Linear modulation

Principles of Communication Prof. V. Venkata Rao
Indian Institute of Technology Madras
4.1
4 CHAPTER 4
Linear Modulation
4.1 Introduction
We use the word modulation to mean the systematic alteration of one
waveform, called the carrier, according to the characteristic of another waveform,
the modulating signal or the message. In Continuous Wave (CW) modulation
schemes, the carrier is a sinusoid. We use ( )c t and ( ),m t to denote the carrier
and the message waveforms respectively.
The three parameters of a sinusoidal carrier that can be varied are:
amplitude, phase and frequency. A given modulation scheme can result in the
variation of one or more of these parameters. Before we look into the details of
various linear modulation schemes, let us understand the need for modulation.
Three basic blocks in any communication system are: 1) transmitter 2) Channel
and 3) Receiver (Fig. 4.1).
Fig. 4.1: A basic communication system
The transmitter puts the information from the source (meant for the
receiver) onto the channel. The channel is the medium connecting the transmitter
and the receiver and the transmitted information travels on this channel until it
reaches the destination. Channels can be of two types: i) wired channels or ii)
wireless channels. Examples of the first type include: twisted pair telephone

4.2
channels, coaxial cables, fiber optic cable etc. Under the wireless category, we
have the following examples: earth’s atmosphere (enabling the propagation of
ground wave and sky wave), satellite channel, sea water etc.
The main disadvantage of wired channels is that they require a man-made
medium to be present between the transmitter and the receiver. Though wired
channels have been put to extensive use, wireless channels are equally (if not
more) important and have found a large number of applications.
In order to make use of the wireless channels, the information is to be
converted into a suitable form, say electromagnetic waves. This is accomplished
with the help of a transmitting antenna. The antenna at the receiver (called the
receiving antenna) converts the received electromagnetic energy to an electrical
signal which is processed by the receiver.
The question is: can we radiate the baseband1
information bearing signal
directly on to the channel?
For efficient radiation, the size of the antenna should be 10λ or more
(preferably around 4λ ), where λ is the wavelength of the signal to be radiated.
Take the case of audio, which has spectral components almost from DC upto 20
kHz. Assume that we are designing the antenna for the mid frequency; that is,10
kHz. Then the length of the antenna that is required, even for the 10λ situation
is,
8
3
4
3 10
3 10
10 10 10
c
f
×
= = ×
⋅ ×
meters, c being the velocity of light.
1
Baseband signals have significant spectral content around DC. Some of the baseband signals
that are of interest to us are: a) Speech b) music and c) video (TV signals).
Approximate spectral widths of these signals are: Speech: 5 kHz, Audio : 20 kHz, Video : 5 MHz

4.3
Even an antenna of the size of 3 km, will not be able to take care of the entire
spectrum of the signal because for the frequency components around 1 kHz, the
length of the antenna would be 100λ . Hence, what is required from the point of
view of efficient radiation is the conversion of the baseband signal into a
narrowband, bandpass signal. Modulation process helps us to accomplish this;
besides, modulation gives rise to some other features which can be exploited for
the purpose of efficient communication. We describe below the advantages of
modulation.
1. Modulation for ease of radiation
Consider again transmission of good quality audio. Assume we choose the
carrier frequency to be 1 MHz. The linear modulation schemes that would be
discussed shortly give rise to a maximum frequency spread (of the modulated
signal) of 40 kHz, the spectrum of the modulated signal extending from (1000 -
20) = 980 kHz to (1000 + 20) = 1020 kHz. If the antenna is designed for 1000
kHz, it can easily take care of the entire range of frequencies involved because
modulation process has rendered the signal into a NBBP signal.
2. Modulation for efficient transmission
Quite a few wireless channels have their own appropriate passbands. For
efficient transmission, it would be necessary to shift the message spectrum into
the passband of the channel intended. Ground wave propagation (from the lower
atmosphere) is possible only up to about 2 MHz. Long distance ionospheric
propagation is possible for frequencies in the range 2 to 30 MHz. Beyond 30
MHz, the propagation is line of sight. Preferred frequencies for satellite
communication are around 3 to 6 GHz. By choosing an appropriate carrier
frequency and modulation technique, it is possible for us to translate the
baseband message spectrum into a suitable slot in the passband of the channel
intended. That is, modulation results in frequency translation.

4.4
3. Modulation for multiplexing
Several message signals can be transmitted on a given channel, by assigning to
each message signal an appropriate slot in the passband of the channel. Take
the example of AM broadcast, used for voice and medium quality music
broadcast. The passband of the channel used to 550 kHz to 1650 kHz. That is,
the width of the passband of the channel that is being used is 1100 kHz. If the
required transmission bandwidth is taken as 10 kHz, then it is possible for us to
multiplex, atleast theoretically, 110 distinct message signals on the channel and
still be able to separate them individually as and when we desire because the
identity of each message is preserved in the frequency domain.
4. Modulation for frequency assignment
Continuing on the broadcast situation, let us assume that each one of the
message signals is being broadcast by a different station. Each station can be
assigned a suitable carrier so that the corresponding program material can be
received by tuning to the station desired.
5. Modulation to improve the signal-to-noise ratio
Certain modulation schemes (notably frequency modulation and phase
modulation) have the feature that they will permit improved signal-to-noise ratio
at the receiver output, provided we are willing to pay the price in terms of
increased transmission bandwidth (Note that the transmitted power need not be
increased). This feature can be taken advantage of when the quality of the
receiver output is very important.
Having understood the need and the potential benefits due to modulation,
let us now get into the details of various linear modulation schemes. The four
important types of linear modulation schemes are
1) Double SideBand, Suppressed Carrier (DSB-SC)
2) Double SideBand, Large Carrier (DSB-LC) (also called conventional AM or
simply AM)

4.5
3) Single SideBand (SSB)
4) Vestigial SideBand (VSB)
We shall begin our discussion with DSB-SC.
4.2 DSB-SC Modulation
4.2.1. Modulation
The DSB-SC is the simplest of the four linear modulation schemes listed
above (simplest in terms of the mathematical description of modulation and
demodulation operations). Consider the scheme shown in Fig. 4.2
Fig. 4.2: DSB-SC modulation scheme
( )m t is a baseband message signal with =M f( ) 0 for ( ),f W c t> is a high
frequency carrier, usually with >>cf W .
DSB-SC modulator is basically a multiplier. Let mg denotes the amplitude
sensitivity (or gain constant) of the modulator, with the units per volt (we assume
that ( )m t and cA are in volts). Then the modulator output ( )s t is,
( ) ( ) ( )( )cosm c cs t g m t A t= ω (4.1a)
For convenience, let 1mg = . Then,
( ) ( ) ( )cosc cs t A m t t= ω (4.1b)

4.6
As DSB-SC modulation involves just the multiplication of the message
signal and the carrier, this scheme is also known as product modulation and can
be shown as in Fig. 4.3.
Fig. 4.3: Product Modulation scheme
The time domain behavior of the DSB-SC signal (with 1cA = ) is shown in
Fig. 4.4(b), for the ( )m t shown in Fig. 4.4(a).

4.7
Fig. 4.4: (a) The message signal
(b) The DSB-SC signal
Note that the carrier undergoes a 180ο
phase reversal at the zero crossings of
( )m t . This is brought out more clearly in the oscillograms, shown in Fig. 4.5 and
Fig. 4.6, where ( )m t is a sinusoidal signal.
With reference to Fig. 4.5, between the points ‘a’ and ‘b’, the carrier in the
DSB-SC signal and the actual carrier (bottom picture) are in phase whereas
between the points ‘b’ and ‘c’, they are 1800
out of phase. Fig. 4.6 is an
expanded version of the central part of the waveforms in Fig. 4.5. Here, we can

4.8
very clearly observe that to the left of ‘b’, both the carriers are in phase whereas
to the right, they are 1800
out of phase.
Fig. 4.5: (top) DSB-SC signal with tone modulation
(bottom) The carrier
Fig. 4.6: Expanded versions of a part of the waveforms in Fig. 4.5
Consider waveforms shown in Fig. 4.7. We have on the top, modulating tone
signal and at the bottom, the corresponding DSB-SC. What do we observe on
the oscilloscope, if we feed the X-plates the tone signal and the Y-plates, the
DSB-SC signal? The result is shown in Fig. 4.8, which can be explained as
follows.

4.9
At the point ‘a’ in Fig. 4.7, the modulating tone is at its maximum and
hence the DSB-SC signal has the maximum value. Point ‘a’ in Fig. 4.8
corresponds to the point ‘a’ in Fig. 4.7. Between the points ‘a’ and ‘b’ in Fig. 4.7,
the tone amplitude decreases (reaching the value zero at point b); hence the
maximum value reached by the DSB-SC signal during each carrier cycle keeps
decreasing. As the X-plates are being fed with the same tone signal, this
decrease will be linear and this corresponds to segment ‘a’ to ‘b’ in Fig. 4.8.
(Note that DSB-SC signal is zero at point b). In the time interval between ‘b’ and
‘c’ of Fig. 4.7, the DSB signal increases and this increase is seen as a straight
line between the points ‘b’ and ‘c’ in Fig. 4.8. Between the points ‘c’ and ‘e’ in Fig.
4.7, the tone amplitude varies from the most negative value to the most positive
value. Correspondingly, the display on the oscilloscope will follow the trace
c d e→ → shown in Fig. 4.8.
Fig. 4.7: (top) modulating signal
(bottom) DSB-SC signal

4.10
Fig. 4.8 Display on the oscilloscope with the following inputs:
X-plates: Tone signal
Y-plates: DSB-SC signal
Taking the Fourier transform of Eq. 4.1(b), we have
( ) ( ) ( )
2
c
c c
A
S f M f f M f f⎡ ⎤= − + +⎣ ⎦ (4.2)
If we ignore the constant
2
cA
on the R.H.S of Eq. (4.2), we see that the
modulation process has simply shifted the message spectrum by cf± . As the
frequency translation of a given spectrum occurs quite often in the study of
modulation and demodulation operations, let us take a closer look at this.
i) Let ( )m t be a real signal with the spectrum ( )M f shown below (Fig.
4.9(a)). Let cf be 100 kHz. Assuming = 1
2
cA
, we have ( )S f as shown in
Fig. 4.9(b).

4.11
Fig. 4.9: Frequency translation (a) baseband spectrum (real signal)
(b) Shifted spectrum.
Note that ( ) ( ) ( )102 kHz
2 kHz 202 kHz
f
S f M M
=
= +
1 0= + 1=
and is the point ‘a’ in Fig. 4.9
ii) Let ( )m t be a complex signal with ( )M f as shown in Fig. 4.10(a). The
corresponding shifted spectrum (with 100cf = kHz) is shown in Fig.
4.10(b)

4.12
Fig. 4.10: Frequency translation (a) Baseband spectrum (complex signal)
(b) Shifted spectrum.
In figures 4.9(b) and 4.10(b), the part that is hatched in red is called the
Upper Sideband (USB) and the one hatched in blue is called the Lower Sideband
(LSB). Any one of these two sidebands has the complete information about the
message signal. As we shall see later, SSB modulation conserves the bandwidth
by transmitting only one sideband and recovering the ( )m t with appropriate
demodulation.
Example 4.1
Consider the scheme shown in Fig. 4.11(a). The ideal HPF has the cutoff
frequency at 10 kHz. Given that f1 10= kHz and f2 15= kHz, let us sketch
( )Y f for the ( )X f given at (b).

4.13
Fig. 4.11: (a) The scheme of example 4.1
(b) The input spectrum, ( )X f
We have ( ) ( ) ( )V f X f f X f f1 1= − + + , which is as shown in Fig. 4.12(a).
The HPF eliminates the spectral components for f 10≤ kHz. Hence ( )W f is
as shown in Fig. 4.12(b).
( ) ( ) ( )Y f W f f W f f2 2= − + + . This is shown in Fig. 4.12(c).

4.14
Fig. 4.12: Spectra at various points in the scheme of Fig. 4.11
4.2.2. Coherent demodulation
The process of demodulation of a DSB-SC signal, at least theoretically, is
quite simple. Let us assume that the transmitted signal ( )s t has been received
without any kind of distortion and is one of the inputs to the demodulator as
shown in Fig. 4.13. That is, the received signal ( ) ( )r t s t= . Also, let us assume
that we are able to generate at the receiving end a replica of the transmitted
carrier (denoted ( ) ( )' cosr c cc t A t= ω in Fig. 4.13) which is the other input to the
demodulator.

4.15
Fig. 4.13: Coherent demodulation of DSB-SC
The demodulation process consists of multiplying these two inputs and lowpass
filtering the product quantity ( )v t .
From Fig. 4.13, we have
( ) ( ) ( )( ) ( )( )'cos cosg c c c cv t d A m t t A t= ω ω
where gd is the gain constant of the multiplier, called the detector gain constant,
in the context of demodulation. For convenience, let us take = 1gd
( ) ( ) ( )2' cosc c cv t A A m t t= ω
( )
( )1 cos 2'
2
c
c c
t
A A m t
⎡ ⎤+ ω⎣ ⎦=
Assuming that =' 2c cA A we have
( ) ( ) ( ) ( )cos 4 cv t m t m t f t= + π (4.3)
The second term on the R.H.S of Eq. 4.3 has the spectrum centered at 2 cf±
and would be eliminated by the lowpass filter following ( )v t . Hence ( )0v t , the
output of the demodulation scheme of Fig. 4.13 is the desired quantity, namely,
( )m t .
Let us illustrate the operation of the detector in the frequency domain. Let
( )m t be real with the spectrum shown in Fig. 4.14(a). Let

4.16
( ) ( ) ( ) ( )2 cos cr t s t m t t= = ω . Then ( ) ( ) ( )c cS f M f f M f f= − + + , shown in
Fig. 4.14(b).
Fig. 4.14: Spectra at various points in the demodulation scheme of Fig. 4.13
(Note that the positive frequency part of ( )S f is shown in red and the negative
frequency part in blue). Assuming ( ) ( ) ( )cos cv t s t t= ω (Fig. 4.13 with cA' 1= ),
then ( ) ( ) ( )
1
2
c cV f S f f S f f⎡ ⎤= − + +⎣ ⎦ . ( )
1
2
cS f f− and ( )
1
2
cS f f+ are shown in

4.17
Fig. 4.14(c) and (d) respectively. ( )0V f is the sum of the outputs of the lowpass
filters shown in Fig. 4.14(c) and (d) which is the desired message spectrum,
( )M f .
From the discussion of the demodulation process so far, it appears that
demodulation of DSB-SC is quite simple. In practice, it is not. In the scheme of
Fig. 4.13, we have assumed that, we have available at the receiver, a carrier
term that is coherent (of the same frequency and phase) with the carrier used to
generate the DSB-SC signal at the transmitter. Hence this demodulation scheme
is known as coherent (or synchronous) demodulation. As the receiver and the
transmitter are, in general, not collocated, the carrier source at the receiver is
different from that used at the transmitter and it is almost impossible to
synchronize two independent sources. Fairly sophisticated circuitry has to be
used at the receiver in order to generate the coherent carrier signal, from an
( )r t that has no carrier component in it. Before we discuss at the generation of
the coherent carrier at the receiver, let us look at the degradation caused to the
demodulated message due to a local carrier that has phase and frequency
differences with the transmitted one.
Case i): Constant phase difference between ( )c t and ( )rc t
Let ( ) ( )cos 2 cc t f t= π and ( ) [ ]cos 2r cc t f t= π + ϕ (the amplitude quantities,
cA and '
cA can be treated as 1)
( ) ( ) ( ) ( )cos cosc cv t m t t t= ω ω + ϕ
( ) ( ) ( ) ( )cos cos cos sin sinc c cm t t t t⎡ ⎤= ω ω ϕ − ω ϕ⎣ ⎦
( ) ( ) ( ) ( )2
cos cos sin cos sinc c cm t t t t⎡ ⎤= ω ϕ − ω ω ϕ⎣ ⎦
( )
( ) ( ) ( )1 cos 2 sin 2
cos sin
2 2
c ct m t t
m t
⎡ ⎤+ ω ω
= ϕ − ϕ⎢ ⎥
⎣ ⎦

4.18
At the output of the LPF, we will have only the term
( ) cos
2
m t ϕ
. That is, the
output of the demodulator, ( )0v t , is proportional to ( ) cosm t ϕ. As long as ϕ
remains a constant, the demodulator output is a scaled version of the actual
message signal. But values of ϕ close to 2π will force the output to near about
zero. When 2ϕ = π we have zero output from the demodulator. This is called
the quadrature null effect of the coherent detector.
Case ii): Constant frequency difference between ( )c t and ( )rc t
Let ( ) ( )cos 2 cc t f t= π and ( ) ( )cos 2r cc t f f t⎡ ⎤= π + ∆⎣ ⎦ . Then,
( ) ( ) ( ) ( )cos 2 cos 2c cv t m t f t f f t⎡ ⎤= π π + ∆⎣ ⎦
By carrying out the analysis similar to case (i) above, we find that
( ) ( ) ( )0 cos 2v t m t f t⎡ ⎤α π ∆⎣ ⎦ (4.4a)
Let us look in some detail the implications of Eq. 4.4(a). For convenience,
let ( ) ( ) ( )0 cos 2v t m t f t⎡ ⎤= π ∆⎣ ⎦ (4.4b)
Assume 100f∆ = Hz and consider the spectral component at 1 kHz in ( )M f .
After demodulation, this gives rise to two spectral components, one at 900 Hz
and the other at 1100 Hz, because
( ) ( ) ( ) ( )3 1
cos 2 10 cos 2 100 cos 2 1100 cos 2 900
2
t t t t⎡ ⎤ ⎡ ⎤ ⎡ ⎤π × π × = π × + π ×⎣ ⎦ ⎣ ⎦⎣ ⎦
The behavior of the sum of these two components is shown in Fig. 4.15.
As can be seen from this figure, the envelope of sum signal (broken red line)
attains the peak value twice in a cycle of the beat frequency f∆ . Also, it goes
through zero twice in a cycle of the beat frequency.

4.19
Fig. 4.15: Time-domain behavior of ( ) ( )3
cos 2 10 cos 2 100t tπ × π ×
Let us examine the effect of frequency offset in the frequency domain. Let ( )M f
be as shown in Fig. 4.16(a). Assume 300f∆ = Hz. Then,
( ) ( ) ( )0
1
2
V f M f f M f f⎡ ⎤= − ∆ + + ∆⎣ ⎦ will be as shown in Fig. 4.16(d), which is
one-half the sum of the spectra shown at (b) and (c). Comparing Fig. 4.16(a) and
(d), we are tempted to surmise that the output of the demodulator is a fairly
distorted version of the actual message signal. A qualitative feeling for this
distortion can be obtained by listening to the speech files that follow.
Introduction
Output 1
Output 2
Output 3
Output 4

4.20
Fig. 4.16: The effect of frequency offset in the demodulation of DSB-SC:
(a) Typical message spectrum, ( )M f
(b) ( )300M f +
(c) ( )300M f −
(d) ( ) ( )M f M f
1
300 300
2
⎡ ⎤+ + −⎣ ⎦

4.21
Example 4.2
In this example, we will show that the DSB-SC signal can be demodulated
with the help of any periodic function ( )p t , as long as ( )p t has a spectral
component at cf , the carrier frequency. This component could be due to the
fundamental or some harmonic of the fundamental.
a) Let ( ) ( ) ( )c cs t A m t tcos= ω . Consider the product ( ) ( )ps t x t where ( )px t
is any periodic signal with the period
c
T
f
0
1
= . That is,
( ) cj n f t
p n
n
x t x e 2
∞
π
= − ∞
= ∑
where nx is the nth
Fourier coefficient. We will show that if x1 0≠ , then it is
possible to extract ( )m t from the product ( ) ( )ps t x t .
b) Let ( )py t be another periodic signal with the period T NT0 0
' = . We will
show that, appropriate filtering of the product ( ) ( )ps t y t , will result in ( )m t .
a) As ( ) π − π⎡ ⎤π = +
⎣ ⎦
2 21
cos 2
2
c cj f t j f t
cf t e e , we have
( ) ( )
( ) ( ) ( )c cj n t j n tc
p n n
n n
A m t
s t x t x e x e
1 1
2
+ ω − ω⎡ ⎤
= +⎢ ⎥
⎣ ⎦
∑ ∑
( ) ( ) ( )c cj n t j n tc
n n
n n
n n
A m t
x x e x x e
1 1
1 1
, ,
1 1
2
+ ω − ω
−
≠ − ≠
⎡ ⎤
⎢ ⎥
= + + +⎢ ⎥
⎢ ⎥
⎣ ⎦
∑ ∑
as [ ]
x x
x
1 1
1Re
2
− +
= , the output, after lowpass filtering would be,
[ ] ( )cx A m t1Re . (We assume that the LPF will reject all the other
spectral components)
b) The product ( ) ( )ps t y t can be written as

4.22
( ) ( )
( ) c c
n n
j f t j f t
c N N
p n n
n n
A m t
s t y t y e y e
2 1 2 1
2
⎛ ⎞ ⎛ ⎞
π + π −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎧ ⎫
⎪ ⎪
= +⎨ ⎬
⎪ ⎪⎩ ⎭
∑ ∑
( ) c c
n n
j f t j f t
c N N
N n N n
n n
n N n N
A m t
y y e y y e
2 1 2 1
, ,2
⎛ ⎞ ⎛ ⎞
π + π −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
−
≠ − ≠
⎧ ⎫
⎪ ⎪
= + + +⎨ ⎬
⎪ ⎪
⎩ ⎭
∑ ∑
We assume that Ny 0≠ . Then, the output of the LPF would be
[ ] ( )N cy A m tRe . (Note that ( ) ( )py t s t has spectral lobes at
c cf f
N N
2
0, , , etc.± ± We assume that the LPF will extract the lobe at f 0=
and reject others).
Example 4.3
Consider the scheme shown in Fig. 4.17. ( )s t is the DSB-SC signal
( ) ( )cm t tcos ω with
( )
f
S f
outside
1 , 99 kHz 101kHz
0 ,
⎧ ≤ ≤⎪
= ⎨
⎪⎩
Let ( )g t be another bandpass signal with
( )
f
G f
outside
1 , 98 kHz 102 kHz
0 ,
⎧ ≤ ≤⎪
= ⎨
⎪⎩
a) We will show that the output ( ) ( )y t m tα .
b) We will show that it would not be possible to recover ( )m t from ( )v t if
( )
⎧ < ≤⎪
= ⎨
⎪⎩
1 , 98.5 kHz 101.5 kHz
0 ,
f
G f
outside

4.23
Fig. 4.17: Scheme of DSB-SC demodulation (example 4.3)
a) Let ( ) ( ) ( )cg t m t f t1 cos 2= π where cf 100 kHz= and
( )
f
M f
outside
1
2 , 2 kHz
0 ,
⎧ ≤⎪
= ⎨
⎪⎩
From ( )S f , we see that
( )
f
M f
outside
2 , 1kHz
0 ,
⎧ ≤⎪
= ⎨
⎪⎩
We have,
( ) ( ) ( ) ( )cv t m t m t t2
1 cos= ω
( ) ( )
( )c t
m t m t1
1 cos 2
2
⎡ ⎤+ ω
= ⎢ ⎥
⎣ ⎦
( ) ( ) ( ) ( )
( )c
m t m t m t m t
t1 1
cos 2
2 2
= + ω
We will assume that the LPF rejects the spectrum around cf2± ,
( ) ( ) ( ) ( )m t m t M f M f1 1
2 2
∗
←⎯→
( ) ( )M f M f1∗ will have a flat spectrum for f 1kHz≤ . By using an ILPF
with cutoff at 1 kHz, we can recover ( )m t from ( )v t .
b) For this case ( )M f1 would be
( )
f
M f
outside
1
2 , 1.5 kHz
0 ,
⎧ ≤⎪
= ⎨
⎪⎩
( ) ( )M f M f1 ∗ will be flat only for f 0.5 kHz≤ . Hence ( )m t cannot be
recovered.

4.24
Exercise 4.1
A signal ( )m t whose spectrum is shown in Fig. 4.18(a) is generated
using the signals ( )m t1 and ( )m t2 . ( )M f1 and ( )M f2 are shown at (b) and
(c) respectively in Fig. 4.18. The signal ( ) ( ) ( )s t m t t5
2 cos 10= π is
transmitted on the channel.
a) Suggest a scheme to obtain ( )m t from ( )m t1 and ( )m t2 .
b) ( )m t1 and ( )m t2 are to be recovered from the received signal
( ) ( )r t s t= . A part of this receiver is shown in Fig. 4.18(d). Complete
the receiver structure by indicating the operations to be performed by the
boxes with the question mark inside.
(a)
(b) (c)
(d)
Fig. 4.18: Proposed receiver structure for the exercise 4.1

4.25
4.2.3 Carrier recovery for coherent demodulation
As explained in detail in sec. 4.2.2, coherent demodulation requires a
carrier at the receiving end that is phase coherent with the transmitted carrier.
Had there been a carrier component in the transmitted signal, it would have been
possible to extract it at the receiving end and use it for demodulation. But the
DSB-SC signal has no such component and other methods have to be devised to
generate a coherent carrier at the receiver. Two methods are in common use for
the carrier recovery (and hence demodulation) from the suppressed carrier
modulation schemes, namely (a) Costas loop and (b) squaring loop.
a) Costas loop: This scheme is shown in Fig. 4.19.
Fig. 4.19: Costas loop
The VCO (Voltage Controlled Oscillator) is a source that produces a periodic
waveform1
whose frequency is controlled by the input voltage ( )ce t . The output
frequency of of the VCO when ( ) 0ce t = is called the free running frequency of
the VCO. The frequency put out by the VCO at any instant depends on the sign
and magnitude of the control voltage, ( )ce t .
1
Here, we shall assume that the VCO output is sinusoidal.

4.26
To understand the loop operation, let us assume that the frequency and
phase of the VCO output are the same as that of the incoming carrier. Then,
( ) ( ) ( )( ) ( )( )1 0cos cosc c cv t A m t t A t= ω ω
( ) ( )2
0 cosc cA A m t t= ω
( )
( )
0
1 2cos
2
c
c
t
A A m t
⎡ ⎤+ ω⎣ ⎦=
The output of the LPF1 is
( )
( )0
2
2
cA A m t
v t = ; that is, ( ) ( )2v t m tα , the desired signal.
Similar analysis shows
( )4 0v t =
Now suppose that VCO develops a small phase offset of ϕ radians. The I-
channel output will remain essentially unchanged but a small voltage will develop
at the output of the Q-channel which will be proportional to ϕsin (If the phase
shift is − ϕ rad, then the Q channel output is proportional to sin− ϕ ). Because of
this, ( )e t is a non-zero quantity given by
( ) ( ) ( ) ( )
2
2 4 0
1
cos sin
4
ce t v t v t A A m t⎡ ⎤= = ϕ ϕ⎣ ⎦
( )
2
0
1
sin2
8
cA A m t⎡ ⎤= ϕ⎣ ⎦
( )e t is input to LPF3, which has very narrow passband (Note that LPF1 and
LPF2 should have a bandwidth of at least W Hz). Hence ( ) 0 sin2ce t C= ϕ
where 0C is the DC value of ( )
2
0
1
8
cA A m t⎡ ⎤⎣ ⎦ . This DC control voltage ensures
that the VCO output is coherent with the carrier used for modulation.
b) Squaring loop
The operation of the squaring loop can be explained with the help of Fig.
4.20.

4.27
Fig. 4.20: Demodulation of DSB-SC using a squaring loop
Let ( ) ( ) ( ) ( )cosc cr t s t A m t t= = ω . Then,
( ) ( ) ( ) ( )
2
2 2
1 cos 2
2
c
c
A
v t r t m t t⎡ ⎤= = + ω⎣ ⎦
( )2
m t will have nonzero DC value which implies its spectrum has an impulse at
= 0f . Because of this, ( )V f will have a discrete spectral component at 2 cf .
( )v t is the input to a very narrowband bandpass filter, with the centre frequency
2 cf . By making the bandwidth of LPF1 very narrow, it is possible to make the
VCO to lock on to the discrete component at 2 cf , present in ( )w t . (The dotted
box enclosing a multiplier, LPF and a VCO, connected in the feedback
configuration shown is called the Phase Locked Loop (PLL)). The VCO output
goes through a factor of two frequency divider, yielding a coherent carrier at its
output. This carrier is used to demodulate the DSB-SC signal. Note that LPF2
must have adequate bandwidth to pass the highest frequency component
present in ( )m t .
Both the Costas loop and squaring loop have one disadvantage, namely,
an 1800
phase ambiguity. Consider the Costas loop; if the input to the loop were

4.28
to be ( ) ( )cosc cA m t t− ω , output of the LPF1 would be ( )0
1
cos
2
cA A m t− ϕ and
that of LPF2 would be ( )0
1
sin
2
cA A m t− ϕ with the result that, ( )e t would be the
same as in the case discussed earlier. Similarly, for the squaring loop, ( )v t would
be the same whether ( ) ( ) ( )cosc cr t A m t t= ω or ( ) ( )cosc cA m t t− ω . Hence
the demodulated output could be either ( )m t or ( )m t− . However, this will not
cause any problem for audio transmission because ( )m t and ( )m t− , would
sound the same to our ears.
Though DSB-SC modulation schemes place the entire transmitted power
into the useful sidebands, the demodulation has to be coherent. The circuit
required to generate a coherent carrier increases the cost of the receiver. If only
a few receivers are to be built for a specific communication need, the cost may
not be a major factor. But in a broadcast situation, there would be a large number
of receivers tuned to a given station and in that scenario, it is better make the
receiver fairly cheap and push the cost up of the transmitter, if required. As will
seen later, the Envelope Detector(ED) is fairly cheap to implement as compared
to a coherent detector. But to make use of ED, the modulated carrier should
carry ( )m t in its envelope. Evidently, DSB-SC does not satisfy this property as
explained below.
Let ( ) ( ) ( )cosc cs t A m t t= ω
Pre-envelope of ( ) ( ) ( ) ci t
cpe
s t s t A m t e ω⎡ ⎤= =
⎣ ⎦
Complex envelope of ( ) ( ) ( )cce
s t s t A m t⎡ ⎤= =⎣ ⎦
Hence the envelope of DSB-SC ( ) ( )pe ce
s t s t= =
( )m tα

4.29
We shall now describe a modulation scheme that has ( )m t as its envelope
which can easily be extracted.
4.3 DSB-LC Modulation (or AM)
By adding a large carrier component to the DSB-SC signal, we will have
DSB-LC, which, for convenience, we shall call simply as AM. By choosing the
carrier component properly, it is possible for us to generate the AM signal such
that it preserves ( )m t in its envelope. Consider the scheme shown in Fig. 4.21.
Fig. 4.21: Generation of an AM signal from a DSB-SC signal
Let ( ) ( ) ( )cosc m cv t A g m t t= ω . Then,
( ) ( ) ( )cosc cAM
s t A t v t⎡ ⎤ = ω +⎣ ⎦
( ) ( )1 cosc m cA g m t t⎡ ⎤= + ω⎣ ⎦ (4.5)
In this section, unless there is confusion, we use ( )s t in place of [ ]( ) AM
s t . We
shall assume that ( )m t has no DC component and ( )( ) ( )( )max min
m t m t= − . Let
( )mg m t be such that ( ) 1mg m t ≤ for all t. Then ( )1 0mg m t⎡ ⎤+ ≥⎣ ⎦ and
( ) AM
s t⎡ ⎤⎣ ⎦ preserves ( )m t in its envelope because
( ) ( )1 cj t
c mpe
s t A g m t e ω
⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦
( ) ( )1c mce
s t A g m t⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦

4.30
As ( )1 0mg m t⎡ ⎤+ ≥⎣ ⎦ , we have
Envelope of [ ]= = +( ) ( ) 1 ( )ce c ms t s t A g m t
The quantity after the DC block is proportional to ( )m t .
If ( )1 mg m t⎡ ⎤+⎣ ⎦ is not nonnegative for all t , then the envelope would be different
from ( )m t . This would be illustrated later with a few time domain waveforms of
the AM signal. Fig. 4.22(b) illustrates the AM waveform for the case
( )1 0mg m t⎡ ⎤+ ≥⎣ ⎦ for all t .
Fig. 4.22: (a) An arbitrary message waveform ( )m t
(b) Corresponding AM waveform

4.31
A few time instants have been marked in both the figures (a) and (b). At
the time instants when ( ) 0m t = , carrier level would be cA which we have
assumed to be 1. Maximum value of the envelope (shown in red broken line)
occurs when ( )m t has the maximum positive value. Similarly, the envelope will
reach its minimum value when ( )m t is the most negative. As can be seen from
the figure, the envelope of ( )s t follows ( )m t in a one-to-one fashion.
Let ( )max
1mg m t x= ≤ . Then ( )s t is said to have (100x) percentage
modulation. For the case of 100% modulation,
( ) ( )max min
1m mg m t g m t⎡ ⎤ ⎡ ⎤= − =⎣ ⎦ ⎣ ⎦ . If ( )max
1mg m t > , then we have over
modulation which results in the envelope distortion. This will be illustrated in the
context of tone modulation, discussed next.
Exercise 4.2
For the waveform ( )m t shown in Fig. 4.23, sketch the AM signal with
the percentage modulation of 40. Assume cA 1= (the figure has to be shown
with reference to ( )m t )
Fig. 4.23: Baseband signal for the exercise 4.2

4.32
4.3.1. Tone Modulation
Let ( )m t to be a tone signal; that is, ( ) ( )cos 2m mm t A f t= π where
m cf f<< . Then Eq. 4.5 becomes
( ) ( ) ( )1 cos cosc m m m cs t A g A t t⎡ ⎤= + ω ω⎣ ⎦ (4.6a)
( ) ( )cos cA t t= ω (4.6b)
Let m mg A = µ . Then for tone modulation,
( ) ( ) ( )1 cos cosc m cs t A t t⎡ ⎤= + µ ω ω⎣ ⎦ (4.7)
µ is called the modulation index or modulation factor. 100µ × is the
percentage modulation. To avoid envelope distortion, we require, µ ≤ 1.
As ( ) ( )1 cosc mA t A t⎡ ⎤= + µ ω⎣ ⎦ , we have
( ) [ ]max
1cA t A⎡ ⎤ = + µ⎣ ⎦
( ) [ ]min
1cA t A⎡ ⎤ = − µ⎣ ⎦
( )
( )
max
min
1
1
A t
A t
⎡ ⎤ + µ⎣ ⎦
=
− µ⎡ ⎤⎣ ⎦
or
( ) ( )
( ) ( )
max min
max min
A t A t
A t A t
⎡ ⎤ ⎡ ⎤−⎣ ⎦ ⎣ ⎦
µ =
⎡ ⎤ ⎡ ⎤+⎣ ⎦ ⎣ ⎦
Fig. 4.24 to 4.26 illustrate the experimentally generated AM waveforms for
µ = 0.5, 1and 1.5 respectively (with 1µ > , we have overmodulation).

4.33
Fig. 4.24: AM with tone modulation ( 0.5µ = )
Fig. 4.25: AM with tone modulation ( 1µ = )

4.34
Fig. 4.26: AM with tone modulation ( 1.5µ = )
Fig. 4.27: Envelope of the AM signal of Fig. 4.26
As can be seen from 4.24 and 4.25, the envelope (shown with a red
broken line) is one-to-one related to the message sinusoid. Note that, for µ = 1,
the carrier amplitude (and hence the envelope) goes to zero, corresponding to
the time-instant when the sinusoid is going through the negative peak. However,
when µ > 1, the one-to-one relationship between the envelope of the modulated
carrier and the modulating tone is no longer maintained. This can be more clearly
seen in Fig. 4.27 which shows the output of the envelope detector when the input

4.35
is the modulated carrier of Fig. 4.26. Notice that the tone signal between ( )t t1 2,
and to the right of t3 of Fig. 4.26 is getting inverted; in other words, the output of
the ED is proportional to ( )+ µ ω1 cos mt which is not equal to ( )+ µ ω1 cos mt ,
when µ > 1.
Fig. 4.28: Oscillogram when the CRO inputs are:
X-plates: tone signal
Y-plates: AM signal with
1
2
µ =
Fig. 4.29: Oscillogram when the CRO inputs are:
X-plates: tone signal
Y-plates: AM signal with ( 1µ = )

4.36
Fig. 4.28 and Fig. 4.29 illustrate the oscillograms when the X-plates of the
CRO are fed with the modulating tone and the Y-plates with the AM signal with
µ = 0.5 and µ = 1 respectively. In Fig. 4.28, A represents the peak-to-peak
value of the carrier at its minimum (that is, [ ]= − µ2 1cA A ) where as B is the
peak-to-peak value of the carrier at its maximum (that is, [ ]= + µ2 1cB A ).
Hence µ can be calculated as
B A
B A
−
µ =
+
In Fig. 4.29, as A = 0 we have µ = 1
4.3.2. Spectra of AM signals
Taking the Fourier transform of Eq. 4.5,
( ) ( ) ( ) ( ) ( )
2 2
c c
c c m c cAM
A A
S f f f f f g M f f M f f⎡ ⎤ ⎡ ⎤ ⎡ ⎤= δ − + δ + + − + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(4.8)
The plot of [ ( )]AMS f is given in Fig. 4.30.
Exercise 4.3
Picture the oscillogram when the X-plates of the CRO are fed with the
modulating tone and the Y-plates with the AM signal with µ = 1.5 .

4.37
Fig. 4.30: (a) Baseband message spectrum ( )M f
(b) Spectrum of the AM signal
Based on Fig. 4.30, we make the following observations:
1) The spectrum has two sidebands, the USB [between cf to cf W+ , and
( )cf W− − to cf− , hatched in red] and the LSB ( cf W− to cf and cf− to
( )cf W− + , hatched in blue).
2) If the baseband signal has bandwidth W , then the AM signal has
bandwidth 2 W . That is, the transmission bandwidth TB , required for the
AM signal is 2 W .
3) Spectrum has discrete components at cf f= ± , indicated by impulses of
area
2
cA
4) In order to avoid the overlap between the positive part and the negative part
of ( )S f , cf W> (In practice, cf W>> , so that ( )s t is a narrowband
signal)

4.38
The discrete components at cf f= ± , do not carry any information and as
such AM does not make efficient use of the transmitted power. Let us illustrate
this taking the example of tone modulation.
Example 4.4
For AM with tone modulation, let us find
Total sideband power
η =
Total power
, as a
function of modulation index µ.
For tone modulation, we have
( ) ( ) ( )1 cos cosc m cs t A t t⎡ ⎤= + µ ω ω⎣ ⎦
Carrier term = ( )cosc cA tω
Carrier Power =
2
2
cA
USB term = ( )cos
2
c
c m
A
t
µ
ω + ω
Power in USB =
2
2
2
2 8
c
2
c
A
A
µ⎛ ⎞
⎜ ⎟ µ⎝ ⎠ =
Power in LSB = Power in USB
Total sideband Power =
2
2 c
c
A
A
22
2
8 4
µµ
× =
Total Power=
( )22 2 2
cc c c
AA A A
22 2 2
1
2 4 2 2 4
+ µ⎛ ⎞µ µ
+ = + =⎜ ⎟
⎝ ⎠
Hence,
( )
c
c
A
A
2 2
2
22 2
4
22
4
µ
µ
η = =
+ µ+ µ
Calculating the value of η for a few value of µ, we have

4.39
µ η
0.25 0.03
0.50 0.11
0.75 0.22
1.0 0.33
As can be seen from the above tabulation, η increases as 1µ → ;
however, even at 1µ = , only 1/3 of the total power is in the sidebands (or side
frequencies), the remaining 2/3 being in the carrier. From this example, we see
that AM is not an efficient modulation scheme, in terms of the utilization of the
transmitted power.
The complex envelope behavior of ( )s t for tone modulation is quite
illustrative.
( ) ( ) ( )c m cs t A t t1 cos cos⎡ ⎤= + µ ω ω⎣ ⎦
( ) ( )c m c mc j j tj t tc
c
A
A e e eRe
2
ω + ω ω − ωω µ⎧ ⎫⎡ ⎤= + +⎨ ⎬⎣ ⎦⎩ ⎭
( ) m mj t j tc c
cce
A A
s t A e e
2 2
ω − ωµ µ
⎡ ⎤ = + +⎣ ⎦ (4.9)
Let us draw a ‘phasor’ diagram, using the carrier quantity as the reference.
The term mj tcA
e
2
ωµ
can be represented as a rotating vector with a magnitude
of cA
2
µ
, rotating counterclockwise at the rate of mf rev/sec. Similarly,
mj tcA
e
2
− ωµ
can be shown as a vector with clockwise rotational speed of mf
rev/sec.

4.40
Fig. 4.31: Phasor diagram for AM with tone modulation
Fig. 4.31 depicts the behavior of all the quantities on the RHS of Eq. 4.9.
From Eq.4.9, we find that the complex envelope is real and is given by
( )c mA t1 cos⎡ ⎤+ µ ω⎣ ⎦ . This can also be seen from the phasor diagram, because
at any given time, the quadrature components of the sideband phasors cancel
out where as the in-phase components add up; the resultant of the sideband
components is collinear with the carrier.
The length of the in-phase component of ( ) ce
s t⎡ ⎤⎣ ⎦ depends on the sign of
the resultant of the sideband phasors. As can be seen from Fig. 4.31 this varies
between the limits [ ]cA 1 − µ to [ ]cA 1 + µ . If the modulation index µ is less than
1, [ ]cA 1 0− µ > and envelope of ( )s t is
( ) ( ) ( )⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + µ ω = + µ ω⎣ ⎦ ⎣ ⎦ ⎣ ⎦1 cos 1 cosc m c mce
s t A t A t
Phasor diagrams such as the one shown in Fig. 4.31 are helpful in the
study of unequal attenuation of the sideband components. We shall illustrate this
with an example.
Example 4.5
Let 1cA = , µ =
1
2
and let the upper sideband be attenuated by a factor of
2. Let us find the expression for the resulting envelope, ( )A t .

4.41
The phasor diagram for this case is shown in Fig. 4.32.
Fig. 4.32: Phasor diagram for an AM signal with unequal sidebands
As can be seen from the figure, the resultant of the sidebands is no longer
collinear with the carrier.
( ) ( ) ( )( ) ( ) ( )( )m m m mce
s t t j t t j t
1 1
1 cos sin cos sin
8 4
⎡ ⎤ = + ω + ω + ω − ω⎣ ⎦
( ) ( )m mt j t
3 1
1 cos sin
8 8
= + ω − ω
( ) ( ) ( )m mA t t t
1
2 2 2
3 1
1 cos sin
8 8
⎡ ⎤⎛ ⎞ ⎛ ⎞
= + ω + ω⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
Evidently, it is not possible for us to recover the message from the above ( )A t .
4.4 Generation of AM and DSB-SC signals
Let ( )x t be the input to an LTI system with the impulse response ( )h t
and let ( )y t be the output. Then,
( ) ( ) ( )y t x t h t= ∗
( ) ( ) ( )Y f X f H f=
That is, an LTI system can only alter a frequency component (either boost or
attenuate), that is present in the input signal. In other words, an LTI system
cannot generate at its output frequency components that are not present in
( )X f . We have already seen that the spectrum of a DSB or AM signal is

4.42
different from that of the carrier and the baseband signal. That is, to generate a
DSB signal or an AM signal, we have to make use of nonlinear or time-varying
systems.
4.4.1 Generation of AM
We shall discuss two methods of generating AM signals, one using a
nonlinear element and the other using an element with time-varying
characteristic.
a) Square law modulator
Consider the scheme shown in Fig. 4.33(a).
Fig. 4.33 (a): A circuit with a nonlinear element
(b): v i− characteristic of the diode in Fig. 4.28(a)
A semiconductor diode, when properly biased has a v i− characteristic that
nonlinear, as shown in Fig. 4.33(b).

4.43
For fairly small variations of v around a suitable operating point, ( )v t2 can be
written as
( ) ( ) ( )v t v t v t2
2 1 1 2 1= α + α (4.10)
where 1α and 2α are constants.
Let ( ) ( ) ( )c cv t A f t m t1
' cos 2= π + . Then,
( ) ( ) ( ) ( ) ( )c c c cv A m t f t m t m t A f tt 2 2 22
2 1 1 2 2
1
2' 1 cos 2 ( ) cos 2'α
= α + π + α + α + α π
α
⎡ ⎤
⎢ ⎥
⎣ ⎦
(4.11)
The first term (on the RHS of Eq. 4.11) is ( )⎡ ⎤⎣ ⎦AM
s t , with the carrier amplitude
c cA A1
'= α and mg 2
1
2α
=
α
.
Now the question is: can we extract ( )⎡ ⎤⎣ ⎦AM
s t from the sum of terms on
the RHS of Eq. 4.11? This can be answered by looking at the spectra of the
various terms constituting v t2( ) . Fig. 4.34 illustrates these spectra (quantities
marked A to E) for the ( )M f of Fig. 4.14(a). The time domain quantities
corresponding to A to E are listed below.
Fig. 4.34: Spectra of the components of ( )v t2 of Eq. 4.11

4.44
Component Spectrum indicated by
(i) ( )c cA f tcos 2π A
(ii) ( ) ( )c cA m t f t2
'2 cos 2α π B
(iii) ( )m t1α C
(iv) ( )m t2
2α D
(v) ( )c cA f t
2
2
2
' cos (2 )α π
E
( )⎡ ⎤⎣ ⎦AM
s t consists of the components (i) and (ii) of the above list. If it is possible
for us to filter out the components (iii), (iv) and (v), then the required AM signal
would be available at the output of the filter. This is possible by placing a BPF
with centre at cf and bandwidth 2W provided ( )cf W W2− > or cf W3> .
Usually, this is not a very stringent requirement. However, this scheme
suffers from a few disadvantages.
i) The required square-law nonlinearity of a given device would be available
only over a small part of the ( )v i− characteristic. Hence, it is possible to
generate only low levels of the desired output.
ii) If cf is of the order of 3W, then we require a BPF with very sharp cut off
characteristics.
b) Switching modulator
In the first method of generation of the AM signals, we have made use of
the nonlinearity of a diode. In the second method discussed below, diode will be
used as a switching element. As such, it acts as a device with time-varying
characteristic, generating the desired AM signals when it is used in the circuit
configuration shown in Fig. 4.35.

4.45
Fig. 4.35: (a) Switching modulator
(b) Switching characteristic of the diode-load combination.
The (ideal) transfer characteristics of the diode-load combination is shown at (b)
in Fig. 4.35. This is explained as follows. We have,
( ) ( ) ( )v t c t m t1 = +
( ) ( )c cA f t m tcos 2= π +
If we assume that ( ) cm t A
max
<< , then the behavior of the diode is governed by
( )c t and can be approximated as
( )
( ) ( )
( )
v t c t
v t
c t
1
2
, 0
0 , 0
⎧ >⎪
⎨
≤⎪⎩
(That is, the diode offers infinite impedance when reverse biased and has zero
impedance, when forward biased. Hence, whether ( )v t1 is switched to the output
or not depends on the carrier cycle)
We can express ( )v t2 as
( ) ( ) ( )pv t v t x t2 1= (4.12)
where ( )px t is the periodic rectangular pulse train of example 1.1. That is,
( )
( ) ( )( )
( ) ( )( )
p
c t c t
x t
c t c t
if
if
1 , 0 positivehalf cycleof
0 , 0 negativehalf cycleof
⎧ >⎪
= ⎨
<⎪⎩
But from example 1.1 ( with =0 ),cf f

4.46
( )
∞
π
= − ∞
= ∑ 2 cj nf t
p n
n
x t x e
where n
n
x c
1
sin
2 2
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
(4.13a)
From Eq. 4.13(a), we find that 0nx = , for n 2, 4= ± ± etc. Combining the terms
− nx and nx , we obtain the trigonometric Fourier series, namely,
( ) ( )
( )
( )
n
p c c
n
x t f t n f t
n
1
2
11 2
cos 2 cos 2 2 1
2 2 1
−∞
=
−
⎡ ⎤= + π + π −⎣ ⎦π −
∑ (4.13b)
From Eq. 4.12 and 4.13(b), we see that ( )v t2 is composed of two components,
namely,
a) The desired quantity: ( ) ( )
⎡ ⎤
+ π⎢ ⎥
π⎣ ⎦
4
1 cos 2
2
c
c
c
A
m t f t
A
b) The undesired terms with
i) Impulses in spectra at c cf f f0, 2 , 4= ± ± etc.
ii) Spectral lobes (same in shape as M(f)) of width 2W, centered at
c cf f0, 3 , 5± ± etc.
As compared to the square law modulator, switching modulator has the following
advantages:
a) Generated AM signals can have larger power levels.
b) Filtering requirements are less stringent because we can separate the
desired AM signal if cf W2> .
However, the disadvantage of the method is that percentage modulation has to
be low in order that the switching characteristics of the diode are controlled only
by the carrier.

4.47
4.4.2. Generation of DSB-SC
a) Product modulator
Generation of a DSB-SC signal involves the multiplication of ( )m t with
( )c cA tcos ω . To generate this signal in the laboratory, any of the commercially
available multipliers can be used. Some of them are:
National: LM 1496
Motorola: MC 1496
Analog Devices: AD 486, AD 632 etc
Signetics: 5596
The power levels that can be generated, the carrier frequencies that can be used
will depend on the IC used. The details can be obtained from the respective
manuals. Generally, only low power levels are possible and that too over a
limited carrier frequency range.
b) Ring modulator
Consider the scheme shown in Fig. 4.36. We assume that the carrier
Fig. 4.36: Ring modulator
signal ( )c t is much larger than ( )m t . Thus ( )c t controls the behavior of diodes
which would be acting as ON-OFF devices. Consider the carrier cycle where the

4.48
terminal 1 is positive and terminal 2 is negative. T1 is an audio frequency
transformer which is essentially an open circuit at the frequencies near about the
carrier. With the polarities assumed for ( )c t , D1, D4 are forward biased, where as
D2, D3 are reverse biased. As a consequence, the voltage at point ‘a’ gets
switched to a' and voltage at point ‘b’ to b' . During the other half cycle of ( )c t ,
D2 and D3 are forward biased where as D1 and D4 are reverse biased. As a
result, the voltage at ‘a’ gets transferred to b' and that at point ‘b’ to a'. This
implies, during, say the positive half cycle of ( )c t , ( )m t is switched to the output
where as, during the negative half cycle, ( )m t− is switched. In other words,
( )v t can be taken as
( ) ( ) ( )pv t m t x t= (4.14)
where ( )px t is square wave as shown in Fig. 4.37.
Fig. 4.37: ( )px t of Eq. 4.14

4.49
Fig. 4.38: (a) A message waveform ( )m t
(b) ( )v t of the ring modulator
Fig. 4.38(b) illustrates the product quantity ( ) ( )pm t x t , for the ( )m t shown in
Fig. 4.38(a). The Fourier series expansion of ( )px t can be written as
( )
( )
( )
n
p c
n
x t n t
n
1
2
1, 3, 5, ...
14
cos
−
=
−
= ω
π
∑ .
When ( )v t is passed through a BPF tuned to cf , the output is the desired DSB-
SC signal, namely, ( ) ( ) ( )cs t m t t
4
cos= ω
π
.
Example 4.6: Generation of DSB-SC
Consider the scheme shown in Fig. 4.39. The non-linear device has the
input-output characteristic given by
( ) ( ) ( )y t a x t a x t3
0 1= +

4.50
Fig. 4.39: The scheme for the example 4.6
Let ( ) ( ) ( )x t A f t m t1cos 2= π + where ( )m t is the message signal. If the
required output ( )s t is a DSB-SC signal with a carrier frequency of 1 MHz, let us
find the value of f1, assuming that a suitable BPF is available.
( ) ( ) ( ) ( ) ( )y t a A f t m t a A f t m t
3
0 1 1 1cos 2 cos 2⎡ ⎤ ⎡ ⎤= π + + π +⎣ ⎦ ⎣ ⎦
1 2
( ) ( ) ( ) ( ) ( ) ( )a A f t m t A f t m t A f t m t3 3 3 2 2 2
1 1 11cos 2 3 cos 2 3 cos 2= π + + π + π⎡ ⎤
⎣ ⎦2
In the equation for the quantity 2 above, the only term on the RHS that
can give rise to the DSB-SC signal is ( ) ( )a A m t f t2 2
1 13 cos 2π .
( ) ( ) ( )
( ){ }f t
a A m t f t a A m t
12 2 2
1 1 1
1 cos 2 2
3 cos 2 3
2
⎡ ⎤+ π⎣ ⎦
π =
Assume that the BPF will pass only the components centered around f12 . Then,
choosing f1 500= kHz, we will have
( ) ( ) ( )c cs t A m t f tcos 2= π
where cA a A2
13= and cf 1= MHz.

4.51
Exercise 4.4
Consider the circuit configuration (called Cowan modulator) shown in Fig.
4.40. Show that the circuit can produce at its output the DSB-SC signal. T1 is the
audio frequency transformer where as T2 and T3 are designed to operate around
the carrier frequency.
Fig. 4.40: Cowan modulator
Exercise 4.5: Balanced Modulator (BM)
Consider the scheme shown in Fig. 4.41. This configuration is usually
called a balanced modulator. Show that the output ( )s t is a DSB-SC signal,
thereby establishing that BM is essentially a multiplier.
Fig. 4.41: Balanced modulator (BM)

4.52
4.5 Envelope Detector
As mentioned earlier, the AM signal, when not over modulated allows the
recovery of ( )m t from its envelope. A good approximation to the ideal envelope
detector can be realized with a fairly simple electronic circuit. This makes the
receiver for AM somewhat simple, there by making AM suitable for broadcast
applications. We shall briefly discuss the operation of the envelope detector,
which is to be found in almost all the AM receivers.
Consider the circuit shown in Fig. 4.42.
Fig. 4.42: The envelope detector circuit
We assume the diode D to be ideal. When it is forward biased, it acts as a short
circuit and thereby, making the capacitor C charge through the source
resistance sR . When D is reverse biased, it acts as an open circuit and C
discharges through the load resistance LR .
As the operation of the detector circuit depends on the charge and
discharge of the capacitor C , we shall explain this operation with the help of Fig.
4.43.

4.53
Fig. 4.43: Envelope detector waveforms
(a) ( )v t1 (before DC block)
(b) ( )outv t (after DC block)
If the time constants sR C and LR C are properly chosen, ( )v t1 follows the
envelope of ( )s t fairly closely. During the conduction cycle of D , C quickly
charges to the peak value of the carrier at that time instant. It will discharge a
little during the next off cycle of the diode. The time constants of the circuit will
control the ripple about the actual envelope. BC is a blocking capacitor and the
final ( )outv t will be proportional to ( )m t , as shown in Fig. 4.43(b). (Note that a
small high frequency ripple, at the carrier frequency could be present on ( )outv t .
For audio transmission, this would not cause any problem, as cf is generally
much higher than the upper limit of the audio frequency range).

4.54
How do we choose the time constants? sR , though not under our control
can be assumed to be fairly small. Values for LR and C can be assigned by us.
During the charging cycle, we want the capacitor to charge to the peak value of
the carrier in as short a time as possible. That is,
1
s
c
R C
f
<< (4.15a)
Discharge time constant should be large enough so that C does not discharge
too much between the positive peaks of the carrier but small enough to be able
follow the maximum rate of change of m(t). This maximum rate depends on W,
the highest frequency in M(f). That is
L
c
R C
f W
1 1
<< << (4.15b)
Too small a value for LR C will make ( )1V t somewhat ragged (sort of saw
tooth ripple on the top) where as, with too large value for LR C , ED fails to follow
the envelope during the periods when ( )m t is decreasing. A more accurate
analysis of the behavior of ED with ( )m t as a tone signal is given in appendix
A4.1.
Example 4.7
Consider the scheme shown in Fig. 4.44. ( )x t is a tone signal given by
( ) ( )x t t4
cos 2 10⎡ ⎤= π ×
⎣ ⎦
and ( ) ( )cc t f tcos 2= π with cf 10= MHz. ( )c t is the
HT of ( )c t . ( )v t , the output of the Balanced Modulator (BM), is applied as input
to an ideal HPF, with cutoff at 10 MHz. We shall find the expression for ( )y t , the
output of an ideal envelope detector.

4.55
It is not too difficult to see that
( ) ( )w t f t61
cos 2 10 10
2
⎡ ⎤= π + ∆
⎣ ⎦
where f 0.01∆ = Hz. That is,
( ) ( ) ( ) ( ) ( )c cw t f t t f t t
1
cos 2 cos sin 2 sin
2
⎡ ⎤⎡ ⎤ ⎡ ⎤= π ∆ ω − π ∆ ω⎣ ⎦ ⎣ ⎦⎣ ⎦
As ( ) ( ) ( )cz t w t f tsin 2⎡ ⎤= + π⎣ ⎦ , we have
( ) ( ) ( ) ( ) ( )c cz t f t t f t t
1 1
cos 2 cos sin 2 1 sin
2 2
⎡ ⎤
⎡ ⎤ ⎡ ⎤= π ∆ ω − π ∆ − ω⎣ ⎦ ⎣ ⎦⎢ ⎥
⎣ ⎦
( )z t represents a narrowband signal with the in phase component
( )f t
1
cos 2
2
⎡ ⎤π ∆⎣ ⎦ and the quadrature component ( )f t
1
sin 2 1
2
⎧ ⎫
⎡ ⎤π ∆ −⎨ ⎬⎣ ⎦
⎩ ⎭
. Hence,
( ) ( ) ( )y t f t f t
1
2 2
21 1
cos 2 sin 2 1
4 2
⎧ ⎫⎪ ⎪⎡ ⎤
⎡ ⎤ ⎡ ⎤= π ∆ + π ∆ −⎨ ⎬⎣ ⎦ ⎣ ⎦⎢ ⎥
⎣ ⎦⎪ ⎪⎩ ⎭
( )t
1
245
sin 2 10
4
⎧ ⎫⎡ ⎤= − π ×⎨ ⎬⎣ ⎦⎩ ⎭
Example 4.8
Consider the scheme shown in Fig.4.45.
Let us find the output ( )y t when,

4.56
a) ( ) ( ) ( )m cx t g m t t1 cos⎡ ⎤= + ω⎣ ⎦ . ( )mg m t 1< and ( )m t is band-limited to
W Hz and the LPF has a bandwidth of W2 . Assume that cf W2>> .
b) ( )x t is a DSB-SC signal; that is ( ) ( ) ( )cx t m t tcos= ω .
a) ( ) ( ) ( ) ( )m cv t x t g m t t
22 2
1 cos⎡ ⎤= = + ω⎣ ⎦
( )
( )c
m
t
g m t
2 1 cos 2
1
2
⎡ ⎤+ ω⎣ ⎦⎡ ⎤= +⎣ ⎦
( ) ( )
( )m m
c
g m t g m t
t
2 2
1 1
cos 2
2 2
⎡ ⎤ ⎡ ⎤+ +⎣ ⎦ ⎣ ⎦= + ω .
The second term on the RHS will be eliminated by the LPF. Hence,
( )
( )mg m t
w t
2
1
2
⎡ ⎤+⎣ ⎦= . As ( )mg m t1 0⎡ ⎤+ ≥⎣ ⎦ , we have
( )
( )mg m t
y t
1
2
⎡ ⎤+⎣ ⎦= .
b) When ( ) ( ) ( )cx t m t tcos= ω , we have
( ) ( ) ( ) ( )
( )c
c
t
v t m t t m t2 2 2 1 cos 2
cos
2
⎡ ⎤+ ω⎣ ⎦= ω =
The output of the LPF would be
( )
( )m t
w t
2
2
=
As the squaring operation removes the information about the sign of the
signal, the output of ( )y t is
( )
( )m t
y t
2
=

4.57
4.6 Theory of Single-Sideband
Assume that from a DSB-SC signal, we have completely suppressed one
of the sidebands, say, the LSB. Let ( ) ( ) ( )
2
c
c cDSB
A
S f M f f M f f⎡ ⎤⎡ ⎤ = − + +⎣ ⎦ ⎣ ⎦
where ( )M f is as shown in Fig. 4.30(a). The resulting spectrum ( )⎡ ⎤⎣ ⎦USB
S f will
be as shown in Fig. 4.47. Can we get back ( )m t from the above signal? The
answer is YES. Let
( ) ( ) ( )⎡ ⎤= × π⎣ ⎦ cos 2 cUSB
v t s t f t
If ( ) ( )⎡ ⎤= ⎣ ⎦
'
USB
S f S f , then ( ) ( ) ( )
1 ' '
2
c cV f S f f S f f⎡ ⎤= − + +⎢ ⎥⎣ ⎦
Exercise 4.6
Consider the waveform ( )m t shown in Fig. 4.46. A DSB-SC is
generated using ( )m t and a suitable high frequency carrier. Sketch the output
of an ideal envelope detector when the input to the detector is the DSB-SC
signal.
Fig. 4.46: ( )m t for the exercise 4.6

4.58
Fig 4.47: Spectrum of the upper sideband signal
By plotting the spectrum of ( )v t and extracting the spectrum for f W≤ , we see
that it is ( )
1
2
cA M f . A similar analysis will show that it is possible to extract
( )m t from ( )⎡ ⎤⎣ ⎦LSB
S f . In other words, with coherent demodulation, it is possible
for us to recover the message signal either from USB or LSB and the
transmission of both the sidebands is not a must. Hence it is possible for us to
conserve transmission bandwidth, provided we are willing to go for the
appropriate demodulation.
Let us now derive the time domain equation for an SSB signal. Let us start
with the two-sided spectrum and then eliminate the unwanted sideband. We shall
retain the upper sideband and try to eliminate the lower sideband. Consider
( ) ( ) ( )sgn
2
c
c c c
A
M f f f f M f f⎡ ⎤− + − −⎣ ⎦
But ( ) ( )
( )
( )
,
sgn
,
c c
c c
c c
M f f f f
f f M f f
M f f f f
⎧ − >⎪
− − = ⎨
− − <⎪⎩
Hence ( ) ( )
( )⎧ − >⎪
⎡ ⎤− + − = ⎨⎣ ⎦ <⎪⎩
,
1 sgn
2 0 ,
c c cc
c c
c
A M f f f fA
M f f f f
f f
That is, the lower sideband has been eliminated from the positive part of the
spectrum.
( ) ( ) 2 cj f t
cM f f m t e π
− ←⎯→

4.59
( ) ( ) ( ) 21
sgn cj f t
c cf f M f f m t e
j
π
− − ←⎯→ − ,
where ( )m t is the HT of ( )m t .
That is, ( ) ( ) ( ) ( ) 21
1 sgn
2 2
cj f tc c
c c
A A
M f f f f m t m t e
j
π⎡ ⎤
⎡ ⎤− + − ←⎯→ −⎢ ⎥⎣ ⎦
⎣ ⎦
(4.16a)
Similarly, ( ) ( )
( )⎧ + < −⎪
⎡ ⎤+ − + = ⎨⎣ ⎦ > −⎪⎩
,1
1 sgn
2 0 ,
c c
c c
c
M f f f f
M f f f f
f f
( ) ( ) ( ) ( )2 2
1 sgn
2 2 2
c cj f t j f tc c c
c c
A A A
M f f f f m t e m t e
j
− π − π
⎡ ⎤+ − + ←⎯→ +⎣ ⎦ (4.16b)
Combining Eq. 4.16(a) and Eq. 4.16(b), we have the time domain equation for
the upper single sideband signal, namely,
( ) ( ) ( ) ( ) ( )cos sinc c cUSB
s t A m t t m t t⎡ ⎤⎡ ⎤ = ω − ω⎣ ⎦ ⎣ ⎦
(4.17)
Assume that the USSB signal is obtained from the DSB-SC signal,
( ) ( )cosc cA m t tω , by filtering out the LSB part. Then,
( ) ( ) ( ) ( ) ( )cos sin
2
c
c cUSB
A
s t m t t m t t⎡ ⎤⎡ ⎤ = ω − ω⎣ ⎦ ⎣ ⎦
(4.18)
A few authors take Eq. 4.18 as representative of the SSB signal. Eq. 4.18
has the feature that the average power of the SSB signal is one-half the average
power of corresponding DSB signal. We shall make use of both Eq. 4.17 and
Eq. 4.18 in our further studies.
By a procedure similar to that outlined above, we can derive a time
domain expression for the LSB signal. The result would be
( ) ( ) ( ) ( ) ( )⎡ ⎤⎡ ⎤ = ω + ω⎣ ⎦ ⎣ ⎦
cos sinc c cLSB
s t A m t t m t t or (4.19a)
( ) ( ) ( ) ( ) ( )cos sin
2
c
c cLSB
A
s t m t t m t t⎡ ⎤⎡ ⎤ = ω + ω⎣ ⎦ ⎣ ⎦
(4.19b)
An SSB signal, whether upper or lower, is also a narrowband bandpass signal.
Eq. 4.18 can be treated as the canonical representation of USB signal with ( )m t
as the in-phase component and ( )m t as the quadrature component. Similarly

4.60
Eq. 4.19 provides the canonical representation of the LSB signal where ( )m t is
the in-phase component and ( )m t− is the quadrature component.
We have already seen that a narrowband signal can also be expressed in
the envelope and phase form. Taking the USB signal, we have for the complex
envelope the quantity ( ) ( )( )2
cA
m t j m t+ . Hence the envelope ( )A t of the USB
signal is
( ) ( ) ( )
22
2
c
USB
A
A t m t m t⎡ ⎤ = +⎣ ⎦ (4.20a)
Similarly for the phase ( )tϕ , we have
( )
( )
( )
arc tan
m t
t
m t
⎡ ⎤
ϕ = ⎢ ⎥
⎢ ⎥⎣ ⎦
(4.20b)
Expressing the USB signal with the envelope-phase form, we have
( ) ( ) ( ) ( )cos cUSB
s t A t t t⎡ ⎤⎡ ⎤ = ω + ϕ⎣ ⎦ ⎣ ⎦ (4.21)
where ( )A t and ( )tϕ are given by Eqs. 4.20(a) and 4.20(b) respectively. The
expression for ( ) LSB
s t⎡ ⎤⎣ ⎦ is identical to Eq. 4.21 but with ( )tϕ given by
( )
( )
( )
tan
m t
t arc
m t
⎡ ⎤
ϕ = −⎢ ⎥
⎢ ⎥⎣ ⎦
(4.22)
That is, ( ) ( ) ( )( )cos cSSB
s t A t t t⎡ ⎤ = ω + ϕ⎣ ⎦ (4.23)
where ( )tϕ is given either by Eq. 4.20(b) or Eq. 4.22. Eq. 4.23 indicates that an
SSB signal has both amplitude and phase variations. (AM and DSB-SC signals
have only the amplitude of the carrier being changed by the message signal.
Note that AM or DSB-SC signals do not have quadrature components.) As such,
SSB signals belong to the category of hybrid amplitude and phase modulation.

4.61
Example 4.9: SSB with tone modulation
As a simple application of the Eqs. 4.18 and 4.19, let ( )m t be ( )cos m tω .
Let us find the SSB signals.
( ) ( ) ( ) ( )m mm t t m t tcos sin= ω ⇒ = ω . Therefore,
( ) ( ) ( ) ( ) ( )cos cos sin sin
2
c
m c m cUSB
A
s t t t t t⎡ ⎤⎡ ⎤ = ω ω − ω ω⎣ ⎦ ⎣ ⎦
( )cos
2
c
c m
A
t⎡ ⎤= ω + ω⎣ ⎦
( ) ( ) ( ) ( ) ( )cos cos sin sin
2
c
m c m cLSB
A
s t t t t t⎡ ⎤⎡ ⎤ = ω ω + ω ω⎣ ⎦ ⎣ ⎦
( )cos
2
c
c m
A
t⎡ ⎤= ω − ω⎣ ⎦
Alternatively,
( ) ( ) ( )cos cosc m cDSB SC
s t A t t
−
⎡ ⎤ = ω ω⎣ ⎦
( ) ( )cos cos
2
c
c m c m
A
t t⎡ ⎤= ω + ω + ω − ω⎣ ⎦
Extracting the USB, we have
( ) ( )cos
2
c
c mUSB
A
s t t⎡ ⎤⎡ ⎤ = ω + ω⎣ ⎦ ⎣ ⎦
If we eliminate the USB, then
( ) ( )cos
2
c
c mLSB
A
s t t⎡ ⎤⎡ ⎤ = ω − ω⎣ ⎦ ⎣ ⎦
Example 4.10
Let ( ) ( ) ( )m t x t y t= where ( )X f and ( )Y f are as shown in Fig. 4.48.
An LSB signal is generated using ( )m t as the message signal. Let us develop
the expression for the SSB signal.

4.62
Fig. 4.48: ( )X f and ( )Y f of example 4.10
Let us take the SSB signal as
( ) ( ) ( ) ( ) ( )c cLSB
s t m t t m t tcos sin⎡ ⎤ = ω + ω⎣ ⎦ (4.24)
We have
( ) ( )x t c t3 3
2 10 sin 2 10⎡ ⎤= × ×
⎣ ⎦
( ) ( ) ( )y t c t t3 2 3
10 sin 500 cos 4 10⎡ ⎤= π ×
⎣ ⎦
( ) ( ) ( )m t x t y t= (4.25a)
What is required is ( )m t , the HT of ( )m t . ( )m t is the product of a lowpass and
a bandpass signal. Hence ( ) ( ) ( )m t x t y t= . (See the note, after example 1.25)
But ( )y t , from the result of example 1.25, is
( ) ( ) ( )y t c t t3 2 3
10 sin 500 sin 4 10⎡ ⎤= π ×
⎣ ⎦
That is,
( ) ( ) ( ) ( )m t c t c t t6 3 2 3
2 10 sin 2 10 sin 500 sin 4 10⎡ ⎤ ⎡ ⎤= × × π ×
⎣ ⎦ ⎣ ⎦
(4.25b)
( ) LSB
s t⎡ ⎤⎣ ⎦ is obtained by using Eq. 4.25 in Eq. 4.24.

4.63
Exercise 4.7
Let ( )M f be as shown in Fig. 4.49. An upper sideband signal is
generated using the signal with this ( )M f . Compute and sketch the spectrum
of the quadrature component of the SSB signal.
Fig. 4.49: Baseband spectrum for the Exercise 4.7
Exercise 4.8
Let ( ) ( )m t c tsin= . Develop the expression for the following:
a) USB signal, in the canonical form.
b) USB signal, in the envelope and phase form
Ans. (b): c
t
c f t
1
sin cos 2
2 4
⎡ ⎤⎛ ⎞ ⎛ ⎞
π +⎜ ⎟ ⎜ ⎟⎢ ⎥
⎝ ⎠ ⎝ ⎠⎣ ⎦
Exercise 4.9
Let ( ) ( ) ( ) ( ) ( )c cs t m t t m t tcos sin= ω − ω
where ( )M f is as shown in Fig. 4.49. Let W 5= kHz and ( )M 0 1= .
a) Sketch the spectrum of (i) ( ) ( )cs t tcos ω and (ii) ( ) ( )cs t tsin ω
b) Show that sum of the spectra of part (a) is proportional to ( )M f
c) Sketch the spectrum of ( ) ( ) ( ) ( )c cs t t s t tcos sinω − ω . Is this related to
( )M f ? Explain.

4.64
4.7 Generation of SSB Signals
We shall consider two broad categories of SSB generation, namely, (i)
frequency discrimination method and (ii) phase discrimination method. The
former is based on the frequency domain description of SSB, whereas the latter
in based on the time-domain description of an SSB signal.
4.7.1 Frequency discrimination method
Conceptually, it is a very simple scheme. First generate a DSB signal and
then filter out the unwanted sideband. This method is depicted in Fig. 4.50.
Fig. 4.50: Frequency discrimination method of SSB generation
( )v t is the DSB-SC signal generated by the product modulator. The BPF is
designed to suppress the unwanted sideband in ( )V f , thereby producing the
desired SSB signal.
As we have already looked at the generation of DSB-SC signals, let us
now look at the filtering problems involved in SSB generation. BPFs with abrupt
pass and stopbands cannot be built. Hence, a practical BPF will have the
magnitude characteristic ( )H f , as shown in Fig. 4.51. As can be seen from the
figure, ( ( )H f is shown only for positive frequencies) a practical filter, besides the
PassBand (PB) and StopBand (SB), also has a TransitionBand (TB), during
which the filter transits from passband to stopband. (The edges of the PB and SB
depend on the attenuation levels used to define these bands. It is a common
practice to define the passband as the frequency interval between the 3-dB

4.65
points. Attenuation requirements for the SB depend on the application. Minimum
attenuation for the SB might be in the range 30 to 50 dB.)
Fig. 4.51: Magnitude characteristic of a practical BPF. Centre frequency 0f=
Because of TB (where the attenuation is not to the desired level), a part of
the undesired sideband may get through the filter. As a rule of the thumb, it is
possible to design a filter if the permitted transitionband is not less than 1% of
center frequency of a bandpass filter. Fortunately, quite a few signals have a
spectral null around DC and if it is possible for us to fit in the transitionband into
this gap, then the desired SSB signal could be generated. In order to accomplish
this, it might become necessary to perform the modulation in more than one
stage. We shall illustrate this with the help of an example.
Example 4.11
Telephone quality speech signal has a spectrum in the range 0.3 to 3.4
kHz. We will suggest a scheme to generate upper sideband signal with a carrier
frequency of 5 MHz. Assume that bandpass filters are available, providing an
attenuation of more than 40 dB in a TB of width 0.01 0f , where 0f is the centre
frequency of the BPF.
Let us look at the generation of the SSB signal in one stage using a carrier
of 5 MHz. When a DSB signal is generated, it will have a spectral null of 600 Hz
centered at 5 MHz. That is, the transitionband is about 0.01 percent of the carrier

4.66
and hence it would not be possible to design such a sideband filter. However, it
would be possible to generate the desired SSB signal using two stages of
modulation.
Consider the scheme shown in Fig. 4.52.
Fig. 4.52: Two stage generation of SSB signal
( )1v t is a DSB-SC signal with the USB occupying the (positive frequency) range
( )1 300cf + Hz to ( )1 3400cf + Hz. The frequency range of the LSB is
( )1 3400cf − Hz to ( )1 300cf − Hz. Let us extract the upper sideband from ( )1v t
with the help of BPF1. Then the centre frequency of BPF1, ( )0 1
f , is
( )
( ) ( )1 1
0 1
300 3400
2
c cf f
f
+ + +
=
( )1 1850cf= +
width of the spectral null around 1 600cf = Hz.
Hence 1 1850
600
100
cf +
≤
or 1 60,000 1850cf ≤ −
58.1≤ kHz
Let us take 1cf as 50 kHz. Let ( )M f be as shown in Fig. 4.53(a). Then ( )2V f will
be as shown in Fig. 4.53(b).

4.67
Fig. 4.53: (a) Message spectrum
(b) The spectrum of the USB signal with 1 50cf = kHz
( )3v t is a DSB-SC signal with the carrier 2cf , ( )2v t being the modulating signal.
Then ( )3V f will be as shown in Fig. 4.54. In this figure,
Fig. 4.54: Spectrum of ( )3v t of Fig. 4.52
( )1 2 53,400cf f= − Hz
( )2 2 50,300cf f= − Hz
( )3 2 50,300cf f= + Hz
( )4 2 53,400cf f= + Hz

4.68
Hence the transition band available to design the sideband filter is
( )3 2 100.6f f− = kHz. With this TB, we can choose centre frequency of BPF2
less than or equal to 10.06 MHz. If we choose 2cf as 4.95 MHz then we will have
the upper sideband occupying frequency range (4.95 + 0.0503) = 5.0003 MHz to
(4.95 + 0.0534) = 5.0034 MHz. This is exactly what would have happened if the
modulation scheme was attempted in one step with 5 MHz as the carrier.
Note: As the spectral occupancy of the USB signal is from 5.0003 MHz to 5.0034
MHz, theoretical centre frequency of the BPF2 is 5.00185. With respect to this
frequency, we have
TB width 100.6
100 2.01 percent
centre freq. 5001.85
= × =
which is about twice the permitted ratio. Hence, it is possible to choose for 1cf a
value lower than 50 kHz.
4.7.2 Phase discrimination method
This method implements Eq. 4.17 or Eq. 4.19(a), to generate the SSB
signal. Consider the scheme shown in Fig. 4.55. This scheme requires two
product modulators, two
2
π
phase shifters and an adder. One of the phase shifter
is actually a Hilbert transformer (HT); it should provide a
2
π
phase shift for all the
components in ( )M f . This is not such an easy circuit to realize. Assuming it is
possible to build the HT, the SSB can be generated for any cf , provided the
product modulators (multipliers) can work at these frequencies.

4.69
Fig. 4.55: SSB generation: Phase discrimination method
Instead of a single wide band phase shifter acting as the HT, it is possible
to have an SSB generator with two Phase Shifting Networks, (PSN), one in each
branch as shown in Fig. 4.56.
Fig. 4.56: An alternate configuration for the phase discrimination scheme
( )1H f and ( )2H f are the phase shifting networks. Let ( ) ( )1
1
j f
H f e
θ
= and
( ) ( )2
2
j f
H f e
θ
= . ( )1 fθ and ( )2 fθ are such that ( ) ( )1 2
2
f f
π
⎡ ⎤θ − θ =⎣ ⎦ for the
frequency range of interest. That is, PSN1 and PSN2 maintain a constant
difference of
2
π
.

4.70
Let us explain the operation of the scheme shown in Fig. 4.56 by taking
( )m t to be a tone signal; that is, ( ) ( )cosm mm t A t= ω . Let
( )1 1
mf f
f
=
θ = θ and ( )2 2
mf f
f
=
θ = θ
where 2 1
2
π
θ = θ + . Then,
( ) ( )1 1cosm mv t A t= ω + θ and ( ) ( )2 2cosm mv t A t= ω + θ .
( ) ( ) ( )3 1cos cosm c m cv t A A t t= ω + θ ω
( ) ( ) ( )4 2cos sinm c m cv t A A t t= ω + θ ω
( )1cos sin
2
m c m cA A t t
π⎛ ⎞
= ω + θ + ω⎜ ⎟
⎝ ⎠
( ) ( )1sin sinm c m cA A t t= − ω + θ ω
( ) ( ) ( )3 4 1cosm c c mv t v t A A t⎡ ⎤+ = ω + ω + θ⎣ ⎦
After coherent demodulation, we will have ( )1cos m tω + θ . We shall assume that
the additional phase shift 1θ which is actually frequency dependent will not cause
any problem after demodulation.
As it is not too difficult to design a Hilbert transformer using digital filter
design techniques, phase shift method of SSB generation is better suited for
digital implementation. For a brief discussion on SSB generation using digital
signal processing, the reader is referred to [1].

4.71
4.8 Demodulation of SSB
SSB signals can be demodulated using coherent demodulation as shown
in Fig. 4.58.
Exercise 4.10
There is a third method of generating the SSB signal, known as Weaver’s
method. This scheme is shown in Fig. 4.57.
Fig. 4.57: Weaver’s method of SSB generation
Let ( )M f be non-zero only in the interval l uf f f≤ ≤ ; and let 1
2
l uf f
f
+
= .
The lowpass filters in the I and Q channels are identical and have a cutoff
frequency of 0
2
u l
c
f f
f
−
= . Assume that 2 0cf f>> . By sketching the spectra at
various points in the above scheme, show that ( )s t is an SSB signal. What is
the actual carrier frequency with respect to which, ( )s t would be an SSB
signal?

4.72
Fig. 4.58: Coherent demodulation of SSB
The received SSB signal is multiplied by the local carrier which is of the same
frequency and phase as the carrier used at the transmitter. From Fig, 4.58, we
have
( ) ( ) ( )' cosc cSSB
v t s t A t⎡ ⎤= × ω⎣ ⎦
( ) ( ) ( ) ( ) ( )'cos sin cos
2
c
c c c c
A
m t t m t t A t⎡ ⎤= ω ± ω × ω
⎣ ⎦
( ) ( ) ( ) ( ) ( )2
'
cos cos sin
2
c c
c c c
A A
m t t m t t t⎡ ⎤= ω ± ω ω
⎣ ⎦
(4.26)
The second term on the RHS of Eq. 4.26 has the spectrum centered at 2 cf±
which will be eliminated by the LPF following ( )v t . The first term of Eq. 4.26 can
be written as,
( )
( )
' 1 cos2
2 2
cc c tA A
m t
⎡ ⎤+ ω
⎢ ⎥
⎣ ⎦
.
As ( ) ( )cm t tcos 2ω has the spectrum centered at 2 cf∓ , even this will be
eliminated by the LPF. Hence ( ) ( )0
'
4
c cA A
v t m t= . That is,
( ) ( )0v t m tα
The difficulty in demodulation is to have a coherent carrier at the receiver. Can
we use a squaring loop or Costas loop to recover the carrier from ( ) SSB
s t⎡ ⎤⎣ ⎦ ?
The answer is NO. Let us look at the squaring loop. From Eq. 4.23,
( ) ( ) ( ) ( )cos cSSB
s t s t A t t t⎡ ⎤ ⎡ ⎤= = ω + ϕ⎣ ⎦ ⎣ ⎦
After squaring, we obtain,

4.73
( ) ( ) ( )2 2 2
cos cs t A t t t⎡ ⎤= ω + ϕ⎣ ⎦
( ) ( )( ){ }2 2
1 cos 2 cA t t t⎡ ⎤= + ω + ϕ⎣ ⎦
As ( )tϕ is a function of time, we do not have a discrete component at 2 cf f=
and hence, carrier acquisition is not possible. It is left as an exercise to show that
Costas loop will not be able to put out ( )m t when the input to the loop is the SSB
signal. Hence, when SSB is used in a communication system, highly stable
crystal oscillators are used both at the transmitter and receiver. If this scheme
does not work (especially, at very high frequencies) a small pilot carrier can be
sent along with the SSB signal. This pilot carrier is recovered at the receiver and
is used for demodulation.
Let us now look at the effects of frequency and phase offset in the carrier
used for demodulation. Let the carrier term at the receiver be
( )' cos 2c cA f f t⎡ ⎤π + ∆⎣ ⎦ . Let the received input to the demodulator (Fig. 4.58) be
( ) ( ) ( ) ( )
1
cos sin
2
c c cA m t t m t t⎡ ⎤ω − ω
⎣ ⎦
Then, ( ) ( ) ( ) ( ) ( ) ( )⎡ ⎤= ω − ω ω + ∆ω
⎣ ⎦
1 ' cos sin cos
2
c c c c cv t A A m t t m t t t
( )0v t , the output of the LPF would be
( ) ( ) ( ) ( ) ( )0
1 ' cos 2 sin 2
4
c cv t A A m t f t m t f t⎡ ⎤α π∆ + π∆
⎣ ⎦
(4.27)
Assume that
( ) ( ) ( ) ( ) ( )0 cos 2 sin 2v t m t f t m t f t= π∆ + π∆ (4.28)
Consider a special case, namely, a frequency component at 1f = kHz in ( )M f
and 100f∆ = Hz. With these values, Eq. 4.28 becomes
( ) ( ) ( ) ( ) ( )3 3
0 cos 2 10 cos 2 100 sin 2 10 sin 2 100v t t t t t= π × π × + π × π ×
( )
2 900 2 900
cos 2 900
2
j t j t
e e
t
π × − π ×
+
= π × = (4.29)

4.74
As this is true of every frequency component in ( )M f , we have the result that
when f∆ is positive and the input is a USB signal, there is an inward shift of
( )M f by f∆ . (We see from Eq. 4.28, 2 1000j t
e π ×
is converted to 2 900j t
e π ×
and
2 1000j t
e− π ×
is converted to 2 900j t
e− π ×
. That is, both the spectral components
have been shifted inward by 100 Hz.) By a similar analysis, we can easily see
that if f∆ is negative and the input is a USB signal, then, after demodulation, the
spectral components in ( )M f would undergo an outward shift by f∆ . In all, we
have four cases to be taken into account and the effects of non-zero f∆ on the
resulting output after demodulation are summarized below.
Case i) 0f∆ > and the input signal is USB: Spectral components in ( )M f will
undergo an inward shift by f∆
Case ii) 0f∆ > and the input signal is LSB: Spectral components in ( )M f will
undergo an outward shift by f∆
Case iii) 0f∆ < and the input signal is USB: Spectral components in ( )M f will
undergo an outward shift by f∆
Case iv) 0f∆ < and the input signal is LSB: Spectral components in ( )M f will
undergo an inward shift by f∆
Let ( )M f be as shown in Fig. 4.59(a). Let 300f∆ = Hz. Then, if the input
is a USB signal, the spectrum of the demodulated output, ( )0V f , would be as
shown in Fig. 4.59(b).

4.75
Fig. 4.59: (a) Baseband message spectrum
(b) Inward spectral shift (after demodulation) of a
1 USB signal. Frequency offset, 300f∆ = Hz

4.76
Example 4.12
Let ( ) ( ) ( ) ( )m t A t A t A t1 2 3cos 200 cos 1100 cos 4000= π + π + π . An
upper sideband signal is generated using this ( )m t . The carrier used for
demodulation had a positive offset of 150 Hz. Let us find the frequencies of the
spectral components after demodulation.
As the received signal is USB and f 0∆ > , there would be an inward shift
of the frequency components by 150 Hz. Spectral components in ( )M f are
shown in Fig. 4.61(a).
Exercise 4.11: Effect of phase error on SSB
Consider the scheme shown in Fig. 4.60. Let
( ) ( ) ( ) ( ) ( )cos sinc cs t m t t m t t⎡ ⎤= ω + ω
⎣ ⎦
. The carrier used to demodulate has
a phase difference of θ with respect to the carrier used for modulation. Show
that ( )0v t has phase distortion (when compared to ( )m t ) by establishing
( )
( )
( )
0
, 0
, 0
j
j
M f e f
V f
M f e f
θ
− θ
⎧ >⎪
= ⎨
<⎪⎩
Fig. 4.60: SSB demodulation with carrier phase difference
Hint: Show that ( ) ( ) ( )v t m t m t0 cos sin⎡ ⎤α θ + θ
⎣ ⎦

4.77
Fig. 4.61: (a) message spectrum
(b) Spectrum after demodulation with frequency error
Note that the negative frequency components have been shown with a broken
line. Let ( )m t be the demodulated output. After demodulation, there would be an
inward shift by 150 Hz and this is shown in Fig. 4.61(b). From this spectrum, we
see that ( )m t' is consisting of components at 50 Hz, 350 Hz and 1850 Hz.
The speech files that follow provide a qualitative measure of the distortion
caused by the frequency offset of the local carrier in the demodulation of SSB
signals. Input to the demodulator is a USB signal.
Introduction
Output 1
Output 2a
Output 2b
Output 2c

4.78
Output 3a
Output 3b
Output 3c
Output 4a
Output 4b
Output 4c
After listening to these speech outputs, one gets the feeling that, for a
given frequency offset, SSB performs better than the DSB. Even in SSB, outward
frequency shift of the original message spectrum ( f∆ negative for USB) has
better clarity than the corresponding inward shift. Of course, voice tends to
become somewhat shrill, which, of course, is expected.

4.79
Exercise 4.12: Quadrature Carrier Multiplexing (QCM)
It is possible to transmit two DSB-SC signals with in a bandwidth of 2 W ,
by using a scheme called QCM, as shown in Fig. 4.62. (QCM is also referred to
as Quadrature Amplitude Modulation (QAM) or simply quadrature modulation.
Using QAM, we are able to achieve the BW efficiency of SSB modulation.)
Fig. 4.62: Quadrature carrier multiplexing scheme
Two message signals ( )1m t and ( )2m t are used to generate two DSB-SC
signals, ( )1v t and ( )2v t respectively. The carriers used in generating ( )1v t and
( )2v t are in phase quadrature. The transmitted signal ( ) ( ) ( )1 2s t v t v t= + . At
the receiver, coherent demodulation is used to recover the two messages ( )1m t
and ( )2m t .
a) Show that ( )1m t and ( )2m t will be recovered by the receiver shown.
b) Let the local carrier have some frequency and phase offset; that is, instead
of ( )2 cos 2 cf tπ , let it be ( )2 cos 2 cf f t⎡ ⎤π + ∆ + ϕ⎣ ⎦ . Then show that the
output of the upper branch is
( ) ( ) ( ) ( ){ }1 2cos 2 sin 2cA m t f t m t f t⎡ ⎤ ⎡ ⎤π ∆ + ϕ − π ∆ + ϕ⎣ ⎦ ⎣ ⎦
where as the output of the lower branch is
( ) ( ) ( ) ( ){ }2 1cos 2 sin 2cA m t f t m t f t⎡ ⎤ ⎡ ⎤π ∆ + ϕ + π ∆ + ϕ⎣ ⎦ ⎣ ⎦
Note: We see from the above result that the carrier phase and frequency have to
be fairly accurate to have proper demodulation; otherwise ( )1m t will interfere with
( )2m t and vice versa. This is called cochannel interference. QAM is used in
color TV for multiplexing the chrominance signals.

4.80
4.9 Vestigial SideBand (VSB) Modulation
One of the widespread applications of VSB has been in the transmission
of picture signals (video signals) in TV broadcast. The video signal has the
characteristic that it has a fairly wide bandwidth (about 5 MHz) with almost no
spectral hole around DC. DSB modulation, though somewhat easy to generate,
requires too much bandwidth (about 10 MHz) where SSB, though bandwidth
efficient, is extremely difficult to generate, as explained below.
With analog circuitry it is very difficult to build the
2
π
phase shifter over a 5
MHz bandwidth; as such phase shift discrimination method is not feasible. To
make use of the frequency discrimination method, we require very sharp cutoff
filters. Such filters have a highly non-linear phase characteristic at the band
edges and spectral components around the cut-off frequencies suffer from phase
distortion (also called group delay distortion). The human eye (unlike the ear)
being fairly sensitive to phase distortion, the quality of the picture would not be
acceptable.
VSB refers to a modulation scheme where in the wanted sideband (either
USB or LSB) is retained almost completely; in addition, a vestige (or a trace) of
the unwanted sideband is added to the wanted sideband. This composite signal
is used for transmitting the information. This vestige of the wanted sideband
makes it possible to come up with a sideband filter that can be implemented in
practice.
4.9.1 Frequency domain description of VSB
Figure 4.63 depicts the scheme of VSB generation. In this figure, ( )v t is a
DSC-SC signal, which is applied as input to a Sideband Filter (SBF), ( )vH f , that
shapes ( )V f so that ( )s t is a VSB signal.

4.81
Fig. 4.63: Generation of VSB using the filtering method
Now the questions that arise are: what is the shape of the SBF and how
do we demodulate such a signal. As coherent demodulation is fairly general, let
us try to demodulate the VSB signal as well, using this method. In the process of
demodulation, we shift the modulated carrier spectrum (bandpass spectrum) up
and down by cf and then extract the relevant baseband. Because of the vestige
of the unwanted sideband, we expect some overlap (in the baseband) of the
shifted spectra. In such a situation, overlap should be such that, ( )M f is
undistorted for f W≤ . In other words, ( ) ( )v c v cH f f H f f− + + should result in
a filter with a rectangular passband within the frequency range ( )toW W− .
With a little intuition, it is not too difficult to think of one such ( )vH f .
Assume that we are retaining the USB almost completely and permitting
the vestige of the LSB. Consider the ( ) ( ) ( )vj f
v vH f H f e
θ
= shown in Fig. 4.64.
We assume the phase characteristic ( )v fθ to be linear over the frequency range
l cf f f W≤ ≤ + with ( ) ( )v c v cf f m2θ = − θ − = − π , where m is an integer.
If a DSB signal is given as input to the above ( )vH f , it will partially
suppress USB (in the frequency range c uf f f≤ ≤ ) and allow the vestige of the
LSB (from l cf f f≤ ≤ ). The demodulation scheme is the same as shown in Fig.
4.13, with the input to the detector being the VSB signal. ( )0V f , the FT of the
output of the detector, is

4.82
( ) ( ) ( ) ( )v c v cV f K M f H f f H f f0 1 ⎡ ⎤= − + +⎣ ⎦ , for f W≤ .
where K1 is the constant of proportionality.
Fig. 4.64: An example of a SBF generating VSB
If ( ) ( ) dj f t
v c v cH f f H f f K e 2
2
− π
⎡ ⎤− + + =⎣ ⎦ , f W≤ ,
where dt determines the slope of the phase characteristic and K2 is a constant,
then ( ) ( )0 dv t K m t t= − , where K K K1 2= .
Let ( ) ( ) ( )
( )vj f
v v v c vH f H f e H f f W f f1,
1, 1, ,
θ
= = − − ≤ ≤
where v c lf f f= − , is the width of the vestige in LSB.
For the ( )vH f shown in Fig. 4.64, ( )1, vH f is as shown in Fig. 4.65.

4.83
Fig. 4.65: ( )v cH f f− for f W≤
(a) magnitude characteristics
(b) phase characteristics
Let ( ) ( ) ( )
( )2,
2, 2,
vj f
v v v cH f H f e H f f
θ
= = + for vf f W− ≤ ≤ . ( )2, vH f is
shown in figure 4.66.

4.84
Fig. 4.66: ( )v cH f f+ for f W≤
(a) magnitude characteristics
(b) phase characteristics
As ( )vH f1, 0= for vf f W≤ ≤ and j m
e 2
1± π
= , we can write
( ) ( ) 2
1, ,dj f t
v c vH f f H f e f W− π
− = ≤ (4.30a)
By a similar reasoning,
( ) ( ) 2
2, ,dj f t
v c vH f f H f e f W− π
+ = ≤ (4.30b)
Therefore,
( ) ( ) ( ) ( ) 2
1, 2, ,dj f t
v c v c v vH f f H f f H f H f e f W− π⎡ ⎤− + + = + ≤
⎣ ⎦
(4.30c)
Summing ( )vH f1, and ( )vH f2, , we have the ideal LPF with unity gain for
f W≤ .
Let us take a closer look at the sideband filter for VSB. The magnitude
characteristic shown in Fig. 4.64(a) can be taken as the sum of ( )
( ) ( )
( )1 2
v vH f H f+

4.85
where ( )
( )1
vH f and ( )
( )2
vH f are shown in Fig. 4.67(a) and (b) respectively.
Fig. 4.67: Decomposition of the ( )vH f of Fig. 4.64
Note that with ( )
( )vH f
1
alone for the sideband filter, we would have generated the
SSB signal. It is the addition of ( )
( )vH f
2
to ( )
( )vH f
1
that gives rise to VSB output.
Consider f 0> ; ( )
( )vH f
2
consists of two straight line segments, one between the
points ( )a b, and the other between the points ( )c d, . Similar is the case for
f 0< . Now the question is: should ( )
( )vH f
2
consist of only straight line
segments? The answer is NO. What is required is that ( )
( )vH f
2
exhibit odd
symmetry about cf± . Two more such characteristics have been shown in fig.
4.68, for f 0> .

4.86
Fig. 4.68: Two more examples of ( )
( )vH f
2
As such, we have a great deal of flexibility in designing ( )vH f . (Subject to some
upper limit, there could be choice in fixing vf , the width of the vestige.) These
features facilitate the design and implementation of a practical sideband filter.
4.9.2 Time domain description of VSB
Let ( )vh t denote the impulse response of the sideband filter, ( )vH f . From
Fig. 4.63, we have
( ) ( ) ( ) ( )vVSB
s t s t h v t d
∞
− ∞
⎡ ⎤ = = τ − τ τ⎣ ⎦ ∫
But ( ) ( ) ( )c cv t A m t tcos= ω . For convenience, let cA 1= .
Then, ( ) ( ) ( ) ( )v cs t h m t t dcos
∞
− ∞
⎡ ⎤= τ − τ ω − τ τ⎣ ⎦∫

4.87
( ) ( ) ( ) ( ) ( ) ( )v c c c ch m t t t dcos cos sin sin
∞
− ∞
⎡ ⎤= τ − τ ω ω τ + ω ω τ τ⎣ ⎦∫
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
v c c
v c c
h m t d t
h m t d t
cos cos
sin sin 4.31
∞
− ∞
∞
− ∞
⎡ ⎤
⎢ ⎥= τ − τ ω τ τ ω
⎢ ⎥
⎣ ⎦
⎡ ⎤
⎢ ⎥− − τ − τ ω τ τ ω
⎢ ⎥
⎣ ⎦
∫
∫
Eq. 4.31 is in the canonical form for the representation of narrowband
signal. Let ( )cm t denote the in-phase component of ( )s t and ( )sm t , the
quadrature component. Then,
( ) ( ) ( ) ( )c v cm t h m t dcos
∞
− ∞
= τ − τ ω τ τ∫ (4.32a)
and ( ) ( ) ( ) ( )s v cm t h m t dsin
∞
− ∞
⎡ ⎤
⎢ ⎥= − τ − τ ω τ τ
⎢ ⎥
⎣ ⎦
∫ (4.32b)
Then, ( ) ( ) ( ) ( )c c s cVSB
s t m t t m tcos sin⎡ ⎤ = ω − ω⎣ ⎦ (4.33)
Eq. 4.33 is the canonical representation of the VSB signal.
Let ( ) ( ) ( )i v ch t h t tcos= ω and ( ) ( ) ( )q v ch t h t tsin= − ω . Then,
( ) ( ) ( )c im t m t h t= ∗ (4.34a)
( ) ( ) ( )s qm t m t h t= ∗ (4.34b)
Taking the FT of Eq. 4.34, we have
( ) ( ) ( )c iM f M f H f= and ( ) ( ) ( )s qM f M f H f=
If ( )M f 0= for f W> , then ( )cM f and ( )sM f are bandlimited to atmost W .
This implies that ( )cm t and ( )sm t are lowpass signals, as required.
As ( ) ( ) ( )i v ch t h t tcos= ω we have,
( )
( ) ( )v c v c
i
H f f H f f
H f
2
− + +
= (4.35a)

4.88
Similarly, ( )
( ) ( )v c v c
q
H f f H f f
H f
j2
+ − −
= (4.35b)
But ( ) ( )v c v cH f f H f f 1⎡ ⎤− + + =⎣ ⎦ , for f W≤ . Hence, ( )
( )
c
m t
m t
2
= .
Let us look at ( )qH f .
From Eq. 4.35(b),
( )
( ) ( )q
v c v c
H f
H f f H f f
j
2
= − − +
Let ( )vH f be as shown in Fig. 4.69.
Fig. 4.69: ( )vH f with vestige in LSB
Fig. 4.70:
( )qH f
j
2
(solid line) for the ( )vH f of Fig. 4.69
Then,
( )qH f
j
2
for f W≤ will be as shown in Fig. 4.70.

4.89
Based on Eq. 4.33, we have another implementation for the generation of
the VSB signal called the phase discrimination method of generating VSB. This
scheme is shown in Fig. 4.71. Plus sign at the last summer will result in a VSB
signal with a vestige in LSB whereas the minus sign will result in a VSB signal
with the vestige in USB for the ( )qH f shown in Fig. 4.70.
Fig. 4.71: Phase discrimination method of VSB generation
Comparing the scheme shown in Fig. 4.71 with that shown in Fig. 4.55 for the
generation of SSB, we find a close resemblance between them. When ( )qh t is a
HT (with the magnitude response equal to 1/2), we have ( ) ( )VSB SSB
s t s t⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ .
If ( )qh t 0= , we have ( ) ( )VSB DSB
s t s t⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ . In other terms, DSB and SSB can
be treated as special cases of VSB. Transmission bandwidth of VSB is,
[ ]T vVSB
B W f= +
With vf 0= , we have the SSB situation where as, vf W= leads to DSB.
Example 4.13
A VSB signal is generated from the DSB-SC signal, ( ) ( )cm t t2 cos ω .
( )M f is as shown in Fig. 4.72(a). If the vestige is as shown in Fig. 4.72(b), let’s

4.90
find the values of all spectral components in the VSB signal for cf f> , assuming
that demodulation in done by multiplying the received VSB signal with
( )c t2cos ω . ( )M f is to be restored to its original values.
Fig. 4.72: (a) Baseband message spectrum used to generate the VSB
(b) Spectral components in the vestige in LSB
Let iα be the magnitude of the spectral component at c if f i, 1, 2, ..., 5+ = in
the VSB spectrum. When the VSB spectrum is shifted to the left by cf , we have
the spectrum shown in Fig. 4.73(a). Similarly, when the VSB spectrum is shifted
to the right by cf , we have the spectrum shown in Fig. 4.73(b).

4.91
Fig. 4.73: Shifted VSB spectra: (a) left shift (b) right shift
From the spectral plots in Fig. 4.73(a) and (b), we have
1 0.1 0.3α + = ; That is 1 0.2α =
2 0.2 0.8α + = ; That is 2 0.6α =
3 4 51, 0.65 and 0.4α = α = α =
Hence the VSB spectrum for f 0> is shown in Fig. 4.74.

4.92
Fig. 4.74: VSB spectrum of example 4.13
4.10 Envelope Detection of VSB+C
As mentioned earlier, one important application of VSB has been in TV
broadcast. As a large number of receivers are involved, it is preferable to make
the detector circuit fairly simple and inexpensive. The envelope of a VSB signal is
not one-to-one related to the message ( )m t . Hence direct envelope detection of
a VSB signal would be of no use. However, if there is a carrier component along
with the VSB signal, that is, VSB+C, then ED might work. We shall now look into
this.
Exercise 4.13
Let ( ) ( ) ( ) ( )m t t t t2 4 cos 200 6 cos 300 5 cos 400= + π + π + π .
Specify the frequency response of a VSB filter that passes the LSB almost
completely and leaves a vestige of the first frequency component in USB. It is
given that the magnitude of the filter at the vestige frequency is
1
8
. Sketch the
spectrum of the VSB signal.

4.93
Let the input to the ED be
( ) ( ) ( ) ( ) ( )c c c c s cA t A m t t m t tcos cos sin⎡ ⎤ω + β ω ± β ω⎣ ⎦
where β is an adjustable scale factor. Then ( )A t , the output of the ED is
( ) ( ) ( )( ){ }c sA t A m t m t
1
2 2 2
1⎡ ⎤= + β + β⎣ ⎦
( )
( )
( )
s
c
m t
A m t
m t
1
2 2
1 1
1
⎧ ⎫⎡ ⎤β⎪ ⎪
⎡ ⎤= + β + ⎢ ⎥⎨ ⎬⎣ ⎦ + β⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
(4.36)
If the distortion component
( )
( )
sm t
m t
1
1
β
<<
+ β
for all t , then the output of the ED
will be ( )cA m t1⎡ ⎤+ β⎣ ⎦ , which, after the DC block, will provide the message
output. This level of the distortion component can be reduced by
i) increasing vf . Note that as vf W→ , we will have the DSB signal; that is,
( )sm t 0= .
ii) decreasing the value of β .
In commercial TV broadcast, vf is about 75 kHz which is about 1/6 of the
width of a full sideband which is about 5 MHz. It has been found that with a
vestige of about 75 kHz, the distortion component is not too bothersome.
It would be instructive to compare the envelope detection of SSB+C with
that of VSB+C. Let the input to the ED be SSC+C, namely,
( ) ( ) ( ) ( ) ( )c c c cA t m t t m t tcos cos sin⎡ ⎤ω + ω ± ω
⎣ ⎦
then, ( ) ( ) ( )cA t A m t m t
1
22 2⎧ ⎫⎡ ⎤⎡ ⎤= + +⎨ ⎬⎣ ⎦ ⎣ ⎦⎩ ⎭

4.94
( ) ( ) ( )
c
c cc
m t m t m t
A
A AA
1
2 22
2
2
1
⎧ ⎫⎡ ⎤⎡ ⎤⎪ ⎪
= + + + ⎢ ⎥⎢ ⎥⎨ ⎬
⎢ ⎥⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭
(4.37)
If
( )
c
m t
A
and
( )
c
m t
A
are very much less than unity, then
( )
c
m t
A
2
⎡ ⎤
⎢ ⎥
⎣ ⎦
and
( )
c
m t
A
2
⎡ ⎤
⎢ ⎥
⎢ ⎥⎣ ⎦
can
be dropped from Eq. 4.37, giving us
( )
( )
c
c
m t
A t A
A
1
22
1
⎡ ⎤
+⎢ ⎥
⎣ ⎦
Retaining only first order term of the binomial expansion,
( )
( )
( )c c
c
m t
A t A A m t
A
1
⎡ ⎤
+ = +⎢ ⎥
⎣ ⎦
Of course, if we ensure ( ) ( ) cm t m t Amax min
= << , then
( )
c
m t
A
can be
neglected in comparison with unity. This may be adequate to make
( )
c
m t
A
much
smaller than unity, most of the time. In any case, it is obvious that ED of SSB+C
results in excessive wastage of the transmitted power. In contrast, ED of AM is
reasonably efficient as the requirement to avoid envelope distortion is
( ) cm t A
max
≤ .
Example 4.14
Consider the SSB+C signal,
( ) ( ) ( ) ( ) ( ) ( )c c c cs t A t m t t m t tcos cos sin= ω + ω + ω
where ( )m t
t2
1
1
=
+
. We will show that, if cA 1>> , then the output of the ED
can be taken as ( )+cA m t .

4.95
As ( )m t
t2
1
1
=
+
, we have ( )
t
m t
t2
1
=
+
. Hence ( )A t , the envelope of
( )s t is,
( ) c
t
A t A
t t
1
2 2 2
2 2
1
1 1
⎡ ⎤⎛ ⎞ ⎛ ⎞
⎢ ⎥= + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥+ +⎝ ⎠ ⎝ ⎠⎣ ⎦
( ) ( )
c
c
A t
A
tt t
1
2
2
2
2 2 2
2 2
21
11 1
⎡ ⎤
⎢ ⎥
= + + +⎢ ⎥
++ +⎢ ⎥
⎣ ⎦
c
c
A
A
t t
1
2
2
2 2
21
1 1
⎡ ⎤
= + +⎢ ⎥
+ +⎣ ⎦
( ) ( )c
c c
A
A t A t
1
2
2 2 2
2 1
1
1 1
⎧ ⎫
⎪ ⎪
= + +⎨ ⎬
+ +⎪ ⎪⎩ ⎭
Neglecting the term
( )+2 2
1
1cA t
, we have
( )
( )c
c
A t A
A t
1
2
2
2
1
1
⎡ ⎤
⎢ ⎥+
⎢ ⎥+
⎣ ⎦
Using the binomial expansion upto the second term, we have
( )
( )c
c
A t A
A t2
1 2
1
2 1
⎡ ⎤
⎢ ⎥+
⎢ ⎥+
⎣ ⎦
cA
t2
1
1
+
+
which is the desired result.

4.96
A note on the linearity of AM, DSB-SC, SSB and VSB: Having discussed
these modulation schemes, let us look at the linearity aspect of these schemes.
By linearity, we imply that the schemes obey the superposition property. This can
be easily verified in the case of DSB-SC, SSB and VSB. Consider DSB-SC.
When message signals ( )m t1 and ( )m t2 are applied separately, the resulting
modulated waveforms are ( ) ( )c cA m t t1 cos ω and ( ) ( )c cA m t t2 cos ω . Let
( ) ( ) ( )m t m t m t1 1 2 2= α + α
where 1α and 2α are constants. Then the modulated carrier is
( ) ( ) ( ) ( ) ( )c c c cA m t t A m t m t t1 1 2 2cos cos⎡ ⎤ω = α + α ω⎣ ⎦ ,
Exercise 4.14
Consider SSB+C with tone modulation. Let
( ) ( ) ( )c c c ms t A t tcos cos⎡ ⎤= ω + β ω +ω⎣ ⎦
Assume that c mf f>> and 1β < .
a) Construct the phasor diagram and develop the expression for the
envelope ( )A t .
b) Let β be such that terms involving m
m, 3β ≥ can be neglected. Show
that
( ) ( ) ( )c m mA t A t t
2 2
1 cos cos 2
4 4
⎡ ⎤β β
+ + β ω − ω⎢ ⎥
⎢ ⎥⎣ ⎦
c) Find the value of β so that second harmonic envelope distortion, that is,
the ratio,
( )
( )
A t
A t
Amplitude of the second harmonic in
Amplitude of the fundamental in
is less than five
percent.

4.97
which establishes the linearity property. Similarly, SSB and VSB can be shown to
be linear. AM is not linear in its strict sense, because if ( )m t is applied as input
to an AM modulator, the output is
( ) ( ) ( ) ( ) ( ){ }c m c c m cA g m t t A g m t m t t1 1 2 21 cos 1 cos⎡ ⎤ ⎡ ⎤+ ω = + α + α ω⎣ ⎦⎣ ⎦
and this is not equal to
( ) ( ) ( ) ( )c m c c m cA g m t t A g m t t1 1 2 21 cos 1 cos⎡ ⎤ ⎡ ⎤+ α ω + + α ω⎣ ⎦ ⎣ ⎦
That is, superposition does not apply to the carrier component. As this is only a
minor deviation, all the four modulation types are put under the category of linear
modulation.
4.11 Superheterodyne Receiver
The important function of a receiver is demodulation; that is, to recover the
message signal from the received modulated waveform. The other functions
which become necessary for the proper reception of a signal are: amplification
and tuning or selective filtering. Amplification becomes necessary because, most
often, the received signal is quite weak and without sufficient amplification, it may
not even be able to drive the receiver circuitry. Tuning becomes important
especially in a broadcast situation because there is more than one station
broadcasting at the same time and the receiver must pick the required station
and reject the inputs from the other (unwanted) stations; tuning also ensures that
out of band noise components do not affect the receiver's performance. Besides
these operations, most of the receivers also incorporate certain other features
such as frequency conversion, automatic gain control etc.
Two of the demodulation methods which we have already discussed are
coherent or synchronous detection and envelope detection. Coherent detection
can be used to demodulate any linear modulation scheme: DSB-SC, DSB-LC,
SSB or VSB. In practice, it is used to demodulate only suppressed carrier

4.98
signals. Envelope detection is mainly used in the demodulation of DSB-LC and
VSB+C signals. We shall now describe the receiver used in AM broadcast.
The receiver for the broadcast AM is of the superheterodyne (or superhet)
variety1
. This is shown schematically in the Fig. 4.75.
Fig. 4.75: Superheterodyne Receiver
The wanted signal ( )s t , along with other signals and noise, is input to the Radio
Frequency (RF) stage of the receiver. The RF section is tuned to cf , the carrier
frequency of the desired signal ( )s t . The bandwidth of the RF stage, RFB , is
relatively broad; hence along with ( )s t , a few adjacent signals are also passed
by it. The next stage in the receiver is the frequency conversion stage consisting
of a mixer and a local oscillator. The local oscillator frequency, LOf tracks the
carrier frequency cf , (with the help of a ganged capacitor) and is usually
( )c IFf f+ , where IFf denotes the Intermediate Frequency (IF). The mixer output
consists of, among others, the frequency components at c IFf f2 + and IFf . The
following stage, called the IF stage, is a tuned amplifier, which rejects all the
other components and produces an output that is centered at IFf . The bandwidth
of the IF stage, IFB , is approximately equal to the transmission bandwidth, TB , of
1
Some of the other applications of superhet are: reception of FM and TV broadcast signals and
RADAR

4.99
the modulation scheme under consideration. For example, if the input signal is of
the double sideband variety, then IFB W2≈
The IF stage constitutes a very important stage in a superheterodyne
receiver. It is a fixed frequency amplifier (it could consist of one or more stages of
amplification) and provides most of the gain of the superhet. Also, as IF TB B= ,
it also rejects the adjacent channels (carrier frequency spacing ensures this).
Next to IF, we have the detector or demodulation stage which removes the IF
carrier and produces the baseband message signal at its output. Finally, the
demodulator output goes through a baseband amplification stage (audio or video,
depending upon the type of the signal) before being applied to the final
transducer (speaker, picture-tube etc.)
The spectral drawings shown in Fig. 4.76 and 4.77 help clarify the action
of a superhet receiver. We shall assume the input to the receiver is a signal with
symmetric sidebands.
Fig. 4.76: Typical spectrum at the input to the RF stage of a superhet
Fig. 4.77: Spectrum at the input of the IF stage of a stage of a superhet

4.100
Fig. 4.76 shows the spectrum of the input signal to the RF stage. As can be
seen, it has the desired signal, plus a few adjacent channels and possibly a
signal with the carrier frequency c c IF LO IF cf f f f f f' '2= + = + ⋅ is called the image
frequency corresponding to cf . If cf ' is allowed to pass through the RF stage,
then it will also give rise to an output at the IF stage which would interfere with
the wanted signal, namely the signal whose carrier frequency is cf . Hence the
main task of the RF section is to pass the frequency components in the range
c Tf B
2
± while rejecting the signal with the spectrum centered at cf ' (image
frequency rejection). If the 3-dB bandwidth, RFB is such that
T RF FB B f12< <
then, ( )RFH f , should be able to provide sufficient attenuation at the image
frequency.
Fig. 4.77 depicts the frequency response characteristic of the IF stage. As
seen from this figure, the IF filter takes care of adjacent channel rejection.
The superheterodyne1
structure results in several practical benefits:
i) Tuning takes place entirely in the front end (RF and mixer stage) so that the
rest of the circuitry (IF stage, detector and the final power amplifier stage)
requires no adjustments to changes in cf .
ii) Separation between cf and IFf eliminates potential instability due to stray
feedback from the amplified output to the receiver's input.
iii) Most of the gain and selectivity is concentrated in the fixed IF stage. IFf is
so selected so that IF IFB f results in a reasonable fractional bandwidth (for
1
The word superheterodyne refers to the operation of the receiver, namely, the incoming signal
at the carrier frequency is heterodyned or mixed with the LO signal whose frequency is higher
than ( )c LO c IFf f f f= + .

4.101
AM broadcast various frequency parameters are given in table 4.1). It has
been possible to build superhets with about 70 dB gain at the IF stage itself.
Table 4.1 Parameters of AM radio
Carrier frequency range 540-1600 kHz
Carrier spacing 10 kHz
Intermediate frequency 455 kHz
IF bandwidth 6-10 kHz
Audio bandwidth 3-5 kHz
An IF of 455 kHz has been arrived at by taking the following points into
consideration.
1) IF must not fall within the tuning range of the receiver. Assume that there is
a station broadcasting with the carrier frequency equal to IFf . This signal,
could directly be picked off by the IF stage (every piece of wire can act as
an antenna). Interference would then result between the desired station and
the station broadcasting at c IFf f= .
2) Too high an IF would result in poor selectivity which implies poor adjacent
channel rejection. Assume that IF was selected to be 2 MHz. With the
required bandwidth of less than 10 kHz, we require very sharp cutoff filters,
which would push up the cost of the receiver.
3) As IF is lowered, image frequency rejection would become poorer. Also,
selectivity of the IF stage may increase; thereby a part of the sidebands
could be lost.
We had mentioned earlier, that LO c IFf f f= + . If we have to obtain the IFf
component after mixing, this is possible even if LO c IFf f f= − . But this causes
the following practical difficulty. Consider the AM situation. If we select
LO c IFf f f= − , then the required range of variation of LOf so as to cover the

4.102
entire AM band of 540-1600 kHz is (540 - 455) = 85 kHz to (1600 - 455) = 1145
kHz. Hence the tuning ratio required is 85 :1145 1:13 . If LO c IFf f f= + , then
the tuning ratio required is 995 : 2055 1: 2. This is much easier to obtain than
the ratio 1:13. With the exception of tuning coils, capacitors and potentiometers,
all the circuitry required for proper reception of AM signals is available in IC chips
(for example, BEL 700).
Example 4.15: Double conversion superhet receiver
In receivers operating in the VHF (30 - 300 MHz) range and meant for
receiving fairly narrowband signals (say telemetry signals), to achieve good
image frequency rejection and selectivity using a single mixer stage is quite
difficult. Hence, receivers have been developed with more than one frequency
conversion stage and more than one IF. We shall now look at the image
frequency problem of a receiver with two mixer stages, usually called double or
dual conversion receiver.
Consider the scheme shown in Fig. 4.78.
Fig. 4.78: Block diagram of a double conversion receiver
In the scheme shown, the first mixer stage has a tunable LO, its output
being 30 MHz below the incoming signal frequency. The second mixer stage has
an LO producing a fixed frequency output at 40 MHz. If the RF stage is tuned to
200 MHz, let us find the possible image frequencies of the incoming signal. We
assume that none of the filters in the cascade are ideal.

4.103
First local oscillator frequency = (200 - 30) = 170 MHz. Hence (170 - 30) =
140 MHz will be an image frequency for the incoming signal w.r.t the first IF. As
the second IF is 10 MHz, a component at 50 MHz would be an image of 30 MHz.
Hence if a frequency component at ( )170 50 220± = or 120 MHz get through
the RF stage, it would interfere with the reception of the wanted carrier at 200
MHz. In other words, the image frequencies are at 220 MHz, 140 MHz and 120
MHz.

4.104
Exercise 4.15
a) For the tuned circuit shown in Fig. 4.79, ( )
( )
( )
V f
H f
I f
0
= is,
( )H f
j f C
R f L
1
1 1
2
2
=
⎛ ⎞⎛ ⎞
+ π −⎜ ⎟ ⎜ ⎟π⎝ ⎠ ⎝ ⎠
( )
R
j R LC
L
2 1
1 1
=
+ ω −
ω
Let f0 be the resonant frequency; then f
LC
0
1
2
=
π
and Q-factor of
filter is
C
Q R
L
=
Fig. 4.79: A tuned circuit with L, C and R
Show that
( )
( )
( ) ( )
V f R R
H f
I f fff f j Q
j Q f f
f f
0
2 2
00
0
0
1
1
= = =
⎡ ⎤ ⎛ ⎞− + −⎜ ⎟⎢ ⎥+
⎝ ⎠⎢ ⎥
⎣ ⎦
and hence
( )
( )
R
H f
Q f2 2
1
=
+ β
where ( )
ff
f
f f
0
0
⎛ ⎞
β = −⎜ ⎟
⎝ ⎠

4.105
b) Let the RF stage in a superhet consist of a simple tuned circuit (TC1)
whose output is input to the mixer stage as shown in Fig. 4.80.
Fig. 4.80: Superhet without RF amplifier
Let λ denote the image frequency rejection, where
( )
( )
c
c
H f
H f '
λ = , cf ' being
the image frequency of cf .
That is, ( )cQ f2 2 '1λ = + β . Calculate the value of λ when the receiver
is tuned to (i) 1.0 MHz and (ii) 20.00 MHz. Assume Q of the resonant
circuit TC1 to be 100.
(Ans.: (i) 138.6λ = (ii) 70.6λ = )
c) It is required to have the value of λ at 20 MHz the same as at 1.00λ =
MHz. For this purpose, the RF stage has been modified to include a
tuned amplifier stage as shown in Fig. 4.81. Calculate the required Q of
the tuned circuit, TC2
Answer: Q of TC2 5.17=
Fig. 4.81: Superhet with RF amplifier

4.106
Appendix A4.1: Analysis of ED with Tone Modulation
Consider ED circuit shown in Fig. 4.42 which is redrawn in Fig. A4.1.
Fig. A4.1: Envelope detector
Let ( ) ( ) ( )m cAM
s t f t f t1 cos 2 cos 2⎡ ⎤ ⎡ ⎤= + µ π π⎣ ⎦ ⎣ ⎦
We assume 1µ < so that ( )mf t1 cos 2⎡ ⎤+ µ π⎣ ⎦ is positive for all t . Then, the
envelope quantity is, ( ) ( )mA t t1 cos⎡ ⎤= + µ ω⎣ ⎦ . We shall derive on upper bound
on µ in terms of LR C and mω so that the detector is able to follow the envelope
for all t .
Consider the waveform ( )v t1 shown in Fig. A4.2. The value of the
envelope ( )A t at t t0= ,
( ) ( )mA t t0 01 cos⎡ ⎤= + µ ω⎣ ⎦

4.107
Fig. A4.2: Waveforms of ED of Fig. A4.1
Let ( ) ( )v t A t1 0 0= . Note that ( )v t1 0 is the voltage across the capacitor at
t t0= . Assuming that the capacitor discharges until the next positive peak in the
carrier cycle, we have,
( ) L cR C f
c
v t v t e
f
1 1
1 0 1 0
1 −⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
From Eq. 4.15(b), we have L cR C f 1>> . Hence
( )( )m
c L c
v t t
f R C f
1 0 0
1 1
1 cos 1
⎛ ⎞ ⎡ ⎤
+ + µ ω −⎜ ⎟ ⎢ ⎥
⎝ ⎠ ⎣ ⎦
m
c c
A t t
f f
0 0
1 1
1 cos
⎧ ⎫⎡ ⎤⎛ ⎞ ⎛ ⎞⎪ ⎪
≤ + = + µ ω +⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭
( ) ( )m m
m m
c c
t t
f f
0 01 cos cos sin sin
⎡ ⎤⎛ ⎞ ⎛ ⎞ω ω
≤ + µ ω − ω⎢ ⎥⎜ ⎟ ⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
Assuming m cf f<< , we approximate
m
c
f
f
cos 2 1
⎡ ⎤
π⎢ ⎥
⎣ ⎦
and m m
c c
f f
f f
2
sin 2
⎡ ⎤ π
π⎢ ⎥
⎣ ⎦
Hence,
( )( ) ( ) ( )m
m m m
L c c
t t t
R C f f
0 0 0
1
1 cos 1 1 cos sin
⎡ ⎤ µω
+ µ ω − ≤ + µ ω − ω⎢ ⎥
⎣ ⎦
That is,

4.108
( )
( )
m
L
m m
t
R C
t
0
0
1 cos
sin
+ µ ω
≤
µω ω
(A4.1a)
or
( )
( )
µω ω
≥
+ µ ω
0
0
sin1
1 cos
m m
L m
t
R C t
(A4.1b)
Rearranging Eq. A4.1(b), we have
( ) ( )m m m
L L
t t
R C R C
0 0
1
sin cos
⎛ ⎞µ
µω ω − ω ≤⎜ ⎟
⎝ ⎠
(A4.2)
Let
m LR C
1 1
tan−
ϕ =
ω
as shown
Then, m
m
LR C
2
2
cos
1
ω
ϕ =
⎛ ⎞
ω + ⎜ ⎟
⎝ ⎠
and, L
m
L
R C
R C
2
2
1
sin
1
ϕ =
⎛ ⎞
ω + ⎜ ⎟
⎝ ⎠
The inequality A4.2 can be written as
( ) ( )m m
L
D t D t
R C
0 0
1
cos sin sin cosµ ϕ ω − µ ϕ ω ≤
where
m
L
D
R C
2
2
1
1
=
⎛ ⎞
ω + ⎜ ⎟
⎝ ⎠
. That is,
( )L mR C D t0sin 1µ ω − ϕ ≤ (A4.3)
The inequality should hold even when ( )m t0sin 1ω − ϕ = . That is,
LR C D 1µ ≤
( )L
m L
R C D R C
2
1 1
1
µ ≤ =
+ ω
or ( )L
m
R C
2
2
11 − µ
≤
ω µ

4.109
Evidently, we will not be able to demodulate, using the ED circuit, a tone
modulated AM signal with 1µ = .
Fig. A4.3 displays the experimentally generated demodulator output (top
waveform) when the time constant LR C is of proper value so as to follow the
envelope for all t (modulating tone is shown at the bottom). Fig. A4.4 is the ED
output (top waveform) when the time constant is too large. As can be seen from
the figure, ED is not able to follow the negative half cycle of the tone fully,
resulting in the clipping of a part of their cycle. (These waveforms are from
Shreya’s experimentor kit.)
Fig. A4.3

4.110
Fig. A4.4
Exercise A4.1
A 1 kHz square wave, switching between the levels 1± V, amplitude
modulates a carrier to a depth of 50 %. The parameters of the carrier are:
cA 1= volt and cf 1= MHz.
a) Sketch the resulting AM signal.
b) Let the signal of (a) be the input of the ED of Fig. A4.1. Sketch the
voltage across the capacitor C for the following cases:
i) LR C 25 sec= µ
ii) LR C m2 sec=

4.111
References
1) Charles Schuler and Mahesh Chugani, Digital Signal Processing: a hands
on approach, Tata McGraw Hill, 2005
Some suggested references
1) A. Bruce Carlson, Paul B. Crilly and Janet C. Rutledge, Communication
systems (4th
ed.), Mc Graw Hill international edition, 2002
2) B. P. Lathi, Modern Digital and Analog Communication Systems, (3rd
ed.)
Oxford University press, 1998
3) John G. Proakis and Masoud Salehi, Communication Systems Engineering,
Prentice Hall international edition, 1994

Linear modulation

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