![Important Formulae to Solve AM 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.](https://image.slidesharecdn.com/communicationengineering-class4-200713130209/85/Communication-Engineering-class-4-10-320.jpg)
![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)](https://image.slidesharecdn.com/communicationengineering-class4-200713130209/85/Communication-Engineering-class-4-15-320.jpg)
![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](https://image.slidesharecdn.com/communicationengineering-class4-200713130209/85/Communication-Engineering-class-4-19-320.jpg)
![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 ]](https://image.slidesharecdn.com/communicationengineering-class4-200713130209/85/Communication-Engineering-class-4-21-320.jpg)
![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 %](https://image.slidesharecdn.com/communicationengineering-class4-200713130209/85/Communication-Engineering-class-4-22-320.jpg)
![Observations:
➢ Double sideband with Suppressed carrier.
➢The sum and difference frequencies are present.
➢Only 66.7% efficiency 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: Power saving can increased by eliminating one side band (USB or
LSB) and carrier as well as.
VDBS-SC (t) =
VmVc
𝟐
[Cos(ωC + ωm)t + Cos(ωC - ωm)t]](https://image.slidesharecdn.com/communicationengineering-class4-200713130209/85/Communication-Engineering-class-4-23-320.jpg)
Communication Engineering class 4
- 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.
- 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. ➢ 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. SAI SIDDHARDHA NARISETTY SANDIREDDY BHAVANA DHEERAJ H PANCHATCHARAM S VAMSI KRISHNA SAMPATH REDDY M MONIKA M REINITA THOMAS NIRANJAN JP KOTA RAGHAVA JAYA SAIRAKESH GOKUL RAJ K SURENDAR G CHARAN KUMAR M V ANISH MATHEW OOMMEN P APARNA T KRISHNA KUMAR K SIVARAM V YASHWANTH N KIRUTHIKA K PRAVEEN KUMAR.S MANDADI SRIHITHA SANDEEP KUMAR K V SHAKTHI A S ABBURI VAMSI KRISHNA BALIREDDY BHANUSREE JALADI VINEETH VISHAL G POOJITHA K PREETHI D KESHAV S SASANK REDDY J KOLA BHAVANI SANKAR APPASAMUDRAM NAGAPRATHIBHA HARSHAVARDHAN M PRAJNA A MOHAMMED KAMIL MADHAV K MANJU PARKAVI G DIDDI SOWMYA SEN SANGARAJU DHARANI MUPPALA MONA SREE LINGAMGUNTA UMESH JAYASHREE S D PRIYADHARSHINI E M PREETHI S NANDHINI S POOJITH ITHA SIRASANAMBETI SOMUNATH REDDY JAISNAVI D HARITHA S MOHAMED YAHYA R SOWMYA V KOKA SAI CHAITANYA KARTHIK BALAJI R HARINI G LANKAMSETTY PREMDEEP SOMU SAMPATH KUMAR REDDY BRINDHA C MATCHA CHANDRA KIRAN NUTHALAPATI BHAVYA KAMAL RAJESHWAR S VIDHYAVARSHINI D NOTI PARAMESWARI SARAN M K JAISREE K G KEERTHIPATI VYSHNAVI PRAVIN V P GOGINENI VENKATA NIKHIL MOHAMMED UMAR S DIVYA RANI R GOKUL R NALIANI VIJAYA DURGA DHEERAJ KUMAR S V S HARSHINI S NIVETHITHA A S SHAMITHA R V RAGAVI V S SAI KRISHNA RAJ MOHAN H O N O R B O A R D H O N O R B O A R D
- 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 • Amplitude Modulation- Derivation and modulation index • Frequency Spectrum – To understand about Bandwidth. • Power Relations – To determine how much power is needed for Txn. • Efficiency – To calculate its efficiency is good or not. • Summary 4 3 2 1 SESSION TOPICS
- 10. Important Formulae to Solve AM 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 DSCFC 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 ]
- 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. Observations: ➢ Double sideband with Suppressed carrier. ➢The sum and difference frequencies are present. ➢Only 66.7% efficiency 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: Power saving can increased by eliminating one side band (USB or LSB) and carrier as well as. VDBS-SC (t) = VmVc 𝟐 [Cos(ωC + ωm)t + Cos(ωC - ωm)t]
- 25. Next Topic • Blooms Taxonomy ➢ Bloom's taxonomy is a powerful tool to help develop learning objectives because it explains the process of learning: ➢ Before you can understand a concept, you must remember it. ➢ To apply a concept you must first understand it. ➢ In order to evaluate a process, you must have analyzed it. Single side band suppressed carrier (SSB-SC)