Angle mod for stdnts 25 apr

Amity School of Engineering & Technology

B.Tech. , IV Semester

Communication Systems

Course Code: BTC 404/BTE 402

Credit Units: 04

Pawan Kumar Bansal

Angle Modulation • Frequency or Phase of carrier is varied in proportion to the

amplitude of modulating signal.

• It has two types

• Phase Modulation:

The phase angle of the carrier signal is varied linearly with the

message (modulating) signal.

• Frequency Modulation :

The frequency of the carrier signal is varied linearly with

the integral of the message (modulating) signal.

• PM and FM are closely related, any variation in frequency will result

in change in phase and vice versa.

Instantaneous Frequency

Instantaneous Angle

PM & FM

Relation Between FM and PM

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-1

FM message

FM: Frequency Deviation

FM: Modulation Index

FM wave equation s(t)

• Q- A single tone FM is

represented by voltage equation as

12cos (6*108t+5sin1250t), determine

following-

• a-fc

• b-fm

• c-modulation index

• d- Max deviation

Solution • Compare with standard equation of FM

• fc =6*108 /2*3.14 =95.5MHz

• fm=1250/2*3.14 =199Hz

• Beta=5

• Deviation=Beta*fm=995 Hz

Numerical Problem?

• A single tone modulating signal

frequency modulates a carrier

find Freqency Deviation?

tCosEtf mm )(

tACostC c)(

mfensitivityFrequencyS

tCosEK

Problem

Solution

Problem

Solution

Problem

Solution

FM types

NarrowBand FM

NBFM: Generation

Phasor Representation(1)

Phasor Representation(2)

)Re( ExpjCos

Bessel Function

Observations

Carson's rule• Carson's rule states that nearly all (~98

percent) of the power of a frequency-

modulated signal lies within a bandwidth

• where , as defined above, is the peak

deviation of the instantaneous frequency

from the center carrier frequency

• This rule gives correct result if modulation

index is greater than six. 39

)/11(2)(2 fffB mT

Problem

• Use Carson’s rule to compare the

bandwidth that would be required to

transmit a baseband signal with a

frequency range from 200Hz to 3KHz

• 1) NBFM with maximum deviation of 4KHz

• 2) WBFM with maximum deviation of

Solution

• For NBFM=2(4+3) =14KHz

• For WBFM=2(73+3) =152KHz

Problem

Solution

FM Generation

• Method types -Direct and indirect

• In Direct method the fc is varied as per

m(t) which is done using VCO(voltage

controlled oscillator.

• In Indirect method m(t) is first used to

produce NBFM signal and frequency

multiplication is next used to increase the

frequency deviation of desired signal.

FM Generation: Direct Method

FM Generation:Indirect Method

Frequency Multiplier

Pre-emphasis and De-emphasis

• In FM Noise has greater effect on higher

modulating frequencies. So if we boost

the level of higher modulating frequency

artificially then we can improve the signal

to Noise ratio (S/N).

• This artificial boosting at transmitter side is

called Pre-emphasis, and reverse

phenomena at Receiver is called De-

emphasis.

• It will help us in better FM reception. 48

Pre-emphasis and De-emphasis

Pre-emphasis• Pre-emphasis refers to boosting the relative amplitudes of the

modulating voltage for higher audio frequencies from 2 to

approximately 15 KHz.

De-emphasis• De-emphasis means attenuating those frequencies by the

amount by which they are boosted.

However pre-emphasis is done at the transmitter and the de-

emphasis is done in the receiver.

The purpose is to improve the signal-to-noise ratio for FM reception.

A time constant of 75µs is specified in the RC or L/Z network for pre-

emphasis and de-emphasis.

Pre-emphasis circuit• At the transmitter, the modulating signal is passed through a simple

network which amplifies the high frequency, components more

than the low-frequency components. The simplest form of such a

circuit is a simple high pass filter of the type shown in fig (a).

Specification dictate a time constant of 75 microseconds (µs)

where t = RC. Any combination of resistor and capacitor (or

resistor and inductor) giving this time constant will be satisfactory.

Such a circuit has a cutoff frequency fco of 2122 Hz. This means

that frequencies higher than 2122 Hz will be linearly enhanced.

The output amplitude increases with frequency at a rate of 6 dB per

octave. (This means that as the frequency doubles, the amplitude

increases 6 dB). The pre-emphasis curve is shown in Fig (b).

This pre-emphasis circuit increases the energy content of the

higher-frequency signals so that they will tend to become stronger

than the high frequency noise components. This improves the

signal to noise ratio and increases intelligibility and fidelity (loyalty).

The pre-emphasis circuit also has an

upper break frequency fu where the

signal enhancement flattens out.

It is usually set at some very high

value beyond the audio range.

An fu of greater than 30KHz is typical.

De-emphasis• To return the frequency response to its normal level, a

de-emphasis circuit is used at the receiver. This is a

simple low-pass filter with a constant of 75 microseconds

(µs) . See figure (c) below.

• It features a cutoff of 2122 Hz and causes signals above this

frequency to be attenuated at the rate of 6bB per octave. The

response curve is shown in Fig (d). As a result, the pre-emphasis

at the transmitter is exactly offset by the de-emphasis circuit in the

receiver, providing a normal frequency response. The combined

effect of pre-emphasis and de-emphasis is to increase the high-

frequency components during transmission so that they will be

stronger and not masked by noise.

Combined Frequency Response

Problem

• A 100 MHz carrier wave has a peak

voltage of 5 volts. the carrier is frequency

modulated by a sinusoidal modulating

signal of frequency 2 KHz such that

frequency deviation is 75 KHz. The

modulated waveform passes through zero

and is increasing at t=0. Determine

expression for modulated carrier

waveform.

Solution

Problem

Solution

Problem

FM Demodulation

FM Demodulation: Block Diagram

FM Demodulation: Circuit Diagram

Thank You

Angle mod for stdnts 25 apr

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