Lec. 4-AM - Part2_DSB-WC.pdf analog comm

Module Leader: Dr. Mohammad Abdellatif
Analog Communication
2023-2024
23ECE12I: Analog
Communication
Modulation Techniques
Amplitude Modulation

Dr. Mohammad Abdellatif
Associate Professor in
Communication and Electronic Department
Faculty of Engineering, The British
University in Egypt
E-mail: mohammad.abdellatif@bue.edu.eg
2

1) Double sideband-suppressed carrier
2) Amplitude modulation (AM)
3) Single sideband (SSB)
4) Vestigial sideband (VSB)
5) Quadrature Amplitude Modulation (QAM)
Amplitude Modulation
3

4
 The (coherent) demodulation of a DSB-SC signal requires
the receiver to possess a carrier signal that is
synchronized with the incoming carrier.
 This requirement is not easy to achieve in practice.
 Because the modulated DSB-SC signal
𝑠𝐷𝑆𝐵−𝑆𝐶 𝑡 = 𝐴𝑐𝑚(𝑡) 𝑐𝑜𝑠 𝜔𝑐𝑡
 may have traveled hundreds of miles and could even
suffer from some unknown Doppler frequency shift, the
bandpass received signal in fact has the typical form of
𝑟 𝑡 = 𝐴𝑐𝑚(𝑡 − 𝑡0) 𝑐𝑜𝑠[ 𝜔𝑐 + ∆𝜔 𝑡 − 𝑡0 ]
 in which ∆𝜔 represents the Doppler effect, 𝑡0 represent
the delay
AM (DSB-WC) Modulation

5
 The alternative to a coherent demodulator is for the
transmitter to send a carrier 𝑨𝒄 𝒄𝒐𝒔 𝝎𝒄𝒕 [along with
the modulated signal 𝑚(𝑡) 𝑐𝑜𝑠 𝜔𝑐𝑡 ] so that there is no
need to generate a coherent local carrier at the
receiver. Which called conventional AM
 In this case, the transmitter needs to transmit at a
much higher power level, which increases its cost as a
trade-off.
𝑠𝐴𝑀 𝑡 = 𝐴𝑐 𝑐𝑜𝑠 𝜔𝑐𝑡 + 𝑚(𝑡) 𝑐𝑜𝑠 𝜔𝑐𝑡
The spectrum
𝑆𝐴𝑀 𝑓 =
1
2
𝑀 𝑓 + 𝑓𝑐 + 𝑀(𝑓 − 𝑓𝑐) +
𝐴𝑐
2
𝛿 𝑓 + 𝑓𝑐 + 𝛿 𝑓 − 𝑓𝑐
Sidebands
Carrier

6
B
-B
AM B.W=2B

7
 If the baseband (message) signal has B bandwidth and the
carrier frequency is fc:
a) The 2 Sidebands (the upper sideband and the lower
sideband) come from the message and appear around the
carrier frequencies ± fc.
b) The necessary condition for undistorted spectrum is that
fc  B, this implies that the sidebands of the AM signal do
not overlap over the frequency domain.
c) Message information exists only in the sidebands.
d) Carrier does not carry any information, in this sense,
carrier components in the spectrum may be seen as waste
of the transmitted power.
e) The double sideband transmission bandwidth of the AM
signal is 2B Hz.

8
= [𝐴𝑐 + 𝑚 𝑡 ] 𝑐𝑜𝑠 𝜔𝑐𝑡
 It is clear that the AM signal is identical to the DSB-SC
signal with 𝑨𝒄 + 𝒎 𝒕 as the modulating signal [instead
of 𝒎(𝒕)].
 The value of 𝑨𝒄 + 𝒎 𝒕 is always chosen to be positive.
 Therefore, to sketch 𝑠𝐴𝑀 𝑡 , we sketch the envelope
𝐴𝑐 + 𝑚 𝑡 and its mirror image − 𝐴𝑐 + 𝑚 𝑡 before
filling in between with the sinusoid of the carrier
frequency 𝑓𝑐 .
 The size of 𝐴𝑐 affects the time domain envelope of the
modulated signal.

9
 𝐴𝑐 + 𝑚 𝑡 ≥ 0 ∀ 𝑡 →
the envelope has the
same shape as 𝑚(𝑡)
 If is not large
enough to satisfy
this condition → the
envelope shape
differs from the
shape of 𝑚(𝑡)
because the negative
part of 𝐴𝑐 + 𝑚 𝑡 is
rectified

10
 This means we can detect the desired signal 𝑚(𝑡) by
detecting the envelope in the first case when 𝐴𝑐 + 𝑚 𝑡 >
0 .
 Such detection is not accurate in the second case.
 We shall see that envelope detection is an extremely
simple and inexpensive operation, which does not require
generation of a local carrier for the demodulation.
 For envelope detection to properly detect m(t), two
conditions must be met:
1) 𝑓𝑐 ≫ 𝐵
2) 𝐴𝑐 + 𝑚 𝑡 > 0

11
 Let ±𝑨𝒎be the maximum and the minimum values of
m(t), respectively
−𝐴𝑚≤ 𝑚(𝑡) ≤ 𝐴𝑚
 Hence, the condition of envelope detection
𝐴𝑐 + 𝑚 𝑡 ≥ 0 ∀𝑡
𝐴𝑐 − 𝐴𝑚 ≥ 0
𝐴𝑐 ≥ 𝐴𝑚

12
 Define the modulation index μ as
𝝁 =
𝑨𝒎
𝑨𝒄
 is also called the modulation factor, or the percentage
modulation
 For envelope detection to be distortion-less, the condition
is 𝐴𝑐 ≥ 𝐴𝑚.
0 ≤ 𝜇 ≤ 1
Note
 Am be the maximum value (amplitude) of | m(t) |, then
𝐴𝑚 = 8

13
Case 1:   1
 Hence, the term [𝐴𝑐 + 𝑚(𝑡) ] is
always POSITIVE.
 Then the envelope will not have
zero crossing
Case 2:   1
 In this case the term [𝐴𝑐 + 𝑚(𝑡) ]
is negative for some t.
 Then the envelope will have zero
crossing

14
 The maximum absolute value of ( × 100) is
referred to the Percentage Modulation
 Consequently,
 Case 1: percentage modulation  100 %.
 Case 2: percentage modulation 100 %.
 Conclusions :
1) The AM wave has a one-to-one correspondence with
the message signal if and only if the percentage
modulation  100%
2) If the percentage modulation exceeds 100%, the
modulated wave suffers from “ envelope distortion ”,
and the wave is said to be “ overmodulated ”.

15
𝐴𝑚𝑎𝑥 = 𝐴𝑐 + 𝐴𝑚
𝐴𝑚𝑖𝑛 = 𝐴𝑐 − 𝐴𝑚
.
𝜇 =
𝐴𝑚𝑎𝑥 − 𝐴𝑚𝑖𝑛
𝐴𝑚𝑎𝑥 +𝐴𝑚𝑖𝑛
=
𝐴𝑐 + 𝐴𝑚 − 𝐴𝑐 − 𝐴𝑚
𝐴𝑐 + 𝐴𝑚 + 𝐴𝑐 − 𝐴𝑚
=
𝑨𝒎
𝑨𝒄
 𝐴𝑚𝑎𝑥 and 𝐴𝑚𝑖𝑛 denote the
maximum and minimum values
of the envelope of the
modulated wave

16
Example 2:
 Sketch 𝑠𝐴𝑀 𝑡 for modulation indices of μ =0.5 and μ =1,
when 𝑚 𝑡 = 𝐴𝑚 𝑐𝑜𝑠 𝜔𝑚𝑡 . This case is referred to as
tone modulation because the modulating signal is a pure
sinusoid (or tone).
Solution
𝑠𝐴𝑀 𝑡 = 𝐴𝑐 + 𝐴𝑚 𝑐𝑜𝑠 𝜔𝑚𝑡 𝑐𝑜𝑠 𝜔𝑐𝑡
= 𝐴𝑐[1 + 𝜇 𝑐𝑜𝑠 𝜔𝑚𝑡 ] 𝑐𝑜𝑠 𝜔𝑐𝑡
𝜇 =
𝐴𝑚
𝐴𝑐
= 0.5
𝐴𝑚 = 0.5𝐴𝑐

17
Example 2: Solution
𝑠𝐴𝑀 𝑡 = 𝐴𝑐 + 𝐴𝑚 𝑐𝑜𝑠 𝜔𝑚𝑡 𝑐𝑜𝑠 𝜔𝑐𝑡
= 𝐴𝑐[1 + 𝜇 𝑐𝑜𝑠 𝜔𝑚𝑡 ] 𝑐𝑜𝑠 𝜔𝑐𝑡
𝜇 =
𝐴𝑚
𝐴𝑐
= 1
𝐴𝑚 = 𝐴𝑐

18
Sideband Power, Carrier Power, and Modulation
Efficiency
 The carrier power
𝑃𝑐 =
𝐴𝑐
2
2
 The sideband power
𝑃𝑠 =
1
2
𝑚2(𝑡)
 The useful message information resides in the sideband
power, whereas the carrier power is used for
convenience of demodulation.
Sidebands
Carrier

19
The total power
𝑃𝑡 = 𝑃𝑐 + 𝑃𝑠
 The power efficiency
𝜂 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑝𝑜𝑤𝑒𝑟
𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑤𝑒𝑟
=
𝑃𝑠
𝑃𝑐 + 𝑃𝑠
=
𝑚2(𝑡)
𝐴𝑐
2
+ 𝑚2(𝑡)
 For the special case of tone modulation,
𝑚 𝑡 = 𝐴𝑚 𝑐𝑜𝑠 𝜔𝑚𝑡 = 𝐴𝑐 𝜇 𝑐𝑜𝑠 𝜔𝑚𝑡
𝑚2(𝑡) =
𝜇𝐴𝑐
2
2

20
𝜂 =
𝜇2
2 + 𝜇2
 with the condition that 0 ≤ 𝜇 ≤ 1 .
 It can be seen that η increases monotonically with 𝜇,
and 𝜂𝑚𝑎𝑥 occurs at 𝜇 = 1, for which
𝜂𝑚𝑎𝑥 = 33%

21
Example 3:
 Single-Tone Modulation Consider a modulating wave
that consists of a single tone or frequency component;
that is,
𝑚 𝑡 = 𝐴𝑚 𝑐𝑜𝑠 𝜔𝑚𝑡
The sinusoidal carrier
𝑐 𝑡 = 𝐴𝑐 𝑐𝑜𝑠 𝜔𝑐𝑡
 AM wave
𝑠𝐴𝑀 𝑡 = 𝐴𝑐[1 + 𝜇 𝑐𝑜𝑠 𝜔𝑚𝑡 ] 𝑐𝑜𝑠 𝜔𝑐𝑡
𝜇 =
𝐴𝑚
𝐴𝑐

22
Example 3
𝑠𝐴𝑀 𝑡
= 𝐴𝑐 𝑐𝑜𝑠 2𝜋𝑓𝑐𝑡 +
1
2
𝜇𝐴𝑐𝑐𝑜𝑠 2𝜋(𝑓𝑐 + 𝑓𝑚)𝑡
+
1
2
𝜇𝐴𝑐𝑐𝑜𝑠 2𝜋(𝑓𝑐 − 𝑓𝑚)𝑡
 The Fourier transform
𝑆𝐴𝑀 𝑓
=
1
2
𝐴𝑐 𝛿 𝑓 − 𝑓𝑐 + 𝛿(𝑓 + 𝑓𝑐)
+
𝜇
4
𝐴𝑐 𝛿 𝑓 − 𝑓𝑐 − 𝑓𝑚 + 𝛿(𝑓 + 𝑓𝑐 + 𝑓𝑚)
+
𝜇
4
𝐴𝑐 𝛿 𝑓 − 𝑓𝑐 + 𝑓𝑚 + 𝛿(𝑓 + 𝑓𝑐 − 𝑓𝑚)

23
Example 3

24
Example 3
 The average power delivered to a 1-ohm resistor by s(t)
is comprised of three components (Using frequency
domain )
2
2
2
2
2
8
1
power
frequency
side
Lower
8
1
power
frequency
side
Upper
2
1
power
Carrier
P
P
P
c
lower
s
c
upper
s
c
c
A
A
A


=
−
=
−
=
−
−
24

25
Example 4:
 Determine power efficiency η and the percentage of the
total power carried by the sidebands of the AM wave
for tone modulation when
(a) μ = 0.5 and
(b) μ = 0.3.

26
Example 4: Solution
𝜂 =
𝜇2
𝜇2 + 2
 For tone modulation with μ = 0.5 → 𝜂 = 11.11%
 only about 11% of the total power is the useful
information power (in sidebands).
 For tone modulation with μ = 0.3 → 𝜂 = 4.3%
 only 4.3% of the total power is the useful information
power (in sidebands).
 Note: increasing μ is preferred because increasing μ
increase the ratio of the useful power (sidebands power)
to the total transmitted power (i.e. decrease the wasted
power).

27
Exercise:
 The radiated output power of a DSB-WC station is 800
watts, the modulation index μ =0.8. Find the power of
the message signal, and the carrier amplitude

28
Example 5:
 A conventional AM waveform 𝑠𝐴𝑀 𝑡 shown. Calculate (a)
the modulation index and (b) the power efficiency.

29
Example 5: Solution
𝐴𝑚𝑎𝑥 = 1.5
𝐴𝑚𝑖𝑛 = 0.5
𝜇 =
𝐴𝑚𝑎𝑥 − 𝐴𝑚𝑖𝑛
𝐴𝑚𝑎𝑥 +𝐴𝑚𝑖𝑛
=
1.5 − 0.5
1.5 + 0.5
= 0.5
𝑃𝑠 =
1
2
𝑚2(𝑡) =
𝐴𝑚
2
2
=
0.52
2
=
1
8
𝜂 =
𝜇2
𝜇2 + 2
0.52
0.52 + 2
11.11%
=
=

Generation of AM Signals
30
 In principle, the generation of AM signals is identical to
any DSB-SC modulation
 Except that an additional carrier component
𝐴𝑐 𝑐𝑜𝑠 𝜔𝑐𝑡 needs to be added to the DSB-SC signal.

AM Generation:
1) Square-law modulator
31
 The block diagram of the Square-law modulator is as
shown
 Square-law device is semiconductor device (nonlinear
device) that has characteristics given by:
where a1and a2 are constants.
 The BPF is used to filter the unwanted signal.
m(t)
+
Accos(2pfct+f)
Square-Law
Device BPF
s(t)
v1(t) v2(t)
)
(
)
(
)
( 2
1
2
1
1
2 t
v
a
t
v
a
t
v +
=
V1(t)
V2(t)

32
 The circuit diagram of the Square-law modulator is
as shown
 Since
 Also, since
 Hence
AM Generation:
1) Square-law modulator (cont’d)
Rℓ
V1(t)
m(t)
Ac cos(2 p fct)
C
L
V2(t)
Nonlinear
device
BPF
)
(
)
(
)
( 2
1
2
1
1
2 t
v
a
t
v
a
t
v +
=
)
(
)
2
cos(
)
(
1 t
m
t
f
A
t
v c
c +
= p
   2
2
1
2 )
(
)
2
cos(
)
(
)
2
cos(
)
( t
m
t
f
A
a
t
m
t
f
A
a
t
v c
c
c
c +
+
+
= p
p

33
AM Generation:
1) Square-law modulator (cont’d)
)
(
)
2
cos(
2
)
(
)
2
(
cos
)
(
)
2
cos(
)
(
2
2
2
2
2
2
1
1
2
t
m
t
f
A
a
t
m
a
t
f
A
a
t
m
a
t
f
A
a
t
v
c
c
c
c
c
c
p
p
p
+
+
+
+
=
Since: ( )
1
)
4
cos(
2
1
)
2
(
cos2
+
= t
f
t
f c
c p
p
So:
( ) )
(
)
(
1
)
4
cos(
2
1
)
(
)
2
cos(
2
)
2
cos(
)
(
1
2
2
2
2
2
1
2
t
m
a
t
m
a
t
f
A
a
t
m
t
f
A
a
t
f
A
a
t
v
c
c
c
c
c
c
+
+
+
+
+
=
p
p
p
( ) )
(
)
(
1
)
4
cos(
2
1
)
2
cos(
)
(
2
1
)
(
1
2
2
2
2
1
2
1
2
t
m
a
t
m
a
t
f
A
a
t
f
t
m
a
a
A
a
t
v
c
c
c
c
+
+
+
+








+
=
p
p
AM wave with
ka=2a2/a1
Unwanted signal that can be
filtered with BPF of center
frequency fc and BW of
twice BW of m(t)

AM Generation:
2) Switching Modulator
34
 The diode acts as ideal switch
 Ac is assumed greater than |m(t)|
 So the diode becomes forward biased
whenever c(t) > 0 and reverse
biased if c(t) < 0.
 The resulting load voltage is:
V2(t) varies periodically between v1(t)
and zero with frequency fc
 Then v2(t) can be expressed as:
𝑣2 𝑡 = 𝑣1 𝑡 𝑤 𝑡
𝑤 𝑡 =
1
2
+
2
𝜋
𝑐𝑜𝑠 𝜔𝑐𝑡 −
1
3
𝑐𝑜𝑠 3𝜔𝑐𝑡 +
1
5
𝑐𝑜𝑠 5𝜔𝑐𝑡 − …





+
=
=
0
)
(
0
0
)
(
)
(
)
2
cos(
)
(
)
( 1
2
t
c
t
c
t
m
t
f
A
t
v
t
v c
c p
m(t)
Rℓ
c(t)=Ac cos(2 p fct)
v1(t) v2(t)
~
BPF

35
AM Modulators:
2) Switching Modulator (cont’d)
So
❑ The first term is an AM wave with ka=4/πAc
❑ The unwanted components that can be removed with a
BPF of center frequency fc and BW of twice BW of m(t).
  





+
+

+
= terms
other
t
f
t
m
t
f
A
t
v c
c
c )
2
cos(
2
2
1
)
(
)
2
cos(
)
(
2 p
p
p
c
c
c
c
f
freq
of
terms
other
component
DC
t
m
t
m
t
f
t
f
A
2
)
(
2
1
)
(
)
2
cos(
2
)
2
cos(
2
1

+
+
+
+
= p
p
p
components
unwanted
t
f
t
m
A
A
t
v c
c
c
+






+
= )
2
cos(
)
(
4
1
2
)
(
2 p
p
From where comes
DC component ???

Demodulation of AM Signals
36
 Like DSB-SC signals, the AM signal can be demodulated
coherently by a locally generated carrier. Coherent, or
synchronous, demodulation
 We shall describe two noncoherent methods of AM
demodulation under the condition of 𝟎 ≤ 𝝁 ≤ 𝟏 :
(1) Rectifier detection and
(2) Envelope detection.

Demodulation of AM Signals:
1) Rectifier detection
37

38
 AM signal is applied to a diode and a resistor circuit.
 The negative part of the AM wave will be removed.
 The output across the resistor is a half-wave rectified
version of the AM signal.
 Which is equivalent to (switching) multiplying AM signal
by 𝑤(𝑡)

39
𝑣𝑅 𝑡 = 𝐴𝑐 + 𝑚 𝑡 𝑐𝑜𝑠 𝜔𝑐𝑡 𝒘(𝒕)
𝑣𝑅 𝑡
= 𝐴𝑐 + 𝑚 𝑡 𝑐𝑜𝑠 𝜔𝑐𝑡
1
2
+
2
𝜋
𝑐𝑜𝑠 𝜔𝑐𝑡 −
1
3
𝑐𝑜𝑠 3𝜔𝑐𝑡 +
1
5
𝑐𝑜𝑠 5𝜔𝑐𝑡 − …
𝑣𝑅 𝑡 =
1
𝜋
𝐴𝑐 + 𝑚 𝑡 + 𝑜𝑡ℎ𝑒𝑟 𝑡𝑒𝑟𝑚𝑠 𝑐𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑎𝑡 ℎ𝑖𝑔ℎ𝑒𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠
 lowpass filter of cutoff B Hz
 The dc term 𝐴𝑐/𝜋 may be blocked by a capacitor to give the
desired output 𝑚(𝑡)/𝜋
Output of LPF Rejected by LPF

2) Envelope Detector
40
 An envelope detector can be constructed using a diode
rectifier followed by a first-order RC filter

41

Charging discharging
Envelope Detector Output
(capacitor voltage)
AM wave
42
* Operation:
❑ At the positive half cycle of sAM(t), diode (D) is forward
biased, Capacitor (C) will charge up rapidly to the peak value of
sAM(t).
❑ When sAM(t) signal voltage falls below the peak, the diode will
be reverse biased and C will discharge through R. The
discharge continues till the next positive half cycle of sAM(t).
❑ When sAM(t) becomes greater than capacitor voltage, charging
will take place again till the new peak value sAM(t) .

43
 The diode conducts again when the input signal voltage
exceeds the voltage across the capacitor. This process
is repeated.
Design Factors :
 Assuming ideal diode characteristics (zero-forward
resistance and infinite resistance in the reverse-biased
state).
 It is important to choose time constant
𝑹𝑪 appropriately so that the capacitor output voltage
follows the envelope of the AM waveform.

44
Design Factors :
 The charging time must be very short (fast charging),
so the time constant must be very small compared to
the period of the carrier frequency.
 The discharging time constant 𝑹𝑪 must be large
enough so that the capacitor discharges slowly
through load resistor 𝑹 between successive positive
peaks of the carrier waveform.
 But if the discharging time get close to 𝑇𝑚 =
1
𝐵
then the
capacitor voltage will not follow the envelope shape.
 So,
𝑅𝐶 ≪
1
𝐵

45
Design Factors :
 However, the discharging time must be small enough
so the envelope detector can track the maximum rate
of change of the message signal.
 But if discharging time is very low (fast discharge)
and get close to 𝑇𝑐 =
1
𝑓𝑐
then the capacitor voltage will
not follow the envelope shape.
𝑅𝐶 ≫
1
𝑓𝑐

46
Design Factors :
 Thus, 𝑅𝐶 should be large compared to 𝑇𝑐 but should be
small compared to 𝑇𝑚 =
1
𝐵
(B is the highest frequency in
𝑚(𝑡)). i.e
1
𝑓𝑐
≪ 𝑅𝐶 ≪
1
𝐵
 This condition can be satisfied since 𝑓𝑐 ≫ 𝐵
Slow discharge (RC too large)
Actual
Envelope
Detected
Envelope
Fast discharge (RC too small)
𝑅𝐶 = 𝑇𝑚 =
1
𝐵
𝑅𝐶 = 𝑇𝑐 =
1
𝑓𝑐

Lec. 4-AM - Part2_DSB-WC.pdf analog comm

More Related Content

Similar to Lec. 4-AM - Part2_DSB-WC.pdf analog comm (20)

Recently uploaded (20)

Lec. 4-AM - Part2_DSB-WC.pdf analog comm