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(1)

COMUNICATION

PRINCIPLES (SEE 3533)

COMUNICATION

PRINCIPLES (SEE 3533)

Associate Professor Dr Mohamad Kamal A Rahim FKE UTM Skudai Johor

H/P 013 7484664

(2)

LECTURER

Assoc. Prof. Dr Mohamad Kamal A Rahim

BSc(Hons) Electrical & Electronic Eng; University of

Strathclyde, U.K; 1985-1987

Master Eng Sc (Communication); University of New

South Wales, Australia; 1991-1992

Phd Electrical Eng.(Antenna design): University of

Birmingham, U.K. 1999- 2003

Email :

mkamal@fke.utm.my

Room : P18

Ext: 36088

Assoc. Prof. Dr Mohamad Kamal A Rahim

BSc(Hons) Electrical & Electronic Eng; University of

Strathclyde, U.K; 1985-1987

Master Eng Sc (Communication); University of New

South Wales, Australia; 1991-1992

Phd Electrical Eng.(Antenna design): University of

Birmingham, U.K. 1999- 2003

Email :

mkamal@fke.utm.my

Room : P18

(3)

REFERENCES

1.

Rozeha, Alias, M. Kamal, M. Rijal, Kamaludin,

Norhafizah,

Prinsip Kejuruteraan Telekomunikasi

,

Fakulti Kej. Elektrik, UTM

 2. Abu Sahmah Supaat, etal, Principles of

Communication, Pearson

3.

B.P.Lathi, “Modern Digital and Analog

Communication System”, 3

rd

Ed., Oxford Univ. Press

4.

Pearson, “Basic Communication Theory”, Prentice

Hall

1.

Rozeha, Alias, M. Kamal, M. Rijal, Kamaludin,

Norhafizah,

Prinsip Kejuruteraan Telekomunikasi

,

Fakulti Kej. Elektrik, UTM

 2. Abu Sahmah Supaat, etal, Principles of

Communication, Pearson

3.

B.P.Lathi, “Modern Digital and Analog

Communication System”, 3

rd

Ed., Oxford Univ. Press

4.

Pearson, “Basic Communication Theory”, Prentice

(4)

COURSE OUTCOMES

At the end of the course the students should be able to:

CO1 Describe basic principles of analog and digital communication systems

CO2 Analyze various types of analog and digital communication transmitters and receivers

CO3 Explain the practical concepts of communication system

CO4 Able to use software tool to understand and solve problems in communication system.

C05 Be able to work in a team and communicate effectively At the end of the course the students should be able to:

CO1 Describe basic principles of analog and digital communication systems

CO2 Analyze various types of analog and digital communication transmitters and receivers

CO3 Explain the practical concepts of communication system

CO4 Able to use software tool to understand and solve problems in communication system.

(5)

PROGRAMME OUTCOMES (POs)

1 Ability to acquire and apply knowledge of mathematics, science and engineering.

2 Ability to analyze and interpret data/graph/circuit.

3 Ability to identify, formulate and solve electrical engineering problems. 4 Ability to work with modern instrumentation, software and hardware. 5 Ability to design a system, component or process to fulfill certain

specifications.

6 Ability to communicate effectively.

7 Ability to function and be productive in a team.

8 Ability to recognize the need for, and to engage in life-long learning. 9 Understand the impact of the work of engineers on society.

10 Understand ethical and professional responsibility.

1 Ability to acquire and apply knowledge of mathematics, science and engineering.

2 Ability to analyze and interpret data/graph/circuit.

3 Ability to identify, formulate and solve electrical engineering problems. 4 Ability to work with modern instrumentation, software and hardware. 5 Ability to design a system, component or process to fulfill certain

specifications.

6 Ability to communicate effectively.

7 Ability to function and be productive in a team.

8 Ability to recognize the need for, and to engage in life-long learning. 9 Understand the impact of the work of engineers on society.

(6)

INTRODUCTION TO

PRINCIPLES COMMUNICATION

SYSTEM (WEEK 1)

INSPIRING CREATIVE AND INNOVATIVE MINDS

INTRODUCTION TO

(7)

INTRODUCTION

• Any RF or Microwave communication system consist of transmitter, channel and receiver

• The transmitter delivers the carrier modulated by information through an antenna

• The receiver recovers the information from the received signal from the antenna

• In this topic various parameters used for transmitter and receiver will be discussed.

7

• Any RF or Microwave communication system consist of transmitter, channel and receiver

• The transmitter delivers the carrier modulated by information through an antenna

• The receiver recovers the information from the received signal from the antenna

(8)

BASIC COMMUNICATION

SYSTEM

Transmitter Channel Receiver

Noise

Modulated

(9)

TRANSMISSION MODE

Simplex – One way transmission

(10)
(11)

APPLICATION SPECTRUM

FREQUENCY

Typical frequencies Application:

AM broadcast band 535-1605 kHz VHF TV (5-6) 76-88 MHz Shortwave radio 3-30 MHz UHF TV (7-13) 174-216 MHz FM broadcast band 88-108 MHz UHF TV (14-83) 470-890 MHz VHF TV (2-4) 54-72 MHz Wireless LAN 2.4 GHz

Microwave ovens 2.45 GHz Bluetooth 2.4 GHz

UWB system – 3.1 to 10.6 GHz Wimax 2.3 GHz, 2.5 GHz and 3.4 GHz Typical frequencies Application:

AM broadcast band 535-1605 kHz VHF TV (5-6) 76-88 MHz Shortwave radio 3-30 MHz UHF TV (7-13) 174-216 MHz FM broadcast band 88-108 MHz UHF TV (14-83) 470-890 MHz VHF TV (2-4) 54-72 MHz Wireless LAN 2.4 GHz

Microwave ovens 2.45 GHz Bluetooth 2.4 GHz

(12)

MICROWAVE BAND

Frequency

(GHz)

Wavelength (cm) IEEE band

1 - 2 30 - 15 L

2 - 4 15 - 7.5 S

4 - 8 7.5 - 3.75 C

4 - 8 7.5 - 3.75 C

8 - 12 3.75 - 2.5 X

12 - 18 2.5 - 1.67 Ku

18 - 27 1.67 - 1.11 K

27 - 40 1.11 - 0.75 Ka

(13)

BANDWIDTH (BW)

The limit between upper frequency and lower frequency

The relationship between bandwidth and quality factor Q is BW = f/Q f – resonance frequency

Q – quality factor

If Q high the BW is small If Q low the BW is wider

The limit between upper frequency and lower frequency

The relationship between bandwidth and quality factor Q is BW = f/Q f – resonance frequency

Q – quality factor

(14)

CLASSIFICATION OF SIGNAL

Deterministic signal

- Can represent as a function of mathematic - e.g. sinusoidal signal or squarewave signal

Random signal

- The signal can’t be predicted and can’t be represented as a function - e.g. noise signal

Analogue Signal – continous signal

Digital signal – discrete or can represent as binary signal

Deterministic signal

- Can represent as a function of mathematic - e.g. sinusoidal signal or squarewave signal

Random signal

- The signal can’t be predicted and can’t be represented as a function - e.g. noise signal

Analogue Signal – continous signal

(15)

DECIBEL (dB)

Is not an absolute value

Network P2

P1 dB = 10 log P2/P1

(16)

dBm dBW dB

dBV dBk

dBm is referred towards 1 milli

dBW or dBV is referred towards watts or voltage dBm is referred towards 1 micro

dBm= 10 log P/1mW

dB = 10 log P/1 W

dBm = 20 log V/1mV

(17)

EXAMPLE

1. Find the ratio between output and input power in dB if the input power is 10 mW while the output is 10 W

2. If the input power is 1 mW

(18)

EXAMPLE

Determine the power ratios in dB for the following input and output powers (i) Pin = 0.001 W, Pout = 0.01 W

(ii) Pin = 0.25 W, Pout = 0.5 W (iii)Pin = 0.0001 W, Pout = 0.1 W (iv)Pin = 0.04 W, Pout = 0.16 W (v) Pin = 0.01 W, Pout = 0.4 W

18

Determine the power ratios in dB for the following input and output powers (i) Pin = 0.001 W, Pout = 0.01 W

(19)
(20)

TYPES OF COMMUNICATION SYSTEM

• Public Switch telephone Network

• Cellular Telephone System

• Internet Communication System

• Radio and TV communication System

• Satellite Communication System

• Microwave Communication System

• Public Switch telephone Network

• Cellular Telephone System

• Internet Communication System

• Radio and TV communication System

• Satellite Communication System

(21)

NOISE IN COMMUNICATION SYSTEM

Electrical energy that interfere with the telecommunication system Divided into TWO

•Internal Noise – cause by the active and passive devices in the circuit

•External Noise – man made noise such as noise motor or lightning Electrical energy that interfere with the telecommunication system

Divided into TWO

•Internal Noise – cause by the active and passive devices in the circuit

(22)

THERMAL NOISE

Internal noise

Due to the movement of atom and electron in the resistive components

According to kinetic theory, the power is proportional to absolute temperature and bandwidth

TB

P

n

Internal noise

Due to the movement of atom and electron in the resistive components

According to kinetic theory, the power is proportional to absolute temperature and bandwidth

kTB

(23)

EQUIVALENT CIRCUIT

The noise received due to the resistive component can be represented by Thevenin and Norton circuit as below

or

Rn R

L

R

Vn RL In R RL

From maximum power transfer theorem, The maximum power can be proved with noise voltage and noise current as

kTBG I

kTBR V

n n

4 4

2 2

 

Vn RL In R RL

(24)

SHOT NOISE

Cause from the active devices in the receiver circuit.

Due to the changes as arandom electron or hole that reached at the collector of atransistor

It can be written as

dc

qBI

I

2

2

qBI

dc

(25)

NATURAL SOURCES

OF RECEIVER NOISE

Two types of noise encounter by receiver

Noise picked up by the antenna – sky noise, earth noise, static or atmospheric noise, galactic noise and man made noise

Noise generated by receiver – Thermal noise, shot noise, flicker or 1/f noise.

The quality of the output signal from the receiver is expressed in term of its signal to noise ratio (SNR)

25

Two types of noise encounter by receiver

Noise picked up by the antenna – sky noise, earth noise, static or atmospheric noise, galactic noise and man made noise

Noise generated by receiver – Thermal noise, shot noise, flicker or 1/f noise.

The quality of the output signal from the receiver is expressed in term of its signal to noise ratio (SNR)

(26)

RECEIVER NOISE FIGURE

The noise figure of a system depends on a number of factors such as losses in the circuit, kind of the solid state device, bias applied and amplification.

The noise factor F of a two port network is defined as

26

The noise figure of a system depends on a number of factors such as losses in the circuit, kind of the solid state device, bias applied and amplification.

The noise factor F of a two port network is defined as

p o

p i

SNR

SNR

F

/ /

(27)

NOISE FIGURE

F, G, Nn

Ni=kTB N

o

Si So

27

So=GSi

The output noise No = GNi + Nn

Substitute F = (SNR)i/p/(SNR)o/p = No/GNi Therefore No = FGNi

(28)

CASCADED NETWORK

F1 G1 F2 G2 F3 G3 FN

GN 1 2 1 2 1 3 1 2

1

1

1

...

..

1

N N

G

G

G

F

G

G

F

G

F

F

F

28 1 2 1 2 1 3 1 2

1

1

1

...

..

1

(29)

EQUIVALENT NOISE TEMP. (T

eff

)

The equivalent noise temperature is defined as Te = (F-1)To

Where To = 290 K (room temperature) and F is in ratio. Therefore F = 1 + Te/To

For cascaded

29

The equivalent noise temperature is defined as Te = (F-1)To

Where To = 290 K (room temperature) and F is in ratio. Therefore F = 1 + Te/To

For cascaded

T1 G1 T2 G2 T3 G3 TN

GN 1 2 1 2 1 3 1 2

1 ... ....

      N N e T GT GTG G G T G

(30)

Quiz 1

Calculate the total noise factor and total effective

noise temperature for the amplifier below

Amplifier 1 (T1 = 5K, G1 = 30 dB)

Amplifier 2 (F2 = 6dB, G2 = 20 dB)

Amplifier 3 (F3 = 12 dB, G3 = 40 dB)

Discuss the significant of changing the position of the

amplifier

Calculate the total noise factor and total effective

noise temperature for the amplifier below

Amplifier 1 (T1 = 5K, G1 = 30 dB)

Amplifier 2 (F2 = 6dB, G2 = 20 dB)

Amplifier 3 (F3 = 12 dB, G3 = 40 dB)

(31)

MODUL 2

ANALOG MODULATION

TECHNIQUE

MODUL 2

(32)

MODULATION

 

t

E

t

v

c

c

cos

c

A proses to shift from low frequency to high frequency or a process of changing one of the characteristic (amplitude, frequency or phase) of a carrier wave.

A carrier wave can be represented by

 

t

E

t

v

c

c

cos

c

A modulating wave can be represented by

 

t

E

t

(33)

WHY NEED MODULATION

PROCESS

• To increase the efficiency of the transmission system • Information can be separated

• Multiplexing process

• Signal to noise ratio increased

•To overcome the limitation of channel transmission • To increase the efficiency of the transmission system • Information can be separated

• Multiplexing process

• Signal to noise ratio increased

(34)

AMPLITUDE MODULATION (AM)

 

t

E

t

v

c c

cos

c

A process of changing the amplitude of the carrier signal according to the instantenous information signal

Let say the carrier signal

Information signal

v

m

 

t

E

m

cos

m

t

 

t

E

m

t

t

v

AM

c

1

cos

m

cos

c

Let say the carrier signal

Information signal

The AM signal can be represented as

t mE

t

mE t

E c c m

m c

c c

c

 cos

2 cos

(35)

AM spectrum can be divided into three

•Carrier frequency

•Upper sideband

•Lower sideband

AM SPECTRUM

Ec

fc-fm fc fc+fm

LSB USB

Ec

(36)

Example

AM signal is obtained from v

m

(t) = 3 cos (2 x

10

3

)t with carrier signal v

c

(t) = 10 cos (2 x

10

6

)t. Determine

1. Modulation index (m =0.3)

2. Upper and lower frequency (USB = 1001 kHz, LSB = 999kHz)

3. Amplitude ratio between sideband and carrier in dB(-16.4dB)

4. The maximum and minimum amplitude of the modulated signal (Emax = 26 V, Emin = 14 V) 5. Draw the spectrum of AM signal

AM signal is obtained from v

m

(t) = 3 cos (2 x

10

3

)t with carrier signal v

c

(t) = 10 cos (2 x

10

6

)t. Determine

1. Modulation index (m =0.3)

2. Upper and lower frequency (USB = 1001 kHz, LSB = 999kHz)

3. Amplitude ratio between sideband and carrier in dB(-16.4dB)

(37)

TYPES OF AMPLITUDE

MODULATION

Divided into FOUR types • AM

•DSBSC •SSBSC •VSB

Divided into FOUR types • AM

(38)

AM POWER

 

t E t mE

t mE

t

v c c m

m c

c c

c

AM  cos2 cos 2 cos

AM signal

Total power transmitted = PUSB + PLSB + Pc

   

 

 

2 1 m2

P PT c

(39)

AM GENERATION

Simple AM generation can be shown in figure below

Id

Vo RL

vm(t)

Vo RL

vc(t)

The non liner diode can be written as

2

d d

d

AV

BV

I

(40)

DSB-SC GENERATION

The DSB-SC can be shown in figure below by using the multipler

vm(t)

vDSBSC(t)=vm(t) x vc(t)

vm(t)

vc(t)

t

t

E

E

t

(41)

SSB-SC GENERATION

vm(t)

vc(t)

LPF

vSSBSC(t)

vc(t)

t

E E t

v m c c m

(42)

SSB-SC GENERATION

+ -vm(t)

vc(t)

VSSB(t)

t

E E t

VSSB( )m c cos cm

-90o

90o

If adder is used

If subtractor is used

t E

E t

VSSB( )m c cos cm

(43)

SIMPLE DETECTOR DSB-SC/SSB-SC

vDSBSC(t) or

LPF

vm(t) vDSBSC(t) or

vc(t)

(44)

TRANSMITTER AM/SSB/DSB

AM vm(t)

vc(t) (560 – 1600 kHz)

DSB/S

SB BPF

10.7 MHz

USB =10 695 MHz LSB = 10.7 MHz

(45)

AM RECEIVER

Mixer IF Amp BPF

12 kHz

AF Amp

vs(t) Antena

vLO(t) ( 955 – 2056 kHz) RF Amp

IF BPF 12 kHz

Detector

(46)

SSB/DSB RECEIVER

Mixer1 IF Amp

BPF

AF Amp

vs(t)

Antenna 4.055 – 2.955 MHz

coherent detector

vLO(t) 13.655 MHz 14.755 MHz

RF Amp IF

vs(t)

(47)

FREQUENCY MODULATION (FM)

A process of changing the frequency carrier according to instantenous of information signal

Let say the carrier signal and information signal is represented by

A process of changing the frequency carrier according to instantenous of information signal

Let say the carrier signal and information signal is represented by

)

(

cos

)

(

t

E

t

V

c c c

)

(cos

)

(

t

E

t

(48)

FREQUENCY MODULATION (FM)

The modulated signal can be represented by

)

sin

(

cos

)

(

t

E

t

t

V

FM

c

c

m

where

Ec – the amplitude of carrier signal

= f / fm – modulation index

m = 2 fm– information frequency

where

Ec – the amplitude of carrier signal

= f / fm – modulation index

(49)

WIDEBAND FM

The equation can be expanded using Bessel function Bessel.

This is called WBFM used in radio broadcasting system

)] sin

( sin sin

] sin

( cos [cos

)

(t E t t t t

VFMc c mc m The equation can be expanded using Bessel function Bessel.

(50)

Wideband FM

 

t E

t t

vfmc cos c sinm

 

t E

t

t

t

t

vfmc cosc cos sinm sinc sin sinm

Dari persamaan isyarat termodulat FM

Isyarat di atas dapat dikembangkan seperti berikut

Fungsi bessel

sinmt

dan sin

sinmt

cos Fungsi bessel

Ia dapat diselesaikan dengan nilai

   

    

 

genap

n n m

o

mt J J n t

sin 2 cos

cos

  

ganjil

n n m

mt J n t

sin 2 sin

(51)

NARROWBAND FM

)

sin

(

cos

)

(

t

E

t

t

V

FM

c

c

m

If small << 0.2 the angle is small and from FM signal

The result is Narrowband FM

t E t E t E t

VNBFM c c c c m c cos( c m) 2 ) ( cos 2 cos ) ( 

(52)

FM POWER

The power signal of FM is constant before or after modulation where

R

E

P

c

T

2

2

From Bessel Function

2

1

2 2 ( )

)

(

 

n n

o

T J J

(53)

SUMMARY OF FM SIGNAL

1.Power after or before modulation is same.

2.Bandwidth will be incresed when the increased 3.If small the bandwith is same with AM signal 4. FM signal has infinite sideband

5.The carrier signal for FM can become zero

Bandwidth = 2 x fm x number of sideband

(54)

QUIZ 3

FM modulated signal is given as

If the carrier signal is given as 1 MHz and the

modulator sensitivity is 4000 rad/s/V

(i)Find the modulating signal (v

m

(t))

(ii) Find the instantenous frequency (f

i

(t))

(iii) Draw the power spectrum

)

6280

sin

5

(

cos

10

)

(

t

t

t

V

FM

c

FM modulated signal is given as

If the carrier signal is given as 1 MHz and the

modulator sensitivity is 4000 rad/s/V

(i)Find the modulating signal (v

m

(t))

(55)

FM GENERATION

(DIRECT METHOD)

FM signal can be generated using varactor diode

The tuned circuit is sued to change the resonance frequency

(56)

FM GENERATION

LC fc 2 1 

Varactor diode is designed to have a bigger range of capacitance The resonance frequency can be written as follow

Let C increased with the increment of C. The resonance frequency will vary with f

 

1/2

2 1 C C L f

fc c

  

1

/

1/2

f

c

C

C

Let C increased with the increment of C. The resonance frequency will vary with f

(57)

WBFM GENERATION

NBFM multiplierX n + multiplierX n

VWBFM(t) down converter

X n

multiplier multiplierX n

Local Oscillator

VLO(t)

(58)

QUIZ 4

X

-+

VNBFM(t)

400 kHz + 24.4 Hz

x32 x96

Vm(t) = 2 cos 2 x 103t

antena

Amplifier

90o

Cos x x t)

Ec sin ct

Phase modulation

Vm(t) = 2 cos 2 x 103t

fLO=11.7792 MHz

Find the carrier frequency and modulation index at the antenna .

(59)

DEMODULATION FM

(DRECT METHOD)

Direct method

To recover the original signal

Using discriminator and envelope detector

vFM(t) vs(t)

 ) (t V m ) sin ( cos )

(t E t t

VFMc c m

) sin ( sin ) sin ( sin ) cos ( ) ( t t E t t t E dt t dV m c c m c m m c c FM      

(60)

DIFFERENTIAL CIRCUIT

V1 V2 V1 V2

V2

C R R

L

V2

1/RC atau L/R V1//2

V1

Differential circuit L and C

s c

V

V

(61)

TUNED CIRCUIT

f2

f1

fin V

o

f1

fc f2

f1

(62)

-FM DEMODULATION

(INDIRECT METHOD)

Indirect method – Phase Lock Loop

Based on comparing signal between VCO and input signal

X LPF Vm

VFM v

e

dc cos

X LPF

VCO VCO o/p at

t + At t

2 + dc

Vcon cos

When Vcon = 0

(63)

PLL ANALYSIS

Knowing that VFM = vi cos it and VVCO = Vo cos ( it + ) The multiplier will produce VFM.VVCO

= ViVo cos it cos ( it+ )

= ½ ViVo [cos(2 it+ ) + cos )

After filter an amplifier is located so that the level of is achieved for Vcon

the output Vcon cos vs

Knowing that VFM = vi cos it and VVCO = Vo cos ( it + ) The multiplier will produce VFM.VVCO

= ViVo cos it cos ( it+ )

= ½ ViVo [cos(2 it+ ) + cos )

After filter an amplifier is located so that the level of is achieved for Vcon

(64)

FM RECEIVER

Mixer IF Amp

BPF

200 kHz

Detector AF Amp

Antenna fRF

88 – 108 MHz

vLO(t) ( 77.3 – 97.3 MHz) RF Amp

IF = 10. 7 MHz BPF

200 kHz vm(t)

deemphasize

(65)

INTERMEDIATE

FREQUENCY (IF)

1. AM Receiver (540 – 1650 kHz) IF = 455 kHz 2. AM SSB Receiver IF = 1.6 – 2.3 MHz

3. FM Receiver (88-108 MHz )IF = 10.7 MHz

4. TV VHF Receiver ( 54 – 223 MHz), IF = 26 – 46 MHz 5. TV UHF Receiver ( 470 -940 MHz) IF = 46 MHz

6. Microwave receiver and RADAR, IF = 30, 60 dan 70 MHz 1. AM Receiver (540 – 1650 kHz) IF = 455 kHz

2. AM SSB Receiver IF = 1.6 – 2.3 MHz

3. FM Receiver (88-108 MHz )IF = 10.7 MHz

4. TV VHF Receiver ( 54 – 223 MHz), IF = 26 – 46 MHz 5. TV UHF Receiver ( 470 -940 MHz) IF = 46 MHz

(66)

STEREO RECEIVER

LPF (0-15 kHz) BPF (23-53 kHz + L+R L-R Composite signal 15 kHz, 23-53 kHz and 19 kHz Adder Subtrac tor 2L 2R BPF (23-53 kHz Pilot 19 kHz x2 38kHz Composite signal 15 kHz, 23-53 kHz and 19

kHz 2R

15 19 23 53 kHz

(67)

MODULE 3

PULSE

MODULATION

MODULE 3

PULSE

(68)

The carrier signal is a pulse signal while the information signal is the analog signal. Three types of pulse modulation

PAM – Pulse Amplitude Modulation (V

Vm)

PULSE MODULATION

PWM – Pulse width modulation

V

m

(69)

PULSE MODULATION

t Pulse signal

vd(t)

t

Information signal

vm(t)

t

PAM

vPAM(t)

t

PWM

vPWM(t)

t

PPM

(70)

SAMPLING

Purpose to change analog signal to discrete form with the sample represent the charactristic of the information

Let saya the pulse signat s(t) is called the sampling signal which represent by

    1 2 cos 2 ) ( n s n s s T nt c T A T A t

s

   1 2 cos 2 ) ( n s n s s T nt c T A T A t

s

(71)

Let say the information signal is represented by m(t) ) ( ). ( )

(t m t s t

v           1 2 cos 2 ). ( ) ( n s n s s T nt c T A T A t m t

v

SAMPLING

         1 2 cos 2 ). ( ) ( n s n s s T nt c T A T A t m t

v

s s s s s s s T t c T A t m T t c T A t m T t c T A t m T A t m t

v( )  ( )  ( )2 1cos 2  ( )2 2cos 4  ( )2 3cos 6

f

3fs

2fs2fs 3fs

fs

fm

(72)

SAMPLING THEOREM

If fs is to big it can avoid with overlaping signal but the bandwidth is becoming big

If fs is to small it can cause the signal overlapping and aliasing

occured. It will overlap with the replica of the sampling signal at nfs

f

3fs

2fs 2fs2 3fs2

fs2

fm

f

3fs

2fs 2fs1 3fs1 fs1

fm

(a) Sampling frequency,fs=fs1

(b) Frekuensi persampelan,fs=fs2

(73)

DIFFERENT SAMPLING

FREQUENCY

t t

Analog, signal m(t)

Sampling signals1(t)

Analog signal m(t)

Sampling signal s2(t)

t t

(a)Sampling period, Ts=T1

T1 T2

t t

Sampling signals1(t)

Sampled signal v1(t)

Sampling signal s2(t)

(74)

NYQUIST THEOREM

This theorem said that the sampling frequency must be greater than the information frequency or equivalent to two times of the maximum

information frequency fmax.

This theorem can be represented by fs> 2 fmax The bandwidth of the sampling frequency 2 fmax

This theorem said that the sampling frequency must be greater than the information frequency or equivalent to two times of the maximum

information frequency fmax.

(75)

PULSE AMPLITUDE

MODULATION (PAM)

dc level PAM Generation

Vm(t)

Vm(t) + dc

Vd(t)

VPAM(t)

LPF

PAM Detection Vd(t)

(76)

PULSE WIDTH

MODULATION (PWM)

V to T converter

VPAM(t) VPWM(t)

PWM Generation

PWM Detection

LPF

(77)

PULSE POSITION

MODULATION (PPM)

d/dt

VPWM(t) VPPM(t)

PPM Generator

PPM Detector

LPF

(78)

PULSE CODE

MODULATION (PCM)

Three main process – Sampling Quantizing

Coding

S/H Quantizer Coding P/S

LPF D/A

S/P Vs(t)

(79)
(80)
(81)

QUANTIZATION NOISE

This noise occur when the sampling volatage is differ from the quantization value

The biggest error occur can happened with ½ quantization size where can be written as

This noise occur when the sampling volatage is differ from the quantization value

The biggest error occur can happened with ½ quantization size where can be written as

dB

n

SNR

Q

1

.

76

6

.

02

(82)
(83)

EXAMPLE

An analog signal has a voltage of 0 – 5 V

(i) Calculate the quantization step when it represent 4, 6 and 8 bit

(ii) What is the voltage value for this sample digit: 10000000, 11100101 (iii) For bandwidth 6 kHz baseband signal , Calculate the minimum

bandwidth for the PCM signal which has 8 and 12 bit quantization An analog signal has a voltage of 0 – 5 V

(i) Calculate the quantization step when it represent 4, 6 and 8 bit

(ii) What is the voltage value for this sample digit: 10000000, 11100101 (iii) For bandwidth 6 kHz baseband signal , Calculate the minimum

(84)

MODULE 4

DIGITAL

RADIO MODULATION

MODULE 4

DIGITAL

(85)

INTRODUCTION

The information signal is the digital signal where the carrier frequency is a sinus signal at high frequency

Advantage - immune to noise, easy to store, regeneration and easy to measure

Disadvantages – wide bandwidth, analog signal must be convereted to digital signal, synchronization process and not suitable with analog

system

The information signal is the digital signal where the carrier frequency is a sinus signal at high frequency

Advantage - immune to noise, easy to store, regeneration and easy to measure

Disadvantages – wide bandwidth, analog signal must be convereted to digital signal, synchronization process and not suitable with analog

(86)
(87)

ASK GENERATION

ASK Generation

Vs(t) LPF VASK(t)

Vc(t)

BPF

Vs(t) VASK(t)

Switch

(88)

ASK DETECTION

VASK(t) Vs(t)

Envelope detector

Coherent detector

VASK(t) Vs(t)

(89)

BIT ERROR RATE (BER)

BER is the ratio between error bit (difference between bit transmitted and received bit) with respect to time range To

The performance can be shown in term of graph BER vs SNR

) ( ) ( density noise spectrum power bit per Energy E SNR  ) ( ) ( density noise spectrum power bit per Energy E SNR

(90)

BIT ERROR RATE ASK

o b e

E erfc

P

4 2

1

The probability of error Pe or BER for the ASK can be represented by:

2

T E E c

b

Eb the average energy for each bit

(91)

FSK GENERATION

0

fc1

V1(t)

VFSK(t)

1

fc1

fc2

(92)

FSK DETECTION

c 

logic

Comparator

c 

VFSK(t) logicComparator

c  

VFSK(t) V

(93)

COHERENT FSK

RECEIVER

VFSK(t)

cos ( c+ )t

Differential Amplifier

cos ( c- )t

+

-Low Pass Filter

multiplier

(94)

BIT ERROR RATE

FSK(ASYNCRONOUS)

        o b E e e

P 2

2 1

The probability for this receiver can be written as

2

T

E

E

c b

Where

o/2 = (AWGN)

Eb the average energy for FSK

(95)

SYNCRONOUS FSK

RECEIVER

VFSK(t)

cos ( c+ )t

Differential amplifier

+ LPF

cos ( c- )t

-Low Pass Filter

(96)

BIT ERROR RATE FSK

(SYNCHRONOUS )

The probabilty of BER for this syhncronous receiver is

         o b

e erfc E

P

6 . 0 2 1          o b

e erfc E

P

6 . 0 2 1 where

Eb average bit energy

o/2 = AWGN double spectrum noise and erfc

(97)

PSK GENERATION

Vs(t)

-1

VPSK(t)

Generation using switch Generation using multiplier

Vs(t) VPSK(t)

LPF -1

fc Vc(t)

(98)

PSK DETECTOR

VPSK(t)

Vs(t)

(99)

BIT ERROR RATE

PSK

         o b e erfc E

P

2 1

The probability for PSK signal at the receiver

)

Eb can be given as this equation

2 2T E E c b  )

(100)

MULTI LEVEL SYSTEM

One symbol is represented by many bits

As an example for 4 symbol can be represented by a combination of 2 bits (00.01.10 and 11) The phase will change from 0 to 3600

For binary PSK the phase change from 0 and 180 For QPSK the phase change 45, 135, 225 and 315 M = 2n

One symbol is represented by many bits

As an example for 4 symbol can be represented by a combination of 2 bits (00.01.10 and 11) The phase will change from 0 to 3600

For binary PSK the phase change from 0 and 180 For QPSK the phase change 45, 135, 225 and 315 M = 2n

(101)

MULTI LEVEL SIGNAL

4 digit baseband level

00 01 10 11 00 01 10 11

00 01 10 11 00 01 10 11

Constelation diagram

01 11

00 10

QPSK (4QAM)

0000 0001 1000 1001

0011 0010

1011 1010

1100 1101 0100 0101

1111 1110 0001 0110

(102)

QUADRATURE PHASE

SHIFT KEYING

v(t)

Low Pass Filter

fs=fb/2

I multiplier

v(t)

fb

Series to Parallel

cos ct

sin ct

Low Pass Filter

Q

(103)

QPSK RECEIVER

Band pass filter

Low pass

filter comparator

cos ct

I QPSK

fb/2 Band

pass filter

Low pass

filter comparator

Paralel to siri

sin ct

Q

fb/2

(104)

MULTI LEVEL

COMPARISON

Modulation Type Bandwidth (C/N) dB

BPSK 1 b/s/Hz 11.5

QPSK 2 b/s/Hz 14.5

8PSK 3 b/s/Hz 19.5

16PSK 4 b/s/Hz 25.5

(105)

COMPARISON BIT

ERROR RATE

Non coherent ASK

10-4

10-3

10-2

SNR

5 10 15 20

Non coherent ASK

Coherent ASK Coherent PSK

DPSK

Coherent FSK

Non coherent FSK 10-7

10-6

10-5

(106)

EXAMPLE

(107)

Example

The following specification of a communication system:

• Input fundamental freq fm = 4 kHz

• Maximum error probability = 1 x 10-9

• Bandwidh allowed = 26 kHz

• Eb= 4.55 x 10-3 Joule

• Noise density = No = 2 x 10-4 volts2/Hz

• For FSK the frequency difference between first and second carrier is 1.5 kHz. Transmission has to include the fundamental frequency and third harmonics

a. Using this information , choose the modulation technique (ASK, FSK, PSK) that suits the requirement. Justify your choice by showing the bandwidth and error probability

b. Based on the answer above, state the minimum storage capacity if the transmission duration is 10 seconds

The following specification of a communication system:

• Input fundamental freq fm = 4 kHz

• Maximum error probability = 1 x 10-9

• Bandwidh allowed = 26 kHz

• Eb= 4.55 x 10-3 Joule

• Noise density = No = 2 x 10-4 volts2/Hz

• For FSK the frequency difference between first and second carrier is 1.5 kHz. Transmission has to include the fundamental frequency and third harmonics

a. Using this information , choose the modulation technique (ASK, FSK, PSK) that suits the requirement. Justify your choice by showing the bandwidth and error probability

(108)

MODULE 5

MULTIPLEXING

TECHNIQUE

MODULE 5

(109)

MULTIPLEXING

Process of transmitting more than two signal simultenously Divided into

•Frquency Division Multiplexing

•Time Division Multiplexing

•Space Divison Multiplexing

•Code Division Multiplexing

frequency

Process of transmitting more than two signal simultenously Divided into

•Frquency Division Multiplexing

•Time Division Multiplexing

•Space Divison Multiplexing

•Code Division Multiplexing time

(110)

FREQUENSI DIVISION

MULTIPLEXING(FDM)

f1

f2

(111)

EXAMPLE FDM

BPF

BPF BPF

vm1(t)

vFDM(t) fc1

vm3(t) vm2(t)

fc2

BPF

vm3(t)

fc3

fc1 fc2 fc3 f

channel

1 channel2 channel3

Bandwidth of FDM depends on the number of sub-carrier used. For N sub-carrier, the bandwidth of FDM is:

m FDM Nf

(112)

TIME DIVISION

MULTIPLEXING(TDM)

Sampel suis Agihan suis

Q F T F R

(113)

samplevm1

sample vm2

samplevm3

commutator

A/D converter

TDM PCM

t1 t2 t3 t

channel 1 channel 2 channel 3

Sampled bitvm1

Sampled Bitvm2

Sampled Bitvm3

channel 1 channel 2 channel 3

Sampled Bitvm1

Sampled Bitvm2

Sampled Bitvm3

Frame period, TF

TDM PCM signal

All sampled bits for one complete cycle of the commutator is known as a

(114)

TDM FOR PAM

AND PCM

PAM dalam TDM

1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8

PCM dalam TDM

0 0 0 1 0 0 1 0 0 0 0 1 0 1 0 0 0 1 1 1 1 0 0 0 1 1 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1 1 1 0 1 0

1 2 3 4 1 2

(115)

SPACE DIVISION

SPACE DIVISION

MULTIPLEXING(SDM)

MULTIPLEXING(SDM)

Transmitter 1 Transmitter 2 Transmitter 3 receiver 1 receiver 2 receiver 3 transmission cable

SDM uses multiple cables

• The channel can still operates eventhough there is a fault in one of the cables.

• Easier maintenance works.

• Increase cable cost and the size of the cables becomes

bigger and entangled

Pemodulatan Digit Antena Tx1 Antena Tx2 Antena Tx3 Antena Rx1 Antena Rx2 Antena Rx3 Transmitter 1 Transmitter 2 Transmitter 3 D S P

Receiver 1

Receiver 2 Receiver 3 D S P transmission medium

SDM uses smart antenna

• The channel can still operates eventhough there is a fault in one of the cables.

• Easier maintenance works.

• Increase cable cost and the size of the cables becomes

(116)

THANK

YOU

Figure

Fig : The position of the sampled bit information signal

References

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