CHAPTER 3

Amplitude
Modulation
3.1 INTRODUCTION TO MODULATION
❑The analog signal to be transmitted is denoted by m(t)
▪ m(t) assumed to be a lowpass signal of bandwidth W
▪ M(f) = 0, for |f| > W
▪ The power content of this signal is denoted by

1 T /2
Pm = lim 
2
m(t ) dt
T → T −T / 2

❑The message signal m(t) is transmitted through the

communication channel by impressing it on a carrier
signal of the form
c(t ) = Ac cos(2f c t + c )

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 INTRODUCTION TO MODULATION Cont…
Ac Carrier amplitude
fc Carrier frequency
c Carrier phase - The value of c depends on the choice of
the time origin
we assume that the time origin is chosen such that c = 0
❑ We say that the message signal m(t) modulates the carrier
signal c(t) in either amplitude, frequency, or phase if after
modulation, the amplitude, frequency, or phase of the
signal become functions of the message signal
❑ Modulation converts the message signal m(t) from
lowpass to bandpass, in the neighborhood of the carrier
frequency fc.
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 3.2 AMPLITUDE MODULATION (AM)
❑ In amplitude modulation, the message signal m(t) is
impressed on the amplitude of the carrier signal
c(t) = Accos(2fct)
▪ This results in a sinusoidal signal whose amplitude is a function
of the message signal m(t)
▪ There are different methods of amplitude modulating the carrier
signal by m(t). Each method results in different spectral
characteristics for the transmitted signal
▪ The different methods are
(a) Double sideband, suppressed-carrier AM (DSB-SC AM)
(b) Conventional double-sideband AM (DSB AM)
(c) Single-sideband AM (SSB AM)
(d) Vestigial-sideband AM (VSB AM)
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3.2.1 Double-Sideband Suppressed-Carrier AM
❑ A double-sideband, suppressed-carrier (DSB-SC) AM signal is
obtained by multiplying the message signal m(t) with the carrier
signal c(t) = Accos(2fct)

❑ DSB-SC Amplitude-modulated signal is

u (t ) = m(t )c(t ) = Ac m(t ) cos(2 f c t )

▪ An example of the message signal m(t), the carrier c(t), and the
modulated signal u (t) are shown in Figure 3.1
▪ This figure shows that a relatively slowly varying message signal m(t)
is changed into a rapidly varying modulated signal u(t), and due to its
rapid changes with time, it contains higher frequency components
▪ At the same time, the modulated signal retains the main characteristics
of the message signal; therefore, it can be used to retrieve the message
signal at the receiver
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Figure 3.1 An example of message, carrier, and DSB-SC
modulated signals

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 DSB-SC AM Waveform

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 Phase reversal at zero crossing

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 Phase reversal at zero crossing

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Spectrum of DSB-SC AM Signal
❑ DSB-SC Amplitude-modulated signal is

u (t ) = m(t )c(t ) = Ac m(t ) cos(2 f c t )

❑ Spectrum of u(t) is it’s Fourier Transform
U(f) =FT{u(t)}= FT{m(t) Ac cos (2πfct)}
= FT{m(t)} * FT{Ac cos (2πfct)}
=M(f) * 0.5 Ac [ δ( f - fc ) + δ( f + fc ) ]
U(f) = 0.5 Ac [ M ( f - fc ) + M( f + fc ) ]
Where M ( f ) is spectrum of m(t) and * is convolution integral
❑ Spectrum of the modulated signal u(t) is
Ac
U ( f ) = [ M ( f − f c ) + M ( f + f c )]
2
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Spectrum of the DSB-SC AM Signal Cont….

Figure 3.2 Magnitude and phase spectra of the message signal m(t) and
the DSB-AM modulated signal u(t)
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 Spectrum of the DSB-SC AM Signal Cont….

❑ Figure 3.2 illustrates the magnitude and phase spectra for M ( f)

and U ( f )
❑ The magnitude of the spectrum of the message signal m(t) has been
translated or shifted in frequency by an amount fc
❑ The bandwidth occupancy, of the amplitude-modulated signal is
2W, whereas the bandwidth of the message signal m(t) is W
❑ The channel bandwidth required to transmit the modulated signal
u(t) is Bc = 2W
❑ The frequency content of the modulated signal u(t) in the
frequency band | f | > fc is called the upper sideband of U(f)
❑ The frequency content in the frequency band | f | < fc is called the
lower sideband of U(f)

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 Spectrum of the DSB-SC AM Signal Cont….

❑ It is important to note that either one of the sidebands of U(f)

contains all the frequencies that are in M ( f )
▪ The frequency content of U( f ) for f > fc corresponds to the frequency
content of M ( f ) for f > 0
▪ The frequency content of U(f) for f < - fc corresponds to the frequency
content of M ( f ) for f < 0
▪ Hence, the upper sideband of U ( f ) contains all the frequencies in
M ( f ) . A similar statement applies to the lower sideband of U ( f )

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 Spectrum of the DSB-SC AM Signal Cont….

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 Spectrum of the DSB-SC AM Signal Cont….

❑The other characteristic of the modulated signal u(t) is

that it does not contain a carrier component
▪ As long as m(t) does not have any DC component, there is no
impulse in U (f) at f = fc
▪ That is, all the transmitted power is contained in the modulating
(message) signal m(t)
❖For this reason, u(t) is called a suppressed-carrier signal
❖Therefore, u(t) is a DSB-SC AM signal.

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 Power Content of DSB-SC Signals
❑The power content of the DSB-SC signal
1 T /2 2
Pu = lim  u (t ) dt
T → T −T / 2

1 T /2 2 2
= lim  Ac m (t ) cos 2 ( 2 f c t ) dt
T →  T −T / 2

Ac2
m (t ) 1 + cos(4 f c t )dt
1 T /2 2
= lim 
2 T → T −T / 2

Ac2
 Pm
2
▪ Pm indicates the power in the message signal m(t) given by

1 T /2
Pm = lim
T → T 
−T / 2
m 2 (t ) dt

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 Power Content of DSB-SC Signals Cont….
▪ The last step follows from the fact that
m2(t) is a slowly varying signal and when
multiplied by cos(4fct), which is a high
frequency sinusoid, the result is a high-
frequency sinusoid with a slowly varying
envelope, as shown in Figure 3.5
▪ Since the envelope is slowly varying, the Figure 3.5 Plot of m2(t)cos(4fct).
positive and the negative halves of each
cycle have almost the same amplitude
(Figure 3.6)
▪ Hence, when they are integrated, they
cancel each other
▪ Thus, the overall integral of
m2(t)cos(4fct) is almost zero

Figure 3.6 This figure shows why the

second term in Power Equation is zero.

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Example 3.2.1: Suppose that the modulating signal m (t) is a
sinusoid of the form m(t) = a cos 2πfmt fm << fc.
(a) Determine the DSB-SC AM signal and its upper and lower
sidebands.
(b) determine the power in the modulated signal and the power in
each of the sidebands.
Solution The DSB-SC AM is expressed in the time domain as

Taking the Fourier transform on both sides

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Figure 3.3 The (magnitude) spectrum of a DSB-SC AM signal for (a) a sinusoidal
message signal and (b) its lower and (c) upper sidebands.

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The lower sideband of u (t) is the signal

Finally, the upper sideband of u (t) is the signal

(b) The message signal m(t) = a cos 2πfmt and power in m(t) is
and

Because of the symmetry of the sidebands, the powers in the upper

and lower sidebands, Pus and P1s, are equal and given by
Pus=Pls=Pu/2 or

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Example 3.2.2: Let the message signal be m(t) = sinc(104t). Determine the DSB-
SC-modulated signal and its bandwidth when the carrier is a sinusoid with a
frequency of 1 MHz.
Solution: The carrier is c(t) =cos(2π106t) and m(t) = sinc(104t).
DSB-SC AM signal is u(t)= c(t) m(t) = cos(2π106t) sinc(104t).
To obtain the bandwidth of the modulated signal, we first need to have the
bandwidth of the message signal.
a f
We know that a sin c (2 W t ) ⎯→
⎯
FT

2W
 2W
--------(1)

Hence M (f) = FT{sinc(104t)} = 10-4 IT ( 10-4 f) ---------(2)

|M(f)|
Comparing equation 1 and 2 we have
2W=10000 and hence band width of the message is
W =5000 Hz
Band width of the modulated signal u(t) is -5000 5000

Bu=2W =10000 Hz
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 Demodulation of DSB-SC AM Signals

❑ Suppose that the DSB-SC AM signal u(t) is transmitted

through an ideal channel (with no channel distortion and no
noise)

❑ Then the received signal is equal to the modulated signal,

r (t ) = u (t ) = m(t )c(t ) = Ac m(t ) cos(2 f c t )

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 Demodulation of DSB-SC AM Signals
r(t) y(t)
Ideal LPF
W

cos(2πfct +ϕ )
❑ Suppose we demodulate the received signal by
1. Multiplying r(t) by a locally generated sinusoid cos(2fct + ),
where  is the phase of the sinusoid
2. We pass the product signal through an ideal lowpass filter with
the bandwidth W
▪ The multiplication of r(t) with cos(2 fct + ) yields
r (t ) cos(2 f c t +  ) = Ac m(t ) cos(2 f c t ) cos(2 f c t +  )
1 1
= Ac m(t ) cos( ) + Ac m(t ) cos(4 f c t +  )
2 2
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 Demodulation of DSB-SC AM Signals cont…
❑ The spectrum of the signal is illustrated in Figure 3.7
❑ Since the frequency content of the message signal m(t) is limited to W Hz,
where W << fc, the lowpass filter can be designed to eliminate the signal
components centered at frequency ±2 fc and to pass the signal components
centered at frequency f = 0 without experiencing distortion
❑ An ideal lowpass filter that accomplishes this objective is also
illustrated in Figure 3.7
Consequently, the output
of the ideal lowpass filter
1
yl (t ) = Ac m(t ) cos( )
2

Figure 3.7 Frequency-domain representation of

the DSB-SC AM demodulation.

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Demodulation of DSB-SC AM Signals cont…
❑ Note that m(t) is multiplied by cos()
▪ Therefore, the power in the demodulated signal is
decreased by a factor of cos2.
▪ Thus, the desired signal is scaled in amplitude by a factor
that depends on the phase  of the locally generated
sinusoid.
1. When   0, the amplitude of the desired signal is reduced
by the factor cos().
2. If  = 45, the amplitude of the desired signal is reduced by
√2 and the power is reduced by a factor of two.
3. If  = 90, the desired signal component vanishes

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Demodulation of DSB-SC AM Signals cont…
❑ The preceding discussion demonstrates the need for a phase-
coherent or synchronous demodulator for recovering the
message signal m(t) from the received signal

❑ That is, the phase  of the locally generated sinusoid should

ideally be equal to 0 (the phase of the received-carrier signal)
❑ A sinusoid that is phase-locked to the phase of the received
carrier can be generated at the receiver in one of two ways

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Demodulation of DSB-SC AM Signals cont…

1. One method is to add a carrier component into the transmitted

signal, as illustrated in Figure 3.8.
▪ We call such a carrier component as " pilot tone."
▪ Its amplitude Ap and its power Ap2 / 2 are selected to be
significantly smaller than those of the modulated signal u(t).

Figure 3.8 Addition of a pilot tone to a DSB-AM signal.

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Demodulation of DSB-SC AM Signals cont…

▪ Thus, the transmitted signal is a double-sideband, but it is

no longer a suppressed carrier signal.

u (t ) = Ac m(t ) cos(2 f c t ) + A p cos (2  f c t )


Carrier component

 
 Ap 
u (t ) = Ac cos (2  f c t ) m(t ) + 
 
Ac 
 DC value 


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Demodulation of DSB-SC AM Signals cont…
▪ At the receiver, a narrowband filter tuned to frequency fc,
filters out the pilot signal component
▪ Its output is used to multiply the received signal, as shown in
Figure 3.9

Figure 3.9 Use of a pilot tone to demodulate a DSB-AM signal.

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 Demodulation of DSB-SC AM Signals cont…
▪ We may show that the presence of the pilot signal results in a
DC component in the demodulated signal
▪ This must be subtracted out in order to recover m(t)

r (t ) = Ac m(t ) cos(2 f c t ) + A p cos (2  f c t )

r (t ) A p cos (2f c t ) = (Ac m(t ) cos (2  f c t ) + A p cos (2  f c t ) )(A p cos (2  f c t ) )
= Ac A p m(t ) cos 2 (2  f c t ) + A p2 cos 2 (2  f c t )
cos(4 f c t )
A p Ac m(t ) A p2 A p2 cos(4 f c t )
= + + +
2 
 2 2
 2
Message component DC component

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 Demodulation of DSB-SC AM Signals cont…

❑Adding a pilot tone to the transmitted signal has a disadvantage

▪ It requires that a certain portion of the transmitted signal power
must be allocated to the transmission of the pilot
❑ As an alternative, we may generate a phase-locked sinusoidal carrier
from the received signal r(t) without the need of a pilot signal
▪ This can be accomplished by the use of a phase-locked loop, as
described in Section 6.4.

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3.2.2 Conventional Amplitude Modulation
❑ A conventional AM signal consists of a large carrier component, in
addition to the double-sideband AM modulated signal
❑The transmitted signal is expressed mathematically as
u (t ) = Ac [1 + m(t )] cos(2 f c t )
= Ac m(t ) cos(2 f c t ) + Ac cos(2 f c t )
    
Double sideband Carrier component

❑The message waveform is constrained to satisfy the condition that

| m(t)|  1
❑ We observe that Acm(t) cos(2fct) is a double-sideband AM signal
and Accos(2fct) is the carrier component
❑ Figure 3.10 illustrates an AM signal in the time domain

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 Conventional Amplitude Modulation cont…

❑As we will see later in this chapter, the existence of this extra carrier
results in a very simple structure for the demodulator
❑That is why commercial AM broadcasting generally employs this
type of modulation

Figure 3.10 A conventional AM signal in the time domain

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 Conventional Amplitude Modulation cont…

❑ As long as |m(t)|  1, the amplitude Ac[1 + m(t)] is always positive

▪ This is the desired condition for conventional DSB AM that
makes it easy to demodulate, as we will describe
❑ On the other hand, if m(t) < -1 for some t , the AM signal is
over modulated and its demodulation is rendered more complex
❑ In practice, m(t) is scaled so that its magnitude is always less than
unity. It is sometimes convenient to express m(t) as

m(t ) = amn (t )
❑ where mn(t) is normalized such that its minimum value is -1
m(t )
mn (t ) =
max m(t )

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Conventional Amplitude Modulation cont…

❑The scale factor a is called the modulation index, which is

generally a constant less than 1
❑Since |mn(t)|  1 and 0 < a < 1, we have 1 + a mn(t) > 0 and now
the modulated signal can be expressed as

𝑢(𝑡) = 𝐴𝑐 [1 + 𝑎 𝑚𝑛 (𝑡)] cos( 2𝜋𝑓𝑐 𝑡)

which will never be over modulated, or

u (t ) = Ac [1 + amn (t )] cos(2 f c t )
= Ac a mn (t ) cos(2 f c t ) + Ac cos(2 f c t )
  
Double sideband Carrier component

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 Spectrum of the Conventional AM Signal
❑Conventional amplitude modulated signal is
u (t ) = Ac [1 + am n (t )] cos(2 f c t )
= Ac a m n (t ) cos(2 f c t ) + Ac cos(2 f c t )
 
L arg e carrier

❑The spectrum of the amplitude-modulated signal u(t) is obtained by taking

Fourier transform on both sides
U ( f ) = FT Ac am n (t ) cos(2 f c t ) + FT Ac cos(2 f c t )

=
Ac a
M n ( f − f c ) + M n ( f + f c ) +  ( f − f c ) +  ( f + f c )
Ac
2 2
❑ M ( f ) is Fourier transform (spectrum) of m(t)

❑A message signal m(t), its spectrum M ( f ) , the corresponding modulated

signal u(t), and its spectrum U(f) are shown in Figure 3.11

❑Obviously, the spectrum of a conventional AM signal occupies a

bandwidth twice the bandwidth of the message signal
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Figure 3.11 Conventional AM in both the time and frequency domain.

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Example 3.2.4: Suppose that the modulating signal m (t) is a sinusoid of the form
the form m(t) = cos (2πfmt) fm << fc. Determine the DSB-AM signal, its upper
and lower sidebands, and its spectrum, assuming a modulation index of a.

Solution: DSB-AM signal is expressed as

The lower-sideband component is

The upper-sideband component is

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The spectrum of the DSB-AM signal u (t) is

Figure 3.12 Spectrum of a DSB-AM signal in Example 3.2.4.

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 Power for the Conventional AM Signal cont..
❑ Conventional amplitude modulated signal is similar to a DSB when
m(t) is substituted with 1 + mn(t)

❑ The power in the DSB-SC modulated signal is found to be

Ac2
Pu = Pm
2
▪ where Pm denotes the power in the message signal

❑ The power in signal 1 + mn(t) is

1 T /2 1 T /2
Pm = lim  [1 + amn (t )] dt = lim 
2
[1 + a 2 mn2 (t )]dt
T → T −T / 2 T → T −T / 2

Pm = 1+ a 2 Pmn

▪ where we have assumed that the average of mn(t) is zero

▪ This is a valid assumption for many signals, including audio signals.

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 Power for the Conventional AM Signal
❑ Hence power in Conventional amplitude modulated signal is
Ac2
Pu = Pm
2
Ac2 Ac2 2
Pu = + a Pmn
2 2
 
Power in carrier Power in message

▪ The first component in the preceding relation applies to the existence of the
carrier, and this component does not carry any information
▪ The second component is the information-carrying comp

❑ Note that the second component is usually much smaller than the first
component (a < 1, |mn(t)| < 1, and for signals with a large dynamic range, Pmn
<< 1)
❑ This shows that the conventional AM systems are far less power efficient than
the DSB-SC systems
❑ The advantage of conventional AM is that it is easily demodulated

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Power for the Conventional AM Signal for sinusoidal input

❑ Message signal is m(t) = cos (2πfmt) fm << fc

AM wave has 3 components

Unmodulated carrier + LSB + USB

Hence total power Pt= Pc + PLSB + PUSB

Ac2 ( Ac a / 2)2 ( Ac a / 2)2
Pt = + +
2
 2  
 2
Power in carrier PLSB PUSB

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Power for the Conventional AM Signal for sinusoidal input
2 2
Ac2 Ac a 2 Ac a 2
Pt = + +
2
 8  8
Power in carrier PLSB PUSB
2
Ac2 Ac a 2
Pt = +
2
 4
Power in carrier PSB

 a2   a2 
2
Ac
    PSB a2
PSB
= 2 
2 2 
=  2 
or =
Pt Ac 
 1 +
a2 

 a2 
1 + 
Pt 1 + a 2
2  2   2 

The maximum value of a=1.

PSB 1
Hence maximum power efficiency = = = 33.33%
Pt 3

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DSB AM of a single tone signal(sinusoidal signal)

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Modulation index of DSB AM of a single tone signal
From the waveform we have the amplitude of the message signal
Vmax − Vmin
Vm = but Vmax and Vmin are maximum and
2 minimum amplitudes of the AM envelope
Vm + Vmin = Ac
respectively
Vmax − Vmin
Ac = + Vmin or
2
Vmax + Vmin
Ac =
2
Vm
Vm = a Ac or a= where a is the modulation index
Ac Hence

Vmax − Vmin
Modulation index a=
Vmax + Vmin

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 DSB AM Waveform

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DSB AM Waveform for different modulation indices

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Example 3.1: The message signal m (t) = 2 cos 400t + 4 sin(500t +π/3 ) modulates
the carrier signal c(t) = A cos(8000 π t), using DSB amplitude modulation.
(a) Find the time-domain representation of the modulated signal
(b) Frequency-domain representation of the modulated signal and plot the
spectrum(Fourier transform) of the modulated signal.
(c) What is the power content of the modulated signal?

Solution: (a) Time-domain representation of the modulated signal is

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(b) Taking Fourier transform of the modulated signal on both sides

Consider

   
2 j j and 2 −j3 −j
e 3
= 2 (− j ) e 3 − e = 2 ( j) e 3
j j
   
−j j −j
=2e 2 e
j
=2e 2 3 3
e

 j
−j
=2e 6 =2e 6

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(b) Taking Fourier transform of the modulated signal on both sides

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(b) Magnitude and Phase plot

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(b ) The power content of the modulated signal is

Where

Hence

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Example 3.4: Suppose the signal x (t) = m (t) + cos 2πfct is applied to a nonlinear
system whose output is y (t) = x (t) + 0.5x2 (t) . Determine and sketch the
spectrum of y (t) when M (f) is as shown in Figure below and W<< fc.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 53

Taking Fourier Transform of y (t) on both sides, we get

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 54

Example 3.5: The modulating signal
m (t) = 2 cos 4000πt + 5 cos 6000πt is multiplied by the carrier c (t)
= 100 cos 2πfct, where fc = 50 kHz.
(a) Obtain the modulated signal and its spectrum.

(b) Plot the spectrum

(c) Determine the power in the modulated signal

(d) Obtain the expression for the upper side band (USB) and
lower side band (LSB) of the modulated signal.

(e) Obtain the spectrum of the upper side band (USB) and
lower side band (LSB) and plot the same

(f) Determine the power in each of the sidebands (USB and

LSB).

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 55

Taking Fourier Transform on both the sides

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 56

Example 3.14: The output signal from an AM modulator is
u (t) = 5 cos 1800πt + 20 cos 2000πt + 5 cos 2200πt.
1. Determine the modulating signal m (t) and the carrier c (t).
2. Determine the modulation index.
3. Determine the ratio of the power in the sidebands to the power in the carrier.
1. Given that

u (t ) = Ac [1 + amn (t )] cos(2 f c t ) --------------(2)

Comparing equations (1) and (2),

The modulating signal is m(t) = cos(2π100t)
The carrier signal is c(t) = 20 cos(2π1000t).

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 57

2. Comparing equations (1) and (2),
The modulation index α= ½
3. Power in the carrier component is
Power in the sidebands is
Ratio of power in the sidebands to the power in the carrier is
Ac2 Ac2 2
Pu = + a Pmn
2
 2
 
Power in carrier Power in message

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 58

 Demodulation of Conventional DSB-AM Signals
❑ The major advantage of conventional AM signal transmission is
the ease in which the signal can be demodulated
❑ There is no need for a synchronous demodulator
❑ Since the message signal m(t) satisfies the condition
|m(t)| < 1, the envelope (amplitude) 1+m(t) > 0
❑ If we rectify the received signal, we eliminate the negative values
without affecting the message signal, as shown in Figure 3.14
❑ The rectified signal is equal to u(t) when u(t) > 0, and it is equal to
zero when u(t) < 0
❑ The message signal is recovered by passing the rectified signal
through a lowpass filter whose bandwidth matches that of the
message signal

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 59

 Demodulation of Conventional DSB-AM Signals cont…
❑ The combination of the rectifier and the lowpass filter is called an
envelope detector

Figure 3.14 Envelope detection of a conventional AM signal.

❑ As previously indicated, conventional DSB-AM signals are easily

demodulated by an envelope detector

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 60

 Demodulation of Conventional DSB-AM Signals cont…

❑ A circuit diagram for an envelope detector is shown in Figure 3.27

❑ It consists of a diode and an RC circuit, which is basically a simple
low pass filter
▪ During the positive half-cycle of the input signal, the diode conducts
and the capacitor charges up to the peak value of the input signal
• When the input falls below the voltage on the capacitor, the diode
becomes reverse-biased and the input disconnects from the output
• During this period, the capacitor discharges slowly through the load
resistor R
▪ On the next cycle of the carrier, the diode again conducts when the
input signal exceeds the voltage across the capacitor
• The capacitor again charges up to the peak value of the input signal
and the process is repeated

Figure 3.27 An envelope detector.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 61

 Demodulation of Conventional DSB-AM Signals cont…

Figure 3.28 Effect of (a) large and (b) small RC values on the performance of the envelope detector.

❑ The time constant RC must be selected to follow the variations in the envelope of
the carrier-modulated signal
▪ If RC is too small, then the output of the filter falls very rapidly after each peak
and will not follow the envelope of the modulated signal closely
• This corresponds to the case where the bandwidth of the lowpass filter is too
large
▪ If RC is too large, then the discharge of the capacitor is too slow and again the
output will not follow the envelope of the modulated signal

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 62

 Demodulation of Conventional DSB-AM Signals cont…

▪ This corresponds to the case where the bandwidth of the lowpass filter is too
small
▪ Effect of large and small RC values Figure 3.28
▪ For good performance of the envelope detector,
1 1
 RC 
fc W
▪ In such a case, the capacitor discharges slowly through the resistor, thus, the
output of the envelope detector, which we denote as m ~ (t ) , closely follows the
message signal
❑ Ideally, the output of the envelope detector is of the form
d (t ) = g1 + g 2 m(t )
▪ where gl represents a DC component and g2 is a gain factor due to the signal
demodulator.
▪ The DC component can be eliminated by passing d(t) through a transformer,
whose output is g2m(t).

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 63

 Application of Conventional AM

❑The simplicity of the demodulator has made conventional DSB-AM a

practical choice for AM-radio broadcasting
▪ Since there are literally billions of radio receivers, an inexpensive
implementation of the demodulator is extremely important
▪ The power inefficiency of conventional AM is justified by the fact that there
are few broadcast transmitters relative to the number of receivers

❑Consequently, it is cost-effective to construct powerful transmitters and

sacrifice power efficiency in order to simplify the signal demodulation at
the receivers

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 64

 3.2.3 Single-Sideband AM ( SSB AM )
❑A DSB-SC AM signal required a channel bandwidth of Bc = 2W Hz for
transmission, where W is the bandwidth of the message signal
▪ However, the two sidebands are redundant

❑We will demonstrate that the transmission of either sideband is sufficient

to reconstruct the message signal m(t) at the receiver
• Thus, we reduce the bandwidth of the transmitted signal to that of the
baseband message signal m(t)
▪ In the appendix at the end of this chapter, we will demonstrate that a single-
sideband (SSB) AM signal is represented mathematically as

u (t ) = Ac m(t ) cos(2 f c t )  Ac mˆ (t ) sin( 2 f c t )

▪ where mˆ (t ) is the Hilbert transform of m(t)
▪ The plus or minus sign determines which sideband we obtain
• The plus sign indicates the lower sideband
• The minus sign indicates the upper sideband

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 65

 3.2.3 Single-Sideband AM ( SSB AM )
❑ Hilbert Transform
▪ The Hilbert transform is unlike many other transforms because it does not
involve a change of domain
▪ The Hilbert transform of a signal x (t) is a signal) whose frequency
components lag the frequency components of x(t) by 90°
▪ Example: Hilbert transform of
x(t) = A cos(2πf0t + θ) is A cos(2 π f0t +θ - 90°) = A sin(2 π f0t + θ). Or
sin  ⎯
HT
→ − cos  and
cos  ⎯
HT
→ sin 
The frequency response of the Impulse response is
1
Hilbert transform is h(t ) =
t
j f 0

H( f )= 0 f =0 = − j sgn ( f )
− j f 0


Latha, Department of ECE,ASE,Bengaluru 3/13/2021 66

 Spectrum of the DSB-SC AM Signal
M(f)

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 67

 Spectrum of the USSB AM Signal
M(f)

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 68

 Spectrum of the LSSB AM Signal Cont….
M(f)

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 69

 Appendix 3A: Derivation of the Expression
for SSB-AM Signal
❑ Let m(t) be a signal with the Fourier transform (spectrum) M ( f)
❑ An upper single-sideband amplitude-modulated signal (USSB AM) is
obtained by eliminating the lower sideband of a DSB amplitude-
modulated signal
❑ Suppose we eliminate the lower sideband of the DSB AM signal, uDSB(t)
= 2Acm(t)cos2fct, by passing it through a highpass filter whose transfer
function is given by
1, | f | f c
H( f ) = 
0, otherwise

▪ H(f) can be written in terms of unit step function as

H ( f ) = u−1 ( f − f c ) + u−1 (− f − f c )

▪ where u-1(.) represents the unit-step function

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 70

 Appendix 3A: Derivation of the Expression
for SSB-AM Signal cont…
❑ Therefore, the spectrum of the USSB-AM signal is given by
U u ( f ) = Ac M ( f − f c )u−1 ( f − f c ) + Ac M ( f + f c )u−1 (− f − f c )
U u ( f ) = Ac M ( f )u−1 ( f ) | f = f − f c + Ac M ( f )u−1 (− f ) | f = f + f c

▪ Taking the inverse Fourier transform of both sides of the above

Equation and using the modulation and convolution properties of the
Fourier transform, we obtain
u u (t ) = Ac m(t )  FT −1 [u −1 ( f )]e j 2f ct + Ac m(t )  FT −1 [u −1 (− f )]e − j 2f ct

▪ Note: From Fourier Transform of unit step function and duality

theorem of the Fourier transform
1 j  1 j 
FT   (t ) +  = u −1 ( f ), FT   (t ) −  = u −1 (− f )
2 2 t  2 2 t 

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 71

 Appendix 3A: Derivation of the Expression
for SSB-AM Signal cont…

To find u-1(f) and u-1(-f) :

1 j  FT
X (−t ) =   (t ) +  ⎯→ x ( f ) = u −1 ( f ),
Duality Theorem: 2 2 t 

If x(t ) ⎯→
FT
X ( f ) then 1
X (t ) =   (t ) −
j  FT
 ⎯→ x (− f ) = u −1 (− f ),
 2 2 t 
X (t ) ⎯→
FT
x (− f ) or
X (−t ) ⎯→
FT
x( f )
1 j 
FT   (t ) −  = u −1 (− f )
We know that 2 2t 

1 1
u (t ) ⎯→  ( ) +
FT
OR
2 j 1 j 
FT   (t ) +  = u −1 ( f ),
1
U( f ) = ( f )+
1 2 2t 
2 j2 f

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 72

 Appendix 3A: Derivation of the Expression
for SSB-AM Signal cont…
❑ After substituting for inverse Fourier transform of u(f) and u(-f) in
u(t), we get
1 j  j 2 f c t 1 j  − j 2 f c t
u u (t ) = Ac m (t )    (t ) + e + Ac m (t )    (t ) − e
2 2 t  2 2 t 

= c  m (t ) + j mˆ (t ) e j 2 f ct + c  m (t ) − j mˆ (t ) e − j 2 f ct
A A
2 2

▪ where we have used the identities

m(t ) *  (t ) = m(t ) , m (t ) *
1
= mˆ (t ) is Hilbert transform of m(t)
t
▪ Using Euler's relations in uu(t) Equation above, we obtain
u u (t ) = Ac m (t ) cos 2 f c t − Ac mˆ (t ) sin 2 f c t
▪ which is the time-domain representation of a USSB-AM signal.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 73

 Appendix 3A: Derivation of the Expression
for SSB-AM Signal cont…
❑ The expression for the LSSB-AM signal can be derived by noting
that
uu (t ) + ul (t ) = u DSB (t )
Ac m(t ) cos 2f c t − Ac mˆ (t ) sin 2f c t + ul (t ) = 2 Ac m(t ) cos 2f c t

❑ Therefore time-domain representation of a LSSB-AM signal is

u l (t ) = Ac m (t ) cos 2 f c t + Ac mˆ (t ) sin 2 f c t
❑Thus, the time-domain representation of a SSB-AM signal can
generally be expressed as

u SSB (t ) = Ac m(t ) cos(2 f c t )  Ac mˆ (t ) sin( 2 f c t )

▪ where the minus sign corresponds to the USSB-AM signal, and the plus sign
corresponds to the LSSB-AM signal
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 74
 Generation of SSB-AM

The SSB-AM signal u(t) may be Another method, illustrated in Figure

generated by using the system 3.16, generates a DSB-SC AM signal
configuration shown in Figure 3.15 and then employs a filter that selects
either the upper sideband or the lower
The method shown in Figure 3.15 sideband of the double-sideband AM
employs a Hilbert-transform filter signal

Figure 3.16 Generation of a single-sideband

Figure 3.15 Generation of a lower AM signal by filtering one of the sidebands of
single-sideband AM signal. a DSB-SC AM signal

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 75

Example 3.2.6:Suppose that the modulating signal is a sinusoid of the form
m(t) = cos 2πfmt, fm << fc . Determine the two possible SSB-AM signals.
Solution : Time-domain representation of a SSB-AM signal can generally be
expressed as
u SSB (t ) = Ac m(t ) cos(2 f c t )  Ac mˆ (t ) sin( 2 f c t )
The Hilbert transform of m(t) is

Hence
If we take - ve sign, we obtain the upper-sideband signal

If we take +ve sign, we obtain the lower-sideband signal

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 76

 Spectra of uu(t) and ul(t)

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 77

 Demodulation of SSB-AM Signal
❑ To recover the message signal m(t) in the received SSB-AM signal,
we require a phase-coherent or synchronous demodulator, as was
the case for DSB-SC AM signals
r(t) y(t)
Ideal LPF
W

▪ For the USSB-AM signal cos(2πfct +ϕ )

r (t ) cos(2 f c t +  ) = u (t ) cos(2 f c t +  )
= 12 Ac m(t ) cos( ) + 12 Ac mˆ (t ) sin( ) + double frequency terms.
▪ By passing above product signal through an ideal lowpass filter, the
double-frequency components are eliminated, leaving us with
yl (t ) = 12 Ac m(t ) cos( ) + 12 Ac mˆ (t ) sin( )
▪ Note that the phase offset not only reduces the amplitude of the
desired signal m(t) by cos, but it also results in an undesirable
sideband signal due to the presence of mˆ (t ) in yl(t)
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 78
 Demodulation of SSB-AM Signal cont…
▪ The latter component was not present in the demodulation of a
DSBSC signal
▪ However, it is a factor that contributes to the distortion of the
demodulated SSB signal
▪ The transmission of a pilot tone at the carrier frequency is a very
effective method for providing a phase-coherent reference signal for
performing synchronous demodulation at the receiver
▪ Thus, the undesirable sideband-signal component is eliminated
▪ However, this means that a portion of the transmitted power must be
allocated to the transmission of the carrier

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 79

 Application and Limitation of SSB-AM
❑ The spectral efficiency of SSB AM makes this modulation method very
attractive for use in voice communications over telephone channels
(wire lines and cables)
❑ The filter method used to select one of the two signal sidebands for
transmission, is particularly difficult to implement
when the message signal m(t) has a large power
concentrated in the vicinity of f = 0 as in X(f)
❑ In such a case, the sideband filter
must have an extremely sharp cutoff
in the vicinity of the carrier in order
to reject the second sideband and it is difficult
to implement such filter
❑ But it is suitable for signals having a small
power concentrated in the vicinity of f = 0
as in Y(f).

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 80

 Spectrum of the LSSB AM Signal Cont….

M(f)

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 81

 3.2.4 Vestigial-Sideband AM
❑ The stringent-frequency response requirements on the
sideband filter in an SSB-AM system can be relaxed by
allowing vestige, which is a portion of the unwanted sideband,
to appear at the output of the modulator
▪ Thus, we simplify the design of the sideband filter at the
cost of a modest increase in the channel bandwidth required
to transmit the signal
▪ The resulting signal is called vestigial-sideband (VSB) AM
• This type of modulation is appropriate for signals that have a
strong low-frequency component, such as video signals
• That is why this type of modulation is used in standard
TV broadcasting

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 82

 Vestigial-Sideband AM cont…
❑ To generate a VSB-AM signal, we begin by generating a DSB-SC
AM signal and passing it through a sideband filter with the
frequency response H( f ), as shown in Figure 3.17

Figure 3.17 Generation of vestigial-sideband AM signal.

The VSB signal may be expressed as

u (t ) = [ Ac m(t ) cos 2f c t ]  h(t ) in time domain,

U( f ) =
Ac
M ( f − f c ) + M ( f + f c )H ( f ) in frequency domain
2
where h(t) and H( f ) are impulse response and frequency response of side band
filter respectively.
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 83
 Vestigial-Sideband AM cont…

• To determine the frequency-response

characteristics of the filter, we will
consider the demodulation of the VSB
signal u(t)

• We multiply u(t) by the carrier

component cos2fct and pass the result
through an ideal lowpass filter, as
shown in Figure 3.18
• Thus, the product signal is

v(t ) = u (t ) cos 2 f c t Figure 3.18 Demodulation of VSB signal.

• or equivalently,

V( f ) =
1
U ( f − f c ) + U ( f + f c )
2

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 84

 Vestigial-Sideband AM cont…
V( f ) =
1
U ( f − f c ) + U ( f + f c ) U( f ) =
Ac
M ( f − f c ) + M ( f + f c )H ( f )
2 2
❑ Substituting for U(f) in V(f) we get
V ( f ) = M ( f − 2 f c ) + M ( f )H ( f − f c ) + M ( f ) + M ( f + 2 f c )H ( f + f c )
Ac Ac
4 4

▪ The lowpass filter rejects the double-frequency terms and passes only
the components in the frequency range | f|W
▪ Hence, the signal spectrum at the output of the ideal lowpass filter is
M ( f ) H ( f − f c ) + H ( f + f c )
Ac
Vl ( f ) =
4

❑ The message signal at the output of the lowpass filter must be

undistorted
▪ Hence, the VSB-filter characteristic must satisfy the condition
H ( f − f c ) + H ( f + f c ) = constant for | f | W

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 85

 Vestigial-Sideband AM cont…
❑We note that H(f) selects the upper sideband and a vestige of the lower
sideband
❑ It has odd symmetry about the carrier frequency fc in the frequency
range fc - fa < f < fc + fa , where fa is a conveniently selected frequency that
is some small fraction of W, i.e., fa << W. Thus, we obtain an undistorted
version of the transmitted signal

Figure 3.19 VSB filter

characteristics

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 86

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 87
The spectrum of the
demodulated signal is Vl ( f ) =
Ac
M ( f ) H ( f − f c ) + H ( f + f c )  =
Ac
M( f )k
4 4

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 88

 Vestigial-Sideband AM cont…
❑Figure 3.20 illustrates the frequency response of a VSB filter that selects the
lower sideband and a vestige of the upper sideband

❑ In practice, the VSB filter is designed to have some specified phase

characteristic. To avoid distortion of the message signal, the VSB filter
should have a linear phase over its passband fc - fa  | f |  fc + W

Figure 3.20 Frequency response of the VSB filter for

selecting the lower sideband of the message signals.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 89

Example 3.2.7: Suppose that the message signal is given as

m(t) = 10 + 4 cos 2πt + 8 cos 4 π t + 10 cos 20 π t.

Specify both the frequency-response characteristics of a VSB filter that passes the
upper sideband and the first frequency component of the lower sideband.

Solution: The spectrum of the DSB-SC AM signal u (t) = m(t) cos2fct is

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 90

Figure 3.21 Frequency-response characteristics of
the VSB filter in Example 3 .2. 7.
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 91
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 92
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 93
Example 3.18: The message signal m(t), whose spectrum is shown in Figure P-3. 1
is passed through the system shown in that figure.

The bandpass filter has a bandwidth of 2W centered at f0, and the lowpass filter has

a bandwidth of W. Plot the spectra of the signals x(t), y1 (t), y2(t), y3 (t), and y4(t).

What are the bandwidths o f these signals?

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 94

 1. The signal x(t) is m(t)+cos(2πf0t).
The spectrum of this signal is X(f ) = M(f )+1/2 [ δ(f −f0)+δ(f +f0) ]
Its bandwidth equals to Wx = f0

2.The signal y1(t) after the Square Law Device is

The spectrum is given by

Its bandwidth is W1 = 2f0

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 95

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 96
 3. The bandpass filter will cut-off the low-frequency components

and the terms with the double frequency components

Thus the spectrum Y2(f ) is given by
the bandwidth of y2(t) is W2 = 2W.

4.The signal y3(t) is y3(t) = 2m(t) cos2(2πf0t) = m(t) + m(t) cos(4πf0t)

And its Spectrum is
bandwidth W3 = f0 + W.
5. The lowpass filter will eliminate the spectral components

so that y4(t) = m(t) with spectrum Y4 = M(f ) and bandwidth W4 = W.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 97

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 98
Latha, Department of ECE,ASE,Bengaluru 3/13/2021
 3.3 Implementation of AM
Modulators and Demodulators
Power-Law Modulation: It is Conventional AM signal generator. A
non linear device such as P-N diode, which has voltage current characteristic
as shown in figure 3.22. Suppose that the voltage input to such a device is
the sum of the message signal m(t) and the carrier Ac cos(2πfct), as illustrated
in Figure 3.23. The nonlinearity will generate a product of the message m(t)
with the carrier, plus additional terms, which can be filtered out to get the
desired Conventional AM signal

Figure 3.22 Voltage-current

characteristic of P-N diode.
Figure 3.23 Block diagram of
power-law AM modulator.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 100

 Implementation of AM
Modulators and Demodulators cont…
❑ Suppose that the nonlinear device has an input-output (square-law)
characteristic of the form
vo (t ) = a1vi (t ) + a v (t )
2
2 i

❑where vi(t) is the input signal, vo(t) is the output signal, and the
parameters (al, a2) are constants
▪ Then, if the input to the nonlinear device is
vi (t ) = m(t ) + Ac cos 2f c t
▪ Its output
vo (t ) = a1[m(t ) + Ac cos 2f ct ] + a2 [m(t ) + Ac cos 2f ct ]2
 2a2 
= a1m(t ) + a2 m (t ) + a2 A cos 2f ct + Ac a1 1 +
2 2
c
2
m(t ) cos 2f ct
 a1 

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 101

 IMPLEMENTATION OF AM
MODULATORS AND DEMODULATORS cont…

❑ The output of the band pass filter with a bandwidth 2W

centered at f = fc yields
 2a2 
u (t ) = Ac a1 1 + m(t ) cos 2f c t
 a1 

 2a 2 
▪ where  m(t )   1 by design
 a1 

▪ Thus, the signal generated by this method is a conventional AM

signal

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 102

 Switching Modulator
❑ Another method for generating a
conventional AM-modulated signal is
by means of a switching modulator
❑ Such a modulator can be implemented
by the system illustrated in Figure
3.24(a)
❑ The sum of the message signal and the
carrier vi (t) are applied to a diode that
has the input-output voltage
characteristic shown in Figure 3.24(b),
where Ac >> m(t)
❑ The output across the load resistor is
simply
v (t ), c(t )  0
vo (t ) =  i
 0, c(t )  0
Figure 3.24 Switching modulator and
periodic switching signal.
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 103
 Switching Modulator cont…
❑ This switching operation may be viewed mathematically as a multiplication of
the input vi(t) with the switching function s(t), i.e.,
vo (t ) = [m(t ) + Ac cos(2 f c t )]s (t ) where s(t) is shown in Figure 3.24(c)
❑ Since s(t) is a periodic function, it is represented in the Fourier series as
1 2  (−1) n −1
s (t ) = +  cos2f c t (2n − 1)
2  n =1 2n − 1
 4 Ac 
vo (t ) = [m(t ) + Ac cos(2 f c t )]s (t ) =
1 + m (t )  cos(2 f c t ) + other terms
 Ac 2 
❑ The desired AM-modulated signal is obtained by passing vo(t) through a
bandpass filter with the center frequency f = fc and the bandwidth 2W
At its output, we have the desired conventional AM signal

❑The output across the load resistor is simply

A  4 
u (t ) = c 1 + m(t ) cos(2 f c t )
2  Ac 

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 104

 Balanced Modulator
❑ A relatively simple method for generating a DSB-SC AM signal is
to use two conventional-AM modulators arranged in the configuration
illustrated in Figure 3.25
▪ For example, we may use two square-law AM modulators as previously
described
▪ Care must be taken to select modulators with approximately identical
characteristics so that the carrier component cancels out at the summing
junction

Figure 3.25 Block diagram of a

balanced modulator.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 105

 Ring Modulator
❑ Another type of modulator for generating a DSB-SC AM signal is
the ring modulator illustrated in Figure 3.26
▪ The switching of the diodes is controlled by a square wave of frequency fc,
denoted as c(t), which is applied to the center taps of the two transformers
▪ When c(t) > 0, the top and bottom diodes conduct, while the two diodes in the
cross-arms are off
▪In this case, the message signal m(t) is multiplied by +1
▪ When c(t) < 0, the diodes in the cross-arms of the ring conduct, while the
other two diodes are switched off
▪In this case, the message signal m(t) is multiplied by -1.
▪ Consequently, the operation of the ring modulator may be described
mathematically as a multiplier of m(t) by the square-wave carrier c(t),

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 106

 Working Principle of Ring Modulator
❑When message signal is zero
▪ When polarity of the carrier is as shown in
figure (a), only D1, D2 conduct while D3,
D4 are off
▪ Current divides equally in the upper and the
lower half of the primary winding of T2
▪ The current in the upper part of the secondary
winding produces a magnetic field that is
equal and opposite to the magnetic field
produced by the lower part of the secondary
winding
▪ Hence magnetic field cancel each other
resulting no out put signal
▪When polarity of the carrier is as shown in
figure (b), only D3, D4 conduct while D1,
D2 are off
▪ In this case also no out put signal is produced
as equal and opposite magnetic fields cancel
each other
▪ Hence Carrier is suppressed

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 107

 Ring Modulator cont…
❑ When low frequency message signal is applied to the primary of the
transformer T1

▪ Modulating signal appears at the secondary of the T1

▪ When the carrier polarity is as in figure (a), then only D1, D2 conduct while
D3, D4 are off
▪ Current flows through D1, primary of T2, D2 and then back to secondary of T1.
▪ When polarity of the carrier is as shown in figure (b), only D3, D4 conduct
while D1, D2 are off
▪ Current flows through D4, primary of T2, D3 and then back to secondary of T1.
▪ But this time current through the primary of T2 is reversed due the connections
of D3 and D4.
▪ This results in 180o phase reversal, if the modulating signal is positive, the
output will be negative with this connection and vice versa.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 108

 Demodulation of DSB-SC AM
❑ The demodulation of a DSB-SC AM signal requires a
synchronous demodulator.
Two different methods are
1. PLL method 2. Costas Loop

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 109

 Demodulation of SSB-AM and VSB-AM signals
❑ The demodulation of SSB-AM and VSB-AM signals requires
the use of a phase-coherent reference.
❑ In both of these two cases the carrier component is transmitted
along with the message
❑ The existence of the carrier component makes it possible to extract a
phase-coherent reference for demodulation in a balanced modulator
❑ In applications such as a TV broadcast, a large carrier component is
transmitted along with the message in the VSB signal in which, it is
possible to recover the message by using an envelope detector.

Figure 3.29 Demodulator for

a SSB-AM signal

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 110

Example 3.3.1 An audio signal of bandwidth W = 5 kHz is modulated
on a carrier of frequency 1 MHz using
conventional AM. Determine the range of values of RC for successful
demodulation of this signal using an envelope detector.
Solution: We must have or

In this case, RC = 10-5 is an appropriate choice

Assume R=10 kΩ then C=10-9 F or 1 nano F

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 111

 3.4 Signal Multiplexing
❑ When we use a message signal m(t) to modulate the amplitude of a
sinusoidal carrier, we translate the message signal frequency by an amount
equal to the carrier frequency fc
❑ If we have two or more message signals to transmit simultaneously over
the communication channel, we can have each message signal modulates a
carrier of a different frequency, where the minimum separation between
two adjacent carriers is either 2W (for DSB AM) or W (for SSB AM),
where W is the bandwidth of each of the message signals
❑ Thus, the various message signals occupy separate frequency bands of the
channel and do not interfere with one another during transmission
❑ Combining separate message signals into a composite signal for
transmission over a common channel is called signal multiplexing
(1) Time-division multiplexing
Time-division multiplexing is usually used to transmit digital
information
(2) Frequency-division multiplexing
Frequency-division multiplexing (FDM) may be used with either analog
or digital signal transmission

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 112

 3.4 Signal Multiplexing
❑ Example (FDM)
Consider three signals m1(t), m2(t), m3(t) having same band width of 4 kHz
m1(t) is modulated with carrier frequency 10 kHz
m2(t) is modulated with carrier frequency 18 kHz
m3(t) is modulated with carrier frequency 26 kHz

m(t) =m1(t) + m2(t) + m3(t) is multiplexed composite signal

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 113

 3.4 .1 Frequency- Division Multiplexing
❑ In FDM, the message signals are separated in frequency

Figure 3.31 Frequency-division multiplexing of multiple signals.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 114

 Frequency- Division Multiplexing cont…
❑ A typical configuration of an FDM system is shown in Figure 3.31
❑ This figure illustrates the frequency-division multiplexing of K message
signals at the transmitter and their demodulation at the receiver
▪ The lowpass filters at the transmitter ensure that the bandwidth of the
message signals is limited to W Hz
▪ Each signal modulates a separate carrier
▪ Hence, K modulators are required
▪Then, the signals from the K modulators are summed and transmitted over
the channel
▪ For SSB and VSB modulation, the modulator outputs are filtered prior to
summing the modulated signals
❑ At the receiver of an FDM system, the signals are usually separated
by passing through a parallel bank of bandpass filters
▪ There, each filter is tuned to one of the carrier frequencies and has a bandwidth
that is wide enough to pass the desired signal
▪ The output of each bandpass filter is demodulated, and each demodulated
signal is fed to a lowpass filter that passes the baseband message signal and
eliminates the double-frequency components

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 115

 Frequency- Division Multiplexing cont…
❑ FDM is widely used in radio and telephone communications
❑ In telephone communications

▪ Each voice-message signal occupies a nominal bandwidth of 4 kHz

▪ The message signal is single-sideband modulated for bandwidth-
efficient transmission
▪ In the first level of multiplexing, 12 signals are stacked in frequency,
with a frequency separation of 4 kHz between adjacent carriers
▪ Thus, a composite 48 kHz channel, called a group channel,
transmits the 12 voice-band signals simultaneously
▪ In the next level of FDM, a number of group channels (typically
five or six) are stacked together in frequency to form a supergroup
channel
▪ Then the composite signal is transmitted over the channel
▪ Higher-order multiplexing is obtained by combining several
supergroup channels
▪ Thus, an FDM hierarchy is employed in telephone communication
systems

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 116

Example: A channel having band width of 52.5kHz and centre
frequency of 50 kHz is available for transmitting message signals.
Answer the following question
1. If message signals having common band widths 3.5 kHz are
used to generate DSBSC-AM. Then what is the maximum number
of message signals that can be transmitted with FDM ? What is the
minimum spacing between the adjacent carrier frequencies?
2. Repeat question 1 if SSB-AM is used
3. Repeat question 1 if VSB-AM with complete upper side band
and 20% of the lower side band is used.
4. Repeat question 1 if a guard band of 875 Hz is used between the
message signals
5. Repeat question 1 if DSBSC-AM with QAM is used in the first
stage of multiplexing
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 117
1. Channel band width = 52.5 kHz
Signal band width = W=3.5 kHz
Since DSBSC-AM is used, then the band width occupied by each
message signal in the composite signal is = 2W =7 kHz
The number of message signals which can be multiplexed is
= 52.5/7 =7.5
Hence the maximum number of message signals which can be
multiplexed is 7
Since DSBSC-AM is used, the minimum spacing between the
adjacent carrier frequencies is = 2W =7 kHz

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 118

2. Channel band width = 52.5 kHz
Signal band width = W=3.5 kHz
Since SSB-AM is used, then the band width occupied by each
message signal in the composite signal is = W =3.5 kHz
The number of message signals which can be multiplexed is
= 52.5/3.5 =15
Hence the maximum number of message signals which can be
multiplexed is 15
Since SSB-AM is used, the minimum spacing between the adjacent
carrier frequencies is = W =3.5 kHz

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 119

3. Channel band width = 52.5 kHz
Signal band width = W=3.5 kHz
Since VSB-AM is used, then the band width occupied by each
message signal in the composite signal is
W =3.5 +0.2(3.5) kHz = 4.2 kHz
The number of message signals which can be multiplexed is
= 52.5/4.2 =12.5
Hence the maximum number of message signals which can be
multiplexed is 12
Since DSBSC-AM is used, the minimum spacing between the
adjacent carrier frequencies is = W =4.2 kHz
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 120
4. Channel band width = 52.5 kHz
Signal band width = W=3.5 kHz
Since DSBSC-AM and a guard band of 875 Hz is used, then the band
width occupied by each message signal in the composite signal is =
W =(3.5 +0.875 ) kHz = 4.375 kHz
The number of message signals which can be multiplexed is
= (52.5+0.875) / 4.2 =12.2
Hence the maximum number of message signals which can be
multiplexed is 12
Since DSBSC-AM is used, the minimum spacing between the
adjacent carrier frequencies is = W =4.375 kHz

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 121

5. Channel band width = 52.5 kHz
Signal band width = W=3.5 kHz
Since DSBSC-AM is used, then the band width occupied by each
message signal in the composite signal is = 2W =7 kHz
The number of message signals which can be multiplexed is
= 52.5/7 =7.5
Hence the maximum number of message signals which can be
multiplexed is 7
With QAM in the first stage we can multiplex 14 message signals
Since DSBSC-AM is used, the minimum spacing between the
adjacent carrier frequencies is = 2W =7 kHz
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 122
 3.4.2 Quadrature Carrier Multiplexing
(QAM)
❑ Another type of multiplexing allows us to transmit two message
signals on the same carrier frequency
▪ This type of multiplexing uses two quadrature carriers, Accos2 fct
and Acsin2 fct
▪ To elaborate, suppose that m1(t) and m2(t) are two separate message
signals to be transmitted over the channel
o The signal ml(t) amplitude modulates the carrier Accos2fct
o The signal m2(t) amplitude modulates the quadrature carrier Acsin2fct
o The two signals are added together and transmitted over the channel
▪ Hence, the transmitted signal is
u (t ) = Ac m1 (t ) cos(2 f c t ) + Ac m 2 (t ) sin( 2 f c t )
▪ Each message signal is transmitted by DSB-SC AM
▪ This type of signal multiplexing is called quadrature-carrier
multiplexing (QAM)
▪ Quadrature-carrier multiplexing results in a bandwidth-efficient
communication system that is comparable in bandwidth efficiency
to SSB AM
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 123
 Quadrature Carrier Multiplexing cont…
❑ Example (QAM)
Consider two signals m1(t) and m2(t) having same band width of 4 kHz
▪ m1(t) is modulated with carrier frequency Accos2 10000t
▪ m2(t) is modulated with carrier frequency Acsin2 10000t

m(t) =m1(t) + m2(t) is QAM signal

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 124

 Quadrature Carrier Multiplexing cont…

Figure 3.32 Quadrature-carrier multiplexing.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 125

 Quadrature Carrier Multiplexing cont…
❑ Figure 3.32 illustrates the modulation and demodulation of the quadrature-carrier
multiplexed signals
❑ As shown, a synchronous demodulator is required at the receiver to separate and
recover the quadrature-carrier modulated signals
❑ Demodulation of m1(t) is done by multiplying u(t) by cos2 fct and then passing
the result through a lowpass filter

u (t ) cos(2 f c t ) = Ac m1 (t ) cos 2 (2 f c t ) + Ac m 2 (t ) cos(2 f c t ) sin( 2 f c t )

Ac A A
= m1 (t ) + c m1 (t ) cos(4 f c t ) + c m 2 (t ) sin( 4 f c t )
2 2 2
◼ This signal has a lowpass component ml(t) and two high-frequency
components
◼ The lowpass component can be separated using a lowpass filter
◼ To demodule m2(t), we can multiply u(t) by sin2 fct and then pass the
product through a lowpass filter

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 126

 3.5 AM Radio Broadcasting
❑ AM radio broadcasting is a familiar form of communication via
analog signal transmission
❑ Commercial AM radio broadcasting utilizes the frequency band
535-1605 kHz for the transmission of voice and music.
❑ The carrier-frequency allocations range from 540-1600 kHz with
10 kHz spacing
❑ Radio stations employ conventional AM for signal transmission
❑ The base band message signal m(t) is limited to a bandwidth of
approximately 5 kHz.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 127

 AM Radio Broadcasting cont…
❑ The receiver most commonly used in AM radio broadcast is called
superheterodyne receiver shown in Figure 3.33.

Figure 3.33 A superheterodyne receiver.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 128

 AM Radio Broadcasting cont…
❑ It consists of a radio-frequency (RF)-tuned amplifier, a mixer, a
local oscillator, an intermediate-frequency (IF) amplifier, an
envelope detector, an audio-frequency amplifier, and a loudspeaker
❑ Tuning for the desired radio frequency is provided by a variable
capacitor, which simultaneously tunes the RF amplifier and the
frequency of the local oscillator
❑ In the superheterodyne receiver, every AM radio signal is
converted to a common intermediate frequency of fIF = 455 kHz.

▪This conversion allows the use of a single-tuned IF amplifier for signals

from any radio station in the frequency band.
▪ The IF amplifier is designed to have a bandwidth of 10 kHz, which
matches the bandwidth of the transmitted signal
▪ The frequency conversion to IF is performed by the combination of the
RF amplifier and the mixer

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 129

 AM Radio Broadcasting cont…
❑The frequency of the local oscillator is
fLO=fc +fIF
where fc is the carrier frequency of the desired AM radio signal
❑ The tuning range of the local oscillator is 995-2055 kHz.
❑ By tuning the RF amplifier to the frequency fc and mixing its
output with the local oscillator frequency fLO = fc + fIF, we
obtain two signal components at the output of the mixer
▪ One is centered at the difference frequency fIF (fc + fIF - fc )
▪ Second is centered at the sum frequency 2fc + fIF (fc + fIF + fc )
❑ Only the first component is passed by the IF amplifier.
❑ At the input to the RF amplifier, we have signals that are picked up
by the antenna from all radio stations.
❑ By limiting the bandwidth of the RF amplifier to the range
Bc <BRF < 2fIF, where Bc is the bandwidth of the AM radio signal
(10 kHz), we can reject the radio signal transmitted at the so-called
image frequency fc’ = fLO + fIF

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 130

 AM Radio Broadcasting cont…
❑ When we mix the local oscillator output cos 2πfLOt with the received
signals

where fc = fLO - fIF and fc’ = fLO + fIF,

❑ The mixer output consists of the two signals

❑ m1(t) represents the desired signal

❑ m2(t) is the signal sent by the radio station transmitting at the carrier
frequency fc’ = fLO + fIF.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 131

 AM Radio Broadcasting cont…
❑ To prevent the signal r2(t) from interfering with the demodulation of the
desired signal r1(t), the RF-amplifier bandwidth is sufficiently narrow so
the image-frequency signal is rejected.

❑ Hence, BRF < 2fIF is the upper limit on the bandwidth of the RF
amplifier.

❑ In spite of this constraint, the bandwidth of the RF amplifier is still

considerably wider than the bandwidth of the IF amplifier.

❑ Thus, the IF amplifier, with its narrow bandwidth, provides signal

rejection from adjacent channels,

❑ RF amplifier provides signal rejection from image channels.

❑ Figure 3.34 illustrates the bandwidths of both the RF and IF amplifiers

and the requirement for rejecting the image-frequency signal.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 132

 AM Radio Broadcasting cont…
Example:
Desired frequency fC=550 (m1(t))
fIF=455, fLO= fc+fIF=1005 Where fLO>fc
When fc and fLO are mixed we get sum and difference frequencies,
sum=fc+fLO=1555 which is rejected by IF amplifier
dif=fLO-fc =455=fIF Which is passed through the IF amplifier
Image frequency is fc’=fLO+fIF=fc+2fIF=1460 (m2(t))
When fc’ and fLO are mixed we get sum and difference frequencies,
sum=fc’+fLO=1460+1005=2465 which is rejected by IF amplifier
dif=fc’-fLO=1460-1005=455=fIF Which is passed through the IF amplifier
The difference between fc and fc’ is 1460-550= 910=2fIF

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 133

 Figure 3.34 Frequency-
response characteristics
of both IF and RF
amplifiers.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 134

 AM Radio Broadcasting cont…
❑ The output of the IF amplifier is passed through an envelope detector,
which produces the desired audio-band message signal m(t).
❑ The output of the envelope detector is amplified, and this amplified
signal drives a loudspeaker.
❑ Automatic volume control (AVC) is provided by a feedback-control
loop, which adjusts the gain of the IF amplifier based on the power level
of the signal at the envelope detector.

3/13/2021 135

Latha, Department of ECE,ASE,Bengaluru

 Effect of noise on Amplitude Modulation
Systems
❑ Here we are going to analyze the performance of linear
modulation systems in the presence of the noise
❑We determine the signal to noise ratio of the output of the receiver
that demodulates the AM signals
❑We compare the result with the effect of noise on an equivalent
base band communication system

AWGN Channel Model

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 136

Effect of noise on Amplitude Modulation Systems
cont...

❑ Effect of Noise on a Baseband System

❑ Effect of Noise on DSB-SC AM
❑ Effect of Noise on SSB-AM
❑ Effect of Noise on Conventional AM

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 137

 Effect of noise on a base band Systems
❑ Since baseband systems serve as a basis for comparison of various
modulation systems, we begin with a noise analysis of a baseband
system.
❑ In this case, there is no carrier demodulation to be performed.
❑ The receiver consists only of an ideal lowpass filter with the
bandwidth W
❑ The noise power at the output of the receiver, for a white noise
input, is
W N
Pn0 =  0
df = N 0W
−W 2

❑ If we denote the received power by PR, the baseband SNR is given

by S P
= R  
 N b N 0W

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 138

Example 6.1.1: Find the SNR in a baseband system with a bandwidth
of 5 kHz and with N0 = 10-14 W/Hz. The transmitter power is 1 ki
lowatt and the channel attenuation is 10-12.
Solution: We have PR = 10-12 PT = 10-12 x 103 = 10-9 Watts. Therefore,

This is equivalent to 10 log10 20 = 13 dB.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 139

 Effect of noise on DSB-SC AM
❑ Transmitted signal : u (t ) = Ac m(t ) cos(2 f ct )
❑ The received signal at the output of the receiver noise-limiting
filter is sum of this signal and filtered noise ie.
r(t)=u(t) + n(t) (filtered noise)
❑ We know that a filtered noise process can be expressed in terms of
its in-phase and quadrature components as

n(t ) = A(t ) cos[2f c t +  (t )]

= A(t ) cos  (t ) cos(2f c t ) − A(t ) sin  (t ) sin( 2f c t )
= nc (t ) cos(2f c t ) − n s (t ) sin( 2f c t )
(where nc(t) is in-phase component and ns(t) is quadrature component)

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 140

 Effect of noise on DSB-SC AM Systems cont...
❑ Received signal (Adding the filtered noise to the modulated signal)

r (t ) = u (t ) + n(t )
= Ac m(t ) cos(2 f c t ) + nc (t ) cos(2 f c t ) − ns (t ) sin (2 f c t )

❑ Demodulation of the received signal is done by first multiplying

r(t) by a locally generated sinusoid cos(2 fct + ), where is the
phase of the sinusoid then passing the product signal through an
ideal lowpass filter having a bandwidth W.
❑ The multiplication of r(t) with cos(2 fct + ) yields

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 141

 Effect of noise on DSB-SC AM Systems cont...
r (t ) cos(2 f c t +  ) = u (t ) cos(2 f c t +  ) + n(t ) cos(2 f c t +  )
= Ac m(t ) cos(2 f c t ) cos(2 f c t +  )
+ nc (t ) cos(2 f c t ) cos(2 f c t +  )
− n s (t ) sin (2 f c t ) cos(2 f c t +  )
= 12 Ac m(t ) cos( ) + 12 Ac m(t ) cos(4 f c t +  )
+ 12 nc (t ) cos( ) + n s (t ) sin ( )
+ 12 nc (t ) cos(4 f c t +  ) − n s (t ) sin (4 f c t +  )

❑ The lowpass filter rejects the double frequency components and

passes only the lowpass components .

y (t ) = 12 Ac m(t ) cos( ) + 12 nc (t ) cos( ) + ns (t ) sin ( )

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 142

 Effect of noise on DSB-SC AM Systems cont...
❑ If a phase-locked loop is employed, then = 0 and the
demodulator is called a coherent or synchronous demodulator
❑ In our analysis in this section, we assume that we are employing a
coherent demodulator
❑ With this assumption,( = 0) we get

y (t ) = 1
2
Ac m(t ) + nc (t )
❑ Therefore, at the receiver output, the message signal and the noise
components are additive and we are able to define a meaningful
SNR. The message signal power is given by
Ac2
Po = PM power PM is the content of the message signal
4

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 143

 Effect of noise on DSB-SC AM Systems cont...
❑ The noise power is given by
1 1
Pn0 = Pnc = Pn
4 4
❑ The power content of n(t) can be found by noting that it is the
result of passing nw(t) through a filter with bandwidth Bc.
❑ Therefore, the power spectral density of n(t) is given by
 N20 | f − f c | W
Sn ( f ) = 
0 otherwise

❑ The noise power is

 N0
Pn =  S n( f )df =  4W = 2WN 0
− 2

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 144

The power in the in-phase and quadrature components in this case is
2N0W which is same as P1
Latha, Dept. of ECE, ASE, Bengaluru 3/13/2021 145
 Effect of noise on DSB-SC AM Systems cont...
❑ Now we can find the output SNR as
Ac2
S P0 PM Ac2 PM
  = =1 = 4

 N  0 Pn0 4 2WN 0 2WN 0

❑ In this case, the received signal power, given by Eq. (3.2.2), is
PR = Ac2PM /2
❑ The output SNR for DSB-SC AM may be expressed as
S PR
  =
 N  0 DSB N 0W
❑ Which is identical to baseband SNR
❑ In DSB-SC AM, the output SNR is the same as the SNR for a
baseband system
❑ DSB-SC AM does not provide any SNR improvement over
a simple baseband communication system

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 146

 Power Content of DSB-SC Signals
❑The power content of the DSB-SC signal
1 T /2 2
Pu = lim  u (t ) dt
T → T −T / 2

1 T /2 2 2
= lim  Ac m (t ) cos 2 ( 2 f c t ) dt
T →  T −T / 2

Ac2
m (t ) 1 + cos(4 f c t )dt
1 T /2 2
= lim 
2 T → T −T / 2

Ac2
 Pm
2
▪ Pm indicates the power in the message signal m(t) given by

1 T /2
Pm = lim
T → T 
−T / 2
m 2 (t ) dt

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 147

 Effect of noise on SSB-AM
❑ SSB modulated signal :
u (t ) = Ac m(t ) cos(2 f c t )  Ac mˆ (t ) sin( 2 f c t )
❑ Input to the demodulator
r (t ) = Ac m(t ) cos(2 f c t )  Ac mˆ (t ) sin( 2 f c t ) + n(t )
= Ac m(t ) cos(2 f c t )  Ac mˆ (t ) sin( 2 f c t ) + nc (t ) cos(2 f c t ) − ns (t ) sin (2 f c t )
= Ac m(t ) + nc (t )cos(2 f c t ) +  Ac mˆ (t ) − ns (t )sin (2 f c t )

❑ Assumption : Demodulation with an ideal phase reference.

❑ Hence, the output of the lowpass filter is the in-phase component
(with a coefficient of ½) of the preceding signal.

y (t ) = 1
2
Ac m(t ) + nc (t )

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 148

 Effect of noise on SSB-AM cont…
❑ Parallel to our discussion of DSB, we have

Ac2
Po = PM
4
1 1 S P0 Ac2 PM
Pn0 = Pnc = Pn   = =
4 4  N  0 Pn0 WN 0
 N
Pn =  S n( f )df = 0  2W = WN 0
− 2

PR = PU = Ac2 PM

S PR S
  = =  
  0 SSB
N N 0W  N b

❑ The signal-to-noise ratio in an SSB system is equivalent to that of

a DSB system.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 149

The power in the filtered noise or bandpass noise in this case is

Latha, Dept. of ECE, ASE, Bengaluru

3/13/2021 150
 Effect of noise on Conventional -AM
The noise analysis of conventional AM is carried out with the
following demodulation Techniques
❑ Synchronous Demodulation
❑ Envelope Demodulation
▪ High Receiver SNR
▪ Low Receiver SNR

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 151

 Effect of noise on Conventional -AM
Synchronous Demodulation
❑ DSB AM signal : u (t ) = Ac [1 + amn (t )] cos(2 f c t )
❑ Received signal at the input to the demodulator
r (t ) = Ac [1 + amn (t )] cos(2 f c t ) + n(t )
= Ac [1 + amn (t )] cos(2 f c t ) + nc (t ) cos(2 f c t ) − ns (t ) sin (2 f c t )
= Ac [1 + amn (t )] + nc (t )cos(2 f c t ) − ns (t ) sin (2 f c t )
❑ a is the modulation index
❑ mn(t) is normalized so that its minimum value is -1
❑ If a synchronous demodulator is employed, the situation is
basically similar to the DSB case, except that we have 1 + amn(t)
instead of m(t).
❑ After mixing and lowpass filtering

y (t ) = 1
2
Ac amn (t ) + nc (t )
Latha, Department of ECE,ASE,Bengaluru 3/13/2021 152
 Effect of noise on Conventional –AM cont…
❑ Received signal power
PR =
Ac2
2

1 + a 2 PM n 
❑ Assumed that the message process is zero mean
❑ Now we can derive the output SNR as

S
1 2 2
A a PM n 2 2
A a PM n 2
a PM n Ac2
1 + a P 
2

  = = =
4 c c 2 Mn

 N  0 AM 1
4 Pnc 2 N 0W 1 + a 2 PM n N 0W
a 2 PM n PR a 2 PM n  S  S
= =   =   
1 + a 2 PM n N 0W 1 + a PM n  N b
2
 N b
❑ η denotes the modulation efficiency
❑ Since a 2
PMn  1+ a 2
PM n , the SNR in conventional AM is always
smaller than the SNR in a baseband system.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 153

 Effect of noise on Conventional –AM cont…

❑ In practical applications, the modulation index a is in the

range of 0.8-0.9.
❑ Power content of the normalized message process depends on
the message source.
❑ Speech signals : Large dynamic range, PM is about 0.1.
❑ The overall loss in SNR, when compared to a baseband system,
is a factor of 0.075 or equivalent to a loss of 11 dB.
❑ The reason for this loss is that a large part of the transmitter
power is used to send the carrier component of the modulated
signal and not the desired signal.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 154

 Effect of noise on Conventional –AM cont…

Envelope Demodulation

❑ To analyze the envelope-detector performance in the presence

of noise, we must use certain approximations.
❑ This is a result of the nonlinear structure of an envelope
detector, which makes an exact analysis difficult.
❑ In this case, the demodulator detects the envelope of the
received signal and the noise process.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 155

 Effect of noise on Conventional –AM cont…

Envelope Demodulation
❑ To analyze the envelope-detector performance in the presence
of noise, we must use certain approximations.
❑ This is a result of the nonlinear structure of an envelope
detector, which makes an exact analysis difficult.
❑ In this case, the demodulator detects the envelope of the
received signal and the noise process.
❑ The input to the envelope detector is
r (t ) = Ac [1 + amn (t )] + nc (t )cos(2 f c t ) − ns (t ) sin (2 f c t )

❑ Therefore, the envelope of r ( t ) is given by

Vr (t ) = Ac [1 + amn (t )] + nc (t )2 + ns2 (t )

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 156

 Effect of noise on Conventional –AM cont…

High Receiver SNR

Vr (t ) = Ac [1 + amn (t )] + nc (t )2 + ns2 (t )
❑ Now we assume that the signal component in r ( t ) is much
stronger than the noise component. Then

❑ Therefore, we have a high probability that

𝑃 𝑛𝑠 (𝑡)) << 𝑃((𝐴𝑐 [1 + 𝑎𝑚𝑛 (𝑡)] + 𝑛𝑐(𝑡))

Vr (t )  Ac [1 + amn (t )] + nc (t )

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 157

 Effect of noise on Conventional –AM cont…
❑ After removing the DC component, we obtain

y (t ) = Ac amn (t ) + nc (t )
❑ which is basically the same as y(t) for the synchronous demodulation
without the ½ coefficient.
❑ This coefficient, of course, has no effect on the final SNR.
❑ So we conclude that, under the assumption of high SNR at the
receiver input, the performance of synchronous and envelope
demodulators is the same.

Low Receiver SNR

❑ However, if the preceding assumption is not true, that is, if we
assume that, at the receiver input, the noise power is much
stronger than the signal power, Then

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 158

 Effect of noise on Conventional –AM cont…
Vr (t ) = Ac [1 + amn (t )] + nc (t )2 + ns2 (t )
= Ac2 [1 + amn (t )]2 + nc2 (t ) + ns2 (t ) + 2 Ac nc (t )[1 + amn (t )]
 
⎯
⎯→ a
( )
n (t ) + n (t ) 1 + 2
2 2 2 Ac nc (t )
(1 + amn (t ) )
 nc (t ) + ns (t )
c s 2

 Ac nc (t ) 
⎯⎯→Vn (t ) 1 + 2
b
(1 + amn (t ) )
 Vn (t ) 
= Vn (t ) + c c (1 + amn (t ) )
A n (t )
Vn (t )
❑ (a) : Ac2 [1 + amn (t )]2 is small compared with the other
components
(b) : nc (t ) + ns (t ) = Vn (t ) is the envelope of the noise
2 2
❑
process

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 159

 Effect of noise on Conventional –AM cont…
❑ Use the approximation
1 +   1 + 2 , for small  , where

=
2 Ac nc (t )
(1 + amn (t ) )
nc (t ) + ns (t )
2 2

Then
Vr (t ) = Vn (t ) +
Ac nc (t )
(1 + amn (t ) )
Vn (t )
❑ We observe that, at the demodulator output, the signal and
the noise components are no longer additive.
❑ In fact, the signal component is multiplied by noise and is
no longer distinguishable.
❑ In this case, no meaningful SNR can be defined.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 160

Example 6.1.2: We assume that the message is a wide-sense stationary
random process M(t) with the autocorrelation function
RM (τ) = 16 sinc2(10,000 τ).
We also know that all the realizations of the message process satisfy
the condition max |m(t)| = 6. We want to transmit this message to a
destination via a channel with 50-dB attenuation and additive white
noise with the power spectral density Sn (f) = N0/2 =10-12 W/Hz. We
also want to achieve an SNR at the modulator output of at least 50 dB
What is the required transmitter power and channel bandwidth if we
employ the following modulation schemes?
1. DSB AM.
2. SSB AM.
3. Conventional AM with a modulation index equal to 0.8.

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 161

Solution: Band width of the message process.
Power spectral density of the message process is

which is nonzero for - 10,000 < f < 10,000; therefore, W = 10,000 Hz

SNR of Base band system:

Since the channel attenuation is 50 dB, it follows that

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 162

• Since the channel attenuation is 50 dB

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 163

Hence

1. For DSBSC-AM we have

Therefore transmitter power

And band width

2. For SSB-AM we have

Therefore transmitter power

And band width

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 164
3. For conventional AM

where
power content of the normalized message signal with
max|m(t)| = 6 is

To find PM we have

Therefore

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 165

 hence

Therefore SNR is

Therefore transmitter power

And band width

Latha, Department of ECE,ASE,Bengaluru 3/13/2021 166

Example 5.4 :Under what conditions can two disjoint events A and B be
independent?

Solution: If two events are mutually exclusive (disjoint) then

P(A∪B) = P(A)∪P(B) which implies that

P(A∩B) =0. If the events are independent, then

P(A∩B) = P(A)∩P(B). Combining these two conditions we obtain that two

disjoint events are independent if

P(A ∩ B) = P(A)P (B) = 0

Thus, at least one of the events should be of zero probability

Latha, Dept. of ECE, ASE, Bengaluru 3/13/2021 167

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