generation of angle modulated signals · atheer alaa sabri 12 demodulation of angle modulated...
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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=∆
The deviation ratio ( ) plays the same rule for arbitrary modulation as the modulation
index plays for sinusoidal modulation. Replacing by and by , we have
= 2( + 1)
Generation of Angle Modulated Signals
A) Narrowband Angle Modulated Signals
The generation of narrowband angle modulated signals is illustrated in the
following figures:
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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B) Wideband Angle Modulated Signals There are two methods of generating wideband (WB) angle modulated; the
indirect method and the direct method.
1) Indirect Method
( ) = [ + ( )]
Then
( ) = [ + ( )]
Use of frequency multiplication normally increases the carrier frequency to an
impractically high value. To avoid this, a frequency conversion is necessary to shift the
spectrum
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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2) Direct Method In the direct method of generating FM signal, the modulating signal directly
controls the carrier frequency. A common method used for generating FM directly is to
vary the inductance or capacitance of a tuned electric oscillator. Any oscillator whose
frequency is controlled by the modulating signal voltage is called a Voltage Controlled
Oscillator (VCO). The main advantage of a direct FM is that large frequency deviations
are possible and thus less frequency multiplication is required. The major disadvantage
is that the carrier frequency tends to drift and so additional circuitry is required for
frequency stabilization.
Example
A given angle modulated signal has a peak frequency deviation of 20 Hz for an
input sinusoid of unit amplitude and a frequency of 50 Hz. Determine the required
frequency multiplication factor, , to produce a peak frequency deviation of 20 KHz
when the input sinusoid has unit amplitude and a frequency of 100 Hz, and the angle
modulated used is (a) FM and (b) PM.
Solution
(a) ∆ = 20 , ∆ = 20
=∆∆
= 1000
(b) ∆ = 20 , ∆ = (20 ) = 40
=∆∆
= 500
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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Demodulation of Angle Modulated Signals
1) Direct Method
Demodulation of an FM signal requires a system that produces an output
proportional to the instantaneous frequency deviation of the input signal. Such a system
is called a frequency discriminator. If the input to an ideal discriminator is an angle
modulated signal
( ) = [ + ( )]
then the output of the discriminator is
( ) =( )
where is the discriminator sensitivity
For FM
( ) = ( )
so ( ) = ( )
The characteristics of an ideal frequency discriminator are
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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The frequency discriminator also can be used to demodulate PM signals. For PM,
( ) is given by
( ) = ( )
( ) =( )
Integration of the discriminator output yields a signal that is proportional to ( ).
A demodulator for PM can therefore be implemented as an FM demodulator followed
by an integrator.
A simple approximation to the ideal discriminator is an ideal differentiator
followed by an envelope detector.
If the input to the differentiator is
( ) = [ + ( )]
Then
, ( ) = − +( )
[ + ( )]
The signal , ( ) is both amplitude and angle modulated. The envelope of , ( ) is
+( )
Then ( ) =
which depends on the instantaneous frequency of the ( ).
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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Example
Determine the instantaneous frequency in Hz for each of the following signals
(a) 10 (200 + /3)
(b) 10 (20 + )
(c) (200 ) [5 (2 )] + (200 ) [5 (2 )]
Solution
(a) ( ) = 200 + /3
= / = 200
The instantaneous frequency of the signal is (100 Hz), which is constant.
(b) ( ) = 20 +
= = 20 + 2 = 2 (10 + )
The instantaneous frequency of the signal is (10 Hz) at (t=0) and increases linearly at
a rate of 1 Hz/sec.
(c) [200 − 5 (2 )]
( ) = 200 − 5 (2 )
= = 200 − 10 (2 ) = 2 [100 − 5 (2 )]
The instantaneous frequency of the signal is (95 Hz) at (t=0) and oscillates
sinusoidally between 95 and 105 Hz.
Example
Consider an angle modulated signal
( ) = 10 [10 + 5 (2 ∗ 10 )]
Find the maximum phase deviation and maximum frequency deviation.
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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Solution
( ) = + ( )
= 10 + 5 (2 ∗ 10 )
( ) = 5 (2 ∗ 10 )
Thus the maximum phase deviation
| ( )| = 5rad.
́ ( ) = 5(2 ∗ 10 ) (2 ∗ 10 )
and the maximum frequency deviation is
∆ = | ́ ( )| = 5(2 ∗ 10 ) rad/sec.
or ∆ = 5 KHz.
Example
An angle modulated signal is described by
( ) = 10 [2 (10 ) + 0.1 {(10 ) }]
(a) if ( ) as a PM signal with = 10, find ( ).
(b) if ( ) as an FM signal with = 10 , find ( ).
Solution
(a) ( ) = + ( )
= 10 [2 (10 ) + 10 ( )]
= 10 [2 (10 ) + 0.1 {(10 ) }]
Thus ( ) = 0.01 {(10 ) }
(b) ( ) = + ∫ ( )
= 10 [2 (10 ) + 0.1 {(10 ) }]
So 10 ∫ ( ) = 0.1 [(10 ) ]
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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Thus
10 ( ) = {0.1 [(10 ) ]}
= 100 [(10 ) ]
( ) = 10 [(10 ) ]
Example
Find maximum and minimum instantaneous frequencies resulting from
modulating a carrier signal with carrier frequency = 100 MHz by the shown ( ) if
a) FM with = 2 × 10 rad/sec/volt is used.
b) PM with = 2 rad/volt is used.
Solution:
1- for FM
= +12
( ) = 10 + 10 ( )
( ) = 10 + 10 [ ( )] = 99.9 MHz
( ) = 10 + 10 [ ( )] = 100.1 MHz
2- for PM
1
-1
t
( )
2 × 10
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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= +2
∙( )
( ) = 10 + 10( )
= 10 − 10 = 99.9MHz
( ) = 10 + 10( )
= 10 + 10 = 100.1MHz
Example
Find the average power in an angle modulated signal with sinusoidal modulation.
Solution
An angle modulated signal with a single tone modulation can be expressed as
( ) = ( ) [( + ) ]
The average power is given by
=12
( ) =12
( ) =12
Example
In a single tone angle modulation, the modulated signal ( ) is
( ) = [ + ( )]
when ≪ 1, we have NB angle modulation. Find the spectrum of this NB angle
modulated signal.
Solution
( ) = [ + ( )]
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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= ( ) [ ( )] − ( ) [ ( )]
when ≪ 1, we can write
[ ( )] ≈ 1
[ ( )] ≈ ( )
Then the NB signal can be approximated by
( ) = ( ) − ( ) ( )
= ( ) −2
[( − ) ] +2
[( + ) ]
We can see that the spectrum of ( ) consists of a carrier line and a pair of side lines
at ± .
Example
Given the angle modulated signal
( ) = 10 [2 (10 ) + 200 {2 (10 ) }]
what is its bandwidth ?
Solution
The instantaneous frequency is
= 2 (10 ) − 4 (10 ) {2 (10 )}
So ∆ = 4 (10 ), = 2 (10 ), and
=∆
=4 (10 )2 (10 )
= 200
= 2( + 1) = 8.04 (10 )rad/sec
Since ≫ 1
≈ 2∆ = 8 (10 )rad/sec
or = 400
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
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Example
Consider an angle modulated signal
( ) = 10 [ + 3 ( )]
= 1 . Calculate the modulation index and find the bandwidth when
(a) is doubled, and
(b) is decreased by one-half.
Solution
1) If ( ) is a PM signal then
( ) = + ( )
= 10 [ + 3 ( )]
⇒ = = 3
We see that the value of is independent of . When = 1
= 2( + 1) = 8
(a) When is doubled, = 3, = 2 , and
= 2(3 + 1)(2) = 16
(b) When is decreased by one-half, = 3, = 0.5 , and
= 2(3 + 1)(0.5) = 4
2) If ( ) is an FM signal then
( ) = + ( )
= 10 [ + 3 ( )]
⇒ = = 3
We see that the value of is inversely proportional to . Thus when = 1
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Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri
20
= 2(3 + 1)(1) = 8
(a) When is doubled, = 3/2, = 2 , and
= 2(3/2 + 1)(2) = 10
(b) When is decreased by one-half, = 6, = 0.5 , and
= 2(6 + 1)(0.5) = 7
Example
A block diagram of an indirect (Armstrong) FM transmitter is as shown
Compute the maximum frequency deviation ∆ of the output of the FM transmitter and
the carrier frequency if = 200 , = 10.8 , ∆ = 25 , = 64, and
= 48.
Solution
∆ = (∆ )( )( ) = (25)(64)(48) = 76.8
= = (64)(200)(10 ) = 12.8
= ± = 12.8 ± 10.8 = 23.62
= = (48)(23.6) = 1132.8(48)(2) = 96