Modulation

page 7.51

Chapter 7, sections 7.10 - 7.14, pp. 322-368 Angle Modulation s(t) = Ac cos[(t)] No Modulation (t) = 2fct + c

s(t) = Ac cos[2fct + c]

Instantaneous Frequency fi(t) = 21 ddt(t) or wi(t) = ddt(t) No Modulation fi(t) = fc

Phase Modulation i(t) = 2fct + kpm(t) s(t) = Ac cos[2fct + kpm(t)]

Frequency Modulation fi(t) = fc + kf m(t) (t) = 2

Zt 0

fi()d = 2fct + 2kf

s(t) = Ac cos[2fct + 2kf |

{z

Zt

(t)

0

Zt 0

m()d] }

62

m()d

page 7.52 Example Sketch FM PM waves for m(t)

kf = 105 kp = 5 fc = 100 MHz

FM

jm(t)j < j

fi (t) = fc + kf m(t) = 108 + 105m(t)

fi min = 108 , 105 = 99:9 MHz

when m(t) = ,1

fi max = 108 + 105 = 100:1 MHz when m(t) = +1 PM

 = 2fct + 2kpm(t) PM for m(t) = FM for dm dt fi = 21 d dt = fc + kpm_ (t)

= 108 + 5m_ (t) = 108 + 5( 102,4

fi min = (108 , 5(m_ (t)max) = 108 , 105 = 99:9MHz 63

page 7.53 Example FM + PM

kf = 108 kp = =2 fc = 100 MHz

FM s(t) = cos(Z(t)) t (t) = 2 fi()d 0 fi = fc + kf m(t) = 108 + 105m(t) (

PM i(t) = 2fct + kpm(t) 8 + 1m = 10 fi = fc + 2kp m_ (t) = 21 d dt 4 _ (t) In nite frequency change and back again in zero time Instead write for PM s(t) = cos[2fct + kpm(t)] = cos[2fct + 2 m(t)] ( 2fct m(t) = ,1 = sin , sin 2fct m(t) = 1

100:1MHz m(t) = +1 99:9MHz m(t) = ,1

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page 7.54

Single Tone Frequency Modulation m(t) = Am cos 2fmt FM s(t) = Ac cos[(t)] fi(t) = fc + kf m(t) = fc + kf Am cos 2fmt = fc + 4 |{z}f cos(2fm t) Frequency Deviation Frequency deviates from center by up to 4f fmax = fc + 4f fmin = fc , 4f

(t) = 2

Zt 0

fi()d

2fi(t) = ddt(t)

f sin 2f t (t) = 2fc t + 4 m f m

= 2fc t + |{z} sin 2fmt modulation index, phase deviation of (t) from 2fct Single tone FM

s(t) = Ac cos[2fct + sin 2fmt] = Ac cos[(t)]

65

page 7.55

Example Find Power Maximum Frequency Deviation 4f for A cos(2f t + sin 2f t) c c m Maximum Phase Deviation 4 Modulation Index

s(t) = Ac cos(2fct + sin 2fmt) = 10 cos(2106t + 0:1 sin 2103t)

fc = 1 MHz fm = 1 KHz fm Ac = 10 f sin 2f t] s(t) = 10 cos[(t)] = Ac cos[2fct + 4 m fm

2 Power = A2c = 50

4f = 0:1

6 102 cos 2 103 t fi = 21 d dt = 10 + |{z} 4f

4f = 100 Hz

alternative:

PM s(t) = Ac cos(2fct + kpm(t)) = 10 cos(2106t + 0:1 sin 2103t) ! 4 = 0:1 Radians f = 100 = 0:1 = 4 fm 1000

s(t) = Ac cos[2fct + sin 2fm t]

" Modulation Index 66

4f = 0:1 fm fm = 103

4f = 102

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

67

page 7.56 Spectrum Analysis of FM - p. 328 Recall FM wave with single tone modulation

s(t) = Ac cos[2fct + sin(2fm(t)] f = 4 fm We will show (p. 328 - 330)

s(t) = Ac

1 X n=,1

Jn( ) cos 2(fc + nfm )t]

1 A c X Jn( )[(f , fc , nfm) + (f + fc + nfm)] S (f ) = 2 n=,1 Z where Jn( ) = 21 exp[j ( sin x , nx)]dx ,

FM Spectrum

n = 0: carrier plus in nite number of sidebands at fc  nfm see notes 7.59 AM spectrum - only one pair of sidebands at fc  fm Narrowband FM

f  1 ! J0( )  1 Jn ( )  0 =4 J1( )  =2 n  2 fm Again only one pair of sidebands for n = 1 at fc  fm

68

page 7.57 Wideband FM

s(t) = Ac cos(2fct + sin(2fmt)] = Ac cos[wct + sin wm t] = Re fAcejwctej sin wmtg

ej sin wm t =

1 X n=,1

Fnejnwmt

Z T=2 ej sin wm te,jnwm tdt Fn = T1 ,T=2  2  Let  = wmt = T t T = 1=fm Z 1 Fn = 2 ej( sin ,n) d = Jn ( ) ,

Evaluate this integral numerically in terms of parameters n and

ej sin wm t =

1 X n=,1

Jn( )ejnwmt

s(t) = Re fAcejwct = Ac = Ac

1 X n=,1

1 X n=,1

1 X n=,1

Jn ( )ejnwmtg

Jn( ) cos[wc + nwm]t Jn( ) cos 2(fc + nfm)t

p. 330 eqn. (7.118)

69

page 7.60 Properties of FM

s(t) = Ac

1 X n=,1

Jn ( ) cos[2(fc + nfm )t]

1 X Jn ( )[(f , fc , nfm ) + (f + fc + nfm )] S (f ) = A2c n=,1

Narrowband FM

J0( )  1

J1( )  2

Jn( ) ' 0; n > 1

Thus for < 0:3

Ac cos[2(f , f )t] c s(t) = Ac cos 2fct + A cos[2  ( f c + fm )t] , c m 2 2

70

page 7.61 e.g. fm = 1000 Hz, 4f = 100 Hz

Narrowband FM (NBFM)

For small modulation index, NBFM is similar to AM, with only one pair of sidebands.

= 4fmf  1 s(t) = Ac cos[(t)] m(t) = Am cos 2fm t s(t) = Ac cos(2fct + 2kf

Zt 0

m()d]

FM wave s(t) = Ac[cos 2fct + sin 2fmt] from p. 73 = Ac cos 2fct cos[ sin 2fm t] ,Ac sin 2fct sin[ sin 2fm t] NBFM  1 ! cos( sin 2fmt)  1 sin( sin 2fm t)  sin 2fm t ! s(t) ' Ac cos 2fct , Ac sin 2fct sin 2fm t = Ac cos 2fct + 21 Ac[cos 2(fc + fm)t , cos 2(fc , fm)t] Recall AM

s(t) = Ac cos 2fct + 12 Ac [cos 2(fc + fm )t + cos 2(fc , fm)t] = Ac[1 +  cos 2fmt] cos 2fct = Ac[1 + m(t)] cos 2fc t

Thus NBFM is similar to, but not the same as AM

71

page 7.62

Transmission Bandwidth of FM Waves - p. 335 In theory FM has in nite bandwidth since there are an in nite # side frequencies Eective Bandwidth For single tone FM Large Small

Bandwidth  24f Bandwidth  2fm

Carsons Rule for bandwidth BT

BT  = 24f + 2fm = 24f (1 + 1= ) Other De nition - 98% BW (1% on each side) Bandwidth beyond which no side frequencies exceed 1% of unmodulated carrier amplitude bandwidth = 2nmaxfm ; nmax such that jJn ( )j < 0:01 Deviation Ratio = for max fm in m(t) Use in Carsons Rule Example: FM 2-way Radio Deviation  5 KHz = 4f Maxfm  3 KHz = fm

BT = 2(4f + fm) = 2(5 + 3) = 16 Hence Notation 16F3 thus channel spacing is 15-30 KHz

72

page 7.64 Example problem Estimate bandwidth of

s(t) = Ac cos(2fct + sin 2fmt) = 10 cos(2106t + 0:1 sin 2103t) Since = 0:1  1 Narrowband Case Bandwidth 2(4f + fm ) = 2 (100 + 1000) = 2200 Hz Read example 9 p. 338 Example Problem 4.6 - PM Spectrum Single Tone

m(t) = Am cos(2fmt) PM s(t) = Ac cos(2fct + kpm(t)] Find spectrum if p = Amkp < 0:3 Solution write as sum of cos and sin

s(t) = Ac cos(2fct + p cos 2fmt]

p = Amkp phase deviation

= Ac cos(2fct) cos( p cos 2fmt) , Ac sin(2fct) sin( p cos 2fmt) If p < 0:3 then cos[ p cos 2fmt] ' 1 and sin[ p cos 2fm t] ' p cos 2fmt

73

page 7.65

s(t) = Ac cos 2fct , pAc sin 2fc t cos 2fmt = Ac cos 2fct , 21 pAc sin[2(fc + fm )t] , 21 pAc sin[2(fc , fm )t] S (f ) = 12 Ac[(f , fc) + (f + fc)] , 41j pAc[(f , fc , fm) , (f + fc + fm )] , 41j pAc[(f , fc + fm) , (f + fc , fm )]

S (f ) has real and imaginary parts example problem 4.7 Choose arbitrary p single tone PM at carrier fc, mod tone fm Apply s(t) to ideal bandpass lter H (f )

74

page 7.66 Generation of FM Waves Indirect - Make narrowband FM, multiply to set WBFM Direct - Modulate carrier directly to get WBFM

s1(t) = A1 cos[2f1t + 1 sin 2fmt] s(t) = Ac cos[2fct + sin 2fmt] = n1 1 fc = nf1 read example 10 p. 342

Direct FM Use varactor or voltage variable capacitor in carrier oscillator

fi(t) =

q1

2 LC (t)

c(t) = Co + 4C cos 2fmt

75

page 7.68

Problem Example f = 10 = 2 FM Wave Deviation 10 KHz - Mod freq. 5 KHz = 4 fm 5 input s(t) at fi1(t), multiply 6x to get fi2(t)

Find 1 Deviation and modulation index at output 2 Frequency separation of adjacent side frequencies Solution

n = 6 fi1(t) = fc + 4f cos(2fmt) fi2(t) = nfc + n4f cos(2fmt) Deviation n4f = 6  10 KHz = 60 KHz Modulation index = nf4f = 605 = 12 m Frequency separation of adjacent side frequencies unchanged at 5 KHz

76

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(t)

= Ac cos 2fct , Ac sin 2fct sin 2fm t = a(t) cos(2fc t + (t)) where =

q

A2c + 2A2c sin2 2fmt

Plot a(t) vs t to nd ratio of max to min part b. Write s(t) in terms of frequency components fc; fc , fm ; fc + fm part c. nd an expression for (t) and expand it

77

page 7.70

Demodulation of FM Waves - p. 346 1 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 fc and one below fc

78

page 7.70A

Frequency Discriminator Ideal dierentiator FM s(t) = Ac cos[2fct + 2kf

Zt

m()d

ds = ,A [2f + 2k m(t)]
sin[2f t + 2k Z t m()d] c c f c f dt 0 If kf m(t)  fc "

#

ds = ,2A f 1 + kf m(t) : sin[2f t + 2k Z t m()d)] c c c f dt fc 0 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 ) = j 2f looks like slope detector

79

page 7.71 Skip 347-350, Instead Consider:

Demodulation of FM using hard limiter and BPF before the dierentiator and envelope detector

s(t) = Ac cos[2fct + 2kf

Zt 0

m()d]

ds = A [2f + 2k m(t)] sin[2f t + 2k Z t m()d] c| c f {z f } dt 0 envelope Dierentiator works correctly only if there are no amplitude variations in s(t) i.e. Ac = constant. if Ac = 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

vi(t) = A(t) cos[(t)]

A(t)  0 envelope (t) = 2fct + 2

v0(t) =

(

)

Zt 0

m()d

+1 A(t) cos(i(t)] > 0 = v () o ,1 A(t) cos(i(t)] < 0

Note:vo() =

80

We can plot v0 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).

v0() = 4 [cos  , 13 cos 3 + 51 cos 5 + : : :] 1 n,1 X = 4 (2,n1), 1 cos[(2n , 1)] n=1 series for square wave, see Text p. 363, notes page 7.13, 7.20.

81

page 7.72

vi[(t)] = 4 fcos (t)] , 13 cos[3(t)] + : : :g Zt = 4 fcos[2fct + 2kf m()d] 0 Zt 1 + 3 cos[6fct + 6kf m()d] + : : :g 0 Bandpass lter at fc to eliminate higher order terms Thus Zt s(t) = 4 cos[2fct + 2kf o m()d] = 4 cos((t)]

which is the desired FM wave with constant amplitude 4=pi.

Practical Frequency Demodulators, Slope Detection

82

page 7.76

Page 364-8 FM Stereo Pilot Carrier 19 KHz =fc Radio Carrier 105.1 MHz = fn

m(t) = [`(t) , r(t)] cos 4fct + [`(t) + r(t)] + cos 2fct

DSBSC with fc = 19 KHz

a) Spectrum

b) If deviation = 75 KHz Bandwidth = 2(4f + fm ) = 2 (75 + 53) = 256 KHz Hence FM stations spaced every 0.2 MHz = 200 KHz c) Receiver Block Diagram

83

page 7.77

LAB 4 Text 7.12 - pp. 353-361 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 Input Phase

s(t) Feedback (Loop) Phase

r(t) e(t) vo(t) VCO ff (t) Output Freq. Feedback Phase

Lab

Text

i(t)

i(t)

cos 2fct + i(t)

sin[2fct + 1(t)]

f (t)

2(t)

sin 2fct + f (t) kc[i(t) , f (t)] e(t) h(t)

cos[2fct + 2(t)] sin[1(t) , 2(t)]

kfvo (t) f (t) = kf

Zt 0

vo(t)dt 2(t) = 2ko

84

Zt 0

v( )d

page 7.78 If loop gain in h(t) is high, then i , f is small and i(t)  f (t)

vo(t) = k1 ddtf f Lab (5.9)

1 d1 v(t) = 2k v dt Text (7.177)

Output voltage proportional to

(

input frequency or deprivative of input phase

Thus have FM demodulation To prove e(t) = ke [i(t) , f (t)] (ignoring constants kc kf etc)

e(t) = = sin  cos = e(t) =

s(t)r(t) cos[2fct + i(t)] sin[2fct + f (t)] sin( , ) + sin( + ) sin[i(t) , f (t)] + sin(4 fct +{zi(t) + f (t)]} | Filter out with h(t)

f To show vo(t)  d df

VCO output frequency ff (t) = kf vo(t) thus VCO output phase f (t) = kf

Zt o

vo(t)dt

For precise proof, see text p. 355-357 to show that 1 vo(t)  ddtf (LAB notation) or v(t)  d dt (TEXT notation)

85

page 7.79

Example Problem 46 p. 398 s(t) PM wave into 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 1 (t) z }| {

s(t) = Ac cos(2fct + 2kf m(t)] 1 d1(t) 2kv dt 1 d 2k m(t) = 2k f v dt = kkf dmdt(t) v

PLL output v(+) =

To reproduce m(t) need to integrate v(t) recall that the transfer function of ideal integrator is H (f ) = j 21f (f 6= 0) since Zt ,1

(f ) = G(f )H (f ) g( )d $ G j 2f

86

page 7.80

Example Problem 42 page 397

s(t) = Ac cos[2fct + 2kf Assume

Zt o

m( )d ]

R  Xc RL  R so that envelope detector does not load lter

Find v2(t). Does this circuit work as an FM demodulator? Solution: cR Filter H (f ) = 1 +j 2jf 2f R c

1 R  Xc ! R  j 2f

c

! j 2fc R  1 ! H (f ) ' j 2fc R

V1(f ) = S (f )H (f ) = CRj 2fS (f ) v1(t) = CR ds dt from properties of Fourier transform Thus v1(t) is proportional to ds=dt which is proportional to m(t)

87