Generation of Angle Modulated Signals · Atheer Alaa Sabri 12 Demodulation of Angle Modulated...

Communication Systems II Second Year Communication Engineering Department Assist. Prof. Dr. Atheer Alaa Sabri

9

=∆

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:


10

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


11

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


12

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


13

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 ( ).


14

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.


15

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 ) ]


16

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


17

= +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

( ) = [ + ( )]


18

= ( ) [ ( )] − ( ) [ ( )]

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


19

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


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

Generation of Angle Modulated Signals · Atheer Alaa Sabri 12 Demodulation of Angle Modulated...

Documents

Transcript of Generation of Angle Modulated Signals · Atheer Alaa Sabri 12 Demodulation of Angle Modulated...