1. Introduction to Communication Prepared By Dr. Hany Taher Modified by: Dr. Mouaaz Nahas

2. Text Book 2

3. Contents 3

4. Contents (Cont.) 4

5. 5

6. 6

7. 7

8. Comm. System (Cont.) ) that converted by input transducer 8

9. Comm. System (Cont.) 9

10. 10

11. Can be eliminated Comm. System (Cont.) 11

12. Cannot be eliminated Comm. System (Cont.) 12

13. .. .. .. Comm. System (Cont.) 13

14. .. Analog & Digital Messages 14

15. Analog & Digital Messages (Cont.) 15

16. .. .. Noise immunity of Digital Signals 16

17. Distorted signal Distorted noisy signal .. Due to channel Noise immunity of Digital Signals (Cont.) 17

18. Noise immunity of Digital Signals (Cont.) .. 18

19. .. .. .. Viability of Regenerative Repeaters in Digital Communication 19

20. Viability of Regenerative Repeaters in Digital Communication (Cont.) .. .. .. 20

21. Signal to noise ratio Signal Bandwidth & SNR A measure of the width of a range of frequencies, measured in hertz (Hz). 21

22. Sinusoidal Waveform 22

23. Modulation 23

24. 24

25. 25

26. Demodulation 26

27. ... 27

28. AM (Cont.) 28

29. AM (Cont.) .. .. LetA=1 .. 29

30. AM (Cont.) .. .. .. 30

31. AM (Cont.) .. .. .. .. 31

32. AM Demodulation 32

33. AM Demodulation (Cont.) 33

34. Example on AM Solution.. 34

35. Example on AM (Cont.) 35

36. Example on AM (Cont.) 36

37. 37

38. Modulators (Cont.) 38

39. Modulators (Cont.) .. 39

40. Modulators (Cont.) .. .. .. 40

41. Modulators (Cont.) .. .. 41

42. Modulators (Cont.) .. 42

43. Modulators (Cont.) .. 43

44. Modulators (Cont.) .. .. .. 44

45. Modulators (Cont.) 45

46. Modulators (Cont.) .. .. .. 46

47. Modulators (Cont.) .. 47

48. .. .. 48

49. Modulators (Cont.) .. .. .. .. .. 49

50. Modulators (Cont.) 50

51. Modulators (Cont.) .. 51

52. .. .. .. 52

53. DSB-SC Demodulation (Cont.) .. .. 53

54. Solution 54

55. Example (Cont.) LPF 55

56. Solution Example: Frequency Mixer 56

57. Example: Frequency Mixer (Cont.) 57

58. Exercises 4.2.1 58

59. Exercises (Cont.) Solution 59

60. Exercises (Cont.) 60

61. Exercises (Cont.) 61

62. Exercises (Cont.) 62

63. Exercises (Cont.) 63

64. Exercises (Cont.) 64

65. Exercises (Cont.) 4.2-4 65

66. Exercises (Cont.) 66

67. Exercises (Cont.) Wanted Unwanted Solution 67

68. Exercises (Cont.) Band pass Filter at w c 68

69. Exercises (Cont.) 69

70. Exercises (Cont.) Solution 70 4.2-5

71. Exercises (Cont.) k 71

72. .. Transmitter may be very far, i.e, thousands of Kms Expensive .. Solution is to use Amplitude modulation: 72

73. AM (Cont.) ..Disadvantage .... More power is needed .... 1. Point to point communications : One transmitter for every receiver. One transmitter for every receiver. Complexity in the receiver is justified. Complexity in the receiver is justified. 2. Broadcast communications: Single transmitter and multi receivers. Single transmitter and multi receivers. Better economically to have one expensive high-power transmitter and simpler less expensive multiple receivers. Better economically to have one expensive high-power transmitter and simpler less expensive multiple receivers. 73

74. AM (Cont.) .. .. .. Remember Compare 74

75. AM (Cont.) 75

76. AM (Cont.) 76

77. AM (Cont.)  Two conditions for the envelop detection of an AM signal are: 1) f c > bandwidth of m(t) 2)A + m(t) ≥ 0 for all t 77  If A + m(t) is not positive for all t, m(t) cannot be recovered from the envelop \ A + m(t)\.  If m(t) ≥ 0 for all t, A=0 already satisfies the condition A + m(t) ≥ 0.  No need to add any carrier because the envelop of the DSB-SC signal m(t) cosω c t is m(t). This signal can be detected by envelop detection.

78. AM (Cont.)  Let, m p the peak amplitude  The modulation index  A< m p µ > 1 (overmodulation) 78

79. Solution 79

80. Example (Cont.) 80

81. Sidebands and AM efficiencies .. .. Time mean square value .. .. 81

82. Sidebands and AM efficiencies (Cont.) .. 82

83. Solution 83 Sidebands and AM efficiencies (Cont.)

84. .. .. 84

85. 85 Generation of AM Signals (Cont.)

86. 86

87. Rectifier Detector (Cont.) 87

88. 88 Rectifier Detector (Cont.)

89. 89 Rectifier Detector (Cont.)

90. 90 Rectifier Detector (Cont.)

91. Envelope Detector 1. First AM positive cycle Capacitor is charged until the peak value 2. The AM signal falls below the peak Diode On Diode is Off Capacitor is discharged into the resistance with time constant ‘RC’ and its voltage starts to decrease from the peak value 91

92. 3. The AM signal now is in its negative cycle Diode is still off Capacitor is still discharged into the resistance and its voltage continues in decreasing from the peak value 4. The AM signal now is going into 2 nd positive cycle Diode is still off until the value of AM signal becomes larger than voltage on the capacitor, at this moment, Diode is on Capacitor is charged until the peak value, and so on. The above steps are repeated again 92 Envelope Detector (Cont.)

93. Ripples  To decrease the ripples, use high ‘RC’ Take Care Very high RC does not accurately follow the envelope T C << RC << T m (1/B) 93 Envelope Detector (Cont.)

94. In DSB (SC and AM), each USB and LSB contains complete information about m(t). So DSB modulation requires twice the bandwidth of baseband to transmit (spectral redundancy). To improve the spectral efficiency, we have two schemes: –Single-Sideband (SSB). –Quadrature Amplitude Modulation (QAM). 94 Bandwidth-Efficient AM

95.  As mentioned earlier, DSB occupies twice of bandwidth of baseband (this is a disadvantage). Solution Two baseband signal on the same carrier frequency, how??!!!!! Two baseband signal on the same carrier frequency, how??!!!!! There is π/2 (radian) phase shift between the two carriers, i.e. Cosine and Sine as example. There is π/2 (radian) phase shift between the two carriers, i.e. Cosine and Sine as example. 95 QAM (or Quadrature Multiplexing) transmits two DSB signals using two carriers of the same frequency but in phase quadrature.

96. QAM (Cont.) Two baseband signals Two baseband signals Carrier signal Carrier signal Phase Shifted Carrier signal Phase Shifted Carrier signal Modulated Signal 96 Phase Shifter

97. QAM (Cont.) In-phase (I) In-phase (I)channel Quadrature (Q) Quadrature (Q)channel 97

98. QAM (Cont.) The last two terms in x 1 (t) form a QAM signal with 2ω c as the carrier frequency. They are suppressed by the low-pass filter yielding m 1 (t). 98 Obtain an expression for x 2 (t)???? Obtain an expression for x 2 (t)????

99. 99 QAM (Cont.) Note that QAM must be totally synchronous. The problem is that any error in phase or frequency of the carrier at the demodulation results in loss or interference between the two signals. Both baseband signals will appear at the filter output instead of m 1 (t) Cochannel interference

100.  DSB has two sidebands, USB and LSB  SSB has one half of the bandwidth of DSB 100

101. SSB (Cont.) 101 SSB signal can be coherently (synchronously) demodulated by multiplying it by cos ω c t (exactly like DSB-SC). Example of USB demodulation is: SSB-SC

102. 102 SSB (Cont.)

103. F -1 103

104. SSB (Cont.) 104

105. SSB (Cont.) Substitute with 105

106. SSB (Cont.) Φ LSB (ω) = M + ( ω + ω c ) + M - ( ω - ω c ) Φ LSB (t) = m + (t) e -jωct + m - (t) e jωct Substitute with 106

107. SSB (Cont.) 107

108. Determination of m h (t) 108

109. SSB (Cont.) Hilbert transform 109

110. SSB (Cont.) 110

111. SSB (Cont.) .. .. 111

112. Solution 112

113. Example (Cont.) 113

114. Example (Cont.) 114

115.  DSB is passed through BPF to eliminate undesired band  Most commonly used  To obtain USB, the filter should pass all components above ω c and attenuate all components below ω c Difficult to design sharp cutoff filter Voice spectrum 300 Hz 115

116. Generation of SSB signals (Cont.) Difficult to build 116

117. Envelope detection of SSB with carrier (SSB+C) If ‘A’ is large enough m(t) can be recovered by envelop detection envelop detection 117

118. Envelope detection of SSB with carrier (SSB+C) (Cont.) 118

119. 119

120. Example (Cont.) 120

121. Solution Diodes conducting The diode resistance is ‘ r’ off state off state The diode resistance is ‘ ∞’  A carrier in positive cycle  A carrier in negative cycle The output voltage is ‘ zero’  The diode acts as a gate with gain ‘2R/(R+r)’ (R/(r+R))φ (t) appears across each resistance ‘R’ 121

122. Solution (Cont.)  The output is, e 0 (t) = (2R/(R+r)) ω (t) m (t)  If e 0 is passed through B.P.F centered at ω C and has 2B band width, the output is ( 4R/ ( π ( R + r ) ) m ( t ) Cos (ω C t ) 122

123. Solution (Cont.)  This circuit can work as a demodulator as follows:  The input is m (t) cos ω C t  LPF is used at the output  The output ( 2R/ ( π ( R + r ) ) m ( t ) 123

124. Solution e 0 (t) = ( 4R/ ( π ( R + r )) m ( t ) cos ω C t e 0 (t) = ( 4R/ ( π ( R + r )) m ( t ) cos ω C t = k m ( t ) cos ω C t = k m ( t ) cos ω C t m ( t ) = sin (ω C t + θ ) e 0 (t)= k sin ( ω C t + θ ) cos ω C t 124

125. Solution (Cont.) e 0 (t) = k/2 (sin ( 2 ω C t + θ ) + sin θ )  The LPF suppresses the sinusoid and transmits only the DC component only the DC component  e o / (t) = k/2 sin θ 125 Phase difference between the two sinusoids

126. Solution Attenuated by LPF Removed by DC blocker 126 ab cd

127. 127

128. Solution 128

129. Solution 129

130. Solution (Cont.) 130

131. Amplitude modulation: Vestigial Sideband (VSB)  Due to the difficulties in generating SSB signals VSB is used  VSB is asymmetric system  VSB is a compromise between DSB and SSB  VSB is easy to generate and its bandwidth is only 25 - 33% greater than SSB 131

132. VSB (Cont.) 132

133. VSB (Cont.) 133

134. VSB (Cont.) 134

135. VSB (Cont.) 135

136. VSB (Cont.) Solution 136

137. Use of VSB in television broadcast 137

138. Carrier Acquisition  In any technique from amplitude modulation techniques, the local oscillator must be in synchronous with the oscillator that is used in transmitter side. Why?  Consider DSB-SC case: .. .. .. 138

139. Carrier Acquisition (Cont.) .. Filtered by LPF .. 139

140. Carrier Acquisition (Cont.) 140 Attenuation to the message .. .. Beating effect Solution Quartz Crystal Oscillator? Difficult to built at high frequencies Phase Locked Loop (PLL)

141. Free running frequency .. .. .. Suppressed by LPF 141

142. PLL (Cont.) 142

143. PLL (Cont.) If the frequencies are different  Suppose the frequency of the input sinusoidal signal is increased from ω c to ω c + k  This means that the incoming signal is  Thus the frequency increase in incoming signal causes θ i to increase to θ i + kt Increasing θ e, and vice versa 143

144. DC component 144

145. Carrier Acquisition in DSB-SC (Cont.) 145

146. 146

147. Carrier Acquisition in SSB-SC (Cont.) 147

148. 148

149. Angle Modulation (Cont.) 149

150. Angle Modulation (Cont.)  Two techniques are possible to transmit m(t) by varying angle θ of a carrier. 150

151. Angle Modulation (Cont.) 151

152. Angle Modulation (Cont.) 152

153. 153

154. 154

155. Solution 155

156. 156 Example (Cont.)

157. 157

158. Example (Cont.) 158

159. Example (Cont.) 159

160. 160 Solution

161. Example (Cont.) 161

162. Example (Cont.) 162 This scheme is called Frequency Shift Keying (FSK) where information digits are transmitted by keying different frequencies.

163. Example (Cont.) 163

164. Example (Cont.) The derivative ṁ (t) is zero except at points of discontinuity of m(t) where impulses of strengths ±2 are present. 164

165. Example (Cont.) 165

166. Example (Cont.) 166 This scheme is called Phase Shift Keying (PSK) where information digits are transmitted by shifting the carrier phase. The phase difference (shift) is π.

167. 167

168. Bandwidth of Angle Modulated Waves (Cont.) .. .. .. 168

169. Bandwidth of Angle Modulated Waves (Cont.) .. .. .. 169

170. .. .. 170

171. Narrowband Angle Modulation (Cont.) Compare with AM modulated signal .. .. 171

172. Generation of NBFM .. 172

173. Generation of NBPM 173

174. Can not be ignored Very complicated analysis 174 In practical FM,

175. WBFM (Cont.) Simple way to analyze the problem m(t)  m(t) is band limited to B Hz  m(t) is approximated by pulses of constant amplitudes (cells),  FM analysis of constant amplitude is easier 175 (Staircase)

176. WBFM (Cont.) m(t) ≈ Pulse interval ≤ 1 / 2B Sec  The FM signal of one of these cells starting at t = t k 1/2B Sec Sinusoidal 176

177. WBFM (Cont.)  The FM spectrum of consists of the sum consists of the sum of the Fourier transforms of the sinusoidal pulses. of the Fourier transforms of the sinusoidal pulses. 177

178. WBFM (Cont.)  The min. and max. amplitude of modulating signals are -m p and m p the min. frequency the max. frequency the Bandwidth 178

179. WBFM (Cont.) .. ..  Do not forget, this value is calculated for  From earlier analysis:  For NBFM, 179 The Peak Frequency Deviation:

180. WBFM (Cont.) 180 For a truly wideband case, A better bandwidth estimate is: The “deviation ratio” plays a role similar to the “modulation index” in AM.

181. .. 181

182. 182

183. Solution 183

184. Example (Cont.) 184

185. Example (Cont.) 185

186. Generation of FM Waves 186 Direct FM Method. Indirect FM Method.

187. NBFM Generation 187 Recall: The output of the this NBFM has some amplitude variations. A nonlinear device designed to limit the amplitude of a bandpass signal can remove most of this distortion.

188. NBFM Generation (Cont.) Bandpass Limiter: used to remove amplitude variations in FM wave. 188 The input-output characteristics of the hard limiter.

189. NBFM Generation (Cont.) Hard limiter input and the corresponding output. 189 Hard limiter output with respect to θ

190. NBFM Generation (Cont.) 190

191. NBFM Generation (Cont.) 191

192. NBFM Generation (Cont.) 192

193. Demodulation of FM Signals The simplest demodulator is an ideal differentiator followed by an envelop detector. 193 envelop

194. Demodulation of FM Signals (Cont.) 194 envelop detector can be used to recover m (t )

195. FM Demodulation using PLL 195 Free running frequency