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L#3: Amplitude Modulation

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26 views17 pages

L#3: Amplitude Modulation

Uploaded by

kumar.ankit.nita
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© © All Rights Reserved
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L#3: Amplitude Modulation

Dr. Jayanta K Rakshit


Conventional Amplitude Modulation (Full AM)
AM is the process of varying the instantaneous amplitude
of carrier signal accordingly with instantaneous amplitude
of message signal.
• If sm(t) is the message signal and
carrier signal, then AM signal is:
S (t )  Ac . cos(ct )  sm (t ). cos(ct )
Conventional Amplitude Modulation (Full AM)
Conventional Amplitude Modulation (Full AM)
Modulating signal (information signal) can also be expressed as:

The amplitude-modulated wave can be expressed as

By substitution
Therefore The full AM signal may be written as

Where,

- modulation index
Using,
Frequency spectrum of Amplitude Modulation
Fourier transform of a cosine signal cos (ωct) consists of
two impulses at ωc and –ωc as

cos(ct )   [ (  c )   (  c )]
So, Ac cos(ct )  Ac [ (  c )   (  c )]
In general, AM wave :

S (t )  Ac . cos(ct )  sm (t ). cos(ct )
sm (t )  Sm ( )

 1 jct
sm (t ). cos(ct )  sm (t )  e  e 
 jct 

2

sm (t ). cos(c t )  S m (  c )  S m (  c )
1
2
Frequency spectrum of Amplitude Modulation

So, the Fourier transform of AM wave is:

S ( )  S m (  c )  S m (  c )  Ac [ (  c )   (  c )
1
2

Considering baseband signal as,

mAc
S ( )  Ac [ (  c )   (  c )   {  (c  m )}   {  (c  m )} 
2
mAc
 {  (c  m )}   {  (c  m )}
2
Frequency spectrum of Amplitude Modulation
S(ω)

πAc

mπAc/2 BW=2fm
LSB
USB

-(ωc+ ωm) ω= -ωc -(ωc- ωm) ω=0 ωc- ωm ω= ωc ωc+ ωm

BW
• Frequency spectrum of AM comprises of:
• Carrier frequency ωc .
• A lower side band whose highest frequency component is present at ωc-ωm
• An upper side band whose highest frequency component is present at ωc+ωm

Because of the two side bands in the frequency spectrum it is often called Double
Sideband -full carrier(DSB-FC)
The information in the base band (information) signal is duplicated in the LSB and USB and
the carrier conveys no information.
Modulation index or percentage of modulation

• m is merely defined as a parameter, which determines


the amount of modulation.
• What is the degree of modulation required to establish a
desirable AM communication link?
Answer is to maintain m<1.0 (m<100%).

• This is important for successful retrieval of the original


transmitted information at the receiver end.

Modulation index (m) is defined as: Am


m
Ac
Modulation index

Amax  Amin
Am 
2
Amax  Amin Amax  Amin
Ac  Amax  Am  Amax  
2 2

Amax  Amin
m
Amax  Amin
Effect of Modulation Index

m=0.5, called under modulation

m=1.0, called 100% modulation

If the amplitude of the


modulating signal is
higher than the carrier
m>1, amplitude, severe
Called over distortion to the
modulation modulated signal will
occurs.
Power distribution in AM

2
Acar ( Ac / 2 ) 2 Ac2
Pc   
R R 2R 2
ASB (mAc / 2 2 ) 2 m 2 Ac2
PLSB / USB   
R R 8R
Power distribution in AM

Now, the information are


contained in AM sidebands only.
Thus the fraction of total power
is used to transmit the
information.

 m12 m22 m32 


For multi tone modulation: Pt  Pc 1          
 2 2 2 

The percentage of total power carried by the sidebands is called the transmission efficiency.

For single tone modulation


Observations
1. If m=0 Pt=Pc no modulation occurs

2. As m PSB ( m P )
2
c , But Pc remain same as it is
4
independent of m

3. If m=1, Pt=1.5Pc and efficiency will be 33.33 %

4. Over-modulation, i.e. Am>Ac , should be avoided


because it will create distortions.

In terms of power efficiency, for m=1 modulation, only 33% power


efficiency is achieved which tells us that only one-third of the
transmitted power carries the useful information.
Numerical Problems
P1. An unmodulated AM transmitted power=100W. Find
AM transmitted power with 100% modulation.
Solution:
PC=100 W, m=1, Pt=?

Pt=(3/2)*100=150

Pt=150 W
Numerical Problems
P2. For an AM signals, total side band power=200 W with
50 % of modulation. Find the total transmitted AM power.

Solution:
m=0.5
Pt=PC+PSB

=0.2

200
  0.2
PT
Pt=1000 W
THANK YOU

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