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

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1 page 7.51 Chapter 7, sections , pp 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) = d(t) dt No Modulation f i (t) =f c Phase Modulation i (t) = 2f c t + k p m(t) s(t) = A c cos[2f c t + k p m(t)] Frequency Modulation f i (t) = f c + k f m(t) (t) = 2 f i ()d =2f c t +2k f m()d s(t) = A c cos[2f c t +2k f m()d] {z } (t) 62

2 page 7.52 Example Sketch FM PM waves for m(t) k f = 1 5 k p = 5 f c = 1 MHz FM f i (t) = f c + k f m(t) jm(t)j < j = m(t) f i min = 1 8, 1 5 =99:9 MHz when m(t) =,1 f i max = = 1:1 MHz when m(t) =+1 PM = 2f c t +2k p m(t) PM for m(t) = FM for dm dt f i = 1 d 2 dt = f c + k p _m(t) = _m(t)=1 8 +5( 2 1,4 f i min = (1 8, 5( _m(t) max )=1 8,1 5 =99:9MHz 63

3 page 7.53 Example FM + PM k f =1 8 k p ==2 f c = 1 MHz FM s(t) = cos((t)) (t) =2 f i ()d f i =f c +k f m(t)= m(t) ( 1:1MHz m(t) =+1 99:9MHz m(t) =,1 PM i (t) =2f c t + k p m(t) f i = f c + k p 2 _m(t) = 1 d 2dt = _m(t) Innite frequency change and back again in zero time Instead write for PM s(t) = cos[2f c t + k p m(t)] = cos[2f c t + ( 2 m(t)] sin 2fc t m(t) =,1 =,sin 2f c t m(t) =1 64

4 page 7.54 Single Tone Frequency Modulation m(t) = A m cos 2f m t FM s(t) = A c cos[(t)] f i (t) = f c + k f m(t) =f c +k f A m cos 2f m t = f c + 4f {z} cos(2f mt) Frequency Deviation Frequency deviates from center by up to 4f f max = f c + 4f f min = f c,4f (t) = 2 2f i (t) = d(t) dt f i ()d (t) = 2f c t + 4f f m sin 2f m t = 2f c t + {z} sin 2f mt modulation index, phase deviation of (t) from 2f c t Single tone FM s(t) = A c cos[2f c t + sin 2f m t] = A c cos[(t)] 65

5 page 7.55 Example Find Power Maximum Frequency Deviation 4f Maximum Phase Deviation 4 Modulation Index for A c cos(2f c t + sin 2f m t) s(t) = A c cos(2f c t + sin 2f m t) = 1 cos(21 6 t +:1 sin 21 3 t) Power = A2 c 2 =5 4f f m =:1 f c = 1 MHz f m = 1 KHz A c =1 s(t) = 1 cos[(t)] = A c cos[2f c t + 4f f m sin 2f m t] f i = 1 2 d dt =16 +1 {z} 2 4f cos 21 3 t 4f = 1 Hz alternative: 4f fm =:1 f m =1 3 4f=1 2 PM s(t) = A c cos(2f c t + k p m(t)) = 1 cos(21 6 t +:1 sin 21 3 t)! 4 = :1 Radians = 4f f m = 1 1 =:1 s(t) = A c cos[2f c t + sin 2f m t] " Modulation Index 66

6 Numerical value of modulation index = Maximum phase deviation for PM only 67

7 page 7.56 Spectrum Analysis of FM - p. 328 Recall FM wave with single tone modulation s(t) = A c cos[2f c t + sin(2f m (t)] = 4f f m We will show (p ) s(t) = A c 1X n=,1 J n () cos 2(f c + nf m )t] S(f) = A c 2 1X n=,1 where J n () = 1 Z exp[j( sin x, nx)]dx 2, J n ()[(f, f c, nf m )+(f+f c +nf m )] FM Spectrum n = : carrier plus innite number of sidebands at f c nf m see notes 7.59 AM spectrum - only one pair of sidebands at f c f m Narrowband FM = 4f 1! J () 1 J n () f m J 1 () =2 n 2 Again only one pair of sidebands for n =1atf c f m 68

8 page 7.57 Wideband FM s(t) = A c cos(2f c t + sin(2f m t)] e j sin wmt = = A c cos[w c t + sin w m t] = Re fa c e jwct e j sin wmt g 1X n=,1 Z T=2 F n e jnwmt F n = 1 T,T=2 2 Let = w m t = t T e j sin wmt e,jnwmt dt T =1=f m F n = 1 Z e j( sin,n) d = J n () 2, Evaluate this integral numerically in terms of parameters n and e j sin wmt = 1X n=,1 J n ()e jnwmt s(t) = Re fa c e jwct 1X = A c 1X n=,1 = A c 1X n=,1 n=,1 J n ()e jnwmt g J n () cos[w c + nw m ]t J n () cos 2(f c + nf m )t p. 33 eqn. (7.118) 69

9 page 7.6 Properties of FM s(t) = A c 1X S(f) = A c 2 n=,1 1X n=,1 J n () cos[2(f c + nf m )t] J n ()[(f, f c, nf m )+(f+f c +nf m )] Narrowband FM J () 1 J 1 () 2 J n () ' ; n >1 Thus for <:3 s(t)=a c cos 2f c t + A c 2 cos[2(f c + f m )t], A c 2 cos[2(f c, f m )t] 7

10 page 7.61 Narrowband FM (NBFM) e.g. f m = 1 Hz, 4f = 1 Hz For small modulation index, NBFM is similar to AM, with only one pair of sidebands. = 4f fm 1 s(t) =A ccos[(t)] m(t) =A m cos 2f m t s(t) =A c cos(2f c t +2k f m()d] FM wave s(t) =A c [cos 2f c t + sin 2f m t] from p. 73 = A c cos 2f c tcos[ sin 2f m t],a c sin 2f c tsin[ sin 2f m t] NBFM 1! cos( sin 2f m t) 1 sin( sin 2f m t) sin 2f m t! s(t) ' A c cos 2f c t, A c sin 2f c tsin 2f m t = A c cos 2f c t A c[cos 2(f c + f m )t, cos 2(f c, f m )t] Recall AM s(t) = A c cos 2f c t A c[cos 2(f c + f m )t + cos 2(f c, f m )t] = A c [1 + cos 2f m t] cos 2f c t = A c [1 + m(t)] cos 2f c t Thus NBFM is similar to, but not the same as AM 71

11 page 7.62 Transmission Bandwidth of FM Waves - p. 335 In theory FM has innite bandwidth since there are an innite # side frequencies Eective Bandwidth For single tone FM Large Small Bandwidth 24f Bandwidth 2f m Carsons Rule for bandwidth B T B T = 24f +2fm =24f(1 + 1=) Other Denition - 98% BW (1% on each side) Bandwidth beyond which no side frequencies exceed 1% of unmodulated carrier amplitude bandwidth = 2n max f m ; n max such that jj n ()j < :1 Deviation Ratio = for max f m in m(t) Use in Carsons Rule Example: FM 2-way Radio Deviation 5 KHz = 4f Maxf m 3 KHz = f m B T =2(4f+f m )=2(5+3)=16 Hence Notation 16F3 thus channel spacing is 15-3 KHz 72

12 page 7.64 Example problem Estimate bandwidth of s(t) = A c cos(2f c t + sin 2f m t) = 1 cos(21 6 t +:1 sin 21 3 t) Since =:11 Narrowband Case Bandwidth 2(4f + f m ) = 2 (1 + 1) = 22 Hz Read example 9 p. 338 Example Problem PM Spectrum Single Tone m(t) = A m cos(2f m t) PM s(t) = A c cos(2f c t + k p m(t)] Find spectrum if p = A m k p < :3 Solution write as sum of cos and sin s(t) =A c cos(2f c t + p cos 2f m t] p = A m k p phase deviation = A c cos(2f c t) cos( p cos 2f m t), A c sin(2f c t) sin( p cos 2f m t) If p < :3 then cos[ p cos 2f m t] ' 1 and sin[ p cos 2f m t] ' p cos 2f m t 73

13 page 7.65 s(t) = A c cos 2f c t, p A c sin 2f c tcos 2f m t = A c cos 2f c t, 1 2 pa c sin[2(f c + f m )t], 1 2 pa c sin[2(f c, f m )t] S(f) = 1 2 A c[(f, f c )+(f+f c )], 1 4j pa c [(f, f c, f m ), (f + f c + f m )], 1 4j pa c [(f, f c + f m ), (f + f c, f m )] S(f) has real and imaginary parts example problem 4.7 Choose arbitrary p single tone PM at carrier f c, mod tone f m Apply s(t) to ideal bandpass lter H(f) 74

14 page 7.66 Generation of FM Waves Indirect - Make narrowband FM, multiply to set WBFM Direct - Modulate carrier directly to get WBFM s 1 (t) = A 1 cos[2f 1 t + 1 sin 2f m t] s(t) = A c cos[2f c t + sin 2f m t] = n 1 1 f c = nf 1 read example 1 p. 342 Direct FM Use varactor or voltage variable capacitor in carrier oscillator f i (t) = 2 q 1 LC(t) c(t) =C o +4Ccos 2f m t 75

15 page 7.68 Problem Example FM Wave Deviation 1 KHz - Mod freq. 5 KHz = 4f f m = 1 5 =2 input s(t) atf i1 (t), multiply 6x to get f i2 (t) Find 1 Deviation and modulation index at output 2 Frequency separation of adjacent side frequencies Solution n = 6 f i1 (t) = f c + 4f cos(2f m t) f i2 (t) = nf c + n4f cos(2f m t) Deviation n4f = 6 1 KHz = 6 KHz Modulation index = n4f f m = 6 5 =12 Frequency separation of adjacent side frequencies unchanged at 5 KHz 76

16 page 7.69 Example Problem 34 page 394 NBFM hints for solution Consider a narrowband FM wave. part a. Find the envelope. s(t) = A c cos 2f c t, A c sin 2f c tsin 2f m t = a(t) cos(2f c t + (t)) where a(t) = q A 2 c + 2 A 2 c sin 2 2f m t Plot a(t) vstto nd ratio of max to min part b. Write s(t) in terms of frequency components f c ;f c,f m ;f c +f m part c. nd an expression for (t) and expand it 77

17 page 7.7 Demodulation of FM Waves - p Discriminator (approximation of ideal dierentiator) 2 Phase locked loop FM Demod Output voltage proportional to input frequency Discriminator In practice two resonant circuits one above f c and one below f c 78

18 page 7.7A Frequency Discriminator Ideal dierentiator FM s(t) = A c cos[2f c t +2k f m()d ds dt =,A c [2f c +2k f m(t)] sin[2f c t +2k f m()d] If k f m(t) f c " # ds dt =,2A cf c 1+ k f m(t) f c :sin[2f c t +2k f m()d)] ds dt looks like an AM signal [1 + m(t)]c(t) Thus it can be detected using envelope detector Ideal dierentiator transfer function H(f) = j2f looks like slope detector 79

19 page 7.71 Skip , Instead Consider: Demodulation of FM - using hard limiter and BPF before the dierentiator and envelope detector s(t) = A c cos[2f c t +2k f m()d] ds dt = A c [2f +2k f m(t) ] sin[2f c t +2k f m()d] {z } envelope Dierentiator works correctly only if there are no amplitude variations in s(t) i.e. A c = constant. if A c = A(t) (time-varying), then envelope of ds dt will include term A(t). To remove amplitude variations use hard limiter and BPF before the dierentiator v i (t) = A(t) cos[(t)] A(t) envelope v (t) = (t) =2f c t +2 ( +1 A(t) cos(i (t)] >,1 A(t) cos( i (t)] < ) = v o () m()d Note:v o () = 8

20 We can plot v as function of instead of as function of t. what follows is proof that limiter output is the desired FM signal with constant amplitude, even if the limiter input contains amplitude variations A(t). v () = 4 [cos, 1 3 cos cos 5 + :::] 5 = 4 1X n=1 (,1) n,1 cos[(2n, 1)] 2n, 1 series for square wave, see Text p. 363, notes page 7.13,

21 page 7.72 v i [(t)] = 4 fcos (t)], 1 cos[3(t)] + :::g 3 = 4 fcos[2f ct +2k f m()d] cos[6f ct +6k f m()d]+:::g Bandpass lter at f c to eliminate higher order terms Thus s(t) = 4 cos[2f ct +2k f m()d] = 4 cos((t)] o which is the desired FM wave with constant amplitude 4=pi. Practical Frequency Demodulators, Slope Detection 82

22 page 7.76 Page FM Stereo Pilot Carrier 19 KHz =f c Radio Carrier 15.1 MHz = f n m(t) = [`(t), r(t)] cos 4f c t DSBSC with f c = 19 KHz + [`(t)+r(t)] + cos 2f c t a) Spectrum b) If deviation = 75 KHz Bandwidth = 2(4f + f m ) = 2 ( ) = 256 KHz Hence FM stations spaced every.2 MHz = 2 KHz c) Receiver Block Diagram 83

23 page 7.77 LAB 4 Text pp Phase-Locked Loop Demodulator Feedback system with components 1. Multiplier (phase comparator) 2. Loop Filter H(f) 3. VCO (Voltage Controlled Oscillator) Notation Dierent in Lab & Text Lab Text Input Phase i (t) i (t) s(t) cos 2f c t + i (t) sin[2f c t + 1 (t)] Feedback (Loop) Phase f (t) 2 (t) r(t) sin 2f c t + f (t) cos[2f c t + 2 (t)] e(t) k c [ i (t), f (t)] sin[ 1 (t), 2 (t)] v o (t) e(t) h(t) f f (t) VCO Output Freq. Feedback Phase f (t) =kf kfv o (t) v o (t)dt 2 (t) =2k o v()d 84

24 page 7.78 If loop gain in h(t) is high, then i, f is small and i (t) f(t) v o (t) = 1 d f v(t)= 1 d 1 k f dt 2k v dt Lab (5.9) Text (7.177) Output voltage proportional to ( input frequency or deprivative of input phase Thus have FM demodulation To prove e(t) =k e [ i (t), f (t)] (ignoring constants k c k f etc) e(t) = s(t)r(t) = cos[2f c t + i (t)] sin[2f c t + f (t)] sin cos = sin(, ) + sin( + ) e(t) = sin[ i (t), f (t)] + sin(4f c t + i (t)+ f (t)] {z } Filter out with h(t) To show v o (t) d f df VCO output frequency f f (t) =k f v o (t) thus VCO output phase f (t) =kf o v o (t)dt For precise proof, see text p to show that v o (t) d f dt (LAB notation) or v(t) d 1 dt (TEXT notation) 85

25 page 7.79 Example Problem 46 p. 398 s(t) PMwaveinto PLL output v(t) into H(f) output m(t). Assuming large loop gain in PLL nd H(f) such that the message signal is reproduced Solution: Write out signals at each point in the system s(t) =Ac cos(2f c t + 1 (t) z } { 2k f m(t)] PLL output v(+) = = 1 2k v 1 2k v d 1 (t) dt d dt 2k fm(t) = k f k v dm(t) dt To reproduce m(t) need to integrate v(t) recall that the transfer function of ideal integrator is H(f) = 1 j2f (f 6= ) since,1 g()d $ G(f) j2f = G(f)H(f) 86

26 page 7.8 Example Problem 42 page 397 s(t) =A c cos[2f c t +2k f m()d] o Assume R X c R L R so that envelope detector does not load lter Find v 2 (t). Does this circuit work as an FM demodulator? Solution: Filter H(f) = j2f cr 1+j2f c R R X c! R 1 j2f c! j2f c R 1! H(f) ' j2f c R V 1 (f) = S(f)H(f) =CRj2fS(f) v 1 (t) = CR ds dt from properties of Fourier transform Thus v 1 (t) is proportional to ds=dt which is proportional to m(t) 87

Angle Modulated Systems

Angle Modulated Systems Angle of carrier signal is changed in accordance with instantaneous amplitude of modulating signal. Two types Frequency Modulation (FM) Phase Modulation (PM) Use Commercial radio