COMUNICATION
PRINCIPLES (SEE 3533)
COMUNICATION
PRINCIPLES (SEE 3533)
Associate Professor Dr Mohamad Kamal A Rahim FKE UTM Skudai Johor
H/P 013 7484664
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
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
rdEd., 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
rdEd., Oxford Univ. Press
4.
Pearson, “Basic Communication Theory”, Prentice
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.
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.
INTRODUCTION TO
PRINCIPLES COMMUNICATION
SYSTEM (WEEK 1)
INSPIRING CREATIVE AND INNOVATIVE MINDS
INTRODUCTION TO
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
BASIC COMMUNICATION
SYSTEM
Transmitter Channel Receiver
Noise
Modulated
TRANSMISSION MODE
Simplex – One way transmission
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
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
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
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
DECIBEL (dB)
Is not an absolute value
Network P2
P1 dB = 10 log P2/P1
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
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
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
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
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
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
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
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
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)
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
/ /
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
CASCADED NETWORK
F1 G1 F2 G2 F3 G3 FN
GN 1 2 1 2 1 3 1 2
1
1
1
...
..
1
N NG
G
G
F
G
G
F
G
F
F
F
28 1 2 1 2 1 3 1 21
1
1
...
..
1
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
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)
MODUL 2
ANALOG MODULATION
TECHNIQUE
MODUL 2
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
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
AMPLITUDE MODULATION (AM)
t
E
t
v
c ccos
cA 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
mcos
mt
t
E
m
t
t
v
AM
c1
cos
mcos
cLet say the carrier signal
Information signal
The AM signal can be represented as
t mE
tmE t
E c c m
m c
c c
c
cos
2 cos
AM spectrum can be divided into three
•Carrier frequency
•Upper sideband
•Lower sideband
AM SPECTRUM
Ec
fc-fm fc fc+fm
LSB USB
Ec
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)
TYPES OF AMPLITUDE
MODULATION
Divided into FOUR types • AM
•DSBSC •SSBSC •VSB
Divided into FOUR types • AM
AM POWER
t E t mE
t mE
tv c c m
m c
c c
c
AM cos 2 cos 2 cos
AM signal
Total power transmitted = PUSB + PLSB + Pc
2 1 m2
P PT c
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
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
SSB-SC GENERATION
vm(t)
vc(t)
LPF
vSSBSC(t)
vc(t)
tE E t
v m c c m
SSB-SC GENERATION
+ -vm(t)
vc(t)
VSSB(t)
tE E t
VSSB( ) m c cos c m
-90o
90o
If adder is used
If subtractor is used
t E
E t
VSSB( ) m c cos c m
SIMPLE DETECTOR DSB-SC/SSB-SC
vDSBSC(t) or
LPF
vm(t) vDSBSC(t) or
vc(t)
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
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
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)
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
FREQUENCY MODULATION (FM)
The modulated signal can be represented by
)
sin
(
cos
)
(
t
E
t
t
V
FM
c
c
mwhere
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
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
VFM c c m c m The equation can be expanded using Bessel function Bessel.
Wideband FM
t E
t t
vfm c cos c sinm
t E
t
t
t
t
vfm c 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
NARROWBAND FM
)
sin
(
cos
)
(
t
E
t
t
V
FM
c
c
mIf 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 ) (
FM POWER
The power signal of FM is constant before or after modulation where
R
E
P
cT
2
2
From Bessel Function
21
2 2 ( )
)
(
n n
o
T J J
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
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))
FM GENERATION
(DIRECT METHOD)
FM signal can be generated using varactor diode
The tuned circuit is sued to change the resonance frequency
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
1/22 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
WBFM GENERATION
NBFM multiplierX n + multiplierX n
VWBFM(t) down converter
X n
multiplier multiplierX n
Local Oscillator
VLO(t)
QUIZ 4
X
-+
VNBFM(t)
400 kHz + 24.4 Hz
x32 x96Vm(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 .
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
VFM c 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
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
TUNED CIRCUIT
f2
f1
fin V
o
f1
fc f2
f1
-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
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
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
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
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 19kHz 2R
15 19 23 53 kHz
MODULE 3
PULSE
MODULATION
MODULE 3
PULSE
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
mPULSE MODULATION
t Pulse signal
vd(t)
t
Information signal
vm(t)
t
PAM
vPAM(t)
t
PWM
vPWM(t)
t
PPM
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 ts
1 2 cos 2 ) ( n s n s s T nt c T A T A t
s
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 tv
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
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
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)
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.
PULSE AMPLITUDE
MODULATION (PAM)
dc level PAM Generation
Vm(t)
Vm(t) + dc
Vd(t)
VPAM(t)
LPF
PAM Detection Vd(t)
PULSE WIDTH
MODULATION (PWM)
V to T converter
VPAM(t) VPWM(t)
PWM Generation
PWM Detection
LPF
PULSE POSITION
MODULATION (PPM)
d/dt
VPWM(t) VPPM(t)
PPM Generator
PPM Detector
LPF
PULSE CODE
MODULATION (PCM)
Three main process – Sampling Quantizing
Coding
S/H Quantizer Coding P/S
LPF D/A
S/P Vs(t)
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
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
MODULE 4
DIGITAL
RADIO MODULATION
MODULE 4
DIGITAL
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
ASK GENERATION
ASK Generation
Vs(t) LPF VASK(t)
Vc(t)
BPF
Vs(t) VASK(t)
Switch
ASK DETECTION
VASK(t) Vs(t)
Envelope detector
Coherent detector
VASK(t) Vs(t)
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
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
FSK GENERATION
0
fc1
V1(t)
VFSK(t)
1
fc1
fc2
FSK DETECTION
c
logic
Comparator
c
VFSK(t) logicComparator
c
VFSK(t) V
COHERENT FSK
RECEIVER
VFSK(t)
cos ( c+ )t
Differential Amplifier
cos ( c- )t
+
-Low Pass Filter
multiplier
BIT ERROR RATE
FSK(ASYNCRONOUS)
o b E e eP 2
2 1
The probability for this receiver can be written as
2
T
E
E
c b
Whereo/2 = (AWGN)
Eb the average energy for FSK
SYNCRONOUS FSK
RECEIVER
VFSK(t)
cos ( c+ )t
Differential amplifier
+ LPF
cos ( c- )t
-Low Pass Filter
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 be erfc E
P
6 . 0 2 1 whereEb average bit energy
o/2 = AWGN double spectrum noise and erfc
PSK GENERATION
Vs(t)
-1
VPSK(t)
Generation using switch Generation using multiplier
Vs(t) VPSK(t)
LPF -1
fc Vc(t)
PSK DETECTOR
VPSK(t)
Vs(t)
BIT ERROR RATE
PSK
o b e erfc EP
2 1
The probability for PSK signal at the receiver
)
Eb can be given as this equation
2 2T E E c b )
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
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
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
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
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
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
EXAMPLE
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
MODULE 5
MULTIPLEXING
TECHNIQUE
MODULE 5
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
FREQUENSI DIVISION
MULTIPLEXING(FDM)
f1
f2
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
TIME DIVISION
MULTIPLEXING(TDM)
Sampel suis Agihan suis
Q F T F R
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
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
SPACE DIVISION
SPACE DIVISION
MULTIPLEXING(SDM)
MULTIPLEXING(SDM)
Transmitter 1 Transmitter 2 Transmitter 3 receiver 1 receiver 2 receiver 3 transmission cableSDM 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