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

Text Book 2

Contents 3

Contents (Cont.) 4

5

6

7

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

Comm. System (Cont.) 9

10

Can be eliminated Comm. System (Cont.) 11

Cannot be eliminated Comm. System (Cont.) 12

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

.. Analog & Digital Messages 14

Analog & Digital Messages (Cont.) 15

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

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

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

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

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

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

Sinusoidal Waveform 22

Modulation 23

24

25

Demodulation 26

... 27

AM (Cont.) 28

AM (Cont.) .. .. LetA=1 .. 29

AM (Cont.) .. .. .. 30

AM (Cont.) .. .. .. .. 31

AM Demodulation 32

AM Demodulation (Cont.) 33

Example on AM Solution.. 34

Example on AM (Cont.) 35

Example on AM (Cont.) 36

37

Modulators (Cont.) 38

Modulators (Cont.) .. 39

Modulators (Cont.) .. .. .. 40

Modulators (Cont.) .. .. 41

Modulators (Cont.) .. 42

Modulators (Cont.) .. 43

Modulators (Cont.) .. .. .. 44

Modulators (Cont.) 45

Modulators (Cont.) .. .. .. 46

Modulators (Cont.) .. 47

.. .. 48

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

Modulators (Cont.) 50

Modulators (Cont.) .. 51

.. .. .. 52

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

Solution 54

Example (Cont.) LPF 55

Solution Example: Frequency Mixer 56

Example: Frequency Mixer (Cont.) 57

Exercises

Exercises (Cont.) Solution 59

Exercises (Cont.) 60

Exercises (Cont.) 61

Exercises (Cont.) 62

Exercises (Cont.) 63

Exercises (Cont.) 64

Exercises (Cont.)

Exercises (Cont.) 66

Exercises (Cont.) Wanted Unwanted Solution 67

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

Exercises (Cont.) 69

Exercises (Cont.) Solution

Exercises (Cont.) k 71

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

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

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

AM (Cont.) 75

AM (Cont.) 76

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.

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

Solution 79

Example (Cont.) 80

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

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

Solution 83 Sidebands and AM efficiencies (Cont.)

.. .. 84

85 Generation of AM Signals (Cont.)

86

Rectifier Detector (Cont.) 87

88 Rectifier Detector (Cont.)

89 Rectifier Detector (Cont.)

90 Rectifier Detector (Cont.)

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

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.)

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.)

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

 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.

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

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

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 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

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

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 SSB (Cont.)

F

SSB (Cont.) 104

SSB (Cont.) Substitute with 105

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

SSB (Cont.) 107

Determination of m h (t) 108

SSB (Cont.) Hilbert transform 109

SSB (Cont.) 110

SSB (Cont.) .. .. 111

Solution 112

Example (Cont.) 113

Example (Cont.) 114

 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

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

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

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

119

Example (Cont.) 120

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

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

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

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

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

Solution Attenuated by LPF Removed by DC blocker 126 ab cd

127

Solution 128

Solution 129

Solution (Cont.) 130

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 % greater than SSB 131

VSB (Cont.) 132

VSB (Cont.) 133

VSB (Cont.) 134

VSB (Cont.) 135

VSB (Cont.) Solution 136

Use of VSB in television broadcast 137

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

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

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

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

PLL (Cont.) 142

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

DC component 144

Carrier Acquisition in DSB-SC (Cont.) 145

146

Carrier Acquisition in SSB-SC (Cont.) 147

148

Angle Modulation (Cont.) 149

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

Angle Modulation (Cont.) 151

Angle Modulation (Cont.) 152

153

154

Solution 155

156 Example (Cont.)

157

Example (Cont.) 158

Example (Cont.) 159

160 Solution

Example (Cont.) 161

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

Example (Cont.) 163

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

Example (Cont.) 165

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

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

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

.. .. 170

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

Generation of NBFM .. 172

Generation of NBPM 173

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

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)

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

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

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

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

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

182

Solution 183

Example (Cont.) 184

Example (Cont.) 185

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

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.

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

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

NBFM Generation (Cont.) 190

NBFM Generation (Cont.) 191

NBFM Generation (Cont.) 192

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

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

FM Demodulation using PLL 195 Free running frequency