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D. Mynbaev, EET 2140 Module 9, Spring 2008 1
NEW YORK CITY COLLEGE of TECHNOLOGYTHE CITY UNIVERSITY OF NEW YORK
DEPARTMENT OF ELECTRICAL ENGINEERING AND
TELECOMMUNICATIONS TECHNOLOGIES
Course : EET 2140 (ET 313) Communications ElectronicsModule 9: Amplitude modulation II
Prepared by: Dr. Djafar K. Mynbaev
Spring 2008
D. Mynbaev, EET 2140 Course notes, Spring 2008 2
Module 9: Amplitude modulation
II.
• Introduction– Review of Quiz # 4.
– Quiz # 5 (AM) will be next week.
• Modulation (review)
• Review of amplitude modulation (AM)
• Transmitting AM signal– Generating AM signal
– Block diagram of AM transmitter
• AM demodulation
Key words
• Modulation
• Amplitude modulation
(AM)
• Transmitting AM signal
– Generating AM signal
– Block diagram of AM
transmitter
• AM demodulation
D. Mynbaev, EET 2140 Course notes, Spring 2008 3
Key words: baseband transmission, broadband
transmission, modulation, amplitude modulation, frequency,
electromagnetic waves, information signal, modulating signal,
modulation index, spectrum, power, sidebands.
Modulation (review)•Baseband transmission and its shortcomings
•The need for modulation
•What is modulation
•Broadband transmission and its advantages
•Carrier and information signals
See slides in Module 8.
D. Mynbaev, EET 2140 Course notes, Spring 2008 4
Baseband transmission, unfortunately, has three
major drawbacks:
• To deliver information, some or all parameters of an
analog signal must be changed. Because of
interference with external electromagnetic waves
(noise), the values of this parameter of a received
analog signal may be distorted; therefore,
information may be difficult to extract from this
signal.
• Only one signal at a time can be sent over a
transmission line because other information signals
have to use the same band of frequencies that is
already occupied by the signal being transmitted.
• Transmission distance is very limited because the
power of an information signal is by its very nature
self-limiting. You will recall that the power delivered
by the wave is proportional to the square of both the
amplitude and the frequency. Therefore, the lower
the signal frequency, the less power is delivered by
this signal.
Mag
nitu
de (
V)
Time (s)
Mag
nitu
de (
V)
Time (s)
Analog transmission
1 2 1 2
Figure 2.15 Signal distortion during analog transmission.
Transmitted signal Received signal
Magnit
ude (
V)
Time (s)1 2
Magnit
ude (
V)
Time (s)1 2Transmission
distance (km)Sent signal Received signal
The need for modulation
v(t)
f (kHz)
Your voiceHer voice
His voice
My voice
40
D. Mynbaev, EET 2140 Course notes, Spring 2008 5
Main problem of a baseband transmission: Baseband transmission can be either
analog or digital. However, in either case the baseband transmission suffers from a
major setback – low transmission speed (capacity). One of the fundamental
principles of communications theory is that transmission speed (capacity), C (bit/s),
is proportional to the frequency of a carrier signal, fC.(Hz).
C (bit/s) ~ fC.(Hz)
Since baseband transmission rely on a low-frequency carrier (remember, in this case
an information and a carrier signal is the same), transmission speed (capacity) of a
baseband transmission is low.
The ultimate solution to the problems of baseband transmission is
shifting a low-frequency information signal to the high-frequency
band. This can be achieved by modulation of an information signal.
Modulation means superimposing a low-frequency information signal
upon a high-frequency carrier signal.
This approach is called modulation by frequency translation.
What is modulation
D. Mynbaev, EET 2140 Course notes, Spring 2008 6
Modulating (information) signal
Carrier signal
Modulated signal
*
What is modulation
D. Mynbaev, EET 2140 Course notes, Spring 2008 7
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Modu-
lator
Demo-
dulator 0
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Modulated signal
Information signal
Carrier signal Carrier signal
Information signal
Transmission link
The scheme of modulation – The block diagram of a broadband system.
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What is modulation
D. Mynbaev, EET 2140 Course notes, Spring 2008 8
Broadband transmission - modulation
Broadband transmission: What it is Broadband transmission occurs at the
band of frequencies broader than the frequencies of an information signal, which
means we need to use modulation by frequency translation.
Since we need to translate transmission frequency by using a carrier signal, the best
candidate for a carrier signal is a cosine (sine) wave. This is because a cosine wave
is a single-frequency signal. (Recall time domain and frequency domain
correspondence.)
Using high-frequency sinusoidal wave as a carrier signal, we will achieve the
following advantages:Main advantage of a broadband transmission: Since broadband
transmission relies on a high-frequency carrier, it provides much higher speed of
transmission. Indeed, the formula C (bit/s) ~ fC.(Hz) proofs that.
Other advantages:
•Efficient transmission: (1) The size of antenna: The size is about 1/10 of the signal wavelength
for 1 kHz we would need 300 km antenna size. For 300 MHz FM transmission we need the antenna
of 1 meter in size. (Read about antenna size in [1], Chapter 15.) and (2) the range of sizes: If we need to
radiate signals for voice (4 KHz) and video, (TV - 6.5 MHz) we need the antennas of different
sizes.
D. Mynbaev, EET 2140 Course notes, Spring 2008 9
Advantages (continued):
• Multiplexing and frequency assignment (interference of signals): Several radio
stations can broadcast their signals over the same area without interference.
• We can choose the correct frequency range to give the best transmission
conditions (e.g., for radio and microwave transmission).
• High-power transmission: Power delivered by the wave is proportional to the
square of both the amplitude and the frequency. Therefore, the higher the
signal frequency, the more power is delivered by this signal with the same
amplitude. (Give an example.) Assignments: See Lecture 8 and the course outline.
Frequency (MHz)
Your voiceHer voice
His voice
My voice
0.0040 50.01 50.02 50.03 50.04
My voiceHis voice
Her voiceYour voice
Am
plit
ude
or P
ower
(any
uni
ts)
D. Mynbaev, EET 2140 Course notes, Spring 2008 10
Amplitude modulation (review)
• Amplitude modulation
– What it is
– Carrier, modulating (information) and modulated signals
– Modulation index
– Carrier and modulating frequencies
– Instantaneous value of AM signal
– Spectrum and bandwidth
– Power distribution
– See slides in Topic 8.
D. Mynbaev, EET 2140 Course notes, Spring 2008 11
The formula of a modulated signal
v(t) = (AC + vM(t)) cos ( Ct)
describes a sinusoidal signal with radian frequency C and amplitude AC + vM(t).
What distinguishes this signal from a regular cosine is that its amplitude varies as
the formula for vM(t) dictates. Note that this variable amplitude of a modulated
signal, AC + vM(t), is referred to as an envelope; thus, the formula for the
instantaneous value of an envelope is vENV(t) = AC + vM(t) = (AC + AM cos Mt).
Using the formula for a modulation index, m = AM / AC , we can rewrite this
formula as follows: vENV(t) = AC (1 + AM / AC cos Mt) = AC (1 + m cos Mt).
Now we can derive the formula for a modulated signal:
vAMt) = (AC + vM(t)) cos ( Ct) = (AC + AM cos Mt) cos Ct.
Again, using a modulation index, we can rewrite this formula as follows:
vAM(t) = AC (1 + AM / AC cos Mt )cos Ct = AC (1 + m cos Mt ) cos Ct
Next slide shows the graphs built by using these formulas.
Amplitude modulation - envelopes
D. Mynbaev, EET 2140 Course notes, Spring 2008 12
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0
20
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1 10
Information signal
Amplitude-modulated signal
Time (µs)
Mag
nit
ude
(V)
Calculated AM signal, information signal, and a positive and a negative envelope.
10
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Positive envelope
Negative envelope
Amplitude modulation - envelopes
D. Mynbaev, EET 2140 Course notes, Spring 2008 13
Observe that the magnitude of an AM signal jumps from maximum to
minimum values very quickly. However, the magnitude of a signal’s
envelope changes slowly, because it follows a change in the magnitude of an
information (modulating) signal. Note, too, that an AM signal has two
envelopes: positive and negative. The positive envelope is given by vENV(t)
= AC + vM(t) = (AC + AM cos Mt); the negative envelope differs from the
positive one in that the negative sign appears at the right side of the formula;
that is, vENV (t) = -(AC + AM cos Mt).
Observe, too, that both envelopes have a period equal to TM = 1/fM. In
other words, the envelopes repeat the change of the information signal (the
positive – in phase and the negative – out of phase).
Slide 26 shows the same AM signal, with the information signal drawn
straight over the graph. Incidentally, this graph was calculated using the
formula v(t) = (AC + AM cos Mt) cos Ct,which is the original presentation
of an AM signal.
Amplitude modulation - envelopes
D. Mynbaev, EET 2140 Course notes, Spring 2008 14
To further explore the formula for an AM signal, we need to recall the simple
trigonometric identity cos A x cos B = ½ (cos (A – B) + cos (A + B)). Making use of
this identity, we obtain:
v (t) = (AC + AM cos Mt) cos Ct = AC cos Ct + ½ (AM cos Mt x cos Ct)
= AC cos Ct + ½ (AM cos ( M - C) t + AM cos ( M + C) t)
You’ll recall that modulation index is defined as m = AM/AC, which results in AM =
mAC. Substitute for AM in the above expression and obtain the explicit formula for a
sinusoidal amplitude-modulated signal:
vAM(t) = AC cos Ct + AM/2 cos ( M - C) t + AM/2 cos ( M + C) t
= AC cos Ct + m AC/2 cos ( M - C) t + m AC/2 cos ( M + C) t
This formula allows us to calculate an amplitude-modulated signal in its entirety
and compute its magnitude at any given time. In other words, this formula allows us
to calculate the instantaneous value of an AM signal.
Amplitude modulation – signal formula
D. Mynbaev, EET 2140 Course notes, Spring 2008 15
Amplitude modulation – frequency spectrum
Frequency spectrum of a sinusoidal AM signal
The formula for an AM signal describes three sinusoidal signals with three
different frequencies: C, ( M - C), and ( M + C). All this means is that
the combination of information and carrier signals produces a new entity—
the modulated signal—that has its own frequency spectrum.
Now we can compare time-domain and frequency-domain presentations of
the same AM signal: In time domain we have an AM signal, given by the
formula:
vAM(t) = AC cos Ct + m AC/2 cos ( M - C) t + m AC/2 cos ( M + C) t
The waveforms of such a signal are shown on pertaining slides. To show its
spectrum (to present this signal in frequency domain), we need to show
three lines with the amplitudes AC and m AC/2 at the frequencies fC, (fM - fC),
and (fM + fC). This presentation is given in the next slide. (See Example in
Topic 8.)
D. Mynbaev, EET 2140 Course notes, Spring 2008 16
Frequency (Hz)
Am
pli
tude
(V)
fCfM - fC fM + fC
AC
m AC/2 m AC/2
BWAM
AM
fM
≈
Frequency spectrum of an AM signal.
0
Amplitude modulation – frequency spectrum
D. Mynbaev, EET 2140 Course notes, Spring 2008 17
Spectral width and frequency spectrum It is necessary to stress here one other
important concept: No source on earth can generate a single-frequency signal. It is
impossible even in theory. Thus, every generator produces a signal that contains a
band (group) of frequencies. The frequency that we refer to as fC, fM - fC, or fM + fC is
nothing more than a peak (central) frequency of appropriate signals. For example, a
carrier signal with peak frequency fC includes, in fact, many other sinusoidal signals
(harmonics) with closely bunched frequencies that gradually deviate from the peak
frequency. The amplitudes of these harmonics become smaller as the value of their
frequencies deviates more from the peak frequency. Therefore, instead of a single
frequency, we need to talk about a frequency band characterized by a bell-shaped
curve. Quite obviously, the more frequencies involved in a particular signal, the
wider the shape of the bell curve representing the signal. This characteristic is
described by the spectral width, Δf, which is the width of the bell curve measured at
half of its maximum amplitude.
Thus, the spectrum of an AM signal contains not three frequencies but three bands
of frequencies. The range of frequencies grouped around fC + fM is called upper
sideband (USB); the range of frequencies grouped around fC - fM is called lower
sideband (LSB). Slide 37 shows these bands.
Amplitude modulation – frequency spectrum
D. Mynbaev, EET 2140 Course notes, Spring 2008 18
0
0.005
0.01
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1
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Frequency (Hz)
Am
pli
tude
(V
)
fCfC - fMfC +
fM
Carrier
Upper
sideband
Lower
sideband
Frequency spectrum of AM signal: sidebands.
BWAM
m AC/2
AC
0
BW
• • •0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
1
fM
Modulating
signal
Amplitude modulation – frequency spectrum
D. Mynbaev, EET 2140 Course notes, Spring 2008 19
1V/V 0V
Y
X
0V
A
B
C
1V 100k Hz
Vc
A BT
G
MULTIPLIER SUMM ER
Carrie r
Modulating Signal
TIN
The screen of what device are looking at?
Is it time domain or frequency domain?
Amplitude modulation – frequency spectrum
D. Mynbaev, EET 2140 Course notes, Spring 2008 20
Amplitude modulation – frequency spectrum
Comments on the computer simulation of an AM signal:
1. Signal generation: Observe that the circuit performs the mathematical operation prescribed by the
formula for an AM signal: v(t) = (AM cos Mt)*(AC cos Ct) + (AC cos Ct).
2. Frequency spectrum: Observe three bands – carrier band, upper sideband and lower sideband.
Bandwidth of an AM signal
What range of frequencies does an AM signal occupy? Looking at the frequency-
spectrum slides, you can see that this range lies between the upper-side
frequency, fC + fM, and the lower-side frequency, fC - fM. This range of
frequencies is called the bandwidth of an AM signal, BWAM, and it is equal to
BWAM = fC + fM – (fC - fM) = 2fM
For instance, the AM bandwidth of the signal shown in Slide 32 is equal to 2 kHz.
Example:
Problem:
What bandwidth is needed for the AM transmission of a signal containing music?
Solution:
To solve this problem, we restrict ourselves to the highest distinguishable
frequency, 20 kHz. Thus, for a rough estimate of this bandwidth, we use
Formula 2.19 and compute:
BWAM = 2fM = 40 kHz .
D. Mynbaev, EET 2140 Course notes, Spring 2008 21
Amplitude modulation – power distribution
Power distribution in AM signal.
To analyze the power distribution in an AM signal, refer to the formula:
vAM(t) = AC cos Ct + m AC/2 cos ( M - C) t + m AC/2 cos ( M + C) t
Power of a sinusoid delivered to a load with impedance R is equal to [1]:
P = (Vrms)2/R = (Vpk/√2)2/R = Vpk2/2R.
For the components of an AM signal, Vpk is equal to AC and AM, respectively.
Thus,
PC = (AC)2/2R
PUSB = PLSB = (mAC/2)2/2R = (m2/4) (AC2/2R ) = (m2/4) PC.
Total power, Pt, of an AM signal is equal to:
Pt = PC + PUSB + PLSB = PC + (m2/4) PC + (m2/4) PC = (1 + (m2/2)) PC
To better interpret this formula, let’s consider a case where the modulation index is
equal to 1. In such a case, PT = PC + 2 PSB = 1.5 PC and PSB = 0.25PC . Thus, PC ≈
0.67 PT and PSB ≈ 0.167 PT . In other words, almost 67% of the total power is
consumed by the carrier signal, which delivers no information, while less than 17%
of the total power is concentrated in the information signal. This is the price we pay
for the advantages of AM transmission.
D. Mynbaev, EET 2140 Course notes, Spring 2008 22
Example (See also Example 5-2 in [2]): Problem:
Calculate the total power and power ratio of the sideband signal to the carrier signal
of the amplitude-modulated signal if AC = 50 V, AM = 30 V and R = 50 Ω.
Solution:
The first step is to apply the formulas PC = AC2/2R and PSB = (AM
2/2)/2R =
(mAC/2)2/2R.. We can easily compute m = AM/AC = 0.6.
Next, we plug the values of the amplitudes of the carrier and sideband signals, AC =
50 V and AM = 30 V, in the formula for PC and obtain: PC = 2500/2R = 25 W and PSB
= PUSB + PLSB = (AM2/2)/2R = 450/100 = 4.5 W.
Now we can compute the required total power and the ratio: Pt = (1 + (m2/2)) PC =
1.18 PC = 29.5 W and PSB / PC = 4.5 (W)/25 (W) = 0.18.
Discussion Thus, the information signal consumes only 18% of the power of the
carrier signal. Again, this is a major drawback to the use of a carrier signal. However, the
picture is not totally bleak. We need to maintain a proper perspective on this matter; that is, we
need to recall, too, all the advantages of broadband transmission. In addition, there are
transmission techniques that help to cope with this problem.
Amplitude modulation – power distribution
D. Mynbaev, EET 2140 Course notes, Spring 2008 23
Quiz # 5 will be next week. Topic: Amplitude modulation.
You must be able to:
• Identify modulating, carrier and modulated signals and their
parameters (amplitudes and periods) in time domain (from
waveforms);
•Identify envelopes of a modulated signal;
•Compute a modulation index by two methods in absolute number
and in percents;
•Compute the amplitudes and the frequencies of upper and lower
sidebands and the carrier;
•Explain the spectrum of an AM signal;
•Compute the bandwidth of an AM signal;
•Compute the power of a carrier signal and upper and lower
sidebands and the total power of an AM signal;
•Comment on the power distribution of an AM signal.
D. Mynbaev, EET 2140 Course notes, Spring 2008 24
Assignments:
1. Reading:
a. Textbook: Pages 68-82, 83-94 and 118-120.
b. Paul Young, Electronic Communication Techniques, 5th Edition, Prentice
Hall, 2004: Section “Amplitude Modulation.”
2. Homework problems: Chapter 2: 23-41 and Chapter 3: ## 1-15.
3. Carefully review the examples given in this lecture.
References:
1. Jeffrey S. Beasley and Garry M. Miller, Modern Electronic Communication, 8th ed., Prentice
Hall, 2005.
2. Paul H. Young, Electronic Communication Techniques, 5th ed., Prentice Hall, 2004.
3. Robert L. Boylestad, Introductory Circuit Analysis, 10th ed., Prentice Hall, 2004.
4. Thomas L. Floyd, Electronic Devices, 7th ed., Prentice Hall, 2005.
5. Richard H. Berube, Learning Electronics Communications Through Experimentation Using
Electronics Workbench Multisim, Prentice Hall, 2002.
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