assuming that mt ( ) = 0sin 0 t, determine the transmission bandwidth of an FM modulated signal with k = 4000 f Solution By Carson s rule: 45

46 Appliation: FM Radio broadasting As with standard AM radio, most FM radio reeivers are of super-heterodyne type Limiter disriminator Typial freq parameters RF arrier range = 88~08 MHz Midband of IF = 0.7MHz IF bandwidth = 00kHz Peak freq. deviation = 75KHz Baseband low-pass filter loudspeaker Audio amplifier with de-emphasis 46

47 Summary Spetrum of sinusoidal FM Wave A S ( f ) = J n( β ) m + m = n [ δ ( f f nf ) + δ ( f + f nf )] Carson rule approximation Universal urve approximation 47

48 Generation of FM waves Diret approah Design an osillator whose frequeny hanges with the input voltage => voltage-ontrolled osillator (VCO) Indiret approah First generate a narrowband FM signal and then hange it to a wideband signal Due to the similarity of onventional AM signals, the generation of a narrowband FM signal is straightforward. 48

49 Generation of Narrow-band FM Consider a narrow band FM wave s t) A os f t φ ( ) [ π + ] ( = t where t ) 0 φ ( t) = πk m( τ dτ f = arrier frequeny k = frequeny sensitivity Given φ (t) << with β 0.3, we may use φ t os sin [ ( )] [ φ ( t) ] φ ( t) Correspondingly, we may approximate s (t) as s ( t) = A os πf t A sin πf t φ ( t) = A os ( ) ( ) ( πft ) πk A sin( πft ) t 0 m( τ ) dτ Narrow-band FW wave 49

50 Narrow-band frequeny modulator m(t) integrator Produt Modulator A sin( πf t) Narrow-band FM wave s (t) phase Carrier wave shifter A os( πft) Next, pass s (t) through a frequeny multiplier Narrow-band FM wave Memoryless nonlinear devie Wideband FM Wave Band-pass filter The input-output relationship of the non-linear devie is: n s t) = as ( t) + a s ( t) + + ans ( t) ( The BPF is used to Pass the FM wave entred at nf and with deviation n f and suppress all other FM spetra 50

51 Example: frequeny multiplier with n = Problem: Consider a square-law devie based frequeny multiplier with Speify the midband freq. and bandwidth of BPF used in the freq. multiplier for the resulting freq. deviation to be twie that at the input of the nonlinear devie Solution: ) ( ) ( ) ( t s a t s a t s + = + = t d m k t f A t s 0 ) ( os ) ( τ τ π π = = t t t t d m k t f A a A a d m k t f A a d m k t f A a d m k t f A a t s ) ( 4 4 os ) ( os ) ( os ) ( os ) ( τ τ π π τ τ π π τ τ π π τ τ π π Removed by BPF with f =f BW > f = 4 f 5

52 Generation of Wideband FM Signal s = + t ( t) A os πft πk m( τ ) dτ 0 Message signal A Integrator os( π ft) Narrow-band phase modulator Crystal-ontrolled osillator Frequeny multiplier s( t) = A + t os πf t πk f m( τ ) dτ 0 Wideband FM signal os( πf l t) BPF Output Mixer: perform up/down onversion to shift the signal to the desired enter freq. f k f f = nf = nk = n f may not be the desired arrier frequeny 5

53 Exerise: A typial FM transmitter Problem: Given the simplified blok diagram of a typial FM transmitter used to transmit audio signals ontaining frequenies in the range 00Hz to 5kHz. Desired FM wave: f = 00MHz, f = 75kHz. Set β = 0. in the narrowband phase modulation to limit harmoni distortion. Speify the two-stage frequeny multiplier fators n and n Message signal Integrator Narrow-band phase modulator Frequeny multiplier n Mixer Frequeny multiplier n FM signal Crystal-ontrolled osillator Crystal-ontrolled osillator 0.MHz? 53

54 Demodulation of FM Balaned Frequeny Disriminator Given FM waves( t) = A + t os πf t πk f m( τ ) dτ 0 d t st () = A π f π k f mt () sin π ft π k f m ( τ) d τ dt Hybrid-modulated wave with AM and FM H Differentiator + envelop detetor = FM demodulator Frequeny disriminator: a freq to amplitude transform devie jπa 0, FM wave ( f f + B / ) ( f + f B / ) Slope iruit H (f) Slope iruit H (f), B / Envelop detetor Envelop detetor B / ( f ) = jπa, f B / f f + B f f f elsewhere / Baseband signal H ( f ) = H( f ) 54

55 Ciruit diagram and frequeny response 55

56 Appliation: FM Radio broadasting As with standard AM radio, most FM radio reeivers are of super-heterodyne type Limiter disriminator Baseband low-pass filter loudspeaker Audio amplifier with de-emphasis Typial freq parameters RF arrier range = 88~08 MHz Midband of IF = 0.7MHz IF bandwidth = 00kHz Peak freq. deviation = 75KHz 56

57 FM Radio Stereo Multiplexing Stereo multiplexing is a form of FDM designed to transmit two separate signals via the same arrier. Widely used in FM broadasting to send two different elements of a program (e.g. voalist and aompanist in an orhestra) so as to give a spatial dimension to its pereption by a listener at the reeiving end 0/4/ m l (t) The sum signal is left unproessed in its baseband form The differene signal and a 38-kHz subarrier produe a DSBSC wave The 9-kHz pilot is inluded as a referene for oherent detetion + + m r (t) m( t) = Frequeny doubler [ ml ( t) + mr ( t) ] + [ m ( t) m ( t) ] m(t) K os( πf t) l r os(4πf t) + K os(πf t) f = 9kHz

58 FM-Stereo Reeiver m(t) Baseband LPF BPF entered at f =38kHz m l (t)+m r (t) Baseband LPF + - m l (t)-m r (t) m l (t) + + m r (t) To two loudspeakers Frequeny doubler Narrow-band filter tuned to f =9kHz 0/4/004 58

59 Think AM vs. FM Compared with AM, FM requires a higher implementation omplexity and a higher bandwidth oupany. What is the advantage of FM then? Why AM radio is mostly for news broadasting while FM radio is mostly for musi program 59

60 Suggested Reading Chapter 3 and Chapter of Fundamentals of Communiations Systems, Pearson Prentie Hall 005, by Proakis & Salehi 60

61 Topis to be Covered Soure Modulator Channel Demodulator Output Amplitude modulation Angle modulation (phase/frequeny) Effet of noise on amplitude modulation Effet of noise on frequeny modulation 6

62 A Benhmark System Baseband system: s m (t) n(t) + LPF No arrier demodulation The reeiver is an ideal LPF with bandwidth W Noise power at the output of the reeiver W N0 Pn = 0 df = N W 0W S PR Baseband SNR is given by = N NW b 0 6

63 Example: Find the SNR in a baseband system with a bandwidth 4 of 5 khz and with N 0 = 0 W/Hz. The transmitter power is kw and the hannel attenuation is 0 Solution: P R = = Watts 9 S PR 0 = = 4 = N NW b 0 0 = 0 log0 0 = 3dB 63

64 Effet of Noise on DSBSC st () n(t) ( t ) 0 n i (t) st () + BPF demod m n ( t) 0 Modulated signal Input to the demodulator rt () = st () + n() t i s() t = Amt ()osωt = Amt ()os wt+ n()os t wt n()sin t wt s Here n i (t) is a Gaussian narrow-band noise S n i N / 0 f f W ( f) = 0 otherwise 64

65 In the demodulator, the reeived signal is first multiplied by a loally generated sinusoid signal r( t)os( wt+ φ) = Amt ( )os wtos( wt+ φ) + n( t)os wtos( wt+ φ) n( t)sin wtos( wt+ φ) s = Amt ( )os φ+ Amt ( )os( wt + φ) + [ n()os t φ+ ns()sin t φ] + + s + [ n( t)os( wt φ) n( t)sin( wt φ) ] Assume oherent detetor, we have φ = 0 65

66 Then the signal is passed through a LPF with bandwidth W Sn ( f) = S ( ) ( ) y() t = [ Amt () + n() t n f f i + Sn f + f i ] where for f W The output SNR an thus be defined as AP m S Po 4 AP = = = m N o Pn WN o P 0 n 4 AP m Sine the reeived power of DSBSC in baseband is P R = The output SNR an be rewritten as S PR S = = N WN N odsb 0 b DBSSC does not provide any SNR improvement over a simple baseband systems 66

67 Effet of Noise on SSB Modulated signal: Input to the demodulator s() t = Amt ()os wt± Amt ()sinwt N0 / f f W / rt () = st () + n Sn ( f ) = i () t where i 0 otherwise = Amt () + n()os t wt+ ± Amt () n()sin t wt [ ] [ ] s Output of LPF: A yt () = mt () + n () t Therefore, the output SNR is AP S P 4 = = = N o Pn WN o Pn 4 m o AP m 0 67

68 But in this ase, Thus, P = AP R m S PR S = = N WN N ossb 0 b SNR in an SSB system is equivalent to that of a DSBSC system 68

69 Effet of Noise on Conventional AM Modulated signal Input to the demodulator rt () = st () + n() t [ ] s() t = A + am()os t w t [ ] With oherent detetor, after mixing and LPF: Removing DC omponent i = A + Aamt () + n()os t wt n()sin t wt s y() t = A + Aam() t + n() t [ ] y() t = Aam() t + n() t [ ] 69

70 Exerise: please show that the output SNR is SNR in onventional AM is always smaller than that in baseband. Hint: the reeived signal power is Modulation effiieny is P = A + ap R m 70

71 Performane of Envelope-Detetor Input to the envelope-detetor Envelope of r(t) [ ] r() t = A + Aamt () + n()os t wt n()sin t wt s [ ] V () t = A + A am() t + n () t + n () t r s If signal omponent is muh stronger than noise V () t A + A am() t + n () t r After removing DC omponent, we obtain y() t = A am() t + n () t At high SNR, performane of oherent detetor and envelop detetor is the same 7

72 Performane of Envelope-Detetor If noise power is muh stronger than the signal power Ignore st term ( ) ( ) V t A am t n t n t A n t am t r() = + () + () + s() + () + () An() t n() t + ns() t V() t n = n() t + ns() t ε An () t + ε + Vn () t + ( + am() t ) Vn () t An () t = Vn () t + ( + am() t ) V () t ( n() t ns() t ) ( am() t ) n The system is operating below the threshold, no meaningful SNR an be defined. 7

73 Exerise Consider that the message is a WSS r.p M(t) with autoorrelation funtion RM ( τ) = 6sin (0000 τ). It is given that mt () = 6. max We want to transmit this message to a destination via a hannel with a 50dB attenuation and additive white noise with PSD. Sn ( f) = N0 / = 0 W/Hz. We also want to ahieve an SNR at the modulator output of at least 50dB. What is the required transmitted power and the hannel bandwidth if we employ the following modulation shemes? DSB-SC SSB AM with modulation index =

74 Topis to be Covered Soure Modulator Channel Demodulator Output Amplitude modulation Angle modulation (phase/frequeny) Effet of noise on amplitude modulation Effet of noise on frequeny modulation 74

75 Effet of Noise in FM The effet of additive noise is desribed by the hanges of frequeny, or the hanges in the zerorossings of the modulated FM single. 75

76 Effet of Noise on Angle Modulation st () + n() t Blok diagram of an angle demodulator w BPF BW=B rt () = st () + nt () yt () Demod LPF BW=W S N o Input to the demodulator is rt () = [ ] = A os ω t+ φ() t + n ()os t ω t n ()sin t ω t s = R ()os( t wt+θ ()) t n n 76

77 Assume that the signal is muh larger than the noise ( θ φ ) rt () A + Rn()os t n() t () t os wt + φ( t) + tg R () n t ( θn φ ) ( θ φ ) R ()sin t () t () t n A + R ()os t () t () t n n The phase term an be further approximated as Rn () t θr() t = φ() t + sin θn() t φ() t A ( ) n () t ns () t R t y () φ () t θ y () t θ () θ () φ () e t A R () n t θ n () t n t t A E n t () 77

78 Therefore, the output of the demodulator is d d Rn () t yt () = θr() t = kfmt () + sin θn() t φ() t πdt πdt A d = kfmt () + Yn() t π dt Desired signal Noise ( ) The noise omponent is inversely proportional to the signal amplitude A. (This is not the ase for AM system) Rn () t Yn() t = sin ( θn() t φ() t ) = [ ns()os t φ() t n()sin t φ() t ] A A = [ ns()os t φ n()sin t φ] (Sine φ() t is slowly varying) A 78

79 The power spetral density of n() is π dt 4π f osφ sinφ Y = n + n 4π A A = ( ) n = A 0 d Y t S ( f) f S ( f) S ( f) n s f f S f A N f B 0 otherwise C S ( ) no f Snf ( f) At the output of LPF, the noise is limited to the freq. range [-W, W] B T f x f x BT -W W f 79

80 Output SNR in FM Now we an determine the output SNR in FM First, the output signal power is s f m The output noise power is P n o P Then, the output SNR is o = k P N = = NW W 0 0 f df W A 3A S Ps 3k 0 f A Pm N o Pn W NW 0 = = o ( max mt ( ) ) 3 3 = β fpm S N b 80

Thus, FM provides a way to trade off bandwidth for transmitted power β Power ontent of the normalized message Having a large B

81 Key Observations Rewrite Inreasing inreases the output SNR, in ontrast to AM Inreasing the bandwidth (by Carson rule ) inreases the output SNR. Thus, FM provides a way to trade off bandwidth for transmitted power β Power ontent of the normalized message Having a large B means having a large noise power. Thus, Inreasing β annot ontinue improving the performane indefinitely due to the threshold effet Inreasing the transmitted power inreases output SNR in both FM and AM systems, but the mehanisms are totally different. 8

82 Threshold Effet There exists a speifi SNR at the input of the demodulator below whih the signal is not distinguishable from the noise S 0 N 0 FM DSB 0 a S i N i 8

83 Comparison of Analog-Modulation Bandwidth effiieny SSB is the most bandwidth effiient, but annot effetively transmit DC VSB is a good ompromise PM/FM are the least favorable systems Power effiieny (refleted in performane with noise) FM provides high noise immunity Conventional AM is the least power effiient Implementation omplexity (transmitter and reeiver) The simplest reeiver struture is onventional AM FM reeivers are also easy to implement DSB-SC and SSB-SC requires oherent detetor and hene is muh more ompliated. 83

84 Think AM vs. FM Compared with AM, FM requires a higher implementation omplexity and a higher bandwidth oupany. What is the advantage of FM then? FM provides high noise immunity. This is why AM radio is mostly for news broadasting while FM radio is mostly for musi program. 84

85 Suggested Reading Chapter of of Fundamentals of Communiations Systems, Pearson Prentie Hall 005, by Proakis & Salehi 85

EE140 Introduction to Communication Systems Lecture 7

EE140 Introduction to Communication Systems Lecture 7 3/4/08 EE40 Introdution to Communiation Systems Leture 7 Instrutor: Prof. Xiliang Luo ShanghaiTeh University, Spring 08 Arhiteture of a (Digital) Communiation System Transmitter Soure A/D onverter Soure