lecture23-24_anglemodulation

8/8/2019 Lecture23-24_AngleModulation

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c(t)

Ac

Angle Modulation

In this type of modulation, the frequency or phase of

carrier is

varied in proportion to the amplitude of the modulating

signal.



i(t) = wct + kpm(t)

where wc = 2irf c.

2. Frequency Modulation:

w i ( t ) = w c +k f m ( t )



kf Am

here, is called frequency deviation (Af ) and

A f

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

Here, k p is phase sensitivity or phase modulationindex.

• Frequency Modulation If m(t) = Am cos(2irf mt) is themessage

signal, then the Frequency modulated signal is given

by

2 7rf i(t) = wc + kf Am cos(2irf mt)

kf Am

9i ( t ) = w c t + s i n ( 2 i r f m t )2 'itf m

is called271 f m

modulation index (3). The Frequency modulated signal isgiven by



s (t)= A

cos

(2irf

t +

/3

sin(

27If

mt)

)

Dep



end

ingon

how

sma

ll /3

isFM

is

eith

er

Narrow



ban

d

FM

(/3

<<

1)or

Wi

de

ba

nd

FM



(/3

1).

–Nar

row

-Ban

d

FM

(NB

FM)

In

N



B

F

M

/3

<

<

1,

th

er

ef

ors(



t)

re

d

u

c

e

s

a

s

fo

llo



w

s:

s (t)

= A

cos

(2ir

f t

+ /

3sin(



2 irf

mt)

)

=

A

c

o

s(



2

i

r

f

t)

c

o

s



f

m

t

)

)

-

A

sin(

2irf t)



sin(

/3

sin(

2irf

mt)

)

Sinc

e, /

3 is

very



sma

ll,

the

abo

ve

equatio

n

red

uce

s to



s (t)

= A

cos

(2ir

f t)

-A /3

sin(

2irf

mt)sin(



2irf

t)



+

t/2Phase shifter

-Asin( ? t)c

A cos( ? t)c

NBFM signalDSB-SC

m(t)

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 ??.

oscillator

Figure 2: Generation of NBFM signal

– 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 aWBFM 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 canbe approximated by sequence of pulses as shown in

Figure

??



m p

T t

F

i

g

u

r

e



3

:

A

p

p

r

ox

i

m

a

ti

o

n

o

f



m

e

s

s

a

g

es

i

g

n

a

l

If

ton

e



mo

dul

ation is

con

side

red,

and

the

pea

k

am

plit

ude

of

th

e

sin



us

oid

ismp

,

th

e

mini

mu

m

an

dma

xi

mu

m

fre



qu

en

cydeviationswillbew -kf

m p

andw+kf

m p

re



spec

tively.

Thespr

eadof pulsesinfrequency

doma



inwil

lbe2

T =471B



as shown in Figure??

Co - km Co + kmc f pf 4irB

Figure 4: Bandwidth calculation of WBFM signal

Therefore, total BW is 2kfm p + 8irB and if

frequency

deviation is considered

1

BW f m = 2 (2k f mp +87rB)



B W f m = 2 ( L f + 2 B )



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 Lf is very small compared to

B. This implies

B fm 4B

But the bandwidth for NBFM is the same as that of

AM

which is 2BA better bandwidth estimate is therefore:

B W f m = 2 ( l X f + B )



B W f m = 2 ( k f m p

271 + B)

This is also called Carson’s Rule



t

t

– Demodulation of FM signals

Let 4fm(t) be an FM signal.

[ ]

f m ( t ) = Acos (w c t + k f m ( o ) d o)0

This signal is passed through a differentiator to get

( )

˙ f m ( t ) = A (wc + k f m ( t ) ) s in wc t + k f m (o ) d o0

If we observe the above equation carefully, it isbothamplitude and frequency modulated.

Hence, to recover the original signal back an

envelope

detector can be used. The envelope takes the



c p

form (see

Figure ??):



Envelope = A (wc + kf m(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??



tS

t A c + k m td/dt

Figure 6: Demodulation of an FM signal

• The analysis for Phase Modulation is identical.

– Analysis of bandwidth in PM



i = w c

+ k p m ' ( t )

m' p =

L w

=k p mp

B

W

p m

=

2 (

l X

f +



B )

B W p m = 2 (kpm' p +

– The

differencebetween FM

and PM is

that the

bandwidth

is independent of signal bandwidth in

FM while it is

strongly

dependent

on signal



bandwidth

in PM. a

aowing to thebandwidth being

dependent on thepeak of the

derivative of m(t) rather than

m(t) itself



Angle Modulation: An Example

• An angle-modulated signal with carrier frequency c = 271 × 106 is described by the equation:

cbEM (t) = 12 cos(wct + 5 sin 1500t + 10 sin

200071t)

1.Determine the power of the modulating signal.

2.What is Lf ?

3.What is,@?

4.Determine Lçb, the phase deviation.5. Estimate the bandwidth of çbEM(t)?

1.P = 122/2 = 72 units

0.Frequency deviation Lf , we need to estimate the

instantaneous frequency:



d

wi= dt

9(t)

=

wc

+ 7,

500

cos

150

0t

+

20,

000

71t

T



h

e

dev

ia

ti

o

nof

th

e

ca

rri

er

is

L

w



=

7,

5

0

0

co

s1

5

0

0t

+

2

0,

0

0



0i

rt.

W

h

e

n

the

tw

o

si

n

us

oi

ds

a



d

d

inp

h

as

e,

the

m

ax

i

mu

m

va

lu

e



wi

ll

be

7,

5

0

0+

2

0,

0

00

Hen

ceLf



=

2ii-= 11 ,193.66Hz3.

= L

f

B

=

11

,

1

9

3

.

6



61000 =11.193

4.

T

h

e

a

n

g

l

e

9

(



t

)

=

w

c

t

+

5

s

i

n



1

5

0

0

t

+

1

0

s

i

n



2

0

0

0

i

r

t)

.

T

h

e

m

ax

i



m

u

ma

n

gl

e

dev

ia

ti

o

n

is

1

5,

w



hi

ch

isth

e

p

h

ase

d

ev

ia

tio

n.0

.

B



E

M

=

2

(

L

f

+

B

)

=



2

4

,

3

8

7.

3

2

H

z

lecture23-24_anglemodulation

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