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Communication Systems (Theory)

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Chapter 2 – Amplitude Modulation

Objective

Upon completion of this chapter you will be able to:

Understand the process of Amplitude Modulation.

Understand different types of Amplitude Modulation.

Design different modulators for Amplitude Modulation.

Understand the different demodulation techniques.

Introduction

Baseband term is used to designate the band of frequencies of the signal delivered by the

source. In telephony, baseband is the audio band i.e. 0 – 3.5 KHz. Baseband Signals are

transmitted without modulation as they have sufficient power at low frequencies so they

cannot be transmitted at radio frequencies but by means of co-axial cables and optical

fibers.

Long haul communication over radio links requires modulation to shift the spectrum to

higher frequencies in order to enable efficient power radiation using antennas of reasonable

dimensions. Modulation is a process by which some characteristics of the carrier signal like

its Amplitude, Frequency or Phase are varied in proportion to message signal.

Amplitude Modulation refers to the process in which amplitude of the carrier wave is varied

with the message signal.

The process of modulation i.e. shifting the signal spectrum to a higher frequency is

illustrated below by an example,

If signal and its Fourier Transform is x t X

By Frequency Shifting Property,

j t j tx t e X and x t e X

o o

Suppose, y t x t cos tc

c cx t j t j t

y t e e2

1y t X X

c c2

Hence,

1F.T.x t cos2 f t X f f X f f

c c c2

The spectrum of x(t) and y(t) are shown below,



In the figures shown above, M(0) represents the magnitude of spectrum at 0 frequency.

As can be seen from the above figures, that spectrum of baseband spectrum is shifted to

higher carrier frequency by multiplication with sinusoidal carrier signal. This is the basic

process of amplitude modulation.

Further, we will see different types of Amplitude Modulation.

DSB-FC (Double Side Band – Full Carrier)

This is also known as ordinary Amplitude Modulation. The following terms are generally used

in regards to Amplitude Modulation,

Baseband signal – message signal m(t)

Carrier signal – c(t)

Modulated signal – s(t)

Here, amplitude of carrier is changing according to message signal.

Assuming, c t A cos tc c

s t A m t cos tc c

A cos t m t cos tc c c

F.T.cos t

c c c

1f f f f

c c2

If we write the spectrum in terms of ‘f’ then we have to divide by ‘2π’



Spectrum of Modulated Signal,

A 1cS f f f f f M f f M f fc c c c2 2

The second term has been derived from Multiplication Theorem of Fourier Transform.

Spectrum of Baseband signal is shown below,

Bandwidth = fm

Spectrum of Modulated Signal looks like as shown below,

Since, the spectrum consists of impulses at carrier frequency and two side bands namely USB

i.e. Upper Side Band and LSB i.e. Lower Side Band so it is called as Double Side Band – Full

Carrier.

B.W. f f f fc m c m

2f Hm z

It means bandwidth of AM signal is twice the bandwidth of message signal.

Note: If message signal m(t) is non – sinusoidal, and contains frequencies f , f , fm1 m2 m3

and

so on and frequency of carrier c(t) is fc

; frequencies in modulated signal will be,

s t f f , f f , f fc m1 c m2 c m3

etc.



Solved Examples

Problem: If a 60kHz carrier is amplitude modulated by a speech band of 300kHz 3kHz .

What is range of frequency for USB and LSB?

Solution: Carrier Frequency, f 60kHzc

Message Frequency, f 300Hz 3kHzm

The lower frequency is mapped to 60 0.3 kHz

The upper frequency is mapped to 60 3 kHz

Assuming a spectrum of message signal, the amplitude modulated spectrum is shown below,

Band pass signal B.W. 59.7 57 63 60.3 5.4kHz

B.W. 3 0.3 2.7kHz

Problem: A 1MHz sinusoidal signal is amplitude modulated by a symmetrical square wave of

100μs, which of the frequency will not be present in output?

a) 1010kHz b) 1020kHz c) 1030kHz d) 990kHz

Solution: If signal is non – sinusoidal and is either square, triangular or saw tooth, then by

half wave symmetry it will contain only add harmonics.

f 1MHzc

T 100 s

Fundamental Frequency, 4f 10 Hzm

Frequency in square wave: f ,3f ,5f ,7fm m m m

Frequency in Modulated Signal,

f f 990kHz, 1010kHzc m

f 3f 1030kHz, 970kHzc m

Ans: 1020 kHz



Problem: A 1000kHz carrier is simultaneously modulated with 300Hz and 2kHz audio sine

wave. Which of the frequencies are not present in output?

a) 998kHz b) 1000.3kHz c) 999.7kHz d)700kHz

Solution: Carrier Frequency, f 1000kHzc

Modulating Signal Frequency, f 0.3kHz1 f 2kHz

2

Frequencies present in output,

1002kHzf fc 1 998kHz

999.7kHzf fc 2 1000.3kHz

Ans: 700kHz

Modulation Index

Modulation Index is defined as the ratio of peak of message signal to the peak of carrier

signal.

Modulated Signal,

s t A m t cos tc c

Modulation Index,

m t

maxma A

c

for sinusoidal signal m t A cos tm m

m t Amax m

so, A

mma A

c

If A A ; m 1; under modulationm c a

If A A ; m 1; over modulationm c a

If A A ; m 1m c a

; critical modulation

Time domain representation of modulated signal for sinusoidal signal

Under-modulation

A Am c

s t A m t cos t m 1c c a



From the above figure it can be observed that shape of message signal remains intact as the

envelope of carrier signal and hence it can easily be detected.

Critical Modulation

A Am c

m 1a

In this case, envelope detector is not an ideal solution because information will be lost at

touching points i.e. the points where the two envelopes meet each other.

Over Modulation

A Am c

m 1a

No information is seen at CRO because positive and negative portion will cancel effect of

each other. Envelope detector cannot be used in this case.

The crossing over of positive and negative envelopes is shown in the figure below,



Calculation of M.I. from modulated waveform

Modulation Index, A

mma A

c

A A A Ac m c mm

a A A A Ac m c m

A Amax minm

a A Amax min

So, maximum and minimum amplitude,

A A 1 mmax C a

A A 1 mmin C a

For eg. if A 10, A 5max min

Modulation Index, 10 5 1

ma 10 5 3

Frequency Domain representation of modulated signal for sinusoidal signal

Modulated Signal, s t A m t cos tc c

let m t A cos tm m

s t A A cos t cos tc m m c

s t A cos t A cos t cos tc c m m c



Ams t A cos t cos t cos t

c c c m c m2

Taking Fourier Transform of the modulated signal,

AmS A

c c c c m c m2

Am

c m c m2

Calculation of power for Amplitude Modulation


A cos t m t cos tc c c

Power in A cos tc c

can be calculated in terms of Fourier Series Coefficients kc ,

c c

c c

j t j tj t j t

A Ae e c cA cos t A e ec c c 2 2 2

Power =

2 2 2

2

k

A A Ac c cc

2 2 2

Power in Modulated Signal,

2 2A m tcPT 2 2

Here, 2m t represents mean value of 2m t

Let m t A cos tm m

2 2 2m t A cos tm m



2A

2 mm t2

Hence,

2 2A Ac mP

T 2 4

2 2A m Ac a mP 1 where m

T a2 2 Ac

2maP P 1

T c 2

This is valid only if message signal is sinusoidal.

2AcP of R 1

c 2

If R 1

Power in Carrier Signal,

2AcP

c 2R

2maP P 1

T c 2

Since, 2P I

2maI I 1

T a 2

Power in Modulated Signal,

2maP P P

T c c2

Here,

2ma P

c2 represents total sideband power.

Since, sideband consists of Upper Sideband and Lower sideband

2maP P

LSB c4

2maP P

USB c4



Average sideband power

2P P mUSB LSB a P

c2 4

If message signal is not given, it is assumed to be sinusoidal. In practical message signal is

never sinusoidal as sinusoidal signals are deterministic so no information is contained in it.

Solved Examples

Problem: If c(t) and a(t) are used to generate an AM signal with Modulation Index (M.I.)=0.5,

What is the ratio of total sideband power to carrier power?

Solution: Total Sideband Power,

2m 1a P Pc c2 8

Total Sideband power 1

carrier power 8

Problem: If equation for AM is s t 10 1 0.5sin2 f t cos2 f tm c

, what is value of average

sideband power?

Solution: Message Signal, m t A sin2 f tm m

Modulated Signal, s t 10 5sin2 f t cos2 f tm c

Modulation Index,m 0.5a

Carrier Power, 100

P 50c 2


2m 1 50a P 50c4 16 16

= 3.125 watt

22 A AA

c mmaxPeak output power2R 2R

Important Formulas

Maximum Amplitude, A A Amax c m

Carrier power

2Ac P

c2R



Total sideband power

2mc P

c2


2mcP P .P

USB LSB c4

Peak output power

2A A

c m

2R

Solved Examples

Problem: A given AM broadcasting station transmitter has a total power of 5W. If carrier is

modulated by a sinusoidal signal of m 0.707a

.

Calculate (i) Average sideband power

(ii) Peak output power

In both cases, assume that antenna is replaced by a 50 resistance.

Solution: Transmitted Power,

2maP P 1

T c 2

Carrier Power,

P 5TP 4Wc 12m 1

a 412


2m 1a P 4c4 8

= 0.5W

Since,

2AcP

c 2R

2A 4 2 50 400c

A 20c

Modulation Index A 1mm

a A 2C

20

A 10 2m 2

Peak output power

2

22 120 1A A2c m 11.65W

2R 50 2



Problem: If modulation index of Am

is changed from 0 to 1, then transmitted power will,

(a) Increase by 50% (b) Decrease by 50%

(c) Increase by 60% (d) Decrease by 60%

Solution: If m 0 ; P Pa T c

If

2m 1am 1 ; P 1 P 1 P 1.5Pa T c c c2 2

change in power 1.5 1 P

c 100Pc

= 50%

Hence, Transmitted Power increases by 50%

Problem: If a carrier has peak amplitude of 10V at a frequency 1MHz. If sinusoidal signal of

frequency 1kHz modulates between 7.5 and 12.5. Then calculate,

(i) Modulation index (ii) Peak output power (iii) Average sideband power

Assume that antenna used has a resistance of 50 .

Solution: Given A 12.5 A 7.5max min

Modulation Index,

A A 12.5 7.5 5 1max minma A A 12.5 7.5 20 4

max min

2 2m A 1 100a cAverage sideband power4 2R 4 16 2 50

1

W 0.015625W64

Peak output power

2 22A A A 1 mc m c a

2R 2R

100 2 21 0.25 1.252 50

= 1.5625W

Transmission efficiency in case of AM

Transmitted Power,

2 2A m tcPT 2 2

Sideband Power,

2m tPSB 2

Transmission efficiency PSB

PT

For sinusoidal signal

2 2 2m A Aa c mP

SB 2 2 4



2AcP

C 2

Efficiency,

2Am 2m

a42 22A 2 mAc am

2 4

Maximum transmission efficiency: m 1a

1100 33.33%

max 3

Only 33.33% power will be utilized while 66.67% power is wasted.

Note: Main disadvantage is that 66.6% power is wasted.

But at the same time, receiver has information about carrier frequency, so there will be no

need to use extra synchronizing circuit, for knowing carrier such as Costas Receiver in case of

suppressed carrier modulation.

Problem: If a message signal is given by 1 1

m t cos t sin t1 12 2

. If m(t) is modulated with

a carrier of frequency c

to generate s t 1 m t cos tc

, then what is the value of

in% ?

Solution: Message Signal

1 2 2 1m t cos t sin t cos t

1 1 12 42 2 2

Hence, A1 1mA A 1 m

m c aA2 2c

Modulation Index, A 1mm

a A 2c

12m 1a 2100% 100% 100% 20%2 1 52 m 2a 2



Problem: For a given AM signal, if output is of the form:

s t Acos 400 t Bcos 380 t Bcos 420 t . If P 100w and 40%c what are

values of A and B?

Solution: Modulated Signal can also be expressed as,

s t Acos400 t B 2cos400 tcos20 t A 2Bcos20 t cos400 t

Comparing with standard equation, s t A A cos t cos tc m m c

Carrier Power,

2AcP 100

c 2

Carrier Amplitude, A A 10 2c

Transmission Efficiency,

2m 2a ma20.4

2 2m m 2a a1

2

Modulation Index, 2

ma 3

2B 2m

a A 3

2B 10

3

Problem: If s t 4cos 1800 t 10cos 2000 t 4cos 2200 t

Calculate (a) ma

(b) Total power PT

(c) (d) Bandwidth

Solution: Combining the first and third term in modulated signal,

s t 10 8cos200 t cos2000 t

Modulation Index, 8

m 0.8a 10

Transmitted Power,

100 0.64P 1 50 1.32 66.0WT 2 2

Transmission Efficiency, 0.64 800

100 24.2%2.64 33

Since, modulated signal requires twice the bandwidth as compared to message signal,

Bandwidth

400B.W. 2f 200Hz

m 2



Amplitude modulation in case of Non – Sinusoidal Signals

Transmitted Power,

2 2A m tcPT 2 2

Here, 2m t is mean square value of message signal m(t). For a periodic message signal this

value can be computed as,

T

12 2m t m t dtT

Where, T is time period of message signal m(t)

Solved Examples

Problem: If c t 2cos tc

and message signal is shown, then calculate

(i) Total power PT

(ii) Efficiency

Solution: Mean Square Value of Message Signal,

TT21 1 22 2m t 2 dt 2 dt

T TT0

2

1 T 1 T2m t 4 4T 2 4 2

=4

2A 2m (t) 4 4cP 4T 2 2 2 2

2m (t)22% 100% 100% 50%

P 4T

Problem: Solve the previous problem if square wave of message signal is replaced by

sawtooth waveform shown below,



Solution: Equation of m(t) in

T T,

2 2

6

m t tT

2T/2 T/2

T/2 T/2

A1 12 2 mm t m t dt t dtT T T

2 23 2A AT 32 m mm t 33 3 3 3T

Since, carrier signal is same so carrier power remains same.

Transmitted Power, 4 3

P 3.5wT 2 2

Efficiency, 3

2 100 42.85%3.5

Problem: Calculate total power and efficiency carrier signal is given as, c t 2cos tc

and

modulation index is m 0.5a . The message signal waveform is shown below,

Solution: Modulation Index, max

m(t) km 0.5

a A 2c

Hence, K=1



Mean Square Value of Message Signal, 2K 12m t

3 3

Total Power,

2 2A m tcPT 2 2

4 1

P 2.167WT 2 6

Transmission Efficiency, 0.16

100% 7.19%2.16

Problem: Figure shows positive envelope of A.M. wave, calculate m anda

.

Solution: Modulation Index, A A

max minma A A

max min

300.5

60

Carrier Amplitude,

A A 45 15max minA 30c 2 2

Peak of modulating Signal, m t A m 30 0.5 15max C a

Mean Square Value of Message Signal,

2

152m t 753

Total Power,

2 2A m tcPT 2 2

900 75

P 487.5WT 2 2

Transmission Efficiency, 37.5

100% 7.69%487.5



Trapezoidal Method

If the modulated signal is plotted in X-Y mode on a CRO then the shape of a trapezoid

appears as shown below,

L L1 2m

a L L1 2

It is generally used for non – periodic signals.

Here, L2

represents the minimum amplitude and L1

represents maximum amplitude.

Solved Examples

Problem: If AM wave pattern for non – periodic, signal in CRO is shown, what is value of

Modulation Index?

Solution: From the figure, minimum amplitude L 02

Modulation Index, L L1 2m

a L L1 2

1

Problem: In trapezoidal method, if ratio of short height to long height is 0.65, what is value

of modulation index?

Solution: Ratio of short height to long height is, L2

L1

Modulation Index,

L21

L L L1 2 1m

a LL L 21 2 1L1

1 0.65 0.350.212

1 0.65 1.65

Multi-tone Message Signal

Multi-tone message signals are those which have multiple frequency components. As for an



example any periodic signal can be resolved into Fourier Series and so it can be treated as

Multi-tone Signal.

Let m t A cos t A cos t A cos t ...................1 m1 2 m2 3 m3


s(t) A A cos t A cos t ........... cos tc 1 m1 2 m2 c

A A1 2s(t) A 1 cos t cos t ........ cos t

c m1 m2 cA Ac c

s t A 1 m cos t m cos t ....... cos tc a1 m1 a2 m2 c

Here, mai

represents modulation index for ith tone

Modulation Index, 2 2 2m m m m ...........a a1 a2 a3

B.W. max 2 ,2 ,2 ,............1 2 3

B.W. 2max , , , ............1 2 3

Frequency spectrum

The frequency spectrum for modulated multi-tone signal is as shown below,

Solved Examples

Problem: If equation of AM wave is given as:

3 3 6s t 10 5cos 2 10 t 2cos 4 10 t cos 2 10 t

Calculate, (i) ma

(ii) PT

(iii) Bandwidth

Solution: Modulation Index for different tones present in the signal are,



5 2

m 0.5 and m 0.2a1 a210 10

Modulation Index, 2 2m 0.5 0.2a 0.29 0.5385

Power Transmitted,

2 2 2A m 0.5385c aP 1 50 1T 2 2 2

= 57.25W

Bandwidth,

3 3B.W. 2max 10 ,2 10 4kHz

Transmission Efficiency,

2

50 0.5385100% 12.66%

2 57.25

Low Level and High Level Modulation

The first figure below shows the block diagram of low level modulation and second figure

shows block diagram shows block diagram for high level modulation

In case of low level, generation of AM takes place in initial stage only and this generated AM

is modified by using a linear device which may be a class A or class B amplifier.

In high level, modulation takes place in final stage of amplifier and modulation circuitry has

to handle high value of power.

These are two basic techniques to implement Amplitude Modulation.

The differences in both types of modulation techniques are listed below,



Low Level Modulation High Level Modulation

Modulation occurs is initial stage of

amplifier

Modulation occurs is final stage of

amplifier

Modulation circuit handles low power Modulation circuit handles high power

Simple circuit Complex Circuit

Low Power Audio Amplifier is required High Power Audio Amplifier is required

Class A amplifier is used Class C amplifier is used

Low efficiency High Efficiency

Low Distortion High Distortion

Generation of AM

There are two basic methods for generation of Amplitude Modulated Wave,

Switching Modulator

Non-Linear Device

Switching Modulator

The input voltage to the circuit is sum of message and

carrier signal.

During positive half cycle of input, Diode is ON

V Vo i

The practical and ideal characteristics of Diode are shown

Below,

Practical characteristics are non-linear due to knee voltage. If knee voltage is neglected the

characteristics become linear.

This diode switching operation can also be assumed as multiplication of input signal by an

infinite square pulse shown below,



Here, the switching pulse is 1 during the positive half cycle so that input and output are

same and switching pulse is 0 during negative half cycle as Diode is OFF so output becomes

zero.

Positive and negative cycles are assumed based on cosine function which is the carrier wave.

The time period of square wave,

p

2T

c

Since, it is a periodic function, it can be represented in terms of Fourier Series

1 2 2p(t) cos t cos3 t ............

c c2 3

The output of switching modulator is then,

1 2 2V A cos t m t cos t cos3 t ....o c c c c2 3

m t 2 22V A cos t A cos t m t cos t .....o c c c c c2

Desired components are (I) and (IV) out of all the components present after multiplication.

If one selects a LPF with cut-off frequency c m

then first and fourth component can

be extracted but in that case 2nd component is also present at the output which is undesired.

Hence, one has to take a BPF between c m

and c m

Here, m

is maximum frequency of message signal.

Modulated Signal is the output of filter,

A 2cs t cos t m t cos tc c2

A 4c 1 m t cos tc2 A

c

4ka A

c

amplitude sensitivity



Modulation Index, m m t .kmaxa a

Using Non-Linear Device

Suppose, Non-Linear Device has the input-output relation as,

2V a v a vo 1 i 2 i

Input Voltage, V m t A cos ti c c

2

V a m t A cos t a m t A cos to 1 c c 2 c c

2 2 2V a m t a A cos t a m t a A cos t 2a A m t cos to 1 1 c c 2 2 c c 2 c c

Out of these terms the desired terms are second and fourth term.

The band of Filter is, ,c m c m

In order to remove the third term from the output of band pass filter, the necessary

condition is,

2m c m

where 2m

is frequency of third term

After passing through a BPF,

V t a A cos t 2a A m t cos to 1 c c 2 c c

2a2a A 1 m t cos t

1 c ca1

Modulation Index, 2a

2M.I. . m tmaxa

1



Solved Examples

Problem: What is Modulation Index of generated A.M. wave where 2V V 0.1V2 i i

and input voltage is V 4cos200 t 2cos2000 ti

?

Solution: Input Voltage, V 4cos200 t 2cos2000 ti

2

V 4cos200 t 2cos2000 t 0.1 4cos200 t 2cos2000 t2

2 2V 4cos200 t 2cos2000 t 1.6cos 200 t 0.4cos 2000 t 1.6cos200 tcos2000 t2

Here, second and fifth term are of interest in Amplitude Modulation. So, we apply a filter of

band pass frequency,

1.6cos200 tcos2000 t = 0.8 cos2200 t cos1800 t

Range of BPF : w w w wc m c m

2000 200 2000 200 1800 2200

At the output of filter we have,

V 2cos2000 t 1.6cos200 tcos2000 ta 2 1 0.8cos200 t cos2000 t

Modulation Index, m 0.8a

Problem: A non – linear device with transfer characteristics

2i 10 2V 0.2Vi i

mA is

applied with a carrier of 1V amplitude and a sinusoidal signal of 0.5V in series. If output of

only frequency components of Amplitude Modulated wave is considered, than what is the

depth of modulation?

Solution: Depth of Modulation or Modulation Index, 2a

2m m tmaxa a

1

0.2

m 2 0.5 0.1a 2

Detection method of Amplitude Modulation

The following three methods are used for demodulation of Amplitude Modulation,

(i) By use of non – linear device/square law method.

(ii) By use of synchronous detections.

(iii) By use of envelope detector.



The input to receiver is the Amplitude Modulated Wave,

s t A 1 k m t cos tc a c

By use of Non – Linear device

Input of Demodulator, s t A cos t A k cos tc c c a c

Assume the input-output relation of demodulator is, 2V a V a V1 1 i 2 i

2

V a A cos t k m t A cos t a A cos t A k m t cos t1 1 c c a c c 2 c c c a c

2 2V a A cos t a A k m t cos t a A cos t1 1 c c 1 c a c 2 c c

2 2 2 2 2 2a A k m t cos t 2a A k m t cos t2 c a c 2 c a c

Since, 22cos t 1 cos2 tc c

This expression can be expressed as,

V a A cos t a A k m t cos t1 1 c c 1 c a c

2a A2 2 2 22 c 1 k m t 2k m t cos2 t k m t cos2 t 2k m t cos ta a c a c a c2

The desired output is the signal proportional to message signal,

Desired output

2a A2 c 2k m t

a2

To extract this term we will Use a LPF with cut – off frequencym

, followed by a capacitor.

Due to presence of 2m t component, there may be harmonic distortion in output. For

neglecting harmonic distortion,

2k m ta 1

2 2k m ta

k m t 2a

It means this method is not suitable for large amplitude message signal, It is generally

preferred if message signal amplitude is less than 1volt.



Solved Examples

Problem: What is a value of maximum Harmonic Distortion possible in non – linear device

output?

Solution: Let m t A cos tm m

2 2 2y t k A cos t 2k A cos ta m m a m m

2 2k Aa m cos2 t 2k A cos t

m a m m2

% Harmonic Distortion

2 2k Aa m 100%

2 2k Aa m

2 2k Aa m 100% 25 k A

a m4

Maximum Harmonic Distortion = 25% max k A 1a m

Problem: Assertion: Square Law Detector are not particularly satisfactory for detection if

modulating signal base amplitude is greater than 1V.

Reason: A square detector H.D. is as high as 25% for completely modulated signal.

Solution: Both Assertion and Reason are correct but reason is not correct explanation of

assertion. Reason will be that is causes harmonic distraction for high amplitude ( 2m t will

be denominating).

Synchronous Detection

Modulated Signal, s t A 1 k m t cos tc a c

Here, synchronization means frequency of carrier and frequency of local oscillator are in

synchronization.

V t s t c t1

A 1 k m t cos t A' cos tc a c c c

2V t A A' 1 k m t cos t1 c c a c

A A

c c 1 k m t 1 cos2 ta c2

A A'

c cV t 1 k m t cos2 t k m t cos2 t1 a c a c2



Output of LPF: A A' A A'

c c c ck m ta2 2

Output of capacitor: A A'

c c k m ta2

The capacitor blocks DC Voltage and so the output is proportional to message signal. So,

there is no distortion in output.

Envelope Detector

For the case of under-modulation, the modulated wave and its envelope are shown below,

Here, if we are able to extract the envelope then message signal can be extracted from the

modulated waveform without the loss of any information.

When modulation index is 1 i.e. the case of critical modulation the modulated waveform and

its envelope are shown below,

But the information is lost when the modulated waveform touches the axis so message

signal cannot be completely recovered.



In the figure shown above, the modulation index is greater than 1 and so it is a case of Over-

Modulation.

Information is lost due to overlapping of positive and negative envelope so envelope

detection will not yield the message signal.

Hence, Envelope detector will work efficiently only when modulation index is less than 1.

Basic Operation

In positive half, diode will be ON and capacitor starts changing through resistance Rs

.

The time constant for charging of capacitor is, R Cs c

.

Here capacitor will charge fast upto peak value and hence should be small.

1

R Cs c f

c

If time constant is small, capacitor charging will be fast.

In negative half, diode becomes off and start discharging through RL

. In this case, R CL

becomes very important.

If R CL

is very small, then there will be fluctuation in the output and high value of ripple

will be present. Due to this there may be some ripples at the output. These spikes/ripples

fluctuations will degrades the performance of output.

1R C

L fc

If R CL

is very high; initially capacitor will be charged upto peak and due to high value of

R CL

. Let voltage discharge upto some value in negative half cycle (during period when

diode is off). Next peak at input having high value will charged capacitor to higher voltage.

During negative half cycle, capacitor discharges very slowly and cannot reach the level of

next peak. If having very small value, voltage is not sufficient to make diode ON and so, it will



miss that peak; so an amount of information will be lost (small peaks) due to large discharge

time this is called as Diagonal clipping.

The capacitor shall be able to discharge by the next peak arrives so that no information is

missed and hence, 1

R C TL m f

m

1 1R C

Lf fc m

{for satisfactory operation}

Solved Examples

Problem: An A.M. signal is detected using an envelope detector carrier frequency are 1MHz

and 2kHz, then what is value of time constant for envelope detector?

(a) 500 s (b) 20 s

(c) 1 s (d) 0.2 s

Solution: For satisfactory detection of message signal,

6 3

1 1RC

10 2 10

1 s RC 500 s

Optimum Time Constant

Optimum value of RC to avoid diagonal clipping in a single tone sinusoidal signal is given by,

21 ma

RC.m

m a

To avoid diagonal clipping, rate of discharge of capacitor voltage should be greater than rate

of decrease of modulating voltage.



Modulated Waveform, s t A 1 m cos t cos tc a m c

Envelope of Modulated Signal is, V A 1 m cos tenvelope c a m

When the capacitor discharges, the capacitor voltage is given by,

ttRCV t V e V 1

c envelope envelope RC

Capacitor should discharge before next peak arrives so no information is lost. Hence,

dVdVenvelopec

dt dt

VdVenvelopec

dt RC

dVenvelope

sin t A m A m sin tm m c a c a m mdt

Venvelope

A m sin tc a m mRC

Thus,

A 1 m cos tc a mRCA m sin t

c a m m

1 m cos ta mRC

m sin ta m m

for maximum value of RC so that time is no diagonal clipping differentiate w.r.t. t;

2m sin t .m sin t m cos t 1 m cos tm a m a m m a m m a m

2m sin t cos t 1 m cos ta m m a m

m cos ta m

1 m cos ta mRC

m sin ta m m

21 ma

2m 1 ma m a



21 ma

RCmax m

a m

Note:

This is valid for single tone sinusoidal signal only.

Here value of time constant depends upon max modulation frequency and modulation

index ma

.

Envelope detector is an Asynchronous detector (since no oscillator is used).

Solved Examples

Problem: If 6s t 10 1 0.5cos 2 500t cos 2 10 t and demodulated by envelope

detector, than what is maximum value time constant to avoid diagonal clipping?

Solution:

21 m1 0.25a

RC 0.55max .m 0.5 2 500

m a

Advantages and Disadvantages of DSB-FC

The biggest disadvantage is that transmitted power is wasted and efficiency is minimum.

One of the advantages is that it has simple modulation and demodulation method.

DSB – SC (Double Side Band - Suppressed Carrier)

Amplitude Modulated signal with DSB-FC is,

s t A cos t m t cos tc c c

If we suppress the carrier i.e. carrier is not present in the output signal. Then, it is known as

Double Side Band with Suppressed Carrier.

s t m t cos tc

Due to carrier term removed the power in Transmitted Signal is,

2maP P

T c 2

Hence, entire power belongs to message signal and no energy is wasted in transmitting the

carrier. So, transmission efficiency % 100%



Solved Examples

Problem: In DSB – FC, if m 1a , what is % saving in transmitted power if:

a) Only carrier is suppressed

b) Carrier and one side band is suppressed.

Solution: For 100% Modulation Index, the transmitted power is

2m 1 3aP P 1 P 1 PT c c c2 2 21

If carrier is suppressed then, transmitted power becomes,

1P PT c22

3 1P Pc c2 2% saving3

Pc2

= 66.67%

Now, if one sideband as well as carrier is suppressed the transmitted power becomes half

1P PT c42

3 1P Pc c2 4% saving 100%3

Pc2

= 83.33%

Problem: Repeat previous question for m 0.5a

Solution: IN DSB-FC transmitted power is,

2m 0.25aP P 1 1 P 1.125PT c c c2 21

With suppressed carrier the transmitted power becomes,

2m 0.25aP P P 0.125PT c c c2 22

1.125 0.125% saving 100% 88.89%

1.125



If one sideband is also suppressed,

0.25P PT c42

1.0.625 0.0625% saving 100% 94.44%

1.0625

So, in order to save more transmitted power:

(i) Carrier should be suppressed

(ii) One sideband should be suppressed.

(iii) Modulation Index should be small

DSB – SC time domain representation

In DSB-SC, whenever message signal crosses zero level i.e. it becomes negative from positive

or vice versa then modulated signal also changes polarity. This can be called as 1800 phase

shift in the sinusoidal carrier wave.

When m(t) is negative. Suppose, m(t) = -g(t)

Then, modulated signal becomes,

0

s t m t cos t g t cos tc c

s t g t cos t 180c

DSB – SC Frequency Domain Representation



Suppression of carrier means first generate DSB – FC and then suppress the carrier.

s t m t cos tc

The spectrum of Modulated Signal is,

1S w M M

c c2

From, the spectrum of modulated wave,

Bandwidth, radBW 2secm

Generation of DSB-SC Signal

There are two means of generating a DSB-SC signal,

Non-Linear Device

Linear Device

Non – Linear Device

The output of Non-Linear Device is,

2V a V a V1 1 i 2 i

The input to Non-Linear Device is sum of message and carrier signal,

V m t A cos ti c c

The output of Non-Linear Device is,

2 2 2V a m t a A cos t a m t A cos t 2A m t cos t1 1 1 c c 2 c c c c

To generate DSB – SC,

Make a 01 so that non – linear device will be simply square law device. Then we can use a

Band Pass Filter of frequency toc m c m

Then, at the output second and fourth terms are present which means DSB-FC is generated.

The carrier is then suppressed to generate DSB-SC signal.



By use of Linear Device

Balanced/Product Modulator

The two signals generated in this modulator are,

s t A m t cos t1 c c

s t A m t cos t2 c c

The output of modulator is,

s t s t 2m t cos t1 2 c

The output of Balanced Modulator is product of Message and Carrier Signal.

Ring Modulator

The input to Ring Modulator is the message signal. The carrier signal controls the switching

of Diodes to transfer the input signal to output.

(i) For m(t) = 0 and D ,D ON1 2

and D ,D OFF3 4

The output is zero.



(ii) For m(t) = 0 and D ,D ON3 4

and D ,D OFF1 2

The output is zero.

(iii) For m t 0 and D ,D ON1 2

and D ,D OFF3 4

Message Signal is transferred to output by means of two transformers with same polarity. So,

output is same as message signal.

(iv) for m t 0 and D ,D ON3 4

and D ,D OFF1 2

Message Signal is connected to output as shown below,

Message Signal is transferred to output with opposite polarity due to cross connection.



Since modulated wave becomes zero when m(t) becomes zero and there is phase reversal

also at m(t) = 0 as shown by circled points in output wave; hence it is DSB – SC.

In ring modulator, if one frequency is f1

and other is f2

, output will have frequency

f f and f f1 2 1 2 and no other frequency will be present.

Multiplication in time domain = convolution in frequency domain and by convolution the

signal gets translated to new frequency so frequency shifts which is happening by this

method. Hence, both methods are generating DSB – SC only.

Detection of DSB – SC

The following methods can be used for detection of message signal from DSB-SC Modulated

Signal.

Switching Modulator

The operation of diode can be replaced by switching waveform shown below,

The Fourier Series of the switching waveform can be expressed as,

1 2 2p(t) cos t cos3 t ............

c c2 3

The output of diode is,

V V t .p to1 i

1 2 2V m t cos t cos t cos3 t ............o1 c c c2 3



1 cos2 t1 2 2cV m t cos t m t m t cos t

o1 c c2 2 3

m t m t1V m t cos t cos2 t ........o1 c c2

To get message signal, one can design a LPF of cut – off frequency practically, it can be taken

as 2m

Synchronous Detection/Coherent Detection

Synchronization between modulating signal and carrier signal,

The carrier multiplied to message signal and the one used for demodulation needs to have

same frequency and hence synchronized to each other.

V m t cos t A cos t1 c c c

A m tc cos 2 t cos

c2

If we use a LPF for detection then the output is,

A m tcV cos

o 2

Quadrature Null Effect

If

2

i.e. the demodulating carrier signal is in quadrature to modulated signal, then

A m t

cV m t cos t A sin t sin 2 t1 c c c c2

After filtering with a Low Pass Filter, there is no message signal recovery at the output. This

effect is called as Quadrature Null Effect.

If 0 , then

A m tcV m t cos t A cos t 1 cos 2 t

1 c c c c2

After filtering with a Low Pass Filter, there is complete message signal recovery at the output.



A m tcV

o 2

In this case, the demodulating signal is in perfect synchronization with the modulating signal.

General Error in case of Synchronous Detection

In general, the signal used for demodulation may have slightly different frequency than the

modulated signal.

Then,

c t A' cos tc c

A' m tcV t m t cos t A' cos t cos 2 t cos t

i c c c c2

After passing through the LPF, output is

'A m tcy t cos t

2

Case 1: 0 and 0

No distortion, no attenuation

m t

y t2

Case 2: 0 and 0

Only attenuation as amplitude is changed.

m t cos

y t2

Case 3: 0 and 0

'A m tcy t cos t

2

Here, attenuation as well as distortion is present in the output.

Case 4: 0 and 0

Only Distortion is present due to time dependent term in the output.

m t cos t

y t2



Costas Receiver

A Costas Loop is a phase locked loop used for recovery of carrier signal from DSB-SC

Modulated Signal.

Modulated Signal, s t m t cos tc

The output y t and y t1 2

will be,

y t A m t cos t cos t1 c c c

A m tc cos 2 t cos

c2

A m tcy t A m t cos t cos t 90 sin 2 t sin

2 c c c c2

When both these signals are passed through LPF then the outputs are,

A m t

cs t cos1 2

A m t

cs t sin2 2

For small value of ; cos 1and sin

A m t

cs t1 2

A m t

cs t2 2

Phase discriminator is a device which generates DC voltage corresponding to phase

difference Φ. This DC voltage will modify phase in value such that VCO phase and input

phase are same so, by Costas Receiver, the receiver circuit will become complex and this is

for only low phase errors.

Solved Examples

Problem: A message signal band – limited to frequency ' f 'm

has power 'P 'm

, what is power

of output shown in figure?



Solution: The power content in the signal m t cos tc

2m tPm 2

Output of multiplier is,

X t m t cos t cos t1 c c

m tcos 2 t cos

c2

The output of Low Pass Filter becomes,

m t

y t cos2

2m t1 2Power cos4 2

P

2mPower of Output cos4

Problem: Determine the output of the system shown below,

Assume 2sin t

m tt

, c t cos 200 t and

sin 198 tn t

t

Solution: Output of Multiplier is, 2

X t sin2 tcos200 t1 t

1

sin202 t sin198 tt

Output of Adder is,

sin202 t sin198 t sin198 t sin202 t

X t n t1 t t t t

Frequency of output, 202

f 101Hzm 2



Since, cut-off frequency f 1Hzc is less than the frequency of message signal. Hence,

y t 0 which means no signal will pass through LPF.

Problem: If the input to a coherent detector is DSB – SC signal plus noise, than what is value

of noise at output?

Assume n t n cos t n sin tI c Q c

and c t cos tc

nI: In-phase component of Noise

nQ

: Quadrature component of Noise

Solution: The input to multiplier is,

s t n t m t cos t n cos t n sin tc I c Q c

The output of multiplier is,

2 2X t m t cos t n cos t n cos tsin t1 c I c Q c c

nm t n m t nQI IX t cos2 t sin2 t

1 c c2 2 2

The output of LPF is,

m t nIy t

2

Hence, in-phase component of noise is present at the output of system.

Types of Envelope

There are three types of envelope,

Pre Envelope

An analytic signal is a complex signal created by adding a signal with its Hilbert Transform in

Quadrature. It is also known as pre-envelope of a signal.

ˆx t x t jx tp

Here, x̂ t is the Hilbert Transform of x(t)



By taking the pre-envelope the negative frequency components are discarded and the real

signal can be easily recovered from complex signal by discarding the imaginary part.

Natural Envelope

The magnitude of pre-envelope of a signal is known as natural envelope of a signal.

Natural Envelope: x t

p

Complex Envelope

Analytic Signals are often shifted in frequency or down converted to 0Hz. It possibly creates

negative frequency components. This helps in reducing the maximum frequency so that

sampling rate also reduces. Complex Envelope is given by,

j t

cx t x t ec p

Solved Examples

Problem: A modulated signal is given by

ats t e cos t t u tc

where andc

are positive constants, calculate:

(i) Pre envelope (ii) Complex envelope (iii) Natural envelope

Solution: Pre-envelope, ˆs t s t js tp

at ats t e cos t t u t jsin t t ep c c

This happens because Hilbert transform of cosine signal is sine signal.

j t

at cs t e u t ep

j tat ce e u t

Complex Envelope,

j tj t j t

at cc cs t s t e e e u t ec p

j tats t e e u t

c

Natural Envelope,

j t

at atcs t e e u t e u tp



Note: In questions, if only envelope is asked, then envelope taken is natural envelope.

If s t s cos t s sin tI c Q c

Natural envelope: 2 2s t s t s sp I Q

Problem: If a modulated signal is given as, x t m t cos tc . Determine natural

envelope of the signal.

Solution: Method-1

Pre-Envelope, x t m t cos t jm t sin tp c c

x t m t cos t jsin tp c c

j t

cm t e

Natural Envelope, x t m tp

Method-2

x t m t cos tcos m t sin tsinc c

m t cos t m t sin tI c Q c

Natural Envelope 2 2x t m t m tp I Q

2m t m t

Problem: For the system shown below,

Determine the output if, sin199 t

n tt

,

2sinn tm t

t

and c t cos200 t

Solution: Output of first multiplier,

1X t m t c t sin 202 t sin 198 t1 t

sin202 t sin198 t sin199 t

m t c t n tt t t

Output of second multiplier,



2sin202 t.cos200 t 2sin199 tcos200 t

c t m t c t n t2t 2t

2sin198 tcos200 t

2t

1c t m t c t n t sin402 t sin2 t sin399 t sin t sin398 t sin2 t

2

Frequency components present in the output,

402f 201Hz1 2

2f 1Hz2 2

399f 199.5Hz3 2

f 0.5Hz4 2

398f 199Hz5 2

Hence, at the output of LPF only 1Hz and 0.5 Hz components are present.

1

y t 2sin2 t sin t2t

Problem: Consider a system shown in figure. Let X(f) and Y(f) denote Fourier Transform of

x(t) and y(t) respectively. If ideal HPF has cut – off frequency of 10kHz. Then positive

frequencies where Y(f) will have spectral peaks are?



Solution: The spectrum of the signal output of first multiplier is,

The output of High Pass Filter will look like,

The output of second multiplier will be,

So, the positive frequency peaks lie in the spectrum at 2kHz and 24kHz.

Alternatively, we can directly solve by observing that input has peaks at -1kHz and 1kHz, so

after first multiplier peaks will lie at 9kHz and 11kHz. After High Pass Filter, only 11kHz and -

11kHz peak passes and after second multiplier -11kHz is shifted to 2kHz and 11kHz is shifted

to 24kHz.

Problem: A DSB – SC signal is to be generated with carrier frequency /fc

=1MHz using a non

– linear device with input-output characteristics 3V a V a Vo 1 i 1 i

where a and ao 1

are

constants. The output of non – linear device can be filtered by an appropriate BPF. Let

V A cos 2 f t m ti c c

where m(t) is message signal. Then value of fc

in MHz is?



Solution: The input-output relation of Non-Linear Device is,

3V a V a Vo o i 1 i

3

V a A cos2 f t m t a A cos2 f t m to o c c 1 c c

3 3 3V a A cos2 f t m t a A cos 2 f t a m to o c c 1 c c 1

2 2 23a m t A cos 2 f t 3a m t A cos2 f t1 c c 1 c c

This can also be expressed as,

3 3 3V a A cos2 f t m t a A cos 2 f t a m to o c c 1 c c 1

1 cos 4 f t2 2c3a m t A 3a m t A cos2 f t

1 c 1 c c2

Desired component for DSB-SC generation,

23a A1 cy t m t cos 4 f t

c2

The DSB-SC for a carrier frequency of f 'c

must be,

m t cos2 f 'tc

On comparing with generated DSB-SC,

4 f 2 f 'c c

f ' 1MHzcf 2.5MHz 250kHz

c 2 2

Problem: A 100MHz carrier of 1V amplitude and 1MHz modulating square of 1V amplitude

are fed to a modulator. If output of modulator is passed through a HPF with cut – off

frequency of 100MHz and output of filter is added to 100MHz signal of 1V amp. And 090

phase shift as shown in figure, than what is resultant of envelope of the signal?



Solution: The input to the modulator is,

6V cos 2 100 10 ti

The output of modulator is,

6 6V cos 2 100 10 t .cos 2 10 t1

=1 6 6cos2 101 10 t cos2 99 10 t2

When it is passed through a HPF, the output is

1 6V cos2 101 10 t

2 2= 1 8 6cos2 10 10 t

2

After the adder the output is,

1 8 6 6y t cos2 10 10 t sin2 10 t2

1 18 6 8 6 6y t cos 2 10 t cos 2 10 t sin 2 10 t sin 2 10 t sin 2 10 t2 2

1 16 6y t cos 2 10 t cos t sin 2 10 t 1 sin tc c2 2

The natural envelope of the signal is,

2 21 1 1 1 16 6 2 2y t cos 2 10 t sin 2 10 t 1 cos t sin t 1 sin t

m m m2 2 4 4 2

Here, 62 10 rad / secm

5 1 6y t sin 2 10 t4 2

Quadrature Amplitude Modulation (QAM)

QAM is a modulation scheme by which we can transmit two message signal by Amplitude

Modulation of two carrier waves which are in phase quadrature to each other. Both the

signals are then summed and transmitted together.

The modulated signal by QAM is of the form,

s t s t s t m t A cos t m t A sin t1 2 1 c c 2 c c

A motivation for the use of quadrature amplitude modulation comes from the fact that a

straight amplitude modulated signal, i.e. double sideband even with a suppressed carrier

occupies twice the bandwidth of the modulating signal. This is very wasteful of the available

frequency spectrum. QAM restores the balance by placing two independent double sideband

suppressed carrier signals in the same spectrum as one ordinary double sideband

suppressed carrier signal.



Generation of QAM

By 900 phase shift we generate a second carrier A sin tc c

in phase quadrature with the

carrier signal A cos tc c

Detection of Message Signal in QAM

The output of first multiplier is,

y t s t .A cos t1 c c

2 2 2A m t cos t A m t sin tcos tc 1 c c 2 c c

21 cos2 t A m t2 c c 2y t A m t sin2 t

1 c 1 c2 2

By the use of LPF, the output is,

2A m tc 1y

01 2



The output of second multiplier is,

y t s t .A sin t1 c c

2 2 2A m t sin tcos t A m t sin tc 1 c c c 2 c

2A m t 1 cos2 t2 c 2 cy t A m t sin2 t

1 c 1 c 2 2

By the use of LPF, the output is,

2A m tc 2y

02 2

In QAM, we transmit two signals using the same Bandwidth and so it leads to efficient

utilization of Bandwidth.

Review of Hilbert Transform

Fourier Transform changes time domain signal to frequency domain but Hilbert transform

changes time domain to time domain signal only but phase is changed.

090 V

0

0

90 V

0

Frequency Domain Representation of this signal is,

j2

j2

e , 0F.D. t

e , 0

j, 0H jsgn

j, 0

The time domain representation of this signal is,

FT1jsgn w

t

Since h t 0 for every value of t, hence h t 0 for t < 0; hence it is non – casual system.

Hilbert Transform of a signal is orthogonal to its Hilbert Transform.

x(t) and x̂ t are orthogonal to each other.

Also, d dH.T.

ˆx t x tdt dt



Solved Examples

Problem: What is Hilbert Transform of x t sin to

?

Solution: The spectrum of signal x(t) is,

Xo oj

Fourier Transform of Hilbert Transform is,

H jsgn

Y X H sgno o

Yo o

Taking Inverse Fourier Transform,

y t cos to

Single Side Band Modulation – Suppressed Carrier (SSB - SC)

The spectrum of different modulation schemes is shown below,

When we use only one of the sidebands instead of two sidebands it is known as Single

Sideband.



In the above spectrum, we have only considered Upper Sideband for SSB Modulation.

In the above spectrum we have considered the lower sideband out of the two sidebands.

SSB – SC requires minimum value of Bandwidth and minimum value of transmitted power.

Since, information is carried out by 2 sidebands and they are images of each other, hence

they carry same information.

Bandwidth is noise – filtering characteristics of the system, means as Bandwidth increases the

effect of Noise increases. The power spectral density of Noise is constant so the higher the

Bandwidth more is the Power contained in Noise and so more is the effect of Noise.

For one to many communication, we will prefer either DSB – SC or DSB – FC since design for

receiver needed is simple for one to one communication, we will use SSB – SC modulation.

SSB can be effectively used for FDM and generally used in point – to – point communication

and mobile communication.

Generation of SSB – SC

To generate SSB-SC, we first create DSB-FC and suppress the carrier in order to generate

DSB-SC and then either upper or lower sideband is suppressed to create SSB-SC.

Let m t A cos tm m

and c t A cos tc c

DSB-FC modulated signal is,

s t A m t cos tc c A cos t A cos t cos t

c c m m c

If the carrier is suppressed,

s t A cos t cos tm m c

Am cos t cos t

c m c m2

A A

m ms t cos t cos tc m c m2 2

The first term has higher frequency and second one has lower frequency. So, first one is

called as Upper Sideband and second term is called as Lower Sideband



For Upper Sideband,

Amy USB cos tcos t sin t sin t

c m c m2

ˆm t m tcos t sin t

c c2 2

Here, m̂ t is Hilbert Transform of m(t)

If ˆm t A cos t and m t A sin tm m m m

For Lower Sideband,

Amy LSB cos tcos t sin t sin t

c m c m2

ˆm t m tcos t sin t

c c2 2

Generation of SSB – SC

There are three methods used for generation of SSB-SC signal,

(1) Frequency discriminator

(2) Phase discriminator

(3) Weaver discriminator

Frequency Discriminator or Filter Method

This method first generate DSB – SC and a filter is used to remove unwanted sidebands.

This filter may be RC, LC or mechanical filter depending on carrier frequency.

This filter may have flat pass band and extremely highly attention outside the pass band.

The frequency generated by this filter may be very low which will be translated to a higher

frequency by use of local oscillator.

This dc filter may be used upto a frequency of 100kHz and is bulky in size which can

create a problem. Hence mechanical filter will be preferred and due to this reason,

mechanical filter are preferred.

Advantages

This method gives side – suppression ratio upto 90 dB which is quite adequate.

Side band filter also help to attenuate carrier if present in Balanced Modulator.

In this case, Pass Band is sufficiently flat and wide.



Disadvantages

They cannot generate SSB at every frequency.

They are bulky in size.

At lower audio frequency, design becomes expensive.

The spectrum at various steps of this modulator is shown below,

Step-1: Output of Multiplier or Balanced Modulator is DSB-SC signal

Step-2: Filter Spectrum is as shown below,

The output of the filter will have the spectrum,

Step-3: This spectrum must be translated to higher carrier frequency by the use of Local

Oscillator. The output will have a spectrum of,



By this method, we can generate SSB-SC in which only upper sideband is present. Similarly,

by changing the signal pass band we can generate the Lower Sideband signal.

Phase shift/Phase Discriminator/Hartley Modulation Method

Here, BM represents Balanced Modulator which multiples message signal with carrier signal.

H.T. represents Hilbert Transform.

The outputs of Balanced Modulators are,

S t cos t cos t1 m c

S t sin t sin t2 m c

These signals can also be represented as,

1S t cos t cos t1 c m c m2

1S t cos t cos t2 c m c m2

The output of summer will be either upper sideband or lower sideband based on whether

signals are added or subtracted.

Addition: y t S S cos t LSB1 2 c m

Subtraction: y t S S cos t USB1 2 c m

Advantages

Bulky filter are replaced by small filter.

It can generate SSB at any frequency.

Low audio frequency may be used for modulation.

There is easy switching from one sideband to other side band.

To generate SSB at high frequency up – conversion and repetitive mixing is not required.



Disadvantages

It requires a complex audio frequency phase shift network because it has to work for low

frequency range.

Sideband suppression depends on phase relationship between two phase shifters else

sideband may not be successfully suppressed.

Output of 2 balanced modulators must be exactly same otherwise correlation is not

complete and desired frequency component is not achieved.

Solved Examples

Problem: In following scheme, if spectrum M(f) of m(t) is shown in the figure, then spectrum

Y(f) of y(t) will be?

Solution: This is a phase discrimination method of generation of SSB-SC and due to addition

of both signals, LSB will be generated.

The DSB-SC spectrum will be as shown below,

When LSB is created the spectrum looks like as shown below,



Weaver Method

In Weaver Method, the band of interest is first translated to zero by modulating with a carrier

wave of frequency lying at the center of band, B

fc1 2

This pair of signals is passed through LPF to remove sidebands that is not centered at zero.

Then it is modulated to desired carrier frequency by a pair of quadrature mixers.

Assume m t cos 2 f tm

The output of first oscillator is, v t cos 2 f t1 1

Due to 900 phase shift, v t sin 2 f t2 1

After the multiplier or Balanced Modulator the output is,

1v t cos 2 f t cos 2 f t cos 2 f f t cos 2 f f t3 m 1 1 m m 12

1v t cos 2 f t sin 2 f t sin 2 f f t sin 2 f f t4 m 1 1 m m 12

After the LPF only the low frequency component is passed through,

1

v t cos 2 f f t5 1 m2

1

v t sin 2 f f t6 1 m2

The output of second oscillator is,

v t cos 2 f t7 2

and v t sin 2 f t8 2



The output of second set of Balanced Modulators is,

1v t cos 2 f f t cos 2 f t9 1 m 22

1v t cos 2 f f f t cos 2 f f f t9 2 1 m 2 1 m4

Similarly,

1v t sin 2 f f t sin 2 f t10 1 m 22

1v t cos 2 f f f t cos 2 f f f t10 2 1 m 2 1 m4

The output of Weaver Modulator is,

1

y t cos 2 f f f t2 1 m2

If B B

f and f f1 2 c2 2

Then,

1y t cos 2 f f t

C m2 which is the Upper Sideband signal and similarly Lower

Sideband signal can be generated.

Comparison of Techniques

Parameter Filter Phase Shift Weaver

900 phase shift Not Required Required Required

SSB Generation Not for all frequency All Frequencies All Frequencies

Up-conversion Required Not Required Not Required

Complexity Less Moderate Highest

Design Aspect Size, Weight and

Frequency

Phase Shift of 900 Symmetry of B.M.

Switching Capacity Not Possible Easy Easy

Detection of SSB – SC

Synchronous Detection



SSB Modulated Signal is,

ˆs t m t cos t m t sin tc c

The output of Balanced Modulator is,

2 ˆy t m t cos t m t sin t cos t1 c c c

ˆm t m t1 cos2 t sin2 t

c c2 2

After Passing through LPF,

m t

y t2

Phase Synchronization problem

SSB Modulated Signal is,


If the carrier input to Balanced Modulator has a phase shift with respect to modulated signal,

c t cos tc

ˆy t m t cos tcos t m t sin tcos t1 c c c c

ˆm t m ty t cos 2 t cos sin 2 t sin1 c c2 2

After passing through LPF,

ˆm t m t

y t cos sin2 2

If

m t

0, y t2

m̂ t; y t

2 2

It means there is no Quadrature Null Effect problem in the SSB – SC due to phase error.

Pilot carrier in SSB



Due to absence of carrier and one side band, transmitted power and Bandwidth are reduced

but at same time due to absence carrier, receiver will become complex. In order to reduce

this problem, a small power carrier signal is directly transmitted from the transmitter so that

these will be synchronization in frequency of local oscillator and modulated signal.

Same function is served by costar receiver in DSB – SC.

Important points about SSB – SC

(1) Due to absence of carrier, a little change in frequency with hamper quality of

transmitted and received signal hence it cannot be used for transmission of good quality

signal such as music but can be used for speech communication.

(2) Bandwidth and transmitted power is reduced in SSB – SC and there is no Quadrature –

Null effect.

Vestigial side – band modulation (VSB - SC)



The first waveform shows the spectrum of DSB-SC modulated signal. The second and

third waveforms represent Upper Sideband and Lower Sideband variation of VSB signal.

B.W. f fm b

Major advantage of SSB is that it reduced bandwidth equal to half as compared to AM or

DSB – FC but SSB signal are very different to generate because it is difficult to desired side

band and eliminate undesired sideband.

For this purpose, filter must have a very sharp cut off frequency and in case information

contains low frequency signal, then generation of SSB be difficult. This difficulty is generate

by a scheme because VSB – SC which is a compromise b/w DSB – SC and SSB – SC.

Mathematical explanation of VSB – SC (Design of VSB filter)

Output of Multiplier, y t m t cos2 f t1 c

Output, y t y t h t m t cos2 f t h t1 c

y t y t h t y t h d

1 1

y t h m t cos 2 f t d

c

y t cos2 f t h m t cos2 f d sin2 f t h m t sin2 f d

c c c c

We can break the output into two components one in-phase component and other is

Quadrature Component.

s t h m t cos2 f d h t cos2 f t m t

I c c

Taking Fourier Transform,

1S f M f . H f f f f fI c c2

1M f . H f f H f f

c c2



For design of VSB – SC filter,

H f f H f f 1c c

, f W

Thus,

M f

S fI 2

and

m t

s tI 2

Quadrature component of output is,

s t h m t sin2 f d m t h t sin2 f tc cQ

1S f M f . H f f H f fQ c c2j

m tQ

s tQ 2

Thus, the output of VSB modulator is,

m̂ tm t Qs t cos t sin tVSB c c2 2



Generation and Detection of VSB

Generation:

S M M .Hc c I

Detection:

Y w S S1 c c

M H HI c I c

Here, we have neglected 2c

components

1H

o H HI c I c

Y Y .H Mo 1 o

= M(w)

Hence, y t m to

Calculation of Envelope in case of SSB – SC

DSB-FC Modulated Signal is represented as,

s t A m t cos tc c

Comparing this to SSB signal, the in-phase and quadrature component of signal are,

s t A m tI c

and s t 0Q

Natural Envelope A m tc

SSB – SC




For USB Signal,

A A

m cs t cos tc m2

A A A A

c m c ms t cos tcos t sin t sin tc m c m2 2

The in-phase and quadrature component are,

A A

c ms cos tI m2

A A

c ms sin tQ m2

A Ac mEnvelope2

So, envelope of SSB – SC is constant. So, envelope detector cannot be used.

SSB – FC

A A

c ms t A cos t cos tc c c m2

A AAm cms t A cos t 1 cos t sin tsin t

c c m c m2 2

The in-phase and quadrature component are,

Ams 1 cos t A

I m c2

Ams sin t

Q m2

2 22 A AA2 2 2c mmEnvelope A 1 cos t A cos t sin tc m m m m4 4

Envelope

2 2A A2 2c m A A A cos tc c m m4

2A2 mA 1 A cos tc m m4

Envelope

2AmA 1 A cos t

c m m4

Envelope of SSB – SC is constant and independent of message.

In case of SSB – FC, envelope is function of message hence it can be detected by envelope

detector.



ISB (Independent side banding)

In this case, different side bands are carrying different information and that is why very useful

for point – to – point communication.

Figure to Merit (FOM)

SNR oF.O.M.SNR c

SNR o Defined as ratio of average power of demodulated signal to average power of

noise both measured at receiver output.

SNR c Defined as ratio of average power of modulating signal to average power of noise

both measured at receiver input.

FOM for DSB – SC

2 2c A PcSNR

o 4 Nm o

2 2c A PcSNR

c 4 Nm o

Here,

c = constant

Ac

= Carrier power

P = power of message signal

m

= message signal Bandwidth

No

2 = Power spectral density of white noise

FOM for DSB – SC = 1

FOM for SSB – SC = 1

In case of SSB – SC, at the output power is just half of DSB – SC, But F.O.M. will remain same.

F.O.M. in both cases is same as transmission efficiency.



FOM for DSB – FC/AM

2 2A K Pc aSNR

0 2 Nm o

2 2A 1 k Pc a

SNRc 2 N

m o

Figure of Merit,

2 2k P Aa mF.O.M. where P

2 21 k Pa

FOM

2ma

22 ma

If 1

m 1 ; F.O.M.a 3 which is same as Transmission Efficiency

Technical names of Amplitude Modulation methods

DSB FC A3E

DSB SC R3E

SSB FC H3E

SSB SC J3E

VSB C3F

ISB B8E

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