! Amplitude of carrier wave varies a mean value in step with the baseband signal m(t)

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1 page 7.1 CHAPTER 7 AMPLITUDE MODULATION Transmit information-bearing (message) or baseband signal (voice-music) through a Communications Channel Baseband = band of frequencies representing the original signal for music 0 Hz - 0,000 Hz, for voice 300-3,400 Hz write the baseband (message) signal m(t) $ M(f) Communications Channel Typical radio frequencies 10 KHz! 300 GHz write c(t) =A c cos(fct) c(t) = Radio Frequency Carrier Wave A c = Carrier Amplitude fc = Carrier Frequency Amplitude Modulation (AM)! Amplitude of carrier wave varies a mean value in step with the baseband signal m(t) s(t) =A c [1 + k a m(t)] cos f c t Mean value A c. 31

2 page 7. Recall a general signal s(t) =a(t) cos[f c t + (t)] For AM a(t) = A c [1 + k a m(t)] (t) = 0 or constant k a = Amplitude Sensitivity Note 1 jk a m(t)j < 1or [1 + k a m(t)] > 0 f c w = bandwidth of m(t) 3

3 page 7.3 AM Signal In Time and Frequency Domain s(t) = A c [1 + k a m(t)] cos f c t s(t) = A c [1 + k a m(t)] (ejfct + e,jfct) s(t) = A c ejfct + A c e,jfct + A ck a m(t)ejfct + A ck a m(t)e,jfct To nd S(f) use: m(t) $ M(f) e jfct $ (f, f c ) e,jfct $ (f + f c ) exp(jf c t)m(t) $ M(f, f c ) exp(,jf c t)m(t) $ M(f + f c ) S(f) = A c [(f, f c)+(f+f c )] + A ck a [M(f, fc)+m(f+f c)] 33

4 page 7.4 s(t) = A c [1 + k a m(t)] cos f c t = A c [1 + k am(t)][exp(jf c t)+exp(,jf c t)] If k a m(t) > 1, then! Overmodulation! Envelope Distortion see Text p. 6 - Fig. 7.1 max value of k a m(t) =Percent Modulation Look at AM signal in frequency domain S(f) = A c [(f, f c)+(f+f c )] + k aa c [M(f, f c)+m(f+f c )] where M(f) =FT of m(t) 34

5 page 7.5 Example 1 p Single Tone Modulation m(t) = A m cos f m t f m = 1 KHz (message frequency) s(t) = A c (1 + k a m(t)] cos f c t s(t) = A c (1 + cos f m t] cos f c t = k a A m < 1 = Modulation Factor (Percentage) Draw s(t) for <1 A max = A c(1 + ) A min A c (1, ) = A max, A min A max + A min Rewrite s(t) using trig identity cos cos = 1 cos(, )+1 cos( + ) with = f c; = f m s(t) = A c [cos f c t + cos (f c + f m )t + cos (f c, f m )t] S(f) = A c [(f, f c)+(f+f c )] + A c 4 [(f, f c, f m )+(f+f c +f m )] + A c 4 ((f, f c + f m )+(f+f c,f m ))] 35

6 page 7.6 [s(t)] = power into 1 js(f)j Carrier Power Positive and Negative Frequencies % Ac = A c Ac Upper Sideband = A c 4 8 Power Lower Sideband Power Ac = A c 4 8 USB + LSB Power Total Power = A c + A c 8 8 A + c A c + A c 8 8 = + Example Modulation (Audio) Carrier % Power Total Audio Power Power Modulation USB + LSB Power Total A c 100 A c 4 ( + ) A c

7 page 7.8 Generation of AM Waves - P. 67 Square Law Modulator - 3 stages: Adder, Non-Linearity (NL), Bandpass Filter (BPF) For this example, we use non-linearity v (t) =a 1 v 1 (t)+a v 1(t) Here v 1 (t) = A c cos f c t + m(t)! v (t) = z } { a 1 A c 1+ a a 1 m(t) cos f c t +a 1 m(t)+a m (t)+a A ccos f c t The rst (bracketed) term is desired AM wave at carrier f c with k a = a a 1 Remaining terms are at baseband or f c and are ltered out by BPF. BPF has center frequency f c bandwidth w. The cos term has components at baseband and at f c because cos = 1 (1 + cos ) with =f ct 37

8 page 7.10 Given v (t) nd V (f) Curve v (t) = a 1 A c cos f c t a a A c m(t) cos f c t b +a 1 m(t) c +a m (t) d +a A c 1 (1 + cos f ct) {z } cos f ct e V (f) = a 1 A c [(f, f c)+(f+f c )] +a A c [M(f, f c )+M(f+f c )] +a 1 M(f) +a FT[m (t)] + a A c [(f)+(f,f c)+(f+f c )] Assignment Text: p. 384, p. 7.1, p. 7.3, p. 69 Ex. 38

9 page 7.11 Generation of AM Waves Continued-P Switching Modulator Same as square law except non linearity isnow:! v (t) = ( v1 (t) v 1 (t)>0 0 v 1 (t)<0 v 1 (t) = A c cos f c t + m(t) = c(t)+m(t) Assume jm(t)ja c Thus v (t) = ( v1 (t) c(t)>0 0 c(t)<0 Thus switch diode on and o at rate f c diode on only when c(t) > 0 v (t) =[A c cosf c t + m(t)]g p (t) where g p (t) is a unipolar square wave atf c. 39

10 page 7.1 Recall Fourier Series ELEC 60 p. 106 f(t) with period T f(t) = a o + 1X n=1 a n cos nt T + b n sin nt T a o = 1 T a n = T b n = T Z T=,T= Z T=,T= Z T=,T= f(t)dt f(t) cos nt T dt f(t) sin nt T dt Consider g p (t) with period T 0 and frequency f c =1=T 0 a o = 1 T o Z To=4,T o=4 a n = T o Z To=4 = 1 n,t o=4 1 dt = 1 T o T o = 1 cos nt T o dt = T o T o n sin nt T o "!!# sin nt o=4, sin, nt o=4 T o T o = 1 n n sin = 8 >< >: n =1; 5; 9 ::: n, n n =3; 7; ::: To 4,To 4 9 >= >; = (,1)n,1 n, 1 n =m,1; m=1;;3;::: 40

11 page 7.13 Relabel m! n g p (t) = 1 + 1X n=1 " # (,1) n,1 (n, n, 1 cos 1)fc t T o with f c =1=T o write g p (t) asfourier cosine series (see ELEC 60) g p (t) = 1 + 1X n=1 (,1) n,1 n, 1 cos[f ct(n, 1)] v (t) = [A c cos f c t + m(t)]g p (t) = [A c cos f c t + m(t)] 1 + cos f ct + ::: = A c cos f ct + m(t) cos f ct + m(t) + ::: = A c 1+ 4 m(t) A c cos f c t A c {z } + ::: {z} = AM wave with k a = 4 A c unwanted components at frequencies removed from f c 41

12 page 7.14 Demodulation of AM Waves Square Law Demodulator No good unless m(t) A c i.e. % modulation very low see P Skip this Envelope Detector - P. 7 4

13 page P Double Sideband Suppressed Carrier Modulation (DSBSC) Carrier c(t) =A c cos f c t does not carry information! waste of power Recall Power in m(t) Power in c(t) = 1 max at 100% modulation 3! Suppress the Carrier AM: s AM (t) = [1 + m(t)]c(t) = c(t)+m(t)c(t) k a =1 DSB: s(t) = m(t)c(t) = A c cos f c tm(t) S(f) = 1 A c[m(f,f c )+M(f,f c )] Phase reversal when m(t) < 0. 43

14 page 7.16 DSBSC s(t) = c(t)m(t) = A c cos f c t m(t) = A c [exp(,jf ct) + exp(jf c t)]m(t) Recall shifting property m(t) $ M(f) exp(jf c t)m(t) $ M(f, f c ) S(f) = A c [M(f+f c)+m(f,f c )] Single tone modulation m(t) = A m cos f m t M(f) = A m ((f, f m)+(f+f m )] M(f, f c ) = A m [(f, f m, f c )+(f+f m,f c )] M(f + f c ) = A m [(f, f m + f c )+(f+f m +f c )] 44

15 page Generation of DSBSC Waves - P. 76 DSBs(t) = c(t)m(t) = A c cos f c t m(t) Balanced modulator s 1 (t) = A c [1 + k a m(t)] cos f c t s (t) = A c [1, k a m(t)] cos f c t s(t) = s 1 (t), s (t) = k a A c cos f c t m(t) = k a c(t)m(t) 45

16 page 7.19 P. 77 To produce DSB signal s(t) =m(t)c(t) Ring modulator (mixer) multiplies m(t) with c(t). For positive half-cycles of c(t) D1 and D3 conduct, D and D4 open point a connected to point b point c connected to point d current through T primary proportional to m(t). For negative half-cycles of c(t) D and D4 conduct, D1 and D3 open point a connected to point d point b connected to point c current through T primary proportional to,m(t). Use square wave c(t) toavoid problems with diode osets. 46

17 page 7.0 Ring Modulator s(t) = c(t)m(t) = 4 1X (,1) n,1 n, 1 cos[f ct(n, 1)] m(t) {z } mod Square Wave Carrier n=1 {z } Compare c(t) here with g p (t) on page 7.1. Here c(t) range,1 to +1, no DC term g p (t) range 0 to 1, with DC term. S(f) =C(f)M(f)= Z 1 C()M(f, )d,1 S(f) = C(f) z } { 4 1X n=1 (,1) n,1 n, 1 [(f, f c (n, 1)) + (f + f c (n, 1))] M(f) For n =1 S(f) = 4 [(f,f c)+(f+f c )] M(f) Z 1 =,1 4 ((, f c)+(+f c )]M(f, )d = 4 [M(f, f c)+m(f+f c )] For n =::: n=3::: 47

18 page 7.0A Superheterodyne Receiver Since f c varies from one signal to the next, the receiver is designed to convert all f c to a xed frequency f IF. This way we can use the same lter and detector for any signal at any f c. RF KHz AM Broadcast LO KHz IF=LO-RF 455 KHz Example Consider CFAX at 1070 KHz Here LO = 155 KHz SUM = 595 IF = LO = RF = 455 KHz DIF = 455 Mixer outputs sum and dierence frequencies thus another station at 1980 KHz (image frequency) will also be received since LO = 155 KHz RF = 1980 KHz RF-LO = 455 KHz Image is ltered out by RF AMP/lter Homodyne receiver uses zero frequency IF, i.e. LO=RF, IF=0 48

19 page 7.0X General I-Q receiver - demodulator applies for any signal, AM, DSB and others. Input signal s(t) = a(t)cos[f IF t + (t)] = x(t) cos f IF t, y(t) sin f IF t Output signals I = x(t) and Q = y(t). Exercise - show this using algebra 49

20 page 7.3 DSBSC Demodulation Costas Receiver same as general I-Q receiver with extra feedback signal Two coherent detectors One LO cos(f c t + ) Other LO sin(f c t + ) Output 1 A c cos m(t) 1 A c sin m(t) Here f c = f = f IF. Phase discriminator output e(t) is a DC control signal e(t) = 1 1 A c cos m(t) A csin m(t) = A c 4 m (t) cos sin = A c 8 m (t) sin The DC control signal is used as a feedback signal to reduce TP11 = cos f c tcos(f c t + ) TP19 = cos f c tsin(f c t + ) TP13 = 50

21 page 7.4 Costas receiver in more detail Costas receiver is same as general I-Q receiver with extra feedback signal e(t) = IQused to adjust oscillator frequency f to be close to f IF. In general f IF 6= f. IF input signal s IF (t)=m(t) cos(f IF t + in ). From table 3 page 66, cos sin = sin(, ) + sin( + ) cos cos = cos(, ) + cos( + ) TP11 = m(t) cos(f IF t + in ) cos(ft + `) = 1 m(t)[cos((f IF, f)t + in, `) + cos((f IF + f)t + in + `)] TP19 = m(t) cos(f IF t + in ) sin(ft + `) = 1 m(t)[sin((f IF, f)t + in, `) + sin((f IF + f)t + ` + in )] TP13 = 1 m(t) cos[(f IF, f)t + in, `) TP1 = 1 m(t) sin((f IF, f)t + in, `) 51

22 page 7.5 Costas receiver continued Dene 4f = f IF, f = `, in e(t) = (TP1) (TP13) = 1 4 m (t) cos(4ft, ) sin(4ft, ) = 1 8 m (t) sin(44ft,) Using sin cos = 1 sin e(t) =0if4f= 0 and =0 If e(t) 6= 0, then VCO will be adjusted so that e(t) =0 After LPF 5Hz e LP (t) = 1 8 sin(44ft,) <m (t)> {z } Average Value = DC Control Signal For small (44ft,)= x; sin x x. Thus e LP (t) ' 44ft, 5

23 page 7.6 P Quadrature Carrier Multiplexing Combine two independent DSBSC modulated waves on the same frequency f c with two modulations m 1 (t) m (t) To achieve this we use two carrier waves at f c but 90 o out of phase s(t) =A c m 1 (t) cos f c t + A c m (t) sin f c t To receive the two independent modulations m 1 (t) m (t) use the general I-Q receiver. The I-Q receiver contains two separate DSBSC demodulators with local oscillators (carriers) 90 o out of phase, cos f c t and sin f c t. For this to work, we need to have the local oscillators to have frequency f c exactly the same as the frequency of s(t). Note that s(t) is in the general format with x(t) =m 1 (t) and y(t) =,m (t). 53

24 page 7.8 Quadrature Carrier Multiplexing - continued We will show that there is no mixing of the two modulations, even though the carrier frequency is the same v(t) = s(t) cos f c t = A c m 1 (t) cos f c tcos f c t + A c m (t) sin f c tcos f c t cos = 1 (1, cos ) cos sin = 1 sin v(t) = 1 A cm 1 (t)[1, cos 4f c t] + 1 A cm (t) sin 4f c t After LPF v o (t) = 1 A cm 1 (t) 54

25 page 7.9 P Single Sideband Modulation In DSBSC, the two sidebands carry redundant information thus we can eliminate one sideband to get SSB-SC usually referred to as SSB. SSB with carrier is called SSB-AM We begin with frequency domain description of SSB 1. Single tone modulation m(t) = cos f m t. General modulation m(t) Looking only at positive frequencies m(t) AM DSB-SC SSB 55

26 page Generation of SSB Waves 1 - Filtering (Frequency Discrimination) - Phase Discrimination (general I-Q transmitter) 1 Filter Method Produce Modulator + Filter Need sharp lter to pass USB and reject LSB e.g 8-pole crystal lter at 10.7 MHz not practical Upconvert to desired carrier frequency Assignment #4P. 388 P. 7.4 # 16, 19a, 0a 56

27 page 7.31A General I-Q transmitter s(t) =x(t) cos f c t, y(t) sin f c t 57

28 page 7.3 text P. 88 Time Domain Description of SSB s(t) = A c [m(t) cos f ct ^m(t) sin f c t], For upper sideband + For lower sideband ^m(t) = Hilbert transform of m(t) see notes p ,90 o Phase shift for all positive frequencies +90 o Phase shift for all negative frequencies H(f) special kind of \lter" Recall H(f) =,j sgn f p h(t) = 1=t ^m(t) = m(t) h(t) = 1 ^M(f) = M(f)H(f) =,j sgn fm(f) Z 1 m(),1 t, d 58

29 page 7.33 Single tone modulation for SSB Example 3 - P. 91 If m(t) = A m cos f m t ^m(t) = A m sin f m t s(t) = A c [m(t) cos f ct ^m(t) sin f c t] = A ca m [cos f m tcos f c t +, sin f m tsin f c t] Recall cos( ) = cos cos, + sin sin, thus s(t) = A ca m cos[(f c f m )t] s(t) is a single tone at frequency f c f m. See also text gure p

30 page 7.34 SSB - Time $ Frequency Domain s(t) = A c [m(t) cos f ct ^m(t) sin f c t] ^M(f) = M(f)H(f) =,jsgn fm(f) p. 7.3 S(f) = A c [M(f) 1 f(f, f c)+(f+f c )g,j sgn fm(f) 1 j f(f,f c),(f+f c )g] = A c 4 [M(f)(f,f c) +M(f)(f+f c ) sgn fm(f)(f,f c ), + sgn fm(f)(f+f c )] = A c 4 [M(f, f c)+m(f+f c ) sgn(f, f c ) M(f, f c ), + sgn(f + f c ) M(f + f c )] = A c M(f, f c); f,f c >0;f >f c similarly S(f) = A c M(f + f c); f+f < >0;f <,f c 60

31 page 7.36 P. 93 Demodulation of SSB c LO (t) = cos f c t s(t) = A c [m(t) cos f ct, +^m(t) sin f c t] v(t) = s(t) cos f c t = A c m(t) cos f ctcos f c t cos = 1 (1 + cos ), + A c ^m(t) cos f ctsin f c t cos sin = 1 sin v(t) = A c 4 {z m(t) + A c } Message Signal 4 m(t) cos, 4f ct + A c 4 ^m(t) sin 4f ct {z } Unwanted Terms This assumes no phase or frequency error in local oscillator C LO (t) 61

32 page 7.37 Demodulation of SSB with Frequency Error C LO (t) = cos (f c + 4f)t s(t) = A c [m(t) cos f ct, +^m(t) sin f c t] v(t) = s(t) cos (f c + 4f)t = A c m(t) cos (f c + 4f)t cos f c t, + A c ^m(t) cos (f c + 4f)t sin f c t cos cos = 1 [cos(, ) + cos( + )] cos sin = 1 [sin( + ), sin(, )] = (f c + 4f)t =f c t, =4ft v(t) = A c m(t)1 cos 4ft, + A c ^m(t) sin 4ft+ Double Frequency Terms After LPF, v 0 (t) is an SSB signal on a low frequency carrier 4f. 6

33 page 7.38 Carrier Plus SSB Wave s(t) = A c cos f c t {z } Carrier + m(t) cos f c t, ^m(t) sin f c t {z } USB = (A c + m(t)] cos f c t, ^m(t) sin f c t = a(t) cos(f c t + ) q where a(t) = [A c +m(t)] +[^m(t)] a(t) = output of envelope detector q = A c +A cm(t)+m (t)+ ^m (t) a(t) = If A c jm(t)jand A c j^m(t)j q A c +A cm(t) recall (1 + x) 1= ' 1+ 1 x = A c s1+ A c m(t) = A c 1+ 1 m(t) =A c +m(t) A c Thus a(t) =m(t)+ DC bias term as long as A c jm(t)j A c j^m(t)j 63

34 page 7.39 Superheterooyne Receiver for USB To tune in # change LO to MHz SIG -LO = = 455 Khz Tunable lter to eliminate image frequency could also choose LO = MHz for Station #1. Question: Can we demodulate LSB with this receiver? 64

35 page 7.40 In practice it is not cost eective to lter at frequencies much greater than 10.7 MHz. This is because the lter skirts cannot be made steep enough. Typical Crystal Filter Spec. Typical Voice Audio signal 300 Hz Hz If 300 Hz passband centered on audio then -60 db attenuation will be at 400 Hz in opposite sideband. 65

36 page 7.50 Summary of modulation types, assuming the message jm(t)j < 1 AM s(t) = [1 + k a m(t)] cos(f c t + ) DSB s(t) = m(t) cos(f c t + ) USB s(t) = m(t) cos(f c t + ), ^m(t) sin(f c t + ) FM Z t s(t) = cos[f c t +k f m()d] = cos[f c t + (t)] = cos[ i (t)] PM s(t) = cos[f c t + k p m(t)] 0 For m(t) = A m cos f m t with A m =1 AM m(t) = [1 + k a cos f m t] cos(f c t + ) DSB s(t) = cos f m tcos(f c t + ) USB s(t) = cos f m tcos(f c t + ),sin f m tsin(f c t + ) FM s(t) = cos(f c t + k f f m sin f m t] PM s(t) = cos[f c t + k p cos f m t 66

page 7.51 Chapter 7, sections , pp Angle Modulation No Modulation (t) =2f c t + c Instantaneous Frequency 2 dt dt No Modulation

page 7.51 Chapter 7, sections 7.1-7.14, pp. 322-368 Angle Modulation s(t) =A c cos[(t)] No Modulation (t) =2f c t + c s(t) =A c cos[2f c t + c ] Instantaneous Frequency f i (t) = 1 d(t) 2 dt or w i (t)