AngleModulation

8/9/2019 AngleModulation

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i(t) = ct + kpm(t)

where c = 2fc.

2. Frequency Modulation:

i(t) = c + kfm(t)

i(t) =

t0

i(t) dt

= 2t0

fi(t) dt +t0

kfm(t) dt

Phase Modulation If m(t) = Am cos(2fmt) is the message

signal, then the phase modulated signal is given by


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s(t) = Ac cos(ct + kpm(t))

Here, kp is phase sensitivity or phase modulation index.

Frequency Modulation If m(t) = Am cos(2fmt) is the messagesignal, then the Frequency modulated signal is given by

2fi(t) = c + kfAm cos(2fmt)

i(t) = ct +kfAm

2fm

sin(2fmt)

here,kfAm

2is called frequency deviation (f) and

f

fmis called

modulation index (). The Frequency modulated signal is

given by


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s(t) = Ac cos(2fct + sin(2fmt))

Depending on how small is FM is either Narrowband

FM(


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The above equation is similar to AM. Hence, for NBFM thebandwidth is same as that of AM i.e.,

2 message bandwidth(2 B).

A NBFM signal is generated as shown in Figure ??.

DSBSC

oscillator

+

m(t)

A cos( t)

Asin( t)

NBFM signal

/2Phase shifter

c

c

Figure 2: Generation of NBFM signal


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Wide-Band FM (WBFM)A WBFM signal has theoritically infinite bandwidth.

Spectrum calculation of WBFM signal is a tedious process.

For, practical applications however the Bandwidth of a

WBFM signal is calculated as follows:Let m(t) be bandlimited to BHz and sampled adequately at

2BHz. If time period T = 1/2B is too small, the signal can

be approximated by sequence of pulses as shown in Figure

??


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t

mp

T

Figure 3: Approximation of message signal

If tone modulation is considered, and the peak amplitude of

the sinusoid is mp, the minimum and maximum frequency

deviations will be c kfmp and c + kfmp respectively.

The spread of pulses in frequency domain will be2T = 4B


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as shown in Figure ??

k m k m +4 B c

pfc f p

Figure 4: Bandwidth calculation of WBFM signal

Therefore, total BW is 2kfmp + 8B and if frequency

deviation is considered

BWfm =1

2(2kfmp + 8B)

BWfm = 2(f + 2B)


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The bandwidth obtained is higher than the actual value.This is due to the staircase approximation of m(t).

The bandwidth needs to be readjusted. For NBFM, kf is

very small an d hence f is very small compared to B.

This implies

Bfm 4B

But the bandwidth for NBFM is the same as that of AMwhich is 2B

A better bandwidth estimate is therefore:

BWfm = 2(f + B)

BWfm = 2(kfmp

2+ B)

This is also called Carsons Rule


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Demodulation of FM signalsLet fm(t) be an FM signal.

fm(t) =

A cos(ct + kft0

m() d)

This signal is passed through a differentiator to get

fm(t) = A (c + kfm(t))sin

ct + kf

t0

m() d

If we observe the above equation carefully, it is both

amplitude and frequency modulated.

Hence, to recover the original signal back an envelope

detector can be used. The envelope takes the form (see

Figure ??):


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Envelope = A (c + kfm(t))

FM signal

Envelope of FM signal

Figure 5: FM signal - both Amplitude and Frequency Modulation

The block diagram of the demodulator is shown in Figure ??


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t( ) ( t)fm fm

.

d/dtDetector

EnvelopeA( + k m(t)) c f

Figure 6: Demodulation of an FM signal

The analysis for Phase Modulation is identical.

Analysis of bandwidth in PM


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i = c + kpm(t)

mp = [m(t)]max

= kpmp

BWpm = 2(f + B)

BWpm = 2(kpm

p

2+ B)

The difference between FM and PM is that the bandwidth

is independent of signal bandwidth in FM while it isstrongly dependent on signal bandwidth in PM. a

aowing to the bandwidth being dependent on the peak of the derivative of

m(t) rather than m(t) itself


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Angle Modulation: An Example An angle-modulated signal with carrier frequency

c = 2 106 is described by the equation:

EM(t) = 12 cos(ct + 5 sin 1500t + 10 sin 2000t)

1. Determine the power of the modulating signal.

2. What is f?

3. What is ?

4. Determine , the phase deviation.

5. Estimate the bandwidth of EM(t)?

1. P = 122/2 = 72 units

2. Frequency deviation f, we need to estimate the

instantaneous frequency:


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i =d

dt(t) = c + 7, 500 cos 1500t + 20, 000t

The deviation of the carrier is

= 7, 500 cos 1500t + 20, 000t. When the two sinusoids

add in phase, the maximum value will be 7, 500 + 20, 000

Hence f =2 = 11, 193.66Hz

3. = fB

= 11,193.661000

= 11.193

4. The angle (t) = ct + 5 sin 1500t + 10 sin 2000t). The

maximum angle deviation is 15, which is the phasedeviation.

5. BEM = 2(f + B) = 24, 387.32 Hz

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