Analogue Modulation (AM)

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COMM 1208 Unit 3 AM

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Analogue Modulation (AM)

1.

Why Modulate?

................................
................................
................................
....................

2

2.

Amplitude Modulation

................................
................................
................................
.......

3

2.1

Time domain

................................
................................
................................
....................

3

2.2

Derivation

................................
................................
................................
........................

4

2.3

Modulation
Index (or Modulation Factor or Depth of Modulation)

...............................

7

2.4

Power in an AM waveform

................................
................................
..............................

9

2.5

Peak Instantaneous Power

................................
................................
.............................

9

3.

Suppressed Carrier Signals and Single Sideband

................................
.....................

10

4.

AM Demodulators

................................
................................
................................
.............

11

4.1

Diode Detector

................................
................................
................................
...............

11

COMM 1208 Unit 3 AM

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1.

Why Modulate?

All audio signals occupy the same frequency band i.e. between 0 and 20 kHz. Before being
broadcast an a
udio signal (speech or music) must be moved, or frequency translated to a
specific frequency range in order to use the available frequency spectrum. To do this the audio
signal (or modulating signal)

modulates

a much higher radio frequency (the carrier
fre
quency). Each audio signal is assigned a carrier

-

defining a channel
-

so that it is possible
for the receiver to discriminate between all the streams of signals coming in.


There are 3 main reasons to modulate a signal on to a high frequency carrier

1.

Aud
io is in the range approx. 30
-

20000

kHz. If an electromagnetic signal with a frequency
of 30 Hz is transmitted it will have a wavelength of (speed of light /frequency) =
300,000/30

km = 10,000 km. To pick up this signal an aerial of size approx. 2,500 km

will
be required

-

impractical. If this signal is used to modulate a carrier of 1 MHz the
wavelength will be 300,000/1,000,000 km

= 300 m, and an aerial of 75 m will suffice. If the
carrier is 100 MHz, the wavelength is 3 m and a 750 cm aerial is sufficie
nt.

2.

A large number of radio transmitters are trying to transmit at the same time. It is
necessary for the receiver to pick up only the wanted signal and to reject the rest. One way
to do to this is to assign a carrier with a known frequency to each transmi
tter, modulate
this carrier with the signal, and then design the receiver to pick up only that known carrier
frequency and reject the rest, using appropriate filtering methods. Then the original signal
is removed from the received carrier. The same concept

is used in carrying a large number
of telephone conversations over a single pair of wires or optical fibre.

3.

Using appropriate modulation techniques it is possible at the receiver to remove a lot of the
noise and other distortions which the transmission me
dium would impose on the signal.

COMM 1208 Unit 3 AM

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2.

Amplitude Modulation

2.1

Time domain

An AM signal is made up of a carrier (with constant frequency) in which its amplitude is
changed (modulated) with respect to the signal (modulating signal) we wish to transmit
(voice, music
, data, binary). In the example below the carrier (a high frequency sine wave) is
being modulated by a lower frequency sine wave. The modulating signal causes the carriers
amplitude to change with time. This resulting shape of the carrier is called
the env
elope. Note
the envelope has the shape of a sine wave.


Figure
2
-
1

AM signal




Figure
2
-
2

Modulating signal (sine wave) and modulated carrier



COMM 1208 Unit 3 AM

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Figure
2
-
3

Modulating signal (
Audio
) and modulated carrier

2.2

Derivation

A carrier is described by


v

=

V
c

Sin (

c

t +

)


To
am
plitude modulate

the carrier its amplitude is changed in accordance with the level of
the audio signal, which is described by


v

=

V
m

Sin (

m

t )


The amplitude of the carrier varies sinusoidally about a mean of V
c
. When the carrier is
modulated its am
plitude is varied with the instantaneous value of the modulating signal. The
amplitude of the variation of the carrier amplitude is V
m

and the angular frequency of the
rate at which the amplitude varies is

m
. The amplitude of the carrier is then:

Carrier
amplitude = V
c

+ V
m

Sin (

m

t )

and the instantaneous value (value at any instant in time)

is

v

=

{V
c

+ V
m

Sin (

m

t )} * Sin (

c

t )






Eqn. 1


=

V
c

Sin (

c

t )

+ V
m

Sin (

m

t )

* Sin (

c

t )


Using Sin A * Sin B = ½ Cos (A
-

B)
-

½
Cos

(A + B
) this becomes

v

=

V
c

Sin (

c

t ) + ½ V
m

Cos ( (

c

-


m
) t )
-

½ V
m

Cos ((

c

+

m
)t)

Eqn. 2


This is a signal made up of 3 signal components



carrier

at




c
(rad/s)



Frequency is

f
c

=

c
/2


Hz



upper side frequency



c

+

m

(rad/s)


Frequency is

(

c

+

m
)/2


= f
m

+ f
c
Hz



lower side frequency



c

-


m


(rad/s)


Frequency is

(

c

-


m
)/2


= f
m

-

f
c
Hz

The
bandwidth

(the difference between the highest and the lowest frequency) is


BW = (

c

+

m

)
-

(

c

-


m
)

=

2 *

m
Rad/s



( =


m
/


Hz)


COMM 1208 Unit 3 AM

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The spectrum
of these signals is shown. This
is described as the signal in the
frequency
domain
, as opposed to the signal in the
time
domain
. In this case the audio signal is made
up of a single frequency.

In this example the
angular frequencies

(expressed in Radians/
sec, or kRad/sec, or
Mrad/sec) are show. In most cases however
the frequency is shown (expressed in Hz, or
kHz, or MHz).


If the audio signal is made up of a range of
frequencies from f1 to f2 (as is normally the
case) rather than a single frequency the

output signal will be a band of frequencies,
contained in



the upper side band (USB), inverted and




the lower side band (LSB), erect.

A broadcast AM station in the Medium Wave
band is usually allocated a frequency slot 9

kHz wide. This means that the carr
iers of stations
in this band are spaced 9 kHz apart.

The maximum amplitude in an AM signal is

V
c

+ V
m .
The minimum amplitude is

V
c

-

V
m.







Figure
2
-
4

Frequency Domain view of Double Sideband


Full Carrier






Example:

A 1 MHz carrier is amplitude modulated by an audio signal which
contains all frequencies in the range 300 Hz to 5 kHz. What are the freque
ncy bands
which are output? What is the output bandwidth? Draw the spectral diagram of these
signals.

Answer:

The carrier is 1 MHz

Amplitude (V)
Angular
Frequency

c

c
+

m

c
-

m
Lower
side
frequency
Carrier
Upper
side
frequency
Bandwidth
= 2 *

m

Spectrum of
audio signal
Carrier
Upper
Sideband
Erect
Lower
Sideband
Inverted
f
c
f1
f2
f
c
+ f1
f
c
+ f2
f
c
- f2
f
c
- f1

COMM 1208 Unit 3 AM

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The Upper Side Band is all frequencies in the range 1,000,300 to 1,005,000 Hz


The Lower Side Band is all frequencies in the

range 995,000 to 999, 700 Hz


The Bandwidth is 1,005,000
-

995,000 = 10,000 Hz = 10 kHz.


Example:

A 1.5 MHz carrier is amplitude modulated by three sinusoidal signals of
frequency 500

Hz, 800 Hz and 1,400

Hz. What are the frequencies in the AM spectrum?

Answer:

Convert all the frequencies to kHz. 1.5 MHz is 1500 kHz. 500 Hz is
0.5

kHz. 800

Hz is 0.800 kHz. 1400 Hz is 1.4

kHz.

The output frequencies are:

1500 kHz,

1500


0.5

kHz,

1500


0.8 kHz

1500


1.4 kHz

or



1500,


1500.5 , 1499.5,

1500.8, 1499
.2,

1501.4 , 1498.6 kHz


Exercise: Draw a diagram showing these frequency bands for the above examples.

COMM 1208 Unit 3 AM

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2.3

Modulation Index (or Modulation Factor or Depth of Modulation)

This is defined as
m =
V
m
V
c



In AM, this quantity, also called modulation d
epth, indicates by how much the modulated
signal varies around its 'original' level. For AM, it relates to the variations in the carrier
amplitude.


So if m = 0.5, the carrier amplitude varies by 50% above and below
its unmodulated level, and
for m

= 1.0 i
t varies by 100%. Modulation depth greater than 100% is generally to be avoided
as it creates distortion.



Using this Eqn. 2 can be re
-
written as

v

= V
c

Sin (

c

t ) + ½

(V
m

Cos ( (

c

-


m
) t )

-

V
m

Cos ((

c

+

m
)t) ) * V
c

/V
c

v

= V
c

{ Sin (

c

t ) + ½

m [ Cos ( (

c

-


m
) t ) + Cos ((

c

+

m
)t) ] }


Eqn. 3


The maximum allowed value of

m

is 1.0. If
this is exceed the envelope of the output
waveform is distorted. This is known as
Over
-
modulation

and should never occur
in practice, because the distorted

envelope
will result in a distorted output sound signal
in the radio receiver. The effect of over
-
modulation can be examined in the
laboratory.


Example:

An AM signal is
represented by the equation

v =

( 15 + 3 Sin( 2


* 5 * 10
3

t) ) *


Sin(

2


* 0.5 *
10
6

t) volts

(i)

What are the values of the carrier
and modulating frequencies?

(ii)

What are the amplitudes of the
carrier and of the upper and lower
side frequencies?

(iii)
What is the modulation index?

(iv)
What is the bandwidth of this
signal?


Answer:

This looks the sam
e as Eqn. 1 above with:



c


(= 2


f
c
)

= 2


* 0.5 * 10
6




m

(= 2


f
m
)

= 2


* 5 * 10
3


V
c

=

15 V



V
m

=

3 V

(i) Therefore the carrier frequency



f
c
is 0.5 * 10
6



= 0.5 MHz

and the modulating frequency



f
m

is 5 * 10
3


= 5 kHz

(iii) The bandwidth





BW = 2 f
m



= 10 kHz

(ii) The modulation index




m = V
m
/V
c



= 3/15 = 0.2


From Eqn. 3 the amplitude of each side frequency is


m* V
c

/2


= 0.2 * 15 /2

= 1.5 V


50% Modulation
-1.5
-1
-0.5
0
0.5
1
1.5
1
9
17
25
33
41
49
57
65
100% modulation
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
1
9
17
25
33
41
49
57
65
150% Modulation
-3
-2
-1
0
1
2
3
1
9
17
25
33
41
49
57
65
Carrier
-1
-0.5
0
0.5
1
1
9
17
25
33
41
49
57
65
Modulating Signal
-1
-0.5
0
0.5
1
1
9
17
25
33
41
49
57
65

COMM 1208 Unit 3 AM

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Variations of modulated signal with percentage modulation are shown below. In each imag
e,
the maximum amplitude is higher than in the previous image. Note that the scale changes
from one image to the next.





Alternative form for modulation index

If an AM signal is being displayed on an oscilloscope it can be difficult to read V
m

and V
c
.
Instead the form for expressing m

can be modified to make it easier to read.

Modulation index : m=


V
m
V
c


=


½
(
V
m

+ V
m
)

-

½
(
V
c

-

V
c

)

½
(
V
c

+ V
c
)

+ ½
(
V
m

-

V
m
)



=

(
V
m

+ V
m
)

-

(
V
c

-

V
c

)
(
V
c

+ V
c
)

+
(
V
m

-

V
m
)


=

(
V
c

+ V
m
)

-

(
V
c

-

V
m

)
(
V
c

+ V
m
)

+
(
V
c

-

V
m
)



=

Max Amplitude
-

Min Amplitude
Max Amplitude + Min Amplitude


It is possible to read the maximum and minimum amplitude of the signal from the
oscilloscope display.

COMM 1208 Unit 3 AM

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2.4

Power in an AM waveform

Assume that the AM signal is dissipated in

a load of R

. The total power dissipated will be
the sum of the powers in all of the components of the signal.

The power in the carrier will be


P
c


=

V
c
2
R


Watts

The power in each of the frequencies is

P
s
=
(
mV
c
/2
)
2
R


=
m
2
4


V
c
2
R



=

m
2
4


P
c

The total power is

P
t


= P
c

+ P
s

+ P
s
= P
c

+ 2 P
s

= P
c
( 1 + 2
m
2
4

) = P
c
( 1 +
m
2
2

)

Watts

The fraction of the power in the carrier is

P
c
P
t


=
1

1 +
m
2
2



The maximum value for m is 1.0. This means that at most only 1/3 of the power in the signal
will be contained in the sidebands. All of the audio information is contained in either one of
the sidebands, so that, in effect, only one sixth of the power (16.
7%) is used to carry
information. The remainder of the signal can in some respects be considered to be redundant!


Example:

A transmitter puts out a total power of 25 Watts of 30% AM signal. How
much power is contained in the carrier and each of the sideba
nds?

Answer:

Total power = 25

= P
c
( 1 +
m
2
2

) = P
c
( 1 +
0.3
2
2

) = P
c

* 1.045

Therefore the carrier power is P
c

= 25/ 1.045 = 23.92 Watts

The total power in the 2 sidebands is 25
-

23.92

= 1.08 W

The power in each sideband is 1.08/2
= .54 W

The fraction of the power in the carrier is 23.92/25 = 0.957, or 95.7%

2.5

Peak Instantaneous Power

The maximum signal voltage is V
c

+ V
m

=

V
c

(1 + m) so that the maximum
instantaneous output power is
V
c
2
R

(1 + m)
2


=

P
c
(1 + m)
2

.

If th
e modulation index is 1.0 the maximum output power will be 4 P
c
. The transmitter must
be designed to carry this level of output power.

COMM 1208 Unit 3 AM

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3.

Suppressed Carrier Signals and Single Sideband

Single Sideband (SSB), is a form of AM


Because most of the output power
from an amplitude modulator is contained in the carrier
there will be major power savings if the carrier can be suppressed. The carrier amplitude and
frequency do not change and so it does not contain any signal information. The suppression of
the carrier
will not cause any of the information in the signal to be lost.

Each sideband is the image of the other and one of them may be suppressed without the loss
of any information.

All of the information is conveyed through the use of a single sideband with no
carrier.


Example:

An AM signal has a depth of modulation of 70%. What is the power
saving if

(a)
the carrier is suppressed and

(b)
the carrier and one sideband is suppressed?

Answer:

The total power in the signal is P
t

= P
c
( 1 +
m
2
2

) .

Therefore

t
he fraction of the total power in the carrier is

1 / ( 1 +
m
2
2

) = 1/(1 + 0.7
2
/2) =

1/1.245 = .803 = 80.3%

(a)
In this case if the carrier is suppressed then the power saving will be 80%, the
transmitter will need to transmit only 20% of the power
it would otherwise need to
transmit.

(b)
If one of the sidebands is suppressed then only half of the remaining power will need
to be transmitted

i.e. 10%.

In this example a transmitter which would have to transmit 10 W of a
full wave

AM
signal will be able to

transmit the same information on 1 W if the carrier and one
sideband are both suppressed.


When the carrier and both side bands are transmitted it is called
Full Wave
transmission.

If only the carrier is suppressed we have
Double Side Band Suppressed Carr
ier
transmission or
DSBSC
for short .

If the carrier and one side band are suppressed we have
Single Side Band

transmission or
SSB
.

Radio receivers for receiving Full Wave signals are cheap to produce but the transmitter must
be capable of transmitting a l
ot of power. It is used for broadcast radio stations in the Medium
Wave band because there will be only one transmitter for a country the size of Ireland but
millions of receivers so that the aim is to keep the receivers as cheap as possible.

Radio receive
rs for SSB or Carrier Suppressed signals are expensive to produce, but the
transmitter need not be capable of outputting a high power level. It is used for ship to shore
communications (e.g. between a fishing boat and the harbour master) or for other one
-
t
o
-
one
communications. In this case there are as many transmitters as receivers, one per boat, so
that there are no major cost savings if the receiver is made a little cheaper, but there will be
major gains if the transmitter can be made cheaper. In additio
n the power on the boat (or
plane, or other vehicle) may be limited and will be needed for lighting and other functions, so
that it is desirable that as little of it as possible is needed for communications.

COMM 1208 Unit 3 AM

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4.

AM Demodulators

4.1

Diode Detector

This is the most

commonly used AM demodulator. It is cheap and reasonably accurate. It is
also used as an integral part of many designs of (older) FM detector. The basic circuit is as
below


It has limitations.



If the time constant R1 *C1 in the
envelope detector

is too

long relative to the
period of the highest frequency modulating
signal it will not be it will not be able to
follow the peaks and troughs of the
envelope giving rise to diagonal clipping. It
is required that
R1*C1

<

[(1

-
m
2
)
1/2
]

/

(m

m
)
where

m

is the hi
ghest frequency component of the modulating
signal and m is the modulation index. This is derived below.



If R1*C1 is too short than there will excessive RF ripple and the output power will be
reduced.



Because the diode is a non linear device there will be
some distortion in the demodulated
signal.


In general R1 C1 must be a lot longer than the period of the carrier and a lot shorter than the
period of the modulating signal.

R1 must be a lot larger than the forward resistance of the diode to maintain detect
or
efficiency. It must also provide matching to the next (audio) stage.


The Diode detector output signal consists of three components

1.

The wanted demodulated audio signal

2.

A DC component proportional to the peak amplitude of the RF signal. This is removed b
y
sending the signal through

a capacitor

C2 (high pass filter). It is also used to provide an
input into Automatic Gain Control.

3.

An unwanted ripple at the carrier frequency and its harmonics. This is blocked from later
stages by using an RC low pass filter

(R2 and C3 in this circuit).

R1
C1
C2
R2
C3
AM
in
Envelope
Detector
DC
Block
LPF
Audio
Out