Lecture 2.pdf communication system UET TAXILA

COMMUNICATION SYSTEMS
EE-312
Course Instructor: Dr. Irfan Arshad
Assistant Professor, EED, UET Taxila
Conventional Amplitude
Modulation

Amplitude Modulation
• Also known as “Linear Modulation”.
• Applications:
- AM radio
- TV video broadcasting (VSB)
- Point to point communications (SSB)

Double Side Band
Suppressed
Carrier
(DSB)
Conventional
AM
Amplitude Modulation
Single side band
(SSB)
Vestigial side
Band
(VSB)

c c
Conventional Amplitude Modulation
Consider a sinusoidal carrier wavec(t)defined by :
c(t) = A cos(2f t)
where: Ac is the carrier amplitude.
fc is the carrier frequency
Let m(t) denotes the message signal (modulating signal).
An amplitude modulated (AM) signal can be describedas :
s(t) = Ac 
1+ kam(t)cos(2fct)
where: ka is the amplitude senstivity of the modulator.

Spectrum of AM Signal
-W 0 W f fc -W fc fc+ W f
-fc -W -fc -fc+ W 0
fc W
S(f)
M(f)
1
BT=2W

Spectrum of AM Signal
f
fc W
BT=2W

Transmission Bandwidth of Conventional AM
AM is Wasteful of Bandwidth

The envelope of s(t) has the same shape as
the baseband signal m(t) provided that two
requirements are satisfied:

(1) The magnitude of kam(t) is less than unity :
kam(t)  1 for all t
As shown in Fig. (b), this condition ensures that
[1 + kam(t)] is always positive.Then
Ac [1+ kam(t)] representsthe envelope.
When the amplitude senstivity of the modulator is
large such that ka m(t)  1, then the carriersignal
becomes over modulated, resulting in carrier
phase reversal whenever [1+ kam(t)] crosses the
zero.
In this case, the modulated signal exhibits envelope
distortion(as in Fig. (c)).

(2)The carrier frequency fc is much greater than
the hightest frequencycomponentW of thesignal
m(t). That is :
fc W
If this condition is not satisfied, the envelope can not
be detected satisfactorily because part of thespectrum
willbe around f = 0 and will be distorted.

Conventional AM: Basic definitions
2


-ve sign
-
Ac[1+ kam(t)]
Envelope

Conventional AM: Average transmitted power
See Appendix
(At the end of this
Lecture)
Assume m(t) has zero mean
[1+ 
[1+ 2k  m(t)  + 
2
P =
1
 C(t) 2

2 2
k m (t) ]
2
2 2
k m (t) ]
2
[1+ kam(t)] 
2
=
=
=
Ac
2
A
A
c
2
c
2
2
a
a
a
avg
c a
Then C(t) = A 
1+ k m(t)
Since s(t) = Ac 

Conventional AM: Modulation efficiency
Modulation Efficiency:
is the ratio between side band power (that conveys information) to
the total power
=


=
 =
side band power
100
1
2
1
2
1
2
1+ k2
m2
(t)
k2
m2
(t)
Total power
2 2 2
A k m (t)
2
A +
2 2 2
A k m (t)
a
a
c a
c
c a


2
2
2
2
[max
]]
k m(t)
2
1
2
1
2
Ac
[1
2
+ max
kam(t)
]
a
c
[A [1+ max
c a
A [1+ k m(t)]]
PEP
P =
1
max C(t) 2
=
=
=
Conventional AM: Peak Envelope Power
Peak Envelope Power of Conventional AM:
is the power that would be obtained if the envelope attains its
maximum value

Remarks on Transmitted Power of Conventional AM
AM is Wasteful of Power

Conventional AM with single tone modulating signal
max
m(t)− min 
m(t)
2
For sinusiodal modulating signal : m(t) = Am cos(2fmt))
 = ka Am
a m
Am − (−Am )
a
Since Modulation index (factor) =  = ka
= k A
Then  = k
2
0   1
Since ka m(t) 1 i.e ka Am cos(2fmt) 1
Then
s(t) = Ac 
1+ cos(2fmt)cos(2fct)
For
single
tone

m(t)= Am cos(2fmt)
m m
M (f )=
Am
(f − f )+(f + f )
2
MODULATING SIGNAL
MODULATED (AM) SIGNAL
CARRIER UPPER SIDEBNAND LOWER SIDEBNAND
s(t)= Ac 
1+  cos(2fmt)cos(2fct)
s(t)= Ac cos(2fct)+ Ac cos(2fct).cos(2fmt)
+ cos(2(fc − fm )t)
c m
c
c
c  ( (f + f )t)
 A
s(t)= ( f t)+ cos 2
2
A cos 2
m
c
m
c − f ))+(f + (f − f ))
+
Ac
(f − (f
m
c
m
c + f ))+ (f + (f + f ))
+
Ac
(f − (f
c
A
S(f )=  fc )+(f + fc )
(f −
4
4
2

AMPLITUDE MODULATED SIGNAL
MODULATING SIGNAL
Ac
2
= 5 V
Am
= 2.5 V
2
BT = 2 fm
4
= 1.25 V
Ac
Am = 5V
ka = 0.1
Ac = 10V
fc = 200 Hz
fm = 10 Hz

2
2 + 2
2
1+
2
 =
 =
 m
 m
m
2
A2
2
A2
A2
=
m
A2
a
a
1+ k2
 m2
(t) 
k2
 m2
(t) 
max
=
1
3

When  =1
For sinusoidal
modulating signal
 = ka Am

]
2
2
2
2
2
2
2
2
m
] +

c
[1
=
A
= Pavg
m
c
[1
2
c c
2 2
2
2
A2
+
A2
Pavg
A
+  (kam(t)) =
A
A
2
2
2
2
2
2
[1+  ]
c
2
Am ]
c
[1
2
c
[1
2
A
+

+ ka max (m(t))] =
= PEP
Am
PPEP
A
P =
A
 = ka Am

Conventional AM with multi-tone modulating signal
Use this avg in the above power formulas and the efficiency :
m2
cos(2f t))+....
c
m2
a m2
m1
a m1
m2 c
c a m1 m1 m2
m1 m1 m2
c
Then 1 = ka Am1 ; 2 = ka Am2 ,...
= A 
1+ k A cos(2f t))+ k A cos(2f t))+....cos(2f t)
cos(2f t))+....)cos(2f t)
= A 1+k (A cos(2f t))+ A
The modulatedsignal : s(t) = Ac 
Multi - tone modulating signal : m(t) = A cos(2f t))+ A
Proof is not required
]
2
avg
avg
A2 2
P = c
[1+
2
]2
2
avg
PEP
A2
P = c
[1+  2
avg
2
avg
2 + 

 =
avg
 = 2
+ 2
+... such that  1
avg 1 2

Example: Conventional AM

 is the modulation
index
, Am =1

Example: Conventional AM (cont.)
A
m(t) = Am cos(2fmt)
m
a m a
s(t) = Ac 
1+ cos(2fmt)cos(2fct)
= 5[1+ 0.5cos(2 2000t)]cos(2 105
t)
 = k A  k =
 =  = 0.5

+ 4 (0.625)2
14

P =  S( f ) 2
= 2 (2.5)2
−
avg
S( f )
f

2
2
A2
PPEP = c
[1+ ]
2
= c
[1+ ]
2
A2
2
Pavg

Appendix:
Power Calculations
The following formulas are valid for
all types of modulations

Appendix: Power Calculations
Normalized average power: (using complex envelope C(t))
2
P =
1
 C(t)
2

avg
Peak envelope power: (using complex envelope C(t))
Modulation Efficiency:
 =
side band power
100
Total power
2
PEP
P =
1
max C(t) 2

Appendix: Power Calculations
Time average:
For Non-periodic signal:
1
T / 2
−T / 2
T →
 x = lim x(t)dt
T
0
0
1
T0
 x = x(t)dt
T
When x(t) is periodic with period To:
x(t) isa real signal
T0 is the signal period

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