1. Transmission / modulation Reception / demodulation
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
ANGLE MODULATION
CHAPTER 3
Part 2 (cont’d)
Transmission / modulation
Reception / demodulation

2. Angle modulated waveforms generally classified as :
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Angle modulated waveforms generally classified as :
Low modulation index
Medium modulation index
High modulation index
Low –index FM systems sometimes called as Narrow Band FM or NBFM

3. Generation of FM Two major FM generation: Direct method:
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Generation of FM
Two major FM generation:
Direct method:
Straight forward, requires a VCO whose oscillation frequency has linear dependence on applied voltage.
Advantage: large frequency deviation
Disadvantage: the carrier frequency tends to drift and must be stabilized.
example circuit:
Reactance modulator
Varactor diode

4. Generation of FM (cont’d)
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Generation of FM (cont’d)
ii) Indirect method:
Frequency-up conversion.
Two ways:
Heterodyne method
Multiplication method
One most popular indirect method is the Armstrong modulator

5. Simple direct FM modulator
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Simple direct FM modulator
Accousticalmechanical energy
Cm varied causes the resonant frequency varied
Tomasi Electronic Communications Systems, 5e
Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.

6. Varactor diode direct FM modulator
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Varactor diode direct FM modulator
Deviate the frequency via capacitances changes
Tomasi Electronic Communications Systems, 5e
Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.

7. JFET reactance modulator
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
JFET reactance modulator
Called as reactance modulator – JFET (active device)
The modulating signal varies with the reactant of Q causes corresponding change in the resonant frequency

8. Armstrong modulator EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Armstrong modulator
Integrator
Balanced modulator
Down converter
Frequency multiplier (x n)
Crystal oscillator
Phase shifter
Vc(t) fc
Vm(t) fm

9. FM Detection/Demodulation
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
FM Detection/Demodulation
FM demodulation
is a process of getting back or regenerate the original modulating signal from the modulated FM signal.
It can be achieved by converting the frequency deviation of FM signal to the variation of equivalent voltage.
The demodulator will produce an output where its instantaneous amplitude is proportional to the instantaneous frequency of the input FM signal.

10. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
FM detection (cont’d)
To detect an FM signal, it is necessary to have a circuit whose output voltage varies linearly with the frequency of the input signal.
The most commonly used demodulator is the PLL demodulator. Can be use to detect either NBFM or WBFM.

11. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
PLL Demodulator
V0(t)
FM input
Phase
detector
Low pass
filter
Amplifier
fVc0
VCO
Vc(t)
The phase detector produces an average output voltage that is linear function of the phase difference between the two input signals. This low frequency component is selected by LPF.

12. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
PLL Demodulator
After amplification, part of the signal is fed back through VCO where it results in frequency modulation of the VCO frequency. When the loop is in lock, the VCO frequency follows or tracks the incoming frequency.
Let instantaneous freq of FM Input,
fi(t)=fc +k1vm(t),
and the VCO output frequency,
f VCO(t)=f0 + k2Vc(t);
f0 is the free running frequency.
For the VCO frequency to track the instantaneous incoming frequency,
fvco = fi

13. PLL Demodulator f0 + k2Vc(t)= fc +k1vm(t)
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
PLL Demodulator
f0 + k2Vc(t)= fc +k1vm(t)
If VCO can be tuned so that fc=f0, then
Where Vc(t) is also taken as the output voltage, which therefore is the demodulated output

14. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Comparison AM and FM
Its the SNR can be increased without increasing transmitted power about 25dB higher than in AM
Certain forms of interference at the receiver are more easily to suppressed, as FM receiver has a limiter which eliminates the amplitude variations and fluctuations.
The modulation process can take place at a low level power stage in the transmitter, thus a low modulating power is needed.
Power content is constant and fixed, and there is no waste of power transmitted
There are guard bands in FM systems allocated by the standardization body, which can reduce interference between the adjacent channels.

15. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Application of FM
used by most of the field VHF portable, mobile and base radios in exploration use today.
It is preferred because of its immunity to noise or interference and at the frequencies used the antennas are of a reasonable size.

16. Summary of angle modulation
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Summary of angle modulation

17. Summary (cont’d) EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Summary (cont’d)

18. Summary (cont’d) Bandwidth:
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Summary (cont’d)
Bandwidth:
Actual minimum bandwidth from Bessel table:
b) Approximate minimum bandwidth using Carson’s rule:

19. Summary (cont’d) Multitone modulation (equation in general):
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Summary (cont’d)
Multitone modulation (equation in general):

20. Summary (cont’d) EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Summary (cont’d)

21. Summary (cont’d)- Comparison NBFM&WBFM
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Summary (cont’d)- Comparison NBFM&WBFM

22. Part 3 Advantages Disadvantages
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
ANGLE MODULATION
Part 3
Advantages
Disadvantages

23. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Advantages
Wideband FM gives significant improvement in the SNR at the output of the RX which proportional to the square of modulation index.
Angle modulation is resistant to propagation-induced selective fading since amplitude variations are unimportant and are removed at the receiver using a limiting circuit.
Angle modulation is very effective in rejecting interference. (minimizes the effect of noise).
Angle modulation allows the use of more efficient transmitter power in information.
Angle modulation is capable of handing a greater dynamic range of modulating signal without distortion than AM.

24. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Disadvantages
Angle modulation requires a transmission bandwidth much larger than the message signal bandwidth.
The capture effect where the wanted signal may be captured by an unwanted signal or noise voltage.
Angle modulation requires more complex and inexpensive circuits than AM.

25. END OF ANGLE MODULATION
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
END OF ANGLE MODULATION

26. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision (Decibel)
Decibel compress the wide range of signals into a smaller group of numbers and make measurements and type of analysis of system performance more convenient.

27. Absolute voltage ratio
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Absolute power ratio
Power gain
Absolute voltage ratio
Voltage gain
AM DSB-FC
AM DSB-SC
AM SSB-FC
AM SSB-SC ?

28. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
If two powers expressed in same units, the ratio can be expressed as:
dB=10 log(10) (P1/P2)
Where P1 = power level 1 (watts) P2 = power level 2 (watts) i.e the reference power
In electronic circuits, to measure a power gain
or loss
Ap(dB)=10log(10) (Pout/Pin)
Where Ap(dB) = Power gain (dB)
Pout/Pin = absolute power gain (unitless)

29. Revision Voltage gain in dB is expressed as
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Voltage gain in dB is expressed as
Av(dB)=20log(10) (Eout/Ein)
Where Av(dB) = Power gain (dB)
Eout = output voltage (volts)
Ein = output voltage (volts)
Eout/Ein = Absolute voltage gain (unitless)

30. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
A dBm is a unit of measurement used to indicate the ratio of power level with respect to a fixed reference level (1mW)
OR…..“decibel relative to 1 milliwatt”
dBm = 10log(P/1mW)
Where P is the power in watts
1mW is the reference voltage

31. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Similarly…..
A dBW is a unit of measurement used to indicate the ratio of power level with respect to a fixed reference level (1W)
OR…..“decibel relative to 1 watt”
dBW = 10log(P/1W)
Where P is the power in watts
1W is the reference voltage

32. Q1) Express the power values as dBW
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Q1) Express the power values as dBW
(a) 1W
(b) 0.73W
(c) 950W
Answer :
dBW=10log(P/1W);
0dBW,-1.37dBW, 29.8dBW.

33. Q2) Express 200W as dBW and as dBm
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Q2) Express 200W as dBW and as dBm
Answer :
dBm=10log(P/1mW);
dBW=10log(P/1W);
23dB,53dBm

34. Q3) Convert the absolute power ratio of 200 to a power gain in dB.
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Q3) Convert the absolute power ratio of 200 to a power gain in dB.
Answer :
Ap(dB) =10log(10) (Pout/Pin)
Ap(dB) =10log(10)(200) = 23dB

35. Q4) Convert the power gain of Ap = 23dB to an absolute power ratio.
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Q4) Convert the power gain of Ap = 23dB to an absolute power ratio.
Answer :
23dB =10log(10) (Pout/Pin)
(Pout/Pin) = 200

36. Av(dB) = 20log(0.1/5)=-34dB. OR 34dB loss
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Q5) A signal is attenuated from 5V to 0.1V. What is the corresponding value in dB?
Answer :
Av(dB) = 20log(0.1/5)=-34dB. OR 34dB loss

37. Ap(dB) = 10 log(P/100)=-30dB, P=0.1W
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Q6) A 100-W signal goes through a circuit with a 30-dB loss. What is the final power value?
Answer :
Ap(dB) = 10 log(P/100)=-30dB, P=0.1W

38. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Q7) For an AM DSBFC wave with a peak unmodulated carrier voltage Vc = 10 Vp,frequency of 100kHz, a load resistor, RL = 10 , frequency of modulating signal of 10kHz and m = 1.0, Determine
Powers of the carrier and the upper and lower sidebands.
ii) Total power of the modulated wave.
iii) Bandwidth of the transmitted wave.
iv) Draw the power and frequency spectrum.

39. Revision Answer the following questions based on information from Q7
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Answer the following questions based on information from Q7
Q8) For the same given values, determine questions (ii)-(iv) for a AM DSB-SC, AM SSB-FC and AM SSB-SC systems.
Q9) Determine also the percentage of power saved in each of the system design.

40. Revision Solution Q7 : DSBFC i) ii)
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Solution Q7 : DSBFC i)
ii)
iii) Bandwidth = 2(fmmax) = 2(10kHz) = 20kHz

41. Revision Solution Q7 :For DSB-SC ii)
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Solution Q7 :For DSB-SC
ii)
iii) Bandwidth = 2(fmmax) = 2(10kHz) = 20kHz
iv)
110kHz
90kHz fc

42. Revision Solution Q8 : For SSB-FC ii) iii) Bandwidth=fmmax=10kHz iv)
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Solution Q8 : For SSB-FC
ii)
iii) Bandwidth=fmmax=10kHz
iv)
100kHz
110kHz
fc-fm

43. Revision Solution Q8 : For SSB-SC ii) iii) Bandwidth=fmmax=10kHz iv)
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Solution Q8 : For SSB-SC
ii)
iii) Bandwidth=fmmax=10kHz
iv)
fc-fm
110kHz fc

44. Revision Solution Q10 :For DSB-SC Solution Q10 :For SSB-FC
EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Revision
Solution Q10 :For DSB-SC
Solution Q10 :For SSB-FC
Solution Q10 :For SSB-SC

45. EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Assignment
An angle modulated signal with carrier frequency as described by the following equation
Determine
(i) The power of the modulated signal (let R=1Ω) ans:50W
(ii) Prove that
(iii) The frequency deviation,Δf ans:12.387kHz
(iv) The deviation ratio, DR ans:12.387Hz
(v) The phase deviation, m ans:15 rad
(vi) Estimate the bandwidth of the modulated signal. ans:26.774kHz