THE CITY UNIVERSITY OF NEW YORK DEPARTMENT OF …websupport1.citytech.cuny.edu/faculty/mseip/files/...DEPARTMENT OF ELECTRICAL ENGINEERING AND TELECOMMUNICATIONS TECHNOLOGIES Course

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


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.


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


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


Modulating (information) signal

Carrier signal

Modulated signal

*

What is modulation


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The scheme of modulation – The block diagram of a broadband system.

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What is modulation


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.


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)


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.


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


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Information signal

Amplitude-modulated signal

Time (µs)

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Calculated AM signal, information signal, and a positive and a negative envelope.

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Positive envelope

Negative envelope



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.



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


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.)


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



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.



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sideband

Frequency spectrum of AM signal: sidebands.

BWAM

m AC/2

AC

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MULTIPLIER SUMM ER

Carrie r

Modulating Signal

TIN

The screen of what device are looking at?

Is it time domain or frequency domain?




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 .


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.


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


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.


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.

Documents

THE CITY UNIVERSITY OF NEW YORK DEPARTMENT OF …websupport1.citytech.cuny.edu/faculty/mseip/files/...DEPARTMENT OF ELECTRICAL ENGINEERING AND TELECOMMUNICATIONS TECHNOLOGIES Course