Angle Modulation 1 06/06/22 H62TLC_RA
Sep 04, 2014
Angle Modulation
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Advantages and Disadvantages (over AM)
– Advantages • good fidelity• good noise immunity• More efficient use of power
– Disadvantages• BW requirement is higher for FM
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SNR is good
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• Applications of FM– TV sound signal– 2 way radio (fixed and mobile)– Satellite and cellular communication
• Applications of PM– Extensively used in data communication– Used as an intermediate step in FM signal
generation
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• Frequency Modulation (FM) and Phase Modulation (PM) are grouped as angle modulation.
• Consider the simple sinewave
where gives the instantaneous phase
• A change in frequency can cause a change in phase. Therefore direct FM is indirect PM and direct PM is indirect FM.
• Note also that
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Frequency Modulation (FM)
• For FM, the carrier’s instantaneous frequency deviation from its unmodulated (resting) value varies in proportion to the instantaneous amplitude of the modulating signal.
• Let the modulating signal be a sine waveem(t) =Em sin mt
• The frequency deviation Δf is proportional to the amplitude of the modulating signal, therefore: fFM(t) = fc + f
fFM(t) = fc + kf em(t) where fFM(t) is the frequency of the modulated signal, fc is the unmodulated carrier frequency
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FM – frequency deviation Frequency deviation proportional to the amplitude of the modulating signal.
Then fFM(t) = fc + kf Em sin mt and kf Em = = peak frequency deviation in Hz hence fFM(t) = fc + sin mt
mf ekf
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kf (Hz/V)= deviation sensitivity of the modulator
em = modulating signal amplitude
f = instantaneous frequency deviation
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Square wave modulating a sine
wave carrier
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Modulated signal in time domain
Change in the carrier’s frequency over time according to the amplitude of the modulating signal
Change in the carrier’s phase over time.
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FM – modulation index
• The FM’s modulation index is defined as, mf = /fm
• Unlike AM , mf can be > 1; there are no limits• Since mf = /fm thus
fFM(t) = fc + mf fm sin mt
• Example 1An FM modulator has kf =30kHz/V and operates at a carrier frequency of 175 MHz. Determine the output frequency for an instantaneous value of the modulating signal equal to 150mV.
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Phase modulation • In PM the phase shift of the carrier is proportional to the
instantaneous amplitude of the modulating signal.m(t)= Ec cos (ct + kpEmcos(mt))
• Similar to FM kp = /em
kp - phase deviation sensitivity, rad/V - phase deviation in radiansem – modulating signal amplitude, volts
• And mp = = kpEm
• where mp is the phase modulation index
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Example 2
The signal vm(t) = 2 cos(22000t) is to be transmitted.
(i) If the signal is to be FM, determine the peak frequency deviation and the modulation index given that kf = 5 kHz/V.
(ii) If the signal is to be PM, determine the peak phase deviation given kp = 2.5 rad/V
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Angle Modulation Spectrum
• FM and PM have similar equation of the form:V(t) = A sin (ct + m sin mt)where m can be mf or mp
When elaborated, V(t) is a series of sinusoids:-
where v(t) = instantaneous amplitude of modulated carrier A = peak amplitude of the carrier Jn = solution to the nth order Bessel function for a given
modulation index (see Bessel table handout)
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An FM wave contains an infinite number of sideband componentsSidebands are separated from the carrier by multiples of the fm, with
decreasing amplitude as the distance from the carrier frequency increases.
Sidebands with amplitude <1% of the total signal voltage are usually ignored.
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Rela
tive
ampl
itude
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• If the unmodulated carrier has an amplitude of 1V, J0 represents the amplitude of the carrier frequency (resting frequency).
• Total power in FM signal is distributed in the carrier and sideband components.
• Total signal voltage and power does not change with modulation.
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• For unmodulated carrier =
• For FM (modulated carrier)
• Where PT = total rms power of FM wave Pjn = rms power in carrier
Pjn = rms power in first, second ….nth set of sidebands Vjn = rms voltage of carrier, first, second…..nth sidebands.
RV
RV
RV
RV
PPPPP
n
N
JJJJ
JJJJT
2222 2.......
22
......
210
210
RV
P rmscT
2
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Angle Mod. Bandwidth
• 2 methods(i) With help of the Bessel’s frequency spectrum
B = 2( n x fm)• n is the number of significant sideband components
(ii) Carson’s rule (an approximation – BW for about 98% of total power)
B = 2( + fm)• δ is the peak frequency deviation • fm is the modulating frequency
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Bandwidth
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Example 3An FM modulator with a modulation index of m=1 modulates the message signal vm = 2sin(2000t) and a carrier signal of Vc = 10sin(106t). The load resistance is 50.
i) How many pairs of significant sidebands are transmitted? ii) Determine the sidebands’ amplitude. iii) Sketch the output frequency spectrum, showing their relative
amplitudesiv) Determine the unmodulated carrier power. v) Determine the total power in the modulated wave.
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Narrowband & Wideband FM
• Theoretically there are no limitations to the m or of the FM signal.• In practice, a trade-off found is between SNR and BW. A good SNR
requires large BW. BW are usually limited by regulations and also Rx design limitations
• With m<1, bandwidth of FM = bandwidth of AM. This is narrow-band FM (NBFM). Eg voice transmission.
• With m>10, we get wideband FM – commercial FM radio, television.
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Note:
• All material in this slide has been compiled/sourced/extracted from various sources; electronic and non electronic media for the purpose of teaching.
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