1. Communications 1 Hebat-Allah M. Mourad Professor Cairo University Faculty of Engineering Electronics & Communications Department

2. Text Book Modern Digital and Analog Communication Systems Third edition B.P.Lathi - Chapters: 4 – 5 – 6 (1 st term) 7 -11 (2 nd term) Communication systems, S. Haykin, John Wiley and Sons inc., 4 th edition

3. Course Contents -Introduction to modulation - Different analog modulation techniques: Amplitude, Frequency and Phase -Transformation from analog to digital: Sampling, Quatization, PCM, DM, ADM -Introduction to satellite and mobile communications

4. Why Modulation? Mainly for two reasons 1 - Practical antenna dimensions light velocity Wavelengthfrequency Dimension is in the order of a quarter wavelength

6. Baseband Power Spectrum

8. Why Modulation? 2- Multiplexing Better utilization of the available frequency band Spectrum Frequency M 1 (f)M 2 (f) M 3 (f) M N (f)

9. Basic Modulation Types s (t) = A (t) cos [Ө (t) ] Ө (t) = ω t + φ (t) ‘ A c cos ω c t ‘ is called un-modulated carrier Analog Modulation Digital Modulation

10. Modulator m(t) Acos(2 πf c t +φ) modulated signal: s(t) Un-modulated carrier

11. Modulation Types (Analog Modulation)

12. Modulation Types (Digital Modulation)

13. Analog Modulation -Different analog modulation techniques - For each type:- - Mathematical presentation * Bandwidth * transmitted power - Modulators - Demodulators - Applications

14. 1- Amplitude Modulation (A.M) Conventional Amplitude Modulation Consider a sinusoidal Carrier wave c(t) the un- modulated carrier. c(t) = A c cos 2 π f c t = A c cos ω c t A c = carrier amplitude f c = carrier frequency A.M is the process in which the amplitude of the carrier wave c(t) is varied about a mean value, linearly with the base band signal

15. 1-Amplitude Modulation s(t) = A c cos 2 π f c t + m(t) cos 2 π f c t = [A c + m(t) ] cos 2 π f c t = modulated signal S(ω) = π A c [ δ(ω + ω c ) + δ(ω - ω c ) ] + (1/2) [ M(ω + ω c ) + M(ω - ω c ) ]

16. 1- Amplitude Modulation s(t)= un-modulated carrier + upper side band (U.S.B) + lower sideband (L.S.B.) Bandwidth (B.W.) = 2W | M(ω) | | S( ω ) | 0 0 W - W - ω c + ω c

18. Condition : A c + m(t) > 0 for all t A c ≥ m(t) min (absolute –ve peak amp.) Define: μ = modulation index = m(t) min / A c 0 ≤ μ ≤ 1 since there is no upper bound on A c and A c ≥ m(t) min The envelope has the same shape of m(t) Over modulation : μ > 1 μ x 100 =percentage modulation

20. 1- Amplitude Modulation Example: Let m(t) = A m cos 2 π f m t single tone signal Then s (t) = A c cos 2 π f c t + A m cos 2 π f m t cos 2 π f c t s (t) = A c [1 + μ cos 2 π f m t ] cos 2 π f c t μ = A m / A c = modulation factor =modulation index

21. 1- Amplitude Modulation | M(ω) | π ω ω mω m - ω m 0 -ωc-ωc ω ω c+ ω mω c+ ω m f c | S( ω ) | πμA c /2 ω c-ω mω c-ω m -ω c+ ω m-ω c+ ω m -ω c-ω m-ω c-ω m πAcπAc - s (t) = A c [1 + μ cos 2 π f m t ] cos 2 π f c t - B.W. = 2 f m - To get the power or energy : divide S(ω) by 2π, square the modulus and add all terms

22. 1- Amplitude Modulation S(ω) = πA c [ (ω - ω c ) + (ω - ω c ) ] + (π A c μ/2) [ (ω +(ω c + ω m )) ] + (π A c μ/2) [ (ω -(ω c + ω m )) ] + (π A c μ/2) [ (ω -(ω c - ω m )) ] + (π A c μ/2) [ (ω +(ω c - ω m )) ]

23. 1- Amplitude Modulation Un-modulated Power = carrier power= A c 2 / 2 U.S.B power = L.S.B. power = μ 2 A c 2 / 8 Total side bands power = μ 2 A c 2 / 4 P t = P c + P s = (A c 2 / 2) + (μ 2 A c 2 / 4) = P c [ 1+ (μ 2 /2) ] η = Side band power = μ 2 total power 2+ μ 2

24. AM Modulators Switching Modulator

25. 1- Switching Modulators By multiplying the signal by any periodic waveform whose fundamental is ω c. This periodic function can be expressed using F.S. as:

26. -Ψ(t) is a periodic pulse train -Ex: ring modulator,bridge modulators (different electronic circuits performing the switching operation)

27. For c >> m(t), the diode acts as a switch controlled mainly by the value of ‘c’ V bb’ (t) = c cosωct + m(t) for c > 0 V bb’ (t) = 0 for c < 0

28. A.M.demodulator 1-Non coherent/ Asynchronous/Envelope Detector

29. (1/W) > RC > T c

30. Non-coherent demodulator

31. Amplitude Modulation Virtues, limitations and modifications of A.M. Advantages: Ease of Modulation and demodulation ( cheap to build the system) Disadvantages - Waste of power. - Waste of B.W. We trade off system complexity to overcome these limitations - DSB-SC - DSB-QAM - SSB - VSB

32. 2- Double Side bands Suppressed Carrier (DSB-SC) The carrier component doesn’t appear in the modulated wave. s(t) = m(t) cos ( 2 ω c t) S(ω) = (1/2) [ M(ω - ω c ) + M(ω - ω c ) ] Total power = side bands power = (1/2) m 2 (t) B.W = 2W

34. DSB-SC Modulators 1- Switching Modulators By multiplying the signal by any periodic waveform whose fundamental is ω c. This periodic function can be expressed using F.S. as:

35. -Ψ(t) is a periodic pulse train -Ex: ring modulator,bridge modulators (different electronic circuits performing the switching operation)

37. DSB-SC Modulators 2- Multiplier modulators x(t)= m(t) cos 2 π f c t X(f) = (½)[ M(f+f c ) + M(f-f c )] X m(t) cos 2 πf c t x(t)

38. DSB-SC Modulators 3- Non linear modulators

40. DSB-SC Demodulators Coherent (synchronous ) detection

42. Main problem : the carrier at the receiver must be synchronized in frequency and phase with the one at the transmitter. Otherwise we can have serious problems in the demodulation as will be seen.

43. DSB-SC Demodulators XL.P.F ~ m(t) cosω c t cos[ (ω c + ∆ ω )t + φ] Frequency shiftPhase shift e o (t)

44. DSB-SC Demodulators e o (t) = m(t) cos ω c t cos[ (ω c + ∆ ω )t + φ] = (1/2) m(t) cos [ (∆ ω t) + φ] + cos [ (2 ω c +∆ ω) t + φ] Second term will be suppressed by the L.P.F. If ∆ ω =0 and φ = 0 e o (t) = (1/2) m(t) ( no frequency or phase error )

45. DSB-SC Demodulators If ∆ω=0 e o (t) = (1/2) m(t) cos φ -If φ = constant, e o (t) is proportional to m(t) -Problems for φ either varying with time or equals to ± (π/2) -The phase error may cause attenuation of the output signal without causing distortion as long as it is constant.

46. DSB-SC Demodulators If φ = 0, ∆ω ≠ 0 e o (t) = (1/2) m(t) cos ∆ω t (Donald Duck) - The output is multiplied by a low frequency sinusoid, this causes attenuation and distortion of the output signal. - This could be solved by using detectors by square law device, or phase locked loops (PLL) (ex: Costas loop)

47. Carrier Acquisition in DSB-SC Signal squaring method S.L.DN.B.F(2f c ) Freq div. by 2 m(t) cos ω c t x(t)

49. DSB-SC Demodulators Costas Receiver Product modulator -90 phase shift L.P.F V.C.O L.P.F Phase Discriminator cos(2πf c t+φ) A c cos(2πfct)m(t) (1/2)A c cos φ m(t) (1/2)A c sin φ m(t) K sin 2 φ

50. DSB-SC Demodulators The L.O. freq. is adjusted to be the same as f c Suppose that the L.O. phase is the same as the carrier wave: ‘I’ o/p = m(t)‘Q’ o/p = 0 Suppose that the L.O. phase drifts from its proper value by a small angle ‘φ’ radians. ‘I’ o/p will be the same and ‘Q’ o/p will be proportional to sin φ ~ φ (for small φ)

51. DSB-SC Demodulators This ‘Q’ channel o/p will have the same polarity as the ‘I’ channel for one direction of L.O. phase drift and opposite polarity for the opposite direction of L.O. phase drift. By combining the ‘I’ and ‘Q’ channel ouputs in a phase discriminator (multiplier followed by L.P.F.) A d.c. control signal is obtained that automatically corrects for local phase error in the V.C.O.

52. 3- DSB quadrature carrier multiplexing (QAM) It means quadrature amplitude modulation. We make use of the orthogonality of the sines and cosines to transmit and receive two different signals simultaneously on the same carrier frequency

53. 3- DSB quadruture carrier multiplexing (QAM) X X X X π/2 ~ ~ + + L.P.F m 1 (t) m 2 (t) cosω c t s(t) cosω c t x1x1 x2x2 x3x3 x4x4

54. 3- DSB quadruture carrier multiplexing (QAM) s(t) =m 1 (t) cos ω c t + m 2 (t) sin ω c t x 1 (t) = s(t) cos ω c t = (1/2) m 1 (t)[1 + cos 2ω c t ] + (1/2) m 2 (t) sin 2ω c t x 3 (t) = (1/2) m 1 (t) x 2 (t) = s(t) sin ω c t = (1/2) m 2 (t)[1 - cos 2ω c t ] + (1/2) m 1 (t) sin 2ω c t x 4 (t) = (1/2) m 2 (t)

55. 4- Single Side Band (SSB) Can be generated by: - Selective Filtering Method - Phase Shift Method Filter Method X S.B. Filter ~ m(t) s(t) S.S.B.

57. 4- Single Side Band (SSB) Phase shift method – Hilbert transform. hilbert transform of s(t) h(t) s(t) |H(ω)| ω ω ӨH(ω)ӨH(ω) π/2 - π/2

58. 4- Single Side Band (SSB) It can be shown that:- s(t) S.S.B = m(t) cos ω c t ± sin ω c t X -π/2 cos ω c t X m(t) sin ω c t m(t) cos ω c t + ± s S.S.B

59. Detection of SSB Coherent Detection y(t) = [ m(t) cos ω c t ± sin ω c t ] cos ω c t = (1/2)m(t) [1+ cos 2ω c t ] ± (1/2) sin 2 ω c t After the L.P.F the output equals (1/2) m(t) Envelope Detection s S.S.B = R(t) cos [ω c t + Ө(t)]

60. Detection of SSB R(t) = Ө(t) = tan -1 The envelope didn’t express the signal but if we have large carrier in phase with the transmitted signal the non-coherent could be used. s S.S.B (t) + A cos ω c t = R(t) cos [ω c t + Ө(t)]

61. Detection of SSB

62. In case of large carrier, the envelope of the S.S.B. signal has the form of m(t) (the base band signal), so the signal can be demodulated by the envelope detector with the condition that A >>> І m(t) І

63. 5- Vestigial Side Band (V.S.B) Vestige means trace. Since S.S.B. is difficult to realize, a compromise between S.S.B. and D.S.B. in spectrum can be obtained using V.S.B. [especially when we have important components at low frequency and for large bandwidth base band signals like T.V.] Most of one side band is passed along with a vestige of the other side band. (B.W ~ 1.125 W) (Reproduced by filters)

65. 5- Vestigial Side Band (V.S.B) What is the shape of H v (f)? X H v (f)X L.P.F A B C DE cos ω c t

66. 5- Vestigial Side Band (V.S.B) Spectrum at A prop. to: M(f) Spectrum at B prop. to: (1/2)[M(f+f c ) + M(f-f c )] Spectrum at C prop. to: H v (f)[M(f+f c ) + M(f-f c )] Spectrum at D prop. to : (1/2) H v (f+f c ) [ M(f+2f c ) + M(f) ] +(1/2) H v (f-f c ) [ M(f) + M(f-2f c ) ] The central lobe of the spectrum at D must equal M(f) (1/2) [ H v (f+f c ) + H v (f-f c ) ] M(f) f f fcfc -f c

67. 5- Vestigial Side Band (V.S.B) What are the conditions on the vestigial filter? 1 – It must have an odd symmetry around f c 2 – 50% response level at f c [ H v (f+f c ) + H v (f-f c ) ] = constant in ‘W’ H v (f) 1 1/2 -f c f c f

68. Comparison of Various AM signals AM and AM_SC - Detectors required for AM are simpler - AM signals are easier to generate - SC requires less power at the transmitter for the same information ( cheaper transmitter) - Effect of fading is must serious in AM because the carrier must maintain a certain strength in relation to the side bands.

69. Comparison of Various AM signals DSB and SSB - SSB needs only half the bandwidth - Fading disturbs the relationship of the two sidebands and causes more serious distortion than in the case of SSB,

70. AM Applications 1- Frequency Division Multiplexing (FDM) If we have ‘N’ signals to be transmitted using AM modulation. All of these are band limited to ‘W’. A1 B1 C1 Mod 1 2 3 + f 1 2 3 Dem 1 2 3 A2 B2 C2 1 2 3 1 2 3 f1f1 f2f2 f3f3

71. Examples: radio, TV, telephone backbone, satellite, …

73. AM Applications Y(f) f f2f2 f1f1 Carrier Demod. 2W at f N 2W at f 2 2W at f 1 Detector ~ B.P.F Detector m1m1 m2m2 mNmN

74. AM Applications Example: Telephone channel multiplexing First Level Mux 1 2 12Basic Group second Level Mux 1 2 5 Super Group Third Level Mux 1 2 10 Master Group North American FDM hierarchy

75. AM Applications All long-haul telephone channels are multiplexed by FDM using SSB. Basic Group consists of 12 FDM SSB voice signals each of BW = 4 KHz. It uses LSB spectra and occupies 60 to 108 KHz. [ alternate group configuration of 12 USB occupies 148 to 196 KHz] Basic group A (LSB)

76. AM Applications Basic Super group consists of 60 channels. It is formed by multiplexing ‘5’ basic groups and it occupies a band of 312 to 552 KHz. [alternate 60 KHz to 300 KHz] Super group 1 (LSB)

77. AM Applications Basic master group consiste of 600 channels. It is formed by multiplexing 10 supergroups.

78. AM Applications 2- Super heterodyne receiver The receiver not only has the task of demodulation but it is also required to perform some other tasks such as:- - Tuning : select the desired signal - Filtering : separate the desired signal from other modulated signals. - Amplification: to compensate for the loss in the signal power Super heterodyne receiver fulfils efficiently all the three functions

80. Basic elements of an AM receiver of the superheterodyne type Audio Amplifier E Mixer R.F. Section I.F. Section Envelop. detector A B CD ~ Local oscillator (f L.O ). Common tuning

81. Spectrum at different points R.F. response at ‘A’ at ‘C’at ‘E’ fcfc f I.F. response A.F. response f I. F 0

82. at ‘A’: [A + m(t)] cos ω c t at ‘B’:[A 1 +a 1 m(t)] cos ω c t cos (ω c + ω I.F )t [A 2 +a 2 m(t)] {cos (2ω c + ω I.F )t +cos ω I.F t } at ‘C’: [A 3 +a 3 m(t)] cos ω I.F t at ‘D’:a 4 m(t)

83. The R.F cannot provide adequate selectivity since f c is high. It rejects a lot of adjacent channel interference and amplifies the signal. This is why we translate it to IF frequency to obtain good selectivity. All selectivity is realized in the IF section. The main role of RF section is image frequency rejection. What is the image frequency?

84. If f c = 1000 KHz, f L.O. = 1000+455 =1455KHz the image frequency = f c +2f I.F. = 1910 KHz. Will also be picked up. Stations that are 2f I.F apart are called image stations and are rejected by RF filter. Up-conversion: f L.O. = f c +f I.F Down conversion: f L.O = f c -f I.F