One gains knowledge from four sources: 1⁄4 from the teacher, 1⁄4 from self-study, 1⁄4 from classmates, and 1⁄4 from gaining experience over time. The document discusses the different sources from which one gains knowledge, with equal portions coming from the teacher, self-study, classmates, and experience over time.

1. One gains ¼ of the knowledge from the Acharya (the teacher), ¼ from
his own self-study and intellect, ¼ from his classmates and the
remaining ¼ is gained as a person becomes matured as time passes.

2. COMMUNICATION ENGINEERING
Mr.R.VIJAY ANANDH

3. Instructions to all Participants
➢ Kindly mute Audio & turnoff Video while entering and till the
session end. Don’t try to record Video.
➢ Get your pen, notebook and calculator along with you.
➢ https://www.slideshare.net/rmkrva/communication-engineering-class-1
➢ https://www.slideshare.net/rmkrva/communication-engineering-class-2
➢ https://www.slideshare.net/rmkrva/communication-engineering-class3.
https://www.slideshare.net/rmkrva/communication-engineering-class-4
➢ Subject clarifications will be given wherever necessary and
during Q&A sessions also.

4. How the classes would be:
➢ Discussion on concepts with Block diagram & Circuit
diagram .
➢ Classes will be interactive with Brain storming
activities, demo , quiz ,small derivations & calculations
➢ Interesting facts about Basic communication
technology.
➢ Group activities off line /online, Assignments will be
given.

5. H
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B
O
A
R
D
H
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N
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B
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A
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D
SASANK REDDY J POOJITH ITHA SIVARAM V
VAMSI KRISHNA SAMPATH
REDDY M
NANDHINI S MOHAMED YAHYA R DHEERAJ KUMAR S V S ANISH MATHEW OOMMEN P
PREETHI S KOLA BHAVANI SANKAR NIVETHITHA A S CHARAN KUMAR M V
SOWMIYA E GOGINENI VENKATA NIKHIL REINITA THOMAS PRAVEEN KUMAR.S
SWATHI N SHAMITHA R V VISHAL G BRINDHA C
KAVANURU JAGAN CHILAKA DEVI SRI PRASAD SOWNDARYA S
APPASAMUDRAM
NAGAPRATHIBHA
DHEERAJ H POOJITHA K SOMU SAMPATH KUMAR REDDY KIRUTHIKA K
MANJU PARKAVI G ZUHAYR A PREETHI D KESHAV S
HARINI G ABBURI VAMSI KRISHNA HARITHA S VIDHYAVARSHINI D
JALADI VINEETH SANDIREDDY BHAVANA VUNNAM MAHITHA SANDEEP KUMAR K V
APARNA T SANGARAJU DHARANI DIVYA RANI R KARTHIK BALAJI R
CHINNI SAI ABHIJITH PRIYADHARSHINI E M KRISHNA KUMAR K GOKUL RAJ K
SHAKTHI A S
RAMAIAHGARI CHENGANNA
BHANUSREE
SOWMYA V PRAJNA A
KEERTHIPATI VYSHNAVI MADHAV K SAI SIDDHARDHA NARISETTY JAISNAVI D
NOTI PARAMESWARI YASHWANTH N MUPPALA MONA SREE KOTA RAGHAVA JAYA SAIRAKESH
RAGAVI V S NIRANJAN JP MONIKA M HARSHINI S
LANKAMSETTY PREMDEEP NANDHINI R JAISREE K G BALIREDDY BHANUSREE
KAVIN RAJ R S BALAKRISHNA GAGANSAI MANDADI SRIHITHA

6. EC8395-COMMUNICATION
ENGINEERING
1 ANALOG MODULATION
2 PULSE MODULATION
3 DIGITAL MODULATION AND TRANSMISSION
4 INFORMATION THEORY AND CODING
5 SPREAD SPECTRUM AND MULTIPLE ACCESS

7. Classification of Modulation System

9. 5
• Single sideband with suppressed
carrier.
• Frequency Spectrum – To understand
about Bandwidth.
• Power Relations
• Efficiency – Power saving
• Summary
4
3
2
1
SESSION TOPICS

10. Important Formulae to Solve AM-DSB-FC problems
ma =
Vm
Vc
% modulation index, = max100
=(
Vm
Vc
)x100
Vmax = Vc+ Vm
Vmin = Vc - Vm
ma =
(Vmax − Vmin)
(Vmax+Vmin )
BW = 2fm
BW = f2 - f1
PC=
Vc
2
2𝑅
PLSB= PUSB =
ma
2
𝟒
PC
PT= PC +PLSB+ PUSB
PT = Pc[ 1 +
ma
2
𝟐
]
VAM (t) = Vc [1+ maCos 2πfmt] Cos 2πfCt
VAM (t) = Vc Cos ωC t +
maVc
𝟐
Cos(ωC + ωm)t +
maVc
𝟐
Cos(ωC - ωm)t
➢ AM Double sideband with FULL
carrier.

14. • For an AM DSBFC wave with a peak unmodulated
carrier voltage Vc = 10 V, a load resistor of RL = 10 
and modulation Co-efficient ma = 1, determine
a) Powers of the carrier and the upper and lower sidebands.
b) Total sideband power.
c) Total power of the modulated wave.
d) Draw the power spectrum.

15. Given,
➢ carrier voltage Vc = 10 V,
➢ Load resistor of RL = 10 
➢ Modulation Co-efficient ma = 1.
PC=
Vc
2
2𝑅
=
102
2 (10)
=
100
2 (10)
= 5𝑊
PLSB= PUSB =
ma
2
𝟒
PC =
12
𝟒
(5)= 1.25W
PT = Pc[ 1 +
ma
2
𝟐
] = 5[ 1 +
12
𝟐
] =5[ 1 + 0.5 ] =5[ 1.5 ] = 7.5 W
Total Side band Power =PLSB+ PUSB = 1.25 W+ 1.25 W = 2.5 W
(a)
(b)
(c)

16. Power spectrum
fc
fLSB
1.25 W
fUSB
5W
1.25 W
Frequency (Hz)
Power (W)
(d)

17. Observations:
➢ AM Double sideband with FULL carrier.
➢The sum and difference frequencies are present.
➢Only 33.3% efficiency is achieved. Because, two-thirds of the power is being
wasted in the carrier, which carries no information.
➢Power consumed by LSB and USB are same.
➢Duplication of Amplitude occur in LSB or USB
Target:
➢ How to improve efficiency?
Answer: To overcome the drawback, unmodulated carrier term can be
eliminated.
VAM (t) = Vc Cos ωC t +
maVc
𝟐
Cos(ωC + ωm)t +
maVc
𝟐
Cos(ωC - ωm)t

18. Double sideband Suppressed carrier
A
B
A→ “message signal crosses Zero”
B → “Ouput signal undergoes phase reversal”
Product Modulator

19. VDBS-SC (t) =
= Vm Cos ωmt . VC Cos ωCt
=
VmVc
𝟐
[Cos(ωC + ωm)t + Cos(ωC - ωm)t]
Note:
Frequency Spectrum of an DSB-SC wave:
BW=(fc+fm)−(fc −fm)
⇒BW=2fm

20. Power Relations in an DSB-SC wave
Note:
Carrier Power: PC=
Vc
2
(rms)
𝑅
Lower side bandUpper side band
VDSB-SC (t) =
VmVc
𝟐
Cos(ωC + ωm)t +
VmVc
𝟐
Cos(ωC - ωm)t
PC=
(Vc/√𝟐) 𝟐
𝑅
=
Vc
2
2𝑅
Power in the Sidebands:
PLSB= PUSB =
Vc
2
(rms)
𝑅
=
(maVc /2
√𝟐
) 𝟐
𝑅
=
ma
2Vc
2
𝟖𝑹
=
ma
2
𝟒
Vc
2
𝟐𝑹
=
ma
2
𝟒
PC

21. If the carrier is suppressed,then the total power transmitted in
DSB-SC AM wave is,
P'T= PLSB+ PUSB
=
ma
2
𝟒
Vc
2
2𝑅
+
ma
2
𝟒
Vc
2
2𝑅
=
ma
2
𝟒
Pc+
ma
2
𝟒
Pc
= Pc [ ma
2
𝟒
+
ma
2
𝟒
]
P'T = Pc[ma
2
𝟐 ]
Note: During transmission, the
carrier term is suppressed.

22. Power savings in DSB-SC wave is calculated as follows:
PDSB-SC =
PT −P′T
PT
=
Pc[ 𝟏+
ma
2
𝟐
] − Pc[ ma
2
𝟐
]
Pc[ 𝟏+
ma
2
𝟐
]
=
𝟏+
ma
2
𝟐
− ma
2
𝟐
𝟏+
ma
2
𝟐
=
𝟏
𝟏+
ma
2
𝟐
=
2
𝟐+ma
2
➢Therefore, the percentage of Power saving is given by,
=
2
𝟐+ma
2
x100
➢If modulation index , ma = 1,for 100% modulation, then the power
saving is given is calculated as, =
2
3
x 100 = 66.7 %

23. Important Formulae to Solve AM-DSB-SC problems
ma =
Vm
Vc
% modulation index, = max100
=(
Vm
Vc
)x100
BW = 2fm
PC=
Vc
2
2𝑅
PLSB= PUSB =
ma
2
𝟒
PC
P'T = Pc[ma
2
𝟐
]
The total power transmitted in DSB-SC AM wave, P'T = Pc[ma
2
𝟐 ]
VDSB-SC (t) =
VmVc
𝟐
Cos(ωC + ωm)t +
VmVc
𝟐
Cos(ωC - ωm)t
➢The percentage of Power saving is given as =
2
𝟐+ma
2
x100

24. • For an AM DSB-SC wave with a peak unmodulated
carrier voltage Vc = 10 V, a load resistor of RL = 10 ,
frequency of modulating signal of 10kHz and modulation
Co-efficient ma = 1, determine
a) Powers of the carrier and the upper and lower sidebands.
b) Total power of the modulated wave.
c) Bandwidth of the transmitted wave.

25. Given,
➢ carrier voltage Vc = 10 V,
➢ Load resistor of RL = 10 
➢ Modulating Signal fm = 10kHz
➢ Modulation Co-efficient ma = 1.
PC=
Vc
2
2𝑅
=
102
2 (10)
=
100
2 (10)
= 5𝑊
P'T = Pc[
ma
2
𝟐
] = 5[
12
𝟐
] =5[ 0.5 ] =5[ 0.5 ] = 2.5 W
(a)
(b)
(c) Bandwidth = 2fm = 2(10kHz) = 20 kHz
Power in Side bands, PLSB= PUSB =
ma
2
𝟒
PC =
12
𝟒
(5)= 1.25W

26. Observations:
➢ Double sideband with Suppressed carrier.
➢The sum and difference frequencies are present.
➢Only 66.7% of saving of power is achieved.
➢Power consumed by LSB and USB are same.
➢The same information is transmitted twice . One in USB and another in LSB.
Target:
➢ How to improve Power Saving?
Answer: Eliminate one side band (USB or LSB) in addition to carrier .
Hence,one side band is enough for transmission as well as
recovering the useful information.
VDBS-SC (t) =
VmVc
𝟐
[Cos(ωC + ωm)t + Cos(ωC - ωm)t]
DSB-SC AM wave is given by,

27. Single side band with Suppressed Carrier
DSB-SC

28. • A SSB modulated wave s(t) is generated using a carrier of frequency ωc and a sinusoidal
modulating wave of frequency ωm . The carrier amplitude is Ac and that of the
modulating wave is Am . Draw a block diagram of a system for generating SSB
modulated wave s′(t) . Define s′(t) assuming that (a) only the upper side-frequency is
transmitted. (b) the lower side-frequency is transmitted.

29. Am = Amplitude of message signal
Ac = Amplitude of carrier signal

30. Frequency Spectrum
BW=(fc+fm)−(fc )
⇒BW= fm
The Bandwidth of SSB-SC is fm ,
which is same as the bandwidth
of the message signal .

31. Total Power saved in SSB-SC is calculated as follows:
Power in SSB-SC is given by,
P'T= PUSB (or) PLSB
P'T =
ma
2
𝟒
Vc
2
2𝑅
=
ma
2
𝟒
Pc
PSSB-SC =
PT − P′T
PT
=
Pc[ 𝟏+
ma
2
𝟐
] − Pc[ ma
2
𝟒
]
Pc[ 𝟏+
ma
2
𝟐
]
=
𝟏+
ma
2
𝟐
− ma
2
𝟒
𝟏+
ma
2
𝟐
=
𝟏+
ma
2
𝟒
𝟏+
ma
2
𝟐
=
𝟒+ma
2
𝟐( 𝟐+ma
2)
=
𝟒+ma
2
𝟒+𝟐ma
2
➢If modulation index , ma = 1,for 100% modulation, then the power saving
is given is calculated as, =
5
6
x 100 = 83.33 %

32. Important Formulae to Solve AM-SSB-SC problems
ma =
Vm
Vc
% modulation index, = max100
=(
Vm
Vc
)x100
BW = fm
PC=
Vc
2
2𝑅
PLSB= PUSB =
ma
2
𝟒
PC
P'T = Pc[ma
2
𝟒
]
The total power transmitted in SSB-SC AM wave, P'T = Pc[
ma
2
𝟒
]
VSSB-SC (t) =
VmVc
𝟐
Cos(ωC + ωm)t OR
VmVc
𝟐
Cos(ωC - ωm)t
➢The percentage of Power saving is given as ==
𝟒+ma
2
𝟒+𝟐ma
2
x100

33. • For an AM SSB-SC wave with a peak unmodulated
carrier voltage Vc = 10 V, a load resistor of RL = 10 ,
frequency of modulating signal of 10kHz and modulation
Co-efficient ma = 1, determine
a) Powers of the carrier and the upper and lower sidebands.
b) Total power of the modulated wave.
c) Bandwidth of the transmitted wave.

34. Given,
➢ carrier voltage Vc = 10 V,
➢ Load resistor of RL = 10 
➢ Modulating Signal fm = 10kHz
➢ Modulation Co-efficient ma = 1.
PC=
Vc
2
2𝑅
=
102
2 (10)
=
100
2 (10)
= 5𝑊
P'T = Pc[
ma
2
𝟒
] = 5[
12
𝟒
] =5[ 0.25 ] = 1.25 W
(a)
(b)
(c) Bandwidth = fm = 10kHz
Power in Side band, PLSB or PUSB =
ma
2
𝟒
PC =
12
𝟒
(5)= 1.25W

35. Points to Discuss:
Advantage of the SSB-SC modulation:
➢ It reduces the bandwidth requirement to half as compared to
DSB-SC modulation.
Disadvantage :
➢SSB-SC are relatively difficult to generate due to the
difficulty in isolating the desired sideband (USB or LSB).
➢ The required filter must have a sharp cut-off characteristic,
particularly when the baseband signal contains extremely
low-frequencies ( Eg. Television and Telegraphic signals).

36. Can you identify the Power spectrum ?
(a) (b) (c)
(d) (a) DSB-FC
(b) DSB-SC
(c) SSB-SC
(d) Vestigial Side band

37. Vestigial Side Band
➢This difficulty in SSB-SC is overcomed by a scheme known as “Vestigial
Side Band” modulation.
➢It is a comprise between SSB-SC and DSB-SC modulation.
➢Instead of removing a side band, a trace or portion or vestige of that
sideband is transmitted along with other sideband is transmitted.
➢ ie., Vestige of LSB + most part of USB
or
Vestige of USB + most part of LSB
(d)

38. Vestigial Side Band

39. Vestigial Side Band
➢This difficulty in SSB-SC is overcomed by a scheme known as “Vestigial
Side Band” modulation,which is a comprise between SSB-SC and DSB-
SC modulation.
The VSB modulated wave in time domain is given by,
S(t) =
Vc
𝟐
𝒎 𝒕 .Cos(2πfC t) ±
Vc
𝟐
𝒎′ 𝒕 .Sin(2πfC t),
Where the plus sign corresponds to the transmission of
a vestige of the upper side band and minus sign
corresponds to the transmission of a vestige of the
Lower side band.

40. Problems to practice
(a)

41. (b)
Problems to practice

42. (c)
Problems to practice

43. ASSIGNMENT QUESTIONS
(a)
(b)

44. • Blooms Taxonomy
• Blooms Taxonomy

45. Next Topic Modulator and Demodulators