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

Introduction to Communication

Prepared By Dr. Hany TaherPrepared By Dr. Hany Taher

Modified by: Dr. Mouaaz NahasModified by: Dr. Mouaaz Nahas

Text Book

2

ContentsContents

3

Contents (Cont.)Contents (Cont.)

4

Contents (Cont.)Contents (Cont.)

5

Comm. System (Cont.)Comm. System (Cont.)

)

that converted by input transducer

8


9


10

Can be eliminatedCan be eliminated


11

Cannot be eliminatedCannot be eliminated


12

.

.

.


13

.

.

Analog & Digital MessagesAnalog & Digital Messages

14

Analog & Digital Messages (Cont.)Analog & Digital Messages (Cont.)

15

.

.

Noise immunity of Digital SignalsNoise immunity of Digital Signals

16

Distorted signalDistorted signal

Distorted noisy signalDistorted noisy signal

.

Due to channelDue to channel

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

17

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

.

18

.

.

.

Viability of Regenerative Repeaters in Viability of Regenerative Repeaters in Digital CommunicationDigital Communication

19

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

.

.

.

20

Signal to noise ratio

Signal Bandwidth & SNRSignal Bandwidth & SNR

A measure of the width of a range of frequencies, measured in hertz (Hz).

21

Sinusoidal WaveformSinusoidal Waveform

22

ModulationModulation

23

DemodulationDemodulation

26

.

.

.27

AM (Cont.)AM (Cont.)

28

AM (Cont.)AM (Cont.).

.

LetLet A=1A=1

.

29


... 30


.

.

.

31

AM DemodulationAM Demodulation

32

AM Demodulation (Cont.)AM Demodulation (Cont.)

33

Example on AMExample on AM

Solution

.

.

34

Example on AM (Cont.)Example on AM (Cont.)

35

Example on AM (Cont.)Example on AM (Cont.)

36

Modulators (Cont.)Modulators (Cont.)

38

Modulators (Cont.)Modulators (Cont.).

39


.

.

40


.

41


.

42


.

43


.

.

44


45


.

.

46


.47

.

.48


.

.

.

.

.

49


50


.

51

.

.

.

52

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

.

53

SolutionSolution

54

Example (Cont.)Example (Cont.)

LPFLPF

55

SolutionSolution

Example: Frequency MixerExample: Frequency Mixer

56

Example: Frequency Mixer (Cont.)Example: Frequency Mixer (Cont.)

57

ExercisesExercises4.2.14.2.1

58

Exercises (Cont.)Exercises (Cont.)SolutionSolution

59

Exercises (Cont.)Exercises (Cont.)

60


61


62


63


64

Exercises (Cont.)Exercises (Cont.)4.2-44.2-4

65


66


Wanted Unwanted

SolutionSolution

67


Band pass Filter at wc

68


69


SolutionSolution

70

4.2-54.2-5


kk

71

.

Transmitter may be very far, i.e, thousands of KmsTransmitter may be very far, i.e, thousands of Kms

Expensive

. Solution is to use Amplitude modulation:

72

AM (Cont.)AM (Cont.). DisadvantageDisadvantage

..

More power is neededMore power is needed

..

1. Point to point communications :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: 2. Broadcast communications:

• Single transmitter and multi receivers. Single transmitter and multi receivers. • Better economically to have one expensive high-power Better economically to have one expensive high-power transmitter and simpler less expensive multiple transmitter and simpler less expensive multiple receivers.receivers. 73


.

. Remember

Compare

74


75


76

AM (Cont.)AM (Cont.)Two conditions for the envelop detection of an Two conditions for the envelop detection of an AM signal are:AM signal are:1)1) ffcc > bandwidth of> bandwidth of m(t) m(t)

2)2)A A ++ m(t) m(t) ≥≥ 0 0 for all for all tt

77

If If A + m(t) A + m(t) is not positive for all is not positive for all t, m(t) t, m(t) cannot be cannot be recovered from the enveloprecovered from the envelop \ A + m(t)\. \ A + m(t)\.

If If m(t) m(t) ≥≥ 0 0 for all for all t, A=0 t, A=0 already satisfies the already satisfies the condition condition A A ++ m(t) m(t) ≥≥ 0. 0. No need to add any carrier because the envelop of No need to add any carrier because the envelop of the DSB-SC signal the DSB-SC signal m(t) m(t) coscosωωcctt is is m(t).m(t). This signal can This signal can

be detected by envelop detection.be detected by envelop detection.

AM (Cont.)AM (Cont.)Let, mLet, mpp the peak amplitude the peak amplitude

The modulation indexThe modulation index

A< mA< mpp µ > 1 (overmodulation)µ > 1 (overmodulation)78

SolutionSolution

79


80

Sidebands and AM efficienciesSidebands and AM efficiencies

.

.

Time mean square value

.

.

81

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

.

82

SolutionSolution

83

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

.

.

84

85

Generation of AM Signals (Cont.)Generation of AM Signals (Cont.)

Rectifier Detector (Cont.)Rectifier Detector (Cont.)

87

88


89


90


Envelope DetectorEnvelope Detector

1. First AM positive cycle1. First AM positive cycle

Capacitor is charged until the peak valueCapacitor is charged until the peak value

2. The AM signal falls below the peak2. The AM signal falls below the peak

Diode OnDiode On

Diode is OffDiode is Off

Capacitor is discharged into the resistance with time Capacitor is discharged into the resistance with time constant ‘RC’ and its voltage starts to decrease from the constant ‘RC’ and its voltage starts to decrease from the peak valuepeak value 91

3. The AM signal now is in its negative cycle3. The AM signal now is in its negative cycle

Diode is still offDiode is still off

Capacitor is Capacitor is still still discharged into the resistance discharged into the resistance and its voltage and its voltage continuescontinues in decreasing from the peak value in decreasing from the peak value

4. The AM signal now is going into 24. The AM signal now is going into 2ndnd positive cycle positive cycle

Diode is still off until the value of AM signal becomes Diode is still off until the value of AM signal becomes larger than voltage on the capacitor, at this moment,larger than voltage on the capacitor, at this moment,Diode is onDiode is on

Capacitor is charged until the peak value, and so Capacitor is charged until the peak value, and so on. The above steps are repeated againon. The above steps are repeated again 92

Envelope Detector (Cont.)Envelope Detector (Cont.)

RipplesRipples

To decrease the ripples, use high ‘RC’To decrease the ripples, use high ‘RC’ Take CareTake CareVery high RC does not accurately follow the Very high RC does not accurately follow the envelope envelope

TTCC << RC << T << RC << Tm m (1/B)(1/B)

93

Envelope Detector (Cont.)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 AMBandwidth-Efficient AM

As mentioned earlier, DSB occupies twice of As mentioned earlier, DSB occupies twice of bandwidth of baseband bandwidth of baseband (this is a disadvantage).(this is a disadvantage).

Solution Solution Two baseband signal on the same carrier frequency, Two baseband signal on the same carrier frequency, how??!!!!!how??!!!!!

There is There is ππ/2 (radian) phase shift between the two /2 (radian) phase shift between the two carriers, i.e. Cosine and Sine as example.carriers, i.e. Cosine and Sine as example.

95

QAM (or Quadrature Multiplexing) transmits two DSB QAM (or Quadrature Multiplexing) transmits two DSB signals using two carriers of the same frequency but in signals using two carriers of the same frequency but in phase quadrature.phase quadrature.

QAM (Cont.)QAM (Cont.)

Two baseband signalsTwo baseband signals

Carrier signalCarrier signal

Phase Shifted Phase Shifted Carrier signalCarrier signal

Modulated SignalModulated Signal

96

Phase Phase ShifterShifter


In-phase (I)In-phase (I)channelchannel

Quadrature (Q)Quadrature (Q)channelchannel

97


The last two terms in xThe last two terms in x11(t) form a QAM signal with 2(t) form a QAM signal with 2ωωcc as as

the carrier frequency. They are suppressed by the low-pass the carrier frequency. They are suppressed by the low-pass filter yielding mfilter yielding m11(t).(t).

98

Obtain an expression for xObtain an expression for x22(t)????(t)????

99

QAM (Cont.)QAM (Cont.)Note that QAM must be totally synchronous.Note that QAM must be totally synchronous.The problem is that any error in phase or frequency of the carrier The problem is that any error in phase or frequency of the carrier at the demodulation results in loss or interference between the two at the demodulation results in loss or interference between the two signals.signals.

Both baseband signals will appear Both baseband signals will appear at the filter output instead of mat the filter output instead of m11(t)(t)

Cochannel interferenceCochannel interference

DSB has two sidebands, USB and LSB DSB has two sidebands, USB and LSB

SSB has one half of the bandwidth of DSBSSB has one half of the bandwidth of DSB

100

SSB (Cont.)SSB (Cont.)

101

SSB signal can be coherently (synchronously) demodulated by SSB signal can be coherently (synchronously) demodulated by multiplying it by cos multiplying it by cos ωωcct (exactly like DSB-SC). Example of t (exactly like DSB-SC). Example of

USB demodulation is:USB demodulation is:

SSB-SCSSB-SC

102



F -1

103


104

SSB (Cont.)SSB (Cont.)Substitute withSubstitute with

105

SSB (Cont.)SSB (Cont.)Φ LSB (ω) = M+ ( ω + ωc ) + M- ( ω - ωc )

ΦLSB (t) = m+ (t) e -jωct + m- (t) e jωct

Substitute withSubstitute with

106


107

Determination of mDetermination of mhh (t) (t)

108


Hilbert transformHilbert transform

109


110


.

.

111

SolutionSolution

112


113


114

DSB is passed through BPF to eliminate undesired bandDSB is passed through BPF to eliminate undesired band

Most commonly usedMost commonly used

To obtain USB, the filter should pass all components To obtain USB, the filter should pass all components above above ωωcc and attenuate all components below and attenuate all components below ωωcc

Difficult to design sharp cutoff filterDifficult to design sharp cutoff filter

Voice spectrumVoice spectrum

300 Hz300 Hz 115

Generation of SSB signals (Cont.)Generation of SSB signals (Cont.)

Difficult to build116

Envelope detection of SSB with Envelope detection of SSB with carrier (SSB+C)carrier (SSB+C)

If ‘A’ is large enough If ‘A’ is large enough m(t)m(t) can be recovered by can be recovered by envelop detection envelop detection

117

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

118


120

SolutionSolution

Diodes conducting Diodes conducting

The diode resistance is ‘ r’ The diode resistance is ‘ r’

off state off state

The diode resistance is ‘ ∞’ The diode resistance is ‘ ∞’

A carrier in positive cycleA carrier in positive cycle

A carrier in negative cycleA carrier in negative cycle

The output voltage is ‘ zero’ The output voltage is ‘ zero’

The diode acts as a gate with gain ‘2R/(R+r)’The diode acts as a gate with gain ‘2R/(R+r)’

(R/(r+R))(R/(r+R))φφ (t) appears across each resistance ‘R’ (t) appears across each resistance ‘R’

121

Solution (Cont.)Solution (Cont.) The output is, eThe output is, e00 (t) = (2R/(R+r)) (t) = (2R/(R+r)) ωω (t) m (t) (t) m (t)

If eIf e00 is passed through B.P.F centered is passed through B.P.F centered

at at ωωCC and has 2B band width, the output is and has 2B band width, the output is ( 4R/ ( ( 4R/ ( ππ ( R + r ) ) m ( t ) Cos ( ( R + r ) ) m ( t ) Cos (ωωCC t ) t ) 122

Solution (Cont.)Solution (Cont.) This circuit can work as a demodulatorThis circuit can work as a demodulatoras follows: as follows:

The input is m (t) cos The input is m (t) cos ωωCC t t

LPF is used at the output LPF is used at the output

The output The output ( 2R/ ( ( 2R/ ( ππ ( R + r ) ) m ( t ) ( R + r ) ) m ( t )

123

SolutionSolution

ee00(t) =(t) = ( 4R/ ( ( 4R/ ( ππ ( R + r )) m ( t ) cos ( R + r )) m ( t ) cos ωωCC t t

== k m ( t ) cos k m ( t ) cos ωωCC t t

m ( t )m ( t ) = sin (= sin (ωωCC t + t + θθ ) )

ee00(t)=(t)= k sin ( k sin ( ωωCC t + t + θθ ) cos ) cos ωωCC t t 124

Solution (Cont.)Solution (Cont.)ee00(t) =(t) = k/2 (sin ( 2 k/2 (sin ( 2 ωωCC t + t + θθ ) + sin ) + sin θθ ) )

The LPF suppresses the sinusoid and transmits The LPF suppresses the sinusoid and transmits only the DC component only the DC component eeoo

//(t) = k/2 sin (t) = k/2 sin θθ

125

Phase difference between the two sinusoids

SolutionSolution

Attenuated by LPF

Removed by DC blocker

126

a b c d

SolutionSolution

128

SolutionSolution

129

Solution (Cont.)Solution (Cont.)

130

Amplitude modulation: Vestigial Amplitude modulation: Vestigial Sideband (VSB)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

VSB (Cont.)VSB (Cont.)

132


133


134


135

VSB (Cont.)VSB (Cont.)SolutionSolution

136

Use of VSB in television broadcastUse of VSB in television broadcast

137

Carrier AcquisitionCarrier 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?Why?Consider DSB-SC case:Consider DSB-SC case:

..

.138

Carrier Acquisition (Cont.)Carrier Acquisition (Cont.)

.

Filtered by LPFFiltered by LPF.

139

Carrier Acquisition (Cont.)Carrier Acquisition (Cont.)

140

Attenuation to the messageAttenuation to the message

.

.

Beating effectBeating effect

Solution

Quartz Crystal Oscillator? Difficult to built at high frequencies

Phase Locked Loop (PLL)

Phase Locked Loop (PLL)Phase Locked Loop (PLL)

Free running frequency Free running frequency ...

Suppressed by LPF Suppressed by LPF

141

PLL (Cont.)PLL (Cont.)

142

PLL (Cont.)PLL (Cont.)If the frequencies are differentIf the frequencies are different

Suppose the frequency of the input sinusoidalSuppose the frequency of the input sinusoidalsignal is increased from signal is increased from ωωcc to to ωωcc + k + k

This means that the incoming signal isThis means that the incoming signal is

Thus the frequency increase in incoming signal Thus the frequency increase in incoming signal causes causes θθii to increase to to increase to θθii + kt + kt

Increasing Increasing θθe, e, and vice versaand vice versa143

DC componentDC component

144

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

145

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

147

Angle Modulation (Cont.)Angle Modulation (Cont.)

149

Angle Modulation (Cont.)Angle Modulation (Cont.) Two techniques are possible to transmit mTwo techniques are possible to transmit m((tt))by varying angle by varying angle θθ of a carrier. of a carrier.

150


151


152

SolutionSolution

155

156



157


158


159

160

SolutionSolution


161


162

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


163


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

164


165


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.

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

.

.

.168

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

.

.

.

169

..

170

Narrowband Angle Modulation (Cont.)Narrowband Angle Modulation (Cont.)

Compare with AM modulated signalCompare with AM modulated signal

.

.

171

Generation of NBFMGeneration of NBFM.

172

Generation of NBPMGeneration of NBPM

173

Can not be ignoredCan not be ignored

Very complicated analysisVery complicated analysis174

In practical FM, In practical FM,

WBFM (Cont.)WBFM (Cont.)Simple way to analyze the problemSimple way to analyze the problem

m(t)m(t)

m(t) is band limited to B Hzm(t) is band limited to B Hz m(t) is approximated by pulses of constant m(t) is approximated by pulses of constant amplitudes (cells), amplitudes (cells),

FM analysis of constant amplitude is easierFM analysis of constant amplitude is easier

175

((StaircaseStaircase))

WBFM (Cont.)WBFM (Cont.)

m(t) ≈m(t) ≈ Pulse interval ≤ 1 / 2B SecPulse interval ≤ 1 / 2B Sec

The FM signal of one of these cells starting at t = tThe FM signal of one of these cells starting at t = tkk

1/2B Sec1/2B Sec

Sinusoidal Sinusoidal

176


The FM spectrum of The FM spectrum of consists of the sumconsists of the sum

of the Fourier transforms of the sinusoidal pulses. of the Fourier transforms of the sinusoidal pulses.

177

WBFM (Cont.)WBFM (Cont.)The min. and max. amplitude of modulating signals The min. and max. amplitude of modulating signals are -mare -mpp and m and mpp

the min. frequencythe min. frequency

the max. frequencythe max. frequencythe Bandwidththe Bandwidth

178

WBFM (Cont.)WBFM (Cont.).

.

Do not forgetDo not forget, this value is calculated for , this value is calculated for

From earlier analysis:From earlier analysis:

For NBFM, For NBFM,

179

The Peak Frequency Deviation:The Peak Frequency Deviation:


180

For a truly wideband case,For a truly wideband case,

A better bandwidth estimate is:A better bandwidth estimate is:

The “deviation ratio” plays a role similar to the The “deviation ratio” plays a role similar to the “modulation index” in AM. “modulation index” in AM.

.

181

SolutionSolution

183


184


185

Generation of FM WavesGeneration of FM Waves

186

• Direct FM Method.

• Indirect FM Method.

NBFM GenerationNBFM 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.)NBFM Generation (Cont.)

• Bandpass Limiter: used to remove amplitude variations in FM wave.

188

• The input-output characteristics of the hard limiter.


• Hard limiter input and the corresponding output.

189

• Hard limiter output with respect to θ


190


191


192

Demodulation of FM SignalsDemodulation of FM Signals

• The simplest demodulator is an ideal differentiator followed by an envelop detector.

193envelop

Demodulation of FM Signals (Cont.)Demodulation of FM Signals (Cont.)

194

envelop detector can be used to recover m (t )

FM Demodulation using PLLFM Demodulation using PLL

195

Free running frequency Free running frequency

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

Documents

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