angle mod for stdnts 25 apr
TRANSCRIPT
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Amity School of Engineering & Technology
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Amity School of Engineering & Technology
B.Tech. , IV Semester
Communication Systems
Course Code: BTC 404/BTE 402
Credit Units: 04
Pawan Kumar Bansal
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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.
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Instantaneous Frequency
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Instantaneous Angle
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PM & FM
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Relation Between FM and PM
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FM
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0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
FM message
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FM
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FM
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FM
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FM: Frequency Deviation
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FM: Modulation Index
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FM wave equation s(t)
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• 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
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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
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Numerical Problem?
• A single tone modulating signal
frequency modulates a carrier
find Freqency Deviation?
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tCosEtf mm )(
tACostC c)(
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mfensitivityFrequencyS
mf
mmf
f
EK
EK
tCosEK
tfK
/
)(*
max
max
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Problem
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Solution
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Problem
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Solution
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Problem
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Solution
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FM types
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NarrowBand FM
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NBFM: Generation
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Phasor Representation(1)
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Phasor Representation(2)
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WBFM
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)Re( ExpjCos
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Bessel Function
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Observations
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Carson's rule• Carson's rule states that nearly all (~98
percent) of the power of a frequency-
modulated signal lies within a bandwidth
of:
• 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
f
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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
using
• 1) NBFM with maximum deviation of 4KHz
• 2) WBFM with maximum deviation of
73KHz
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Solution
• For NBFM=2(4+3) =14KHz
• For WBFM=2(73+3) =152KHz
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Problem
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Solution
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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.
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FM Generation: Direct Method
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FM Generation:Indirect Method
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Frequency Multiplier
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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
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Pre-emphasis and De-emphasis
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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.
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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.
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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.
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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.
•
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• 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.
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Combined Frequency Response
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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.
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Solution
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Problem
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Solution
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Problem
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Problem
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Problem
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FM Demodulation
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FM Demodulation: Block Diagram
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FM Demodulation: Circuit Diagram
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Thank You
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