Analog & Digital Communications

By
KASULA RAGHU
Assistant Professor
Dept. of E.C.E.,MGIT

08/05/2021 KASULA RAGHU 1

 UNIT- II

Angle Modulation
 Contents

Basic concepts of Phase Modulation, Frequency Modulation: Single tone frequency

modulation, Spectrum Analysis of Sinusoidal FM Wave using Bessel functions,
Narrow band FM, Wide band FM, Constant Average Power, Transmission bandwidth
of FM Wave - Generation of FM Signal- Armstrong Method, Detection of FM Signal:
Balanced slope detector, Phase locked loop, Comparison of FM and AM., Concept of
Pre-emphasis and de-emphasis.
 Basic Definitions

• Let ϴi(t) Denotes the angle of a modulated sinusoidal carrier, which is a

function of the message. we express the resulting angle-modulated wave as

s(t) = Ac Cos[ϴi(t)] ………….1

Where Ac is the carrier amplitude.

A complete oscillation occurs whenever ϴi(t) changes by 2П radians. If

ϴi(t) increases monotonically with time, the Average frequency in Hz.
Over an interval from ‘t’ to ‘t+Δt’, is given by
 fΔt(t)= ϴi(t+Δt)
2ПΔt
− ϴi(t)

We can now define instantaneous frequency of the angle modulated wave s(t) by

f i(t)= 𝐿𝑖𝑚 f Δt (t)

Δt 0

ϴi(t+Δt) − ϴi(t)
f i(t)= 𝐿𝑖𝑚 2ПΔt
Δt 0

1 𝑑ϴi(t)
fi(t) = 2П 𝑑𝑡

According to s(t) = Ac Cos[ϴi(t)] Angle Modulated wave s(t) is a rotating phasor of length
Ac and Angle ϴi(t). The Angular velocity of such a phasor is 𝑑ϴi(t)/dt, In the simple case of
an unmodulated carrier. The angle ϴi(t) is
 ϴi(t) = 2 Пfc t + φc
And the corresponding phasor rotates with a constant angular velocity
equal to 2 Пf c . The constant φc is the value of ϴi(t) at t=0

We can have Two kinds of Angle Modulations

1) Frequency Modulation
2) Phase Modulation
 Phase Modulation
• Phase Modulation (PM) is that form of angle modulation in which the
angle ϴi(t) is varied Linearly with the baseband signal m(t),as shown
by
ϴi(t) = 2Пfc t + Kpm(t)
The term 2Пfc t represents Unmodulated Carrier, Kp represents Phase
sensitivity of the modulator expressed in Radians/Volt

s(t) = Ac Cos[2Пfc t + Kpm(t)] …… Equation of Phase Modulation.

 Frequency Modulation
• Frequency Modulation (FM) is that form of angle modulation in which the instantaneous
frequency f i(t) is varied linearly with the baseband signal m(t), as shown by

f i(t) = fc + Kf m(t)

The term fc represents the frequency of unmodulated Carrier and the constant Kf
Represents the frequency sensitivity of the modulator, expressed in Hertz/Volt

We Know that

1 𝑑ϴi(t)
fi(t) = 2П 𝑑𝑡
 𝑡
• ϴi(t) = 2 Пfc t + 2 П Kf ‫׬‬0 𝑚 𝑡 𝑑𝑡

The Frequency Modulated wave therefore described by

𝐭
s(t) = Ac Cos[2 Пfc t + 2 П Kf ‫𝐦 𝟎׬‬ 𝐭 𝐝𝐭] ……Equation of Frequency Modulation.
• s(t) = Ac Cos[2Пfc t + Kpm(t)] Equation of Phase Modulation.
𝐭
• s(t) = Ac Cos[2 Пfc t + 2 П Kf ‫𝐦 𝟎׬‬ 𝐭 𝐝𝐭] Equation of Frequency Modulation.
FM wave
FM vs PM Wave
Single Tone Angle Modulation (PM)
Let a message signal m(t ) = Am cos 2 f mt
Single -Tone modulation of Frequency Modulation

𝐭
s(t) = Ac Cos[2 Пfc t + 2 П Kf ‫]𝐭𝐝 𝐭 𝐦 𝟎׬‬ 𝑤ℎ𝑒𝑟𝑒 𝑚(𝑡) = 𝐴𝑚 cos 2 𝜋𝑓𝑚 𝑡

s(t) = Ac Cos[ϴi(t)]

i
 i

SFM(t) = Ac cos[2Пfc t + β f sin(2Пƒmt)]

β is Modulation Index

 When it is Small or less than 1 Radian then it is called as Narrow Band FM

 When it is Greater or More than 1 Radian then it is called as Wide Band FM
Brief Summary
 SFM(t) = Ac cos[2Пfc t + β f sin(2Пƒmt)]

sPM (t ) = Ac cos ct  k p m(t )

 t
sFM (t ) = Ac cos ct  2 k f  m( ) d 
  
Narrowband F.M
Narrowband F.M.  f   / 2 (  f  0.2)

 Ac  Ac
s AM (t ) = Ac cos ct  cos(c  m )t  cos(c  m )t
2 2
 Frequency spectrum of NBFM
The time domain expression for NBFM is

The frequency spectrum of Narrow Band Frequency

Modulation is represented by

Ac Ac  f
S FM ( f ) = [ ( f  f c )   ( f  f c )  [ ( f  f c  f m )   ( f  f c  f m )
2 4
Ac  f
 [ ( f  fc  f m )   ( f  f c  f m )
4
Spectrum of AM and NBFM
 Bandwidth of NBFM

The time domain expression for NBFM is

It resembles the AM signal and hence bandwidth is twice of the message

signal frequency

BW= 2fm
Power of NBFM
•Pt = Pc + Power of USB + Power of LSB (as in AM)

𝐴𝑐 2 𝛽 2 𝐴𝑐 2 𝛽 2 𝐴𝑐 2
•Pt = + +
2 8 8
𝐴𝑐 2 𝛽 2 𝐴𝑐 2
•Pt = +
2 4
𝑃𝑐𝛽 2
•Pt = 𝑃𝑐+
2

𝛽2
•𝑃𝑡 = 𝑃𝑐(1 + )
2
 sPM (t ) = Ac cos ct  k p m(t )

sFM (t ) = Ac cos ct  2 k f  m( ) d 

t
  
 NBFM for Base Band Signal
t
sFM (t ) = Ac cos[ct  k f  m(t ) dt ]
 t
= Ac cos c t  cos k f  m(t ) dt
 
t
For | k f  m(t ) dt | 1
 t
 Ac sin c t  sin k f  m(t ) dt
 
𝑡 𝑡 𝑡
cos 𝑘𝑓 න 𝑚(𝑡) 𝑑𝑡 ≃ 1 and sin 𝑘𝑓 න 𝑚(𝑡) 𝑑𝑡 ≃ 𝑘𝑓 න 𝑚(𝑡) 𝑑𝑡
−∞ −∞ −∞

𝑡
𝑆𝑁𝐵𝐹𝑀 (𝑡) = 𝐴𝑐 cos 𝜔𝑐 𝑡 − 𝑘𝑓 න 𝑚(𝑡) 𝑑𝑡 sin 𝜔𝑐 𝑡
−∞
t
sNBFM (t ) = Ac cos ct  k f Ac m '(t )sin ct where m '(t ) = 

m(t ) dt

𝐴𝑐 𝐴𝑐
𝑆𝑁𝐵𝐹𝑀 (𝑓) = 𝛿(𝑓 − 𝑓𝑐 ) + 𝛿(𝑓 + 𝑓𝑐 ) + 𝑘𝑓 𝑀′(𝑓 − 𝑓𝑐 ) + 𝑀′(𝑓 + 𝑓𝑐 )
2 2 2𝑗
 NBPM for Base Band Signal
sPM (t ) = Ac cos[ct  k p m(t )]
sPM (t ) = Ac cos ct  cos k p m(t )  sin ct  sin k p m(t )
For | k p m(t ) | 1 , cos |𝑘𝑝 𝑚(𝑡)| ≃ 1 and sin |𝑘𝑝 𝑚(𝑡)|1 ≃ 𝑘𝑝 𝑚(𝑡)

s NBPM (t ) = Ac cos ct  k p m(t ) sin ct 

= Ac cos ct  Ac k p m(t ) sin ct

𝐴𝑐
𝑆𝑁𝐵𝑃𝑀 (𝑓) = 𝛿(𝑓 − 𝑓𝑐 ) + 𝛿(𝑓 + 𝑓𝑐 )
2
𝐴𝑐 𝑀′(𝑓 − 𝑓𝑐 ) + 𝑀′(𝑓 + 𝑓𝑐 )
+ 𝑘𝑝
2 2𝑗
 Comparison
Similarities:
• Both have the same modulated bandwidth 2W, where W is the highest modulating
signal frequency.

• The sideband spectrum for FM has a phase shift of 900 with respect to the carrier,
whereas that of AM is in-phase with the carrier.
Differences:

• In an AM signal, the oscillation frequency is constant, and the amplitude varies

with time, whereas in an FM signal, the amplitude stays constant, and frequency
varies with time.
 sPM (t ) = Ac cos ct  k p m(t )

sFM (t ) = Ac cos ct  2 k f  m( ) d 

t
  
Wideband FM
 f 1
 The above Eq. can be rewritten as

j2Πƒc t jβfsin(2 Π ƒmt)

sFM(t) = Re{Ac e e }
For simplicity, the modulation index of FM has been considered as β
instead of βƒ. Since sin(2 Π ƒmt) is a periodic signal with fundamental
jβfsin(2 Π ƒmt)
period T = 1⁄ƒm, the complex expontial e is also periodic
with the same fundamental period. Therefore, this complex exponential
can be expanded in Fourier series representation as
 The above Eq. can be rewritten as

j2Πƒc t jβfsin(2 Π ƒmt)

sFM(t) = Re{Ac e e }

j2Πƒc t
SFM(t) = Ac Re{Ś(t) e }

Modulation index of FM has been considered as β instead of βƒ.

Since sin(2 Π ƒmt) is a periodic signal with fundamental period T = 1⁄ƒm,
jβfsin(2 Π ƒmt)
the complex expontial e is also periodic with the same
fundamental period.
Therefore, this complex exponential can be expanded in Fourier series
representation as
Ś(t) =

Where T = 1⁄ƒm

In the Above Equation Substitute

𝑥 𝑑𝑥
𝑡= ; d𝑡 = 2Π𝑓𝑚
2Π𝑓𝑚
𝑇
𝑖𝑓 ′𝑡′ 𝑖𝑠 − 𝑥 = −П
2
𝑇 𝑥=П
𝑖𝑓 ′𝑡′ 𝑖𝑠 +
2
 
s FM (t ) = Ac 
n=
J n (  f ) cos(c  nf m )t

Spectrum of WBFM

∞
𝐴𝑐
𝑆𝐹𝑀 (𝑓) = ෍ 𝐽𝑛 (𝛽𝑓 ) 𝛿(𝑓 − 𝑓𝑐 − 𝑛𝑓𝑚 ) + 𝛿(𝑓 + 𝑓𝑐 + 𝑛𝑓𝑚 )
2
𝑛=−∞

Ac
The carrier with amplitude J 0 ( f )
2
 Wideband FM…
∞

𝑠𝐹𝑀 (𝑡) = 𝐴𝑐 ෍ 𝐽𝑛 (𝛽𝑓 ) cos2Π( 𝑓𝑐 + 𝑛𝑓𝑚 )𝑡 ;

𝑛=−∞

Ac 
S FM ( f ) =  J n (  f ) [ ( f  f ct  nf m )   ( f  f ct  nf m ) 
2 n=
Ac
The carrier with amplitude = J 0 ( f )
2
𝑠𝐹𝑀 (𝑡) = 𝐴𝑐 [𝐽0 (𝛽𝑓 ) cos 2Π𝑓𝑐 𝑡
−𝐽1 (𝛽𝑓 ) cos2Π( 𝑓𝑐 − 𝑓𝑚 ) − cos2Π( 𝑓𝑐 + 𝑓𝑚 )
+𝐽2 (𝛽𝑓 ) cos2Π( 𝑓𝑐 − 2𝑓𝑚 ) − cos2Π( 𝑓𝑐 + 2𝑓𝑚 )
−𝐽3 (𝛽𝑓 ) cos2Π( 𝑓𝑐 − 3𝑓𝑚 ) − cos2Π( 𝑓𝑐 + 3𝑓𝑚 )
+. . . . . . . . ]
• 𝐽0 (𝛽𝑓 ) is Carrier Component
• 𝐽1 (𝛽𝑓 ) First Set of Side Frequencies
• 𝐽2 (𝛽𝑓 ) Second Set of Side Frequencies
• 𝐽3 (𝛽𝑓 ) Third Set of Side Frequencies and So on……….

Ac
The carrier with amplitude J 0 ( f )
2
A set of side frequencies specified simultaneously on the either
side of the carrier at a frequency separation of
f m , 2 f m ,... , nf m
 Properties of Bessel Functions
𝐽−𝑛 (𝛽𝑓 ) 𝑓𝑜𝑟 𝑛 − 𝑒𝑣𝑒𝑛
• 1) 𝐽𝑛 (𝛽𝑓 ) = ൝
−𝐽−𝑛 (𝛽𝑓 ), 𝑓𝑜𝑟 𝑛 − 𝑜𝑑𝑑

𝑓𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 𝑉𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝛽

β𝑛 𝐽0 𝛽𝑓 ~ 1
• 2) 𝐽𝑛 𝛽𝑓 = 𝑛 𝐽1 𝛽𝑓 ~ 𝛽/2
2 𝑛! 𝐽𝑛 𝛽𝑓 ~ 0

∞ 2
• 3) σ𝑛=−∞ 𝐽𝑛 𝛽𝑓 = 1
The Bessel functions
Table of Bessel functions of the first kind
 ∞
𝐴𝑐
𝑆𝐹𝑀 (𝑓) = ෍ 𝐽𝑛 (𝛽𝑓 ) 𝛿(𝑓 − (𝑓𝑐 +𝑛𝑓𝑚 )) + 𝛿(𝑓 + (𝑓𝑐 +𝑛𝑓𝑚 ))
2
𝑛=−∞

𝐴𝑐
n=0 𝑆𝐹𝑀 𝑓 = 𝐽0 (𝛽𝑓 ) 𝛿(𝑓 − 𝑓𝑐 ) + 𝛿(𝑓 + 𝑓𝑐 )
2
𝐴𝑐
n=1 𝑆𝐹𝑀 𝑓 = 𝐽1 (𝛽𝑓 ) 𝛿 𝑓 − (𝑓𝑐 +𝑓𝑚 ) + 𝛿 (𝑓 + (𝑓𝑐 +𝑓𝑚 ))
2
𝐴𝑐
n=-1 𝑆𝐹𝑀 𝑓 = 2 𝐽−1 (𝛽𝑓 ) 𝛿 𝑓 − (𝑓𝑐 −𝑓𝑚 ) + 𝛿 (𝑓 + (𝑓𝑐 −𝑓𝑚 ))
𝐴
= − 2𝑐 𝐽1 (𝛽𝑓 ) 𝛿 𝑓 − (𝑓𝑐 −𝑓𝑚 ) + 𝛿 (𝑓 + (𝑓𝑐 −𝑓𝑚 ))
𝐴
n=2 𝑆𝐹𝑀 𝑓 = 2𝑐 𝐽2 (𝛽𝑓 ) 𝛿 𝑓 − (𝑓𝑐 +2𝑓𝑚 ) + 𝛿 (𝑓 + (𝑓𝑐 +2𝑓𝑚 ))
𝐴𝑐
n=-2 𝑆𝐹𝑀 𝑓 = 2 𝐽−2 𝛽𝑓 𝛿 𝑓 − (𝑓𝑐 −2𝑓𝑚 ) + 𝛿 (𝑓 + (𝑓𝑐 −2𝑓𝑚 ))
𝐴
= 2𝑐 𝐽2 (𝛽𝑓 ) 𝛿 𝑓 − (𝑓𝑐 −2𝑓𝑚 ) + 𝛿 (𝑓 + (𝑓𝑐 −2𝑓𝑚 ))
and So On……………….
 Spectrum Analysis
1) Spectrum consists of Carrier and Infinite No. of Side bands.

2) Theoretical BW of FM is Infinite.

3) The Spacing between the spectral components is equal to ‘fm’ message

frequency.

4) In the Spectrum (fc+fm) and (fc-fm) are called as 1st order side bands
(fc+2fm) and (fc-2fm) are called as 2nd order side bands and so on….
Hence Spectrum contains infinite order of side bands.

5) The Magnitude of spectrum components depends on the Bessel function

values. But these values gradually decreases as ‘n’ increases. So, the
magnitude of higher order frequencies are negligible.
 Spectrum Analysis contd…..
6) The Carrier magnitude in the spectrum depends on the modulation index.
The Bessel function 𝐽0 𝛽𝑓 = 0 for 𝛽 = 2.4,5.5,8.6,11.8 … …

7) For the Above Values of ′𝛽’ the carrier magnitude in the spectrum is Zero and
the efficiency is 100%
i.e. Carrier power is suppressed and the Total power is given to side bands only.
 𝜟𝒇 𝒌𝒇𝑨𝒎
𝜷𝒇 = =
𝒇𝒎 𝒇𝒎
Case-1 : Case 1: Fix fm and vary A m

Let Kf = 1Khz/v of the Modulator & fm = 1Khz

𝟏𝑲𝒉𝒛
𝜟𝒇 ∗𝟏𝒗
𝒗
If Am= 1v 𝜷𝒇 = 𝒇𝒎
= 𝟏𝑲𝒉𝒛
= 1 ; 0.77(fc), 0.44(fc+fm), 0.11(fc+2fm)
-
𝟏𝑲𝒉𝒛
𝜟𝒇 ∗𝟐𝒗 + +
𝒗
If Am= 2v 𝜷𝒇 = = = 2 ; 0.22(fc), 0.57(fc fm), 0.35(fc 2fm), 0.12(fc
𝒇𝒎 𝟏𝑲𝒉𝒛 − −
+
3fm)
−
𝟏𝑲𝒉𝒛
𝜟𝒇 ∗𝟑𝒗 + + +
𝒗
If Am= 3v 𝜷𝒇 = = = 3 ; fc, (fc fm), (fc 2fm), (fc 3fm), (fc
𝒇𝒎 𝟏𝑲𝒉𝒛 − − −
+ +
−
𝟒fm), (fc −
𝟓fm),
• 𝜷𝒇 =1

2 𝜟𝒇
• 𝜷𝒇 =2

2 𝜟𝒇
• 𝜷𝒇 =3

2 𝜟𝒇
 𝜟𝒇 𝒌𝒇𝑨𝒎
• Case-2 : Fix A m and vary fm 𝜷𝒇 = =
𝒇𝒎 𝒇𝒎

• Let Kf = 1Khz/v of the Modulator & fm = 1Khz

• If 𝜷𝒇 = 𝟏 𝒇𝒎 = 𝜟𝒇 ; (fc), (fc+fm),

𝜟𝒇 𝜟𝒇 + +
• If 𝜷𝒇 = 𝟐 𝟐= ( = 𝒇𝒎 ) ; (fc), (fc 500Hz), (fc 1Khz)
𝒇𝒎 𝟐 − −

𝜟𝒇 𝜟𝒇 + +
• If 𝜷𝒇 = 𝟓 𝟓= 𝒇𝒎
( 𝟓
= 𝒇𝒎 ) ; fc, (fc −
200Hz), (fc −
𝟒00Hz),
+ +
(fc − 𝟔00Hz), (fc − 𝟖𝟎𝟎𝐇𝐳),
+
(fc 𝟏𝐊𝐇𝐳),
−
• 𝜷𝒇 =1

2 𝜟𝒇

• 𝜷𝒇 =2

2 𝜟𝒇
• 𝜷𝒇 =5

2 𝜟𝒇
 Carson’s Rule
The Carson rule state that the approximate bandwidth necessary to transmit
an angle modulated wave as twice the sum of the frequency deviation and
the highest modulating signal frequency.

B.W𝐹𝑀 = 2𝑛𝑓𝑚 n= 1+ 𝛽𝑓
1
B.W𝐹𝑀 = 2𝑓𝑚 1 + 𝛽𝑓 = 2 (Δ𝑓 + 𝑓𝑚 ) = 2Δ𝑓(1 + )
𝛽𝑓

Approximates the 98% of the Total power & Carson’s BW is less than actual
Tx BW
Universal Curve for FM Transmission Bandwidth
Carson’s rule is simple but unfortunately it does not always provide a good
estimate of the transmission bandwidth, in particular, for the wideband frequency
modulation.
 Total Power of WBFM
• 𝑠𝐹𝑀 (𝑡) = 𝐴𝑐 σ∞
𝑛=−∞ 𝐽𝑛 (𝛽𝑓 ) cos2Π( 𝑓𝑐 + 𝑛𝑓𝑚 )𝑡 ;

𝐴𝑐 σ∞
𝑛=−∞ 𝐽𝑛 (𝛽𝑓 ) 2
Pt = ( )
√2
𝐴𝑐 2 2 𝐴𝑐 2 2 𝐴𝑐 2 2 𝐴𝑐 2 2 𝐴𝑐 2
Pt = 𝐽0 𝛽𝑓 + 𝐽1 𝛽𝑓 + 𝐽1 𝛽𝑓 + 𝐽2 𝛽𝑓 + 𝐽2 2 𝛽𝑓 + ⋯
2𝑅 2𝑅 2𝑅 2𝑅 2𝑅

𝐴𝑐 2 2 𝐴𝑐 2 2 𝐴𝑐 2 2
Pt =
2𝑅 0
𝐽 𝛽𝑓 + 2 ∗ 2𝑅 1
𝐽 𝛽𝑓 + 2 ∗ 2𝑅 2
𝐽 𝛽𝑓 + ⋯
Carrier Power 1st Order Side Bands Power 2nd Order Side Bands Power
 𝐴𝑐 2
Pt = [ 𝐽0 2 𝛽𝑓 + 𝐽1 2 𝛽𝑓 + 𝐽1 2 𝛽𝑓 + 𝐽2 2 𝛽𝑓 + 𝐽2 2 𝛽𝑓 + ⋯ ]
2𝑅

𝐴𝑐 2 ∞ 2
• Pt = (σ𝑛=−∞ 𝐽𝑛 𝛽𝑓 )
2𝑅

𝐴𝑐 2
• Pt =
2𝑅

• Note:
1) Same as Unmodulated Carrier Power
2) Independent of Modulation Index But in AM & DSB-
SC, SSB-Sc & NBFM it is Dependent on Modulation
Index
 Observations
In AM: Carrier and first two sidebands.
In WBFM: Carrier and infinite number of sidebands.
The J coefficients represent the amplitude of a particular pair of sidebands.
Thus, the modulation index determine the number of sideband components
have significant amplitudes.
 As No. of sidebands increases
f
f = , fm    f 
fm

In AM   PSB  
, Pt 
but in FM as  f  Total power is constant, but B.W 
 Observations…
In AM, the BW = 2 fm , but in FM the BW is determined by fm and f

In FM, the amplitude of the carrier component is constant, whereas in

AM, the carrier component does not remain constant. The J coefficients J0
is a function of f . The overall amplitude of the FM wave remains
constant.
The carrier component of the FM wave disappear completely at f = 2.4, 5.5,
8.6, 11.8 etc.

First carrier null, second carrier null . . .

 Band Width of a Sinusoidally Modulated FM Signal
NBFM: B.W = 2fm
WBFM: The bandwidth of FM wave depends on both modulating frequency
and modulation index.
f
f = , f m    f  No. of sidebands increases
fm
hence BW increases.
The BW depends on . BW = 2  f Hz
f
Effect of the Modulation Index on Bandwidth
f
f =
fm
 f  0.2 B.WNBFM = 2 f m

 f 1 BW = 2f = 2 f f m

B.W = 2(nf m )
The Carson’s rule
1
B.W𝐹𝑀 = 2𝑓𝑚 1 + 𝛽𝑓 = 2 (Δ𝑓 + 𝑓𝑚 ) = 2Δ𝑓(1 + )
𝛽𝑓
Generation of NBFM
Block Diagram for generation of NBFM
Generation of Wide Band FM
•Direct Method {Not there in Syllabus}
•Indirect Method (or) Armstrong Method

Detection of FM
•Single Slope
•Balanced Slope
•Phase Locked Loop(PLL)***
Indirect Method

Base Band Signal Frequency WBFM Signal

Integrator NBFM Multiplier
Xn

Crystal
Controlled
Oscillator
 𝑡
𝑠1 (𝑡) = 𝐴𝑐 cos 2𝜋𝑓1 𝑡 + 2𝜋𝑘𝑓 න 𝑚(𝑡) 𝑑𝑡
0
Output of NBFM
𝑠1 (𝑡) = 𝐴𝑐 cos 2𝜋𝑓1 𝑡 + β f sin(2Пƒmt)] (β f < 0.3)

Output of Frequency Multiplier

𝑠1 (𝑡) = 𝐴𝑐 cos 2𝜋𝑛𝑓1 𝑡 + nβ f sin(2Пƒmt)]

𝑠1 (𝑡) = 𝐴𝑐 cos 2𝜋𝑓𝑐 𝑡 + β f sin(2Пƒmt)] (β f > 1)

FM Demodulators
 Basic Idea

Vout (t ) = kd f

where Vout (t ) = demodulated output signal (Volts)

kd = demodulator transfer function (Volts per Hertz)

 f = difference between the input frequency and the

center frequency of the demodulator (Hertz).
1. Slope Detector
 Slope Detector . . .
Advantages: The only advantage of the basic slope
detector circuit is its simplicity.
Limitations:
(i) The range of linear slope of tuned circuit is quite
small.
(ii) The detector also responds to spurious amplitude
variations of the input FM.
These drawbacks are overcome by using balanced
slope detector.
2. Balanced Slope Detector
Advantages:
(i) This circuit is more efficient than simple slope
detector.
(ii) It has better linearity than the simple slope detector.
Limitations:
(i) Even though linearity is good, it is not good enough.
(ii) This circuit is difficult to tune since the three tuned
circuits are to be tuned at different frequencies, and
(iii) Amplitude limiting is not provided.
 Limitations of direct methods of FM generation

Difficult to obtain a high order of stability in carrier frequency because tank

circuit consists of L and C.

The crystal oscillator can be used for carrier frequency stability, but
frequency deviation is limited.

The non linearity produces a frequency variation due to harmonics of the

modulating signal hence there are distortions in the output FM signal.
Phase-Locked Loop (PLL) as FM Demodulator
𝑡
• s(t) = 𝐴𝑐 cos 2𝜋𝑓𝑐 𝑡 + 2𝜋Kf ‫׬‬0 𝑚 𝑡 𝑑𝑡
• s(t) = Ac cos[2𝜋ƒct + ∅1(t)]
Let the VCO output be defined by
𝑡

𝑟𝑣𝑐𝑜 (𝑡) = 𝐴𝑣 Sin 2𝜋𝑓𝑐 𝑡 + 2𝜋Kv න 𝑣 𝑡 𝑑𝑡

0

𝑡
where ∅2(t) = 2𝜋Kv ‫׬‬0 𝑣 𝑡 𝑑𝑡

PLL Demodulator Kv is the Frequency Sensitivity of the

VCO in Hz/V
We Assume,
1) The Frequency of the VCO Output is Precisely set at Unmodulated Carrier
Frequency fc
2) VCO Output has 90 ֩ Phase shift w.r.t the Unmodulated Carrier Frequency
FM Wave s(t) and VCO Output r(t) is applied to the Multiplier Producing
two Components
1) High Frequency component
𝑘𝑚 𝐴𝑐 𝐴𝑣 sin 4𝜋𝑓𝑐 𝑡 + φ1 (𝑡) + φ2 (𝑡)
2) Low Frequency component is Obtained at the Output of Multiplier
e(t) = 𝑘𝑚 𝐴𝑐 𝐴𝑣 sin φ1 𝑡 − φ2 (𝑡)
e(t) = 𝑘𝑚 𝐴𝑐 𝐴𝑣 sin φ𝑒 𝑡 = K sin φ𝑒 𝑡
where, φ𝑒 𝑡 = φ1 𝑡 − φ2 (𝑡) (Phase Error)
Because φ𝑒 𝑡 is Small, sin φ𝑒 𝑡 = φ𝑒 𝑡
𝑡
φ𝑒 𝑡 = φ1 𝑡 − 2𝜋Kv ‫׬‬0 𝑣 𝑡 𝑑𝑡
where, φ𝑒 𝑡 = φ1 𝑡 − φ2 (𝑡) (Phase Error)

Φ1(t) Φe(t) Loop Filter V(t)

+ g(t)
-

Φ2(t) 𝑡

2𝜋Kv න 𝑣 𝑡 𝑑𝑡
0

𝑡
φ𝑒 𝑡 = φ1 𝑡 − 2𝜋Kv ‫׬‬0 𝑣 𝑡 𝑑𝑡
 𝑡
φ𝑒 𝑡 + 2𝜋Kv ‫׬‬0 𝑣 𝑡 𝑑𝑡 = φ1 𝑡 … . . Eq1
Differentiating ………Eq1
𝑑φ𝑒 𝑡 𝑑φ1 𝑡
+ 2𝜋Kv 𝑣 𝑡 = … . 𝐸𝑞2
𝑑𝑡 𝑑𝑡
∝
where, 𝑣 𝑡 =‫׬‬−∝ φ𝑒 τ 𝑔 𝑡 − τ 𝑑 τ = φ𝑒 𝑡 *g(t) in T.D.
V(f) = φ𝑒 𝑓 G 𝑓 in F.D.
Finding the F.T. of the Equation-2
j2𝜋f φ𝑒 𝑓 + 2𝜋Kv φ𝑒 𝑓 G 𝑓 = j2𝜋f φ1 𝑓
 jfφ𝑒 𝑓 + Kv φ𝑒 𝑓 G 𝑓 = jf φ1 𝑓
𝐾𝑣
jf [1+ G 𝑓 ] φ𝑒 𝑓 = jf φ1 𝑓
𝑗𝑓
φ1 𝑓
φ𝑒 𝑓 =
𝐾𝑣
1+ 𝐺(𝑓)
𝑗𝑓
Equation of the Controlled Voltage to the VCO is 𝑣 𝑓 = φ𝑒 𝑓 𝐺 𝑓
𝐾𝑣
Now the Design of G(f) is Such That Mod. Of 𝐺 𝑓 𝑖𝑓 ≫ 1
𝑗𝑓

𝑗𝑓2𝜋
𝑣 𝑓 = φ
2𝜋𝑘𝑣 1
𝑓 …eq3 We Know, ∅1(t)=
𝑡
2𝜋Kf ‫׬‬0 𝑚 𝑡 𝑑𝑡
𝑑
Taking Inverse F.T of Eq3 φ1 𝑡 = 2𝜋kf 𝑚 𝑡
𝑑𝑡
1 𝑑
𝑣 𝑡 = φ 𝑡
2𝜋𝑘𝑣 𝑑𝑡 1
 1
𝑣 𝑡 = 2𝜋Kf 𝑚 𝑡
2𝜋𝐾𝑣

Kf
𝑣 𝑡 = 𝑚 𝑡
𝐾𝑣

Since The Control voltage of the VCO is Proportional

to the Message Signal ,Hence V(t) is the Demodulated
Signal.
PREEMPHASIS AND DEEMPHASIS NETWORKS

Figure ;(a) Pre emphasis network. (b) Frequency response of preemphasis network
Figure (a) Deemphasis network. (b) Frequency response of Deemphasis network.
 Comparison of AM and FM:
S.No AM FM
1 Pt= Pc(1+μ2) Pt= Pc
2
2 AM Requires More Power Less Power
3 Power Varies with Modulation Independent of Modulation Index
Index
4 33.33% (Max. Efficiency) 33.33% (Max. Efficiency) when β=
2.4,5.5,8.6,11.2..
5 BW = 2fm BW = 2(1+β)fm
6 BW is Low Very High
7 BW is Independent of Mod. Index BW Varies with Modulation Index
8 AM Receiver is Less Complex More Complex
 Comparison of AM and FM contd…

S.No AM FM
9 Effect of Noise is More in AM Effect of Noise is Less in FM
10 550KHz to 1650KHz 88MHz to 108MHz
11 Intermediate Frequency =455KHz Intermediate Frequency =10.7MHz
12 Practical BW = 10KHz Practical BW = 200KHz
13 μ= 1 Β= 5
14 fm = 5KHz fm = 15KHz
15 Ionospheric Propagation(NON-LOS) Line of Sight Propagation(LOS)
16 Area of Coverage is More Limited Because of LOS
17 Frequency Reuse is Not Possible Frequency Reuse is Possible
1) A sinusoidal modulating waveform of amplitude 5 V and a frequency
of 2 KHz is applied to FM generator, which has a frequency sensitivity
of 40 Hz/volt. Calculate the frequency deviation, modulation index, and
bandwidth.
Given, Am=5V , fm=2KHz , kf=40Hz/volt
Δf=kf *Am
Δf=40×5=200Hz
β=Δf / fm ; β=200/2000= 0.1
Here, the value of modulation index, β is 0.1, which is less than one.
Hence, it is Narrow Band FM.
BW=2fm= 2×2K=4KHz
Therefore, the bandwidth of Narrow Band FM wave is 4KHz.
2) An FM wave is given by s(t)=20cos(8π×106t+9sin(2π×103t)).
Calculate the frequency deviation, bandwidth, and power of FM wave.
Sol: Given, the equation of an FM wave as
s(t)=20cos(8π×106t+9sin(2π×103t)).
We know the standard equation of an FM wave as
• s(t)=Ac cos(2πfct+βsin(2πfmt)) comparing the above two equations.
Ac=20V ,fc=4×106Hz=4MHz ,fm=1×103Hz=1KHz , β=9
Here, the value of modulation index is greater than one. Hence, it
is Wide Band FM.
We know the formula for modulation index as β=Δf/fm
Δf =βfm = Δf =9×1KHz =9KHz
The formula for Bandwidth of Wide Band FM wave is
BW=2(β+1)fm
BW=2(9+1)1K=20KHz
Formula for power of FM wave is Pc=Ac2/2R
Assume, R=1Ω and substitute Ac value in the above equation.
P=(20)2/2(1)=200W
Therefore, the power of FM wave is 200 watts.
3) Consider a Narrow band Signal S(t)=10cos(2π×106t+0.4sin(2π×103t)) is
passed through Two Frequency Multipliers connected in cascade and N1= 2 and
N2= 5. Determine the carrier frequency. Modulation index & BW at the output
of 1st and 2nd Multipliers.
Solution:

N1 = 2 N2 = 5
NBFM NBFM WBFM

fc= 1MHz fc= 2MHz fc= 10MHz

β= 0.4 β= 0.8 β= 4
fm=1KHz fm=1KHz fm=1KHz
BW= 2KHz BW= 10KHz
4) Consider a Angle Modulated Signal S(t)=10cos(2π×106t+ 8 sin(4π×103t)).
Assuming the given signal as FM
i) Determine the Modulation Index, Frequency Deviation, BW & Power
ii) Repeat the above calculations if the message signal is Doubled.
Solution:
a) Δf =βfm = 8×2KHz =16KHz
b) BW=2(β+1)fm = 2(8+1)2KHz = 36KHz
c) P=(10)2/2(1)=50W
a) fm is Doubled ,Hence Mod. Index is Halved
b) β= Δf/fm = 16/4KHz = 4 ; b) BW=2(β+1)fm = 2(4+1)4KHz = 40KHz
c) P=(10)2/2(1)=50W
5) In an FM System the audio frequency is 500Hz and Audio Voltage is 2.4V, the
Deviation is 4.8KHz a) find the Expression.
b) If the audio frequency voltage is increased to 7.2V. What is the new Deviation
c) If the Audio frequency voltage is raised to 10V, While the audio frequency is
dropped to 200Hz. What is the deviation and also calculate Modulation Index for All
the above cases and write the expression for Modulated waveform.
• Solution:

a) fm= 500Hz ; Δf= 4.8KHz ; Am= 2.4V; β=?

β= Δf/ fm = 4.8KHz/ 500Hz
=9.6
s(t)=AcCos(2π×fct+9.6sin(2π×500t)).
b) Am= 7.2V; fm= 500Hz ; Δf= ? β=? ;
Kf is Constant for a Modulator (from (a))
K f = Δf/ fm = 4.8KHz/ 500Hz= 2KHz
Δf = Kf *fm = 2KHz *7.2V= 14.4KHz
β= 14.4KHz/500Hz = 28.8
s(t)=AcCos(2π×fct+28.8sin(2π×500t))

c) Am= 10V; fm= 200Hz ; Δf= ? β=? ; Kf = 2KHz

Δf = Kf *fm = 2KHz *10V= 20KHz
β= 20KHz/200Hz = 100
s(t)=AcCos(2π×fct+100sin(2π×200t))
6) A FM Transmitter is Operated at a carrier of 100MHz with a carrier
voltage of 8V the modulating signal has an amplitude of 3V and
frequency 6KHz resulting in a frequency deviation of 60KHz. Write
the voltage equations for the following Conditions.
a) Original Values
b) Audio Amplitude increased to 4V
c) Audio Changed to 2V and 3KHz
Solution:
a) Ac = 8V; fc= 100MHz ; Am= 3V; fm=6KHz; Δf=60KHz
β= Δf/ fm = 60KHz / 6KHz = 10
s(t)= 8Cos(2π×100*106t+10sin(2π×6*103t)).
b) Ac = 8V; fc= 100MHz ; Am= 3V; fm=6KHz; Δf=60KHz
Δf = kf*Am = 20KHz*4= 80KHz
β= Δf/ fm = 80KHz/ 6KHz = 13.33
s(t)= 8Cos(2π×100*106t+13.33sin(2π×6*103t)).

b) Ac = 8V; fc= 100MHz ; Am= 3V; fm=6KHz; Δf=60KHz

Δf = kf*Am = 20KHz*2= 40KHz
β= Δf/ fm = 40KHz/ 3KHz = 13.33
s(t)= 8Cos(2π×100*106t+13.33sin(2π×6*103t)).
7) An FM Wave is described by s(t)=20cos(3π×108t+10cos(2000πt)). Find the
Approximate Bandwidth of an FM Wave.
Solution:
β= 10 fm=1KHz;
BW= 2(1+β)fm = 2(1+10)1KHz = 22KHz
8) Find the number of Channels that can be accommodated in the FM Band
with frequency deviation of 75KHz and message signal frequency equal to
15KHz.
Solution:
fm=15KHz; Δf= 75KHz; (FM Range: 88MHz- 108MHz)
Total BW= (108MHz-88MHz)=20MHz
Number of Signals = Total BW/Signal BW = 20MHz/180KHz = 110
Signal BW= 2(Δf +fm) = 2(75KHz+15KHz) =180KHz
9) For an FM modulator with peak frequency deviation 10 KHz, a
modulating signal frequency 10 KHz, peak modulating signal
amplitude 10 V, and a 500 KHz carrier, determine
a) Actual minimum bandwidth from the Bessel table.
b) Plot the output frequency spectrum for the Bessel approximation.
c) Approximation minimum bandwidth using Carson’s rule.
d) Comment the results.
• From the Bessel table for  f = 1, J 0 = 0.77, J1 = 0.44, J 2 = 0.11, and J3 = 0.02

Solution:
(d) Comments: The bandwidth from Carson’s rule is
less than the actual minimum bandwidth required to
pass all the significant sideband sets as defined by the
Bessel table.
Therefore, a system that was design using Carson’s
rule would have a narrower bandwidth and their poor
performance than a system designed using the Bessel
table.
For modulation indices above 5, Carson’s rule is a
close approximation to the actual bandwidth required.
Beyond the Syllabus
Direct Generation of WBFM
Varactor Diode Modulator (Direct Method)
The capacitance of a
varactor diode is inversely
proportional to the reversed
biased voltage amplitude.
1
fi (t ) = C (t ) = C0  Cv
2 ( L1  L2 )C (t ) where

For m(t) , the capacitance C (t ) = C0  kc m(t )

f0 1
f i (t ) = where f0 =
kc 2 C0 ( L1  L2 )
1 m(t )
C0

 
fi (t ) = f0 1 
kc
m(t )  fi (t ) = f0 1  k f m(t ) 
 2C0 
3. Foster-Seeley Discriminator (Phase Discriminator)
fin = f o fin  f o fin  f o
Advantages:
• Tuning procedure is simpler than balanced slope detector, because it
contains only two tuned circuits, and both are tuned to the same
frequency .
• Better linearity, because the operation of the circuit is dependent
more on the primary to secondary phase relationship which is very
much linear.

Limitations:
It does not provide amplitude limiting. So, in the presence of noise
or any other spurious amplitude variations, the demodulator output
respond to them and produce errors.
 4. Ratio Detector
Similar to the Foster-Seeley discriminator .
(i) The direction of diode is reversed.
(ii) A large capacitance Cs is included in the circuit.
(iii) The output is taken different locations.
Advantages:
• Easy to align.
• Good linearity due to linear phase relationship between primary
and secondary.
• Amplitude limiting is provided inherently. Hence additional
limiter is not required.
Example
Audio 100Hz  15KHz

f1 fc = 100MHz
f = 75KHz
f1 f m = 100Hz
Let 1 = 0.2 1 = ,
fm Then
To produce f = 75KHz
f1 = 20Hz
Frequency Multiplication f 75KHz
= = 3750
f1 20Hz
Two stages n1n2 = 3750
Audio 100Hz  15KHz

fc = 100MHz
f2 f = 75KHz
fc
Two stages n1n2 = 3750 f 2  n1 f1 =  n1 = 75, n2 = 50
n2
Thank You