Điện tư cho Truyền thông

7/27/2019 in t cho Truyn thng

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Eecc f Cca

Trn Quang Vinh VNU - 2014 Sde 11


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Course Outline

2


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:

Jacb Ma,, Ed. McGaH

, , .

:

Ee e Maee, , Ed. Dd

Chhe Me, , Ed. Lae

Jea Aa, , Ed. Dd

Ta Te Lag, , Ed. Ma

Bdee, , Ed. Rad ( Teche de ge)

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Eece 15%

Mde Ea 25%

Fa Ea 60%

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Introduction for

Information Transmission System

1

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1. Electronic system for information transmission

Transmitter Input signal s(t), different frequency spectrums

Baseband processor (noise filter, amplification, coder, decoder, )

Modulator (carrier, mixer)

Output amplifier to a transmission channel (line, antenna)

Transmission channel and noise

Receiver

Demodulator (detector)

Baseband processor

Output signal s(t).

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Presentation of signal in the time domain

s(t)

t0

T = 2A

-A

() = ( + )() = (+T)

+>

2

f

S(f)

f0+ f1 f0+ f2

f

average

minmaxminmax

ff

)ff()ff(


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Meg a da ehd hch e chae baddh a.F eae, a caa cabe ha a baddh f 100' f Mh. Baebad eech a afe H

Multiplexing

Modulation and frequency shift

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Th a eea 'eage' be aaed f baebad, hee he ae a he ae feec bad, adace b eag a f he ec.

A eae f FDM badca ad (g ae LW, ed ae MW, .)

Frequency Division Multiplexing FDM


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TDM ahe f f eg baed ag hch a daeche. I TDM, ae f eea aage eage b, each eaed , ae aed a eece, .. he ae cc adacee .

Time Division Multiplexing TDM


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26

Transmitter

Receiver

X

X


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AM radio transmitter and receiver with dual frequency shift.


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1. Electronic Mixer

(analog multiplier)

2. Filters (low-pass, high-pass, band-pass, )

3. Oscillator and voltage-controlled oscillator

4. Amplifiers

5. Add and substrate analog signals

f0 ()=V.(2f0. ) () f= f0+ K ()

Ae() ()=A . e()

VCO

e1()

e2()

()= +e1() e2()+

e1()

e2()

() = e1() . e2()

Circuits for modulation and frequency shift


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)]sin()[sin(

2

1)cos().sin(

)]cos()[cos(2

1)sin().sin(

)]cos()[cos(2

1

)cos().cos(

bababa

bababa

bababa

++=

+=

++=

)cos())(sin(

)sin())(cos(

xdx

xd

xdx

xd

=

=

Some trigonometric function :

Non-linear circuit elements and mathematic tools


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Fourier transform :

)(.)2()()(

)(.)()().()(*)(

)().()(*)()(.)(

)(11)(

)()(

)()()()(

0

2 0

fXjffYFTdt

xdty

fYfXFTdtyXtytx

dfyXfYfXFTtytx

fFTFTt

ffXFTetx

fYbfXaFTtybtxa

n

n

n

tfj

==

=

=

++

+

+

Some properties of the Fourier transform :


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Phin tc thc chia lm 3 bng ln: Sng mt t ( Ground ware ), sng tri ( Skyware ) v sng truyn theo ng tm mt ( light of sight ) LOS.

Slan truyn sng in ttrn knh truyn

Sng mt t c di tn s< 2 MHz c khuynh hng truyn theo chu vi tri t, c

dng phbin trong cc i AM. y sphsng a phng theo ng cong mt tv tn hiu truyn trn ng chn tri thy c.

sbc xc hiu qu, antenna cn di hn 1/10 bc sng.V d: Vi sng mang fC = 10KHz, bc sng l: = C/fC = ( 3.108/ )/104H = 3.104 Nhvy, mt anten di t nht 3.000m bc xc hiu qumt sng in t10KHz!

5. Propagation of RF waves

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Sng tri c di tn s 2 30 MHz.Struyn ca sng ny da vo sphn xtng ion( ion sphere - tng in ly ) v mt t. Nh, c thtruyn mt khong rt xa.

struyn sng, gc phn xv shao ht tn hiu ti mt im phn xtng ion ty thucvo f, vo thi gian trong ngy, theo ma v stc ng ca vt en mt tri.Ban ngy, g bhp thu, c rt t sng trli tri t.

Ban m, xy ra hin tng khc xtng bc. Do sphn xnhiu ln gia tng ion vmt t, sng tri truyn i rt xa. V th, c nhng sng tri pht ra tnhng i xa bn kiatri t vn c ththu c trn bng sng ngn.

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Sng truyn thng LOS cho di tn s > 30 MHz., sng in ttruyn theo ngthng, c rt t sng bkhc xbi tng ion. Sng struyn ngang qua tng ny. Tnh cht c dng cho thng tin vtinh.Cch truyn LOS bt li cho vic truyn thng tin gia 2 trm mt t, khi m ng i tnhiu phi trn ng chn tri. cong mt t schn ng truyn LOS.

Anten pht cn phi t trn cao, sao cho anten thuphi thy c n.

d2 + 2 = ( + h )2

d2 = 2h + h2 h2


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A transmission line is a device designed to guide electrical energy from one point to another. It

is used, for example, to transfer the output RF energy of a transmitter to an antenna. This

energy will not travel through normal electrical wire without great losses.

Although the antenna can be connected directly to the transmitter, the antenna is usually

located some distance away from the transmitter.

A transmission line is used to connect the transmitter and the antenna.

The transmission line has a single purpose for both the transmitter and the antenna. This

purpose is to transfer the energy output of the transmitter to the antenna with the least possible

power loss. How well this is done depends on the special physical and electrical characteristics

(impedance and resistance) of the transmission line.

6. Transmission line

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Two-Wire Open Line Two-Wire Ribbon Type Line Twisted Pair

Shielded Pair Rigid (Air) Coaxial Line Flexible (Solid) Coaxial Lines

Waveguides

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Characteristic impedance, Z0, is the ratio of E to I at every point along the line. For maximum

transfer of electrical power, the characteristic impedance and load impedance must be matched.

The velocity at which a wave travels over a given length of transmission line can be found by

using the formula:

A transmission line that is not terminated in its characteristic impedance is said to be finite.

When DC is applied to an open-ended line, the voltage is reflected back from the open end

without any change in polarity, amplitude, or shape. Current is reflected back with the same

amplitude and shape but with opposite polarity.

When DC is applied to a short-circuited line, the current is reflected back with the same

amplitude and polarity. The voltage is reflected back with the same amplitude but with

opposite polarity.

When AC is applied to an open-end line, voltage is always reflected back in phase with the

incident wave and current is reflected back out of phase.

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7 A


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After an RF signal has been generated in a transmitter, some means must be used to

radiate this signal through space to a receiver. The device that does this job is the antenna.

The transmitter signal energy is sent into space by a transmitting antenna; the RF signal isthen picked up from space by a receiving antenna.

The cea aea a cdc, e f cdc, ha adae ece eecagec ae eeg. A dea aea ha a defe egh ad af daee, ad cee aed ace.

The aea be abe adae effce he e ed b heae aed. A effce ag aea hae eacde. The de ae deeed b he ag feece. Thede f he eceg aea ae cca f eae adfeece. Hee, a he feec f he gabeg eceed ceae, hedeg ad aa f he eceg aea bece e cca.

M acca ag aea ae dded bac cafca, he(hafae) aea ad Mac (aeae) aea.

7. Antennas

Introduction

37

Antenna


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38

Antenna


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Curent and voltage distribution on antenna

39

Antenna


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Standing waves of current and voltage on an antenna

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Antenna


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A half-wave antenna (referred to as a dipole, Hertz, or doublet) consists of two lengths of wire

rod, or tubing, each one-quarter wavelength long at a certain frequency. It is the basic unit

from which many complex antennas are constructed. The half-wave antenna operates

independently of ground; therefore, it may be installed far above the surface of the earth or

other absorbing bodies.For a dipole, the current is maximum at the center and minimum at the ends. Voltage is

minimum at the center and maximum at the ends, as shown in figure.

A half-wave antenna

Development of

Vertical and

Horizontal Pattern

41

Antenna


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A one-half wavelength antenna is the shortest antenna that can be used in free space. If we cut a

half-wave antenna in half and then ground one end, we will have a grounded quarter-wave or

Marconi antenna. This antenna will resonate at the same frequency as the ungrounded half-wave

antenna. Quarter-wave antennas are widely used in the military. Most mobile transmitting and

receiving antennas are quarter-wave antennas.

Quarter wave antennas

Mobile Antennas

Grounded Quarter-Wave Antenna Image

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2

43

1 Components


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Passive devices

()

()

A B

()

() A B

()

()

() = () () = ()/() = ()

We hae: () = d()/d

he () = C d()/d

= =

= 1/

1. Components

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Active devices

FET and BJT transistors

Operational amplifier

da

ce

gae

Nchae e

ee

bae

cec

+

V

V+

Ve

Ve

VSAd(f)0

0

0

0

11

+

=

+

=

j

A

f

fj

AA ddd

Ace a fe ee:

fT0 gf

Ad

0

Addb

f0LM741 TL081

Ad0 . f0 = fT

45

2 Frequency Mixing circuits


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2. Frequency Mixing circuits

()

()

() = .().()?

Mixing x(t) and y(t)multiplying x(t) by y(t)

:() =.()+.()2+.()

3

Linear region: vout(t) =A.vin(t)

() =.()

()

()

() =

.()+.()2

+.()3

()

()

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Need to be minimized (intermodulation terms)

nonlinear

device

y(t)

x(t) vin(t)vout(t)

multiplication

Consequence :

Any nonlinear device

can act as a mixer

Using a non-linear device to design mixers

47

If inis a signal with large amplitude:


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exp

lnx(t)

vs(t) =K.x(t).y(t)

lny(t)

cant be used at frequency higher than few MHz

very sensitive to temperature variations

works only for positives signals

BUT !

s(t) = RIs.exp(-e(t)/VT)

+

-e(t)

R

s(t)

V

I

+

-e(t)

Rs(t)

i = Is[exp(v/VT)-1]

V

s(t) = -VTln(e(t)/(RIs))

ln function : logarithmic circuit based on op-amp exp function : exponential circuit based on op-amp

Analogic multiplier based on op-amp

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Application domain : high frequencies ( < 2 GHz)

identical transistors

vs= RC((i4-i3)+(i6-i5))

x = va-vb

y = vd-vc

+

)TU

xexp(1

2i6i

++++

====

)TU

xexp(1

1i

4i++++

====

)TU

yexp(1

EI2i

++++

====

)TU

yexp(1

EI1i

++++

====

)TU

xexp(1

1i3i ++++

====

)TU

xexp(1

2i5i

++++

====

)vv).(vv(U4

IRv cdba2

T

ECs ====

The Gilbert cell

T5

T2T1

T3 T6T4

IE

vc

VCC

RC RCvs

vd

va vb

i3

i1 i2

i4 i5 i6

i3+i5 i4+i6

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Application domain :very high frequencies ( > 1 GHz)

RF(t)=LO(t).sign(HF(t))

RF(t)

t

HF(t) RF(t)

LO(t)

D1

D2D3

D4

HF(t)

t

LO(t)

t

HF(t)=A.sin(2f0.t) LO(t)=B.sin(2f.t)

A

D

B

C

N M

Diode ring mixer

Amplitude spectrum of RF wave.

f0 3f0 7f0f5f0

RF(f)

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3. Frequency filter


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First-order passive filters

51

q y


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First-order active filters using op-amp

52

4. Oscillator


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Conditions for generation of an electronic oscillation

Sinusoidal oscillators

T be abe aae he cag cd Bac feedbac de ed. Heehe ca bc: A afe hch cdeed be debad

ad a feedbac e ha a feec eece a h fge.

The Barkhausen oscillation criteria is the condition that the system will be self-generating

and will perform continuous oscillation:

53

Sinusoidal Oscillator


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Armstrong generator LC

Three-point generators LC

They are known as the Colpits oscillator and theHartley oscillator.

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Crystal Oscillator

High Quality factor Q = 104 105 The relative unstable is very small:

The crystal has two resonance frequencies: a

series resonance and parallel at S and P.

for s= j, we have the crystal impedance:

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RC oscillator at low-frequency

Phase-shift oscillator

Wien-bridge oscillator

The basic structure of the phase-shift

oscillator consists of a negative gain amplifier

(-K) with a three-section (third-order) RC

ladder network in the feedback. The circuit

will oscillate at the frequency for which the

phase shift of the RC network is 180 . Only atthis frequency will the total phase shift

around the loop be 0 or 360 .

56

Relaxation Oscillator


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Relaxation oscillators

Astable multivibrator

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Mono-stable multivibrator (One-shot)

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Bistable multivibrator (Latch)

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The VCO has its oscillation frequency controlled by a voltage input.

The frequency of oscillation is varied by the applied DC voltage, while modulating

signals may also be fed into the VCO to cause frequency modulation (FM) or phase

modulation (PM); a VCO with digital pulse output may similarly have its repetition rate

(FSK, PSK) or pulse width modulated (PWM). Voltage-controlled harmonic oscillators generate a sinus waveform.

Voltage-controlled relaxation oscillators can generate a sawtooth or triangular

waveform.

KVCO: gain

Center frequency. 10 GHz or higher

The Tuning range, 2

1

Variation in output phase and frequency as a

result of noise on the control line is important.

To minimize the effect, the VCO gain must be

minimized (in conflict with the tuning range)

Tuning linearity

Voltage-controlled oscillator (VCO)

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VCO


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LC sinusoidal VCOs using varicap

Relaxation VCO using IC - 55561


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A phase-locked loop (PLL) is an electronic circuit that consist of a phase detector, a low-

pass filter, and a voltage-controlled oscillator (VCO)

Phase-locked Loop (PLL)

The PLL can be used in a wide variety of applications:1. Frequency synthesis.

2. Frequency demodulation (detection).

The Phase Detector (PD) compares the phases of the input and output signals,

generating an error that varies with the VCO frequency until the phases are aligned,i.e. the loop is locked.

The PD output consists of a DC component (desirable) and high frequency

components (undesirable).

The PD output is therefore filtered by a low-pass filter (LPD).

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A

3

63

3.1. Introduction


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The carriersignalc(t), in continuous modulation is the sinusoidal wave.

The modulating signal m(t)is the baseband signal (message signal) varies one of three

parameters of carrier to leads to 3 basic types schemes created output modulated signal

o(t)known as Amplitude Modulation AM, Frequency Modulation FM, and Phase

Modulation PM.

These types of modulation are carrier/continuous wave modulation

Frequency & Phase Modulation are also known as Angle Modulation.

64

Example single tone modulation

Introduction to analog modulation


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m(t) is a sinusoidal wave which is wanted to transmit:

c(t) is also a sinusoidal in the high-frequency range:

where (t) is the instantaneous phase , (t) = ct + (t)

(t) is the deviation to the phase.

0(t) is the output modulated signal :

where K(t) amplitude of the modulated signal.

If m(t) creates a change on:

- K(t) Amplitude Modulation (AM)- (t) Phase Modulation (PM)- fi(t) Frequency Modulation (FM)- (t) and K(t) at the same time Angular modulation (QAM)

Example, single-tone modulation

65

Introduction to analog modulation


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Glancing signals in the time domain

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3.2. Amplitude Modulation


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Outline

1. Idc AM

2. Dbe SdebadSeed Cae Mda (DSBSC)3. Dbe Sdebad F Cae Ade Mda (DSBAM)

4. Sge Sdebad Mda (SSB)

5. QadaeAde Mda (QAM)

6. Feec ec hfg

67

1. Introduction


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Amplitude Modulation (AM) allows the baseband signal to vary the amplitude of a

carrier signal in accordance with the instantaneous voltage of the baseband signal.

In the case of single-tone transmission:

Modulating signal is the sinusoidal wave:

Carrieris the sinusoidal wave: where c>> m

Modulatedoutput signal 0is:

There are 4 kinds of Amplitude Modulation techniques, namely:

Double Sideband-Full Carrier maplitude modulation (DSBAM or AM)

Carrier + Upper Sideband + Lower Sideband

Double Sideband-Suppressed Carrier (DSBSC)

Upper Sideband + Lower Sideband Single Sideband (SSB)

Only one Sideband (Upper Sideband or Lower Sideband)

Quadrature Amplitude Modulation(QAM)

Transmit simultaneously two message signals in one channel.

68

M d l ti i d ( d l ti f t )

Introduction


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Modulation index m (modulating factor) :

AM waveform in time domain

= 1

< 1

> 1

69

Frequency spectrum by the trigonometric identity:

Introduction


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Thus, as the carrierfcis modulated by an information signalfm,

new signals at different frequencies are generated as part of the

process. These new frequencies are called side frequencies.

( ) ( )1

2 = + +

( ) ( )2 2

= + + +

Sidebands of complex modulating signal

If the modulating signal is a complex wave, such as voice or video, a whole range of

frequencies modulate the carrier, and thus a whole range of sidebands are generated.

The upper sideband fUSBand lower sideband fLSBare computed as:

The total bandwidth is simply the difference between the upper and lower sideband

frequencies:

70

2. Double Sideband Suppressed Carrier modulation ( DSB-SC)


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- 0(t) and c(t) processes the same frequency and are in phase

- k has the dimension volt

-1

amplitude of 0(t) is directly proportional to m(t)

Signal

(u.a.

)

mt (rd)

m(t)

320

Temporal evolution of 0(t) for an arbitrary signal m(t)

just a voltage multipier !

It is a balanced modulator

- 0(t) follows m(t) when the m(t) signal is positive

and m(t) for negative values

This modulation with suppressed carrier is not

directly used but this is the bases for more evoluated

modulations

m(t)

DSB-SC modulator

71

Time domain and frequency spectrum

DSB-SC


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Time domain and frequency spectrum

72

Waveform and spectrum of the single-tone AM wave.

Spectrum of complex baseband signal and its AM waves spectrum

Circuits for DSB-SC modulator

DSB-SC


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73

DSB-SC


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Mda

The cae e ae cdeab hghe

feec ad ade ha hedag ga.

The cae e ae ed a a ce ffad ad eee ba f he dde.

The cae he dde ff ad a a

hgh ae f eed.

The dde ac e che ha cec hedag ga a he ecda f he a f .

74

DSB-AM

3. Double Sideband Full Carrier modulation (DSB-AM)


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( ) ( )( ) ( )

1 1

1 1

= =

+

Thus, is the modulation index

In fact, the message signal is not only the sinusoidal wave, so m should be presented by an

arbitrary function -1 e(t) 1 :

with

DSB-AM modulator

75

Time domain

DSBAM


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Frequency domain and Bandwidth requirement for DSB-AM

In general, the message signal m(t) will not be a single 'sine' wave, but a band of frequencies

extending up toBHz as shown.

Signal(u.a.

)

mt (rad)

320

a) m < 1

1 + m e(t)

-(1 + m e(t))

Signal(u.a.

)

mt (rad)

320

b) m = 1

Signal(u.a.

)

mt (rad)

320

c) m > 1

Time domain

76

Power relations in DSB-AM waveform

DSBAM


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In radio transmission, the AM signal is amplified by a power amplifier and fed to the antenna

with a characteristic impedance that is ideally, but not necessarily, almost pure resistance. The

AM signal is really a composite of several signal voltages, namely, the carrier and the two

sidebands, and each of these signals produces power in the antenna. The a aede he f he cae e ad he e he debad USB

ad LSB:

he e a AM ga dbed ad cacaed b gg bac he ga AMea:

hee he f e he cae, he ecd e he e debad, ad he hde he e debad.

F e caca, ae be ed f he age. Ce f ea b ddg he ea ae b g b 0.707. The cae ad debad

age ae he

77

The power in the carrier and sidebands can be calculated by using the power formula2

DSBAM


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where P = V2

/R is the output power, Vis the rms output voltage, andRis the resistive part ofthe load impedance, which is usually an antenna.

Use the coefficients on the sine and cosine terms above in the power formula:

If e ee he debad e e f he cae e, he a e bece

Finally, we get a handy formula for computing the total power in an AM signal when

the carrier power and the percentage of modulation are known:

For 100% modulation (m = 1), he a debad e aa ehaf ha f hecae e. Whe he eceage f da e ha he 100, hee ch e e he debad.

78

Power with e(t)?

DSBAM


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)t(eofpowerthebeingPlet e

tcos)t(me(V)t(v cco += 1

>===1 fading

- Good security due to not knowing the carrier frequency limitary used.

Introduction

83

SSB - AM

SSB AM modulator


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The first step in generating an SSB signal is to create a DSBSC wave with no power is wasted

on the carrier. The second step is to filter out the wanted sideband (USB or LSB).

h () = c , e a e:

( ) ( ) ( )( ) ( )( )

= + + +

The SSB filter removes the LSB (say) and the output is

( ) ( ) ( )( )

= + +

For SSBSC, output signal = ( ) ( )( )2

= +

()

()

()

84

Single-tone SSB-AM

SSB - AM


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Modulating

voltage

Caeeedaef DSBSC

Sgedebadaef SSB

85

Sideband spectrum of SSB-AM

SSB - AM


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86

Power in SSB

SSB - AM


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he a e he DSB ga

Hece, f ad ae , he cae e ad e e debad a bedeeed. Aeae, ce SSB ga =

he he e SSB ga (Naed Aeage Pe)

2 2 2

12 4 4

= + = + +

( ) ( ) ( )( )2

= + +

22 2 2

2 2 82 2

= + = +

=2 2

2 8

+

87

SSB d l t i th filt i th d

SSB - AM


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SSB modulator using the filtering method

Baace

da

C02 10 MH

Baace

da

Fe Fe

C01100 H

300 3400 H

100.3 103.4 H 10.1003 10.1034 MH

Eag e de bad f a eed cae ga.

T eae:

A eehe ca h a he e debad:

0= 60 H; = 300 3400 H.

h hghe cae feec, bece e dffc de e eee fe eee.

= 600 H / 10 MH f 0 0.006%

0

88

SSB modulator using the phase shift method

SSB - AM


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SSB modulator using the phase-shift method

Baaceda

9090

Baaceda

+

Mc

M

Vc c

Vcc c

E0c(c)

E0/2[c(c) c (c+)]

E0/2[c(c) + c (c+)]

89

Applications of DSB and SSB

B th DSB d SSB t h i id l d i i ti SSB i l till


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Both DSB and SSB techniques are widely used in communication. SSB signals are still

used in some two-way radios. Two-way SSB communication is used in marine applications,

in the military, and by hobbyists known as radio amateurs (hams). DSB signals are used in

FM and TV broadcasting to transmit two-channel stereo signals and to transmit the color

information for a TV picture.

An unusual form of AM is that used in TV broadcasting.

Vestigial sideband transmission of a TV picture signal.

90

5. Quadrature Amplitude Modulation (QAM)


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QAM (quadrature amplitude modulation) is a method of combining two AM

signals into a single channel, thereby doubling the effective bandwidth. QAM is

used also with pulse amplitude modulation (PAM) in digital systems, especially in

wireless applications.

In a QAM signal, there are two carriers, each having the same frequency but

differing in phase by 90 degrees (one quarter of a cycle, from which the term

quadrature arises). One signal is called the I signal, and the other is called the Q

signal.

Mathematically, one of the signals can be represented by a sine wave, and the other

by a cosine wave.

The two modulated carriers are combined at the source for transmission. At the

destination, the carriers are separated, the data is extracted from each, and then the

data is combined into the original modulating information.

91

QAM Modulator

QAM


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s(t) = V1e1(t) cos(ct) + V2e2(t) sin(ct)

+

+

Sec

92

Example, a QAM mono-tone modulation

Cae :


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ttVm

ttVV

tV

tVVV

tttVVttVVtVts

cc

cc

cccc

ccccccc

2sincos2

2coscos2

2cos2

cos22

sincoscoscoscoscos)(

22

11

11

22

2

11

2

1

++++=

++=

ttVV

tV

ttVV

tVV

ttVVtttVVttVts

cc

cc

ccc

cccccccc

2sincos2

2sin2

2coscos2

cos2

sincossincoscossincos)(

11

22

22

2

22112

++=

++=

Vc(1

+L+R)

Vc(L R)

Cae :

Meage ga 1:

Meage ga 2:

93

Demodulator

QAM


94/272

/2

sr(t)

pr(t) = A1cos(2f0t + )

s1(t)

s2(t)

d1(t) = e1(t) + ...X

X d2(t) = e2(t) + ...

k

k

( )

( )

++=

++=

)sin(sin)(cos)()(

)cos(sin)(cos)()(

010220112

010220111

tkAtteVtteVts

tkAtteVtteVts

+=

=

cos2

)(sin2

)()(

sin2

)(cos

2

)()(

2211112

2211111

teVkAteVkAtd

teVkAteVkAtd

94

6. Frequency shifting

The bac ea ed gedebad da fac a

Frequency shift


95/272

The bac ea ed ge debad da fac af f .

SSB da ee efeed a , , .

The fe.

Bada fe baddh: ea ha f he daed ga ()ed a .

95

Due to frequency translation performed by the mixer : We may set

Frequency shift


96/272

The band-pass filter rejects the unwanted frequency and keeps the desired

one.

Mixing is a linear operation operation.

96

Frequency shift structure

Frequency shift


97/272

da

BFca HF ca

Fe bad

a

I

ga

Modulation with a first frequency

- Shift of the spectra to the adapted channel with a mixer

(the second operation is realized via a second multiplier)

97

Frequency shift of a modulated signal with a sine carrier

Frequency shift


98/272

k

c(t) = C1cos(1t)

s(t)

B is the bandwidth of s(t)

( ) ( ) ( ) ( )( ) ( ) ( )( )1 0 1 0 12 2 2 22

= + + + +

f+B+fHfB fH-B+fH

|S(f)|

+B-fB-B+fB0

If |f0-f1| > B spectrum of s(t) is formed with two componantes :

- one centered at : fB= |f0-f1| shifted to the lower frequencies

- the second centered on : fH= f0-f1shifted to the HF frequencies

The desired band is then selected with a band-pass filter

98

Amplitude modulation and Frequency shift

Frequency shift


99/272

e()

8.8

196156 & 196

15

20

176

0 5 20 25 156 196

20

191 196 201

15

AM modulation is between 440 and 490 kHz and the carrier waves are ranging

from 30 kHz to 3 MHz.

Deg f a a e h ade da ad e 8.8

99

Thee ae 2 a ehd f AM Deda:

3.3. AM Demodulation Analog modulation


100/272

Eee ched deec

Sched chee deda.

100

1. Envelope or non-synchroned detection


101/272

101

R i d i l|A(f)| baseband

s(t) C (1+m e(t)) cos( t) with m A k

Envelope Detection


102/272

R Cs(t) u(t)

demodulation : the ideal case m < 102

1

2

1

fRC

FM >>>>

Received signal :

fFM0-FM

R Cs(t) u(t)

B

u1(t)

u(t)

C0(1+m)

s(t)

t

C0(1-m)

-C0(1-m)

-C0(1+m)

tBm C0

-Bm C0

s(t) = C0(1+m e(t)) cos(ct) with m = A0k

102

Envelope Detection


103/272

M

2

0 Fm2

m1RCf2

1


104/272

Chaacec.

The dde chaacec f he f () = + + + ..., hee

..DSBAM ga( ) ( ) = +

104

.. ( )( ) ( ) ( )( ) ( )( )2

+ + + +

Envelope Detection


105/272

=

=

=

( ) ( ) ( ) ( ) ( )22 22

+ + + + +

( ) ( ) ( ) ( ) ( )22 1 1

2 22 2

+ + + + +

( ) ( ) ( ) ( )

( )2

2 222

2 2 2 2

+ + + + + +

'LPF' ee ce.

Output signal = ( )2

2

+ + ..he ca ()

105

Example,

Demodulation of an AM signal by a crystal radio


106/272

106

2. Synchronised or Coherent Detection

Synchronous Detection

Detector structureVx(t) ( )AM


107/272

m(t)

Vx(t) sout(t)

k

c(t) = Vccos(ct)Local oscillator

(synchronous)

AM

Analysing this for a DSBAM input : AM input = ( )( ) ( ) +

Vx= AM input LO =

= =

( ) ( )2 + ( ) ( ) ( ) = +

( ) ( )1 1

22 2

+ +

( ) ( ) ( )

( )2 22 2 2 2

+ + +

The AM input has been 'split into two' 'half' has moved or shifted up to :

and half shifted down to baseband, and( )

( ) ( )2 2 22

+

2

Vc ( )2

tm

107



108/272

The signal spectrum

from 'modulator to Vx'

108

Analysis

Sch deec hd e f bh ga f DSBAM ad DSBSC

Detect for AM input signal of DSB full carrier



109/272

Detect for AM input signal of DSB full carrier

The ea f DSB

dhed cae eed cae be e.

hee Ca f cae (DSBAM),

Hece, = AM I LO

Sce

( ) ( ) +

( )( ) ( ) ( )( ) = + + +

( ) ( )1

2 = + +

( )( ) ( )( ) ( )( )2

+ = + + + + + +

( )( )( ) ( )2

2 2

= + + + + +

( )( ) ( ) ( ) ( )( ) ( ) ( )2 22 2 2 2

= + + + + + + + + +

109

The LPF h a cff feec fcH ee he ce a 2c(.e.

ce abe c) ad hece( )

( )

= + + +



110/272

Ob, f ad e hae, a e

Consider now if is equivalent to a few Hz offset from the ideal LO:

The output, if speech and processed by the human brain may be intelligible, but wouldinclude a low frequency 'buzz' at , and the message amplitude would fluctuate. Therequirement = 0 is necessary for DSBAM.

( )2 2

= + + +

0= 0=

( )2 2

= +

( ) ( ) ( )2 2

= +

Consider now that = 0 but 0, i.e. the frequency is correct at cbut there is aphase offset. Now we have

'c()' cae fadg (.e. ade edc) f he .

( ) ( )

( )2 2

= +

110

The 'VC' ce a, b cde f (),



111/272

f (900), .e.

(1800),f .e.

The hae e f = a be a be f eech c, b a bea be f h e f da ed dedae PRK

Hee, he a be ha a ceae ad he ga egh

ge eae (fade) ad a he e

2

= 0=

2

cos

( )0=

2

cos

2

tm=Vout

2

=

( )( ) ( )tm=cos

2

tm=Vout

2

( ) 1=cos

2

111



112/272

Thus the requirement for = 0 and = 0 is a 'strong' requirement for DSB amplitudemodulation.

112

The ea f SSB h a cae deedg VC

Detect for AM input signal of SSB



113/272

.e. ag

Hence

=

( ) ( )2

+ +

( ) ( ) =

( ) ( ) ( )( )2

= + + + +

( )( ) ( )

( )( ) ( )( )

22 2

24 4

+ + + + +

+ + + + +

113


114/272

The Local Oscillator (LO) must be synchronised or coherent, i.e.at the same frequency andin phase, with the carrier in the AM input signal. If the LO is not perfectly synchronised to

Requirement for carrier recovery



115/272

-f f0ff + FMf - FM

|D(f)|

the carrier, it will be mixed between high and low frequencies after passing low-pas filter.

If the AM input contains a small or large component of the carrier frequency, the LO may be

derived from the AM input as shown below.

Conclusion :

Carrier recovering is required

115

Carrier recovering using PLL


Systems without the emission of the carrier wave. Recover the carrier wave is then needed :

Phase lock loop (PLL) or Quadratic filter.

Xsr(t) u(t)


116/272

When the PPL is stabilized both frequencies at the input of the phase comparator are equal

)()(0

tavftfi

+=

)(cos)( tAtyy

=

cos)(

2

1aVtav

dt

d==

++=

+

t

yy constduumeAakAtAakA

t

0

00)(

42

)(tanln

)2

2cos()(0

+= tfYty

VCO

dt

df

dt

dtf

i

2

10

2

1)( +=

=

0ff

c


117/272

0 2fm 2f02f0 - 2fm 2f0 + 2fm

Va(f)4fm

f

mf

fQ

2

0=

PLLZero level

comparator:2 pr(t)

sinus => square

w(t)

k

sr(t)

Centr 2f0

Q grand

v(t)w(t)

( ) ( )( ) ( )( )tf4costf4cos1tf4cos14

kA)t(v 0mm

2r +++=sr(t) = A cos(2fmt) cos(2f0t)

y(t), 2f0

117

* General case : e(t)



118/272

0 2Fm 2f0

Va(f)

f2FM

2f0- 2Fm 2f0+ 2Fm

PLL

Centered on 2f0

s(t) :2 pr(t)Zero level

comparator

sinus => square

Demodulated

AM output

118

Outline

3.4. Angle modulation Analog modulation


119/272

1. Idc

2. Phae da

3. Feec da

4. FM deec

119

1. Introduction


120/272

Phase and frequency modulation are examples of the angle modulation.

Before 1980, the main purpose of phase modulation is the generation of

frequency modulated wave for FM generators of commercial purpose. In the last 25 yeas, the phase modulation becomes the important thing for the

transmission of digital data.

Frequency modulation also to be applied for digital transmitters and is one of

two modulation forms which have signification in the century of 21.

120

Phase modulation

Structure of a phase modulator consists of the modulating input signal and the

carrier. The output phase modulated signal has a constant amplitude with a phase angle

i i l h li d f h d l i i l

Introduction


121/272

)tsinMktcos(Vv mpccout ++= 0

he haedaed age.Msinmt is the sinusoidal modulating signal c is the carrier frequency ca deeed b he cc aaee

When the maximum value of Vmis applied to the modulator, the largest phase deviationsoccur.

Then mp = kpMmax is called thephase modulation index.

varies proportional to the amplitude of the modulating signal.

121

zero phase signal

Advancing phase signal A chage hae a a chage

feec. Dffeeag he age

aae hae h eec

Introduction


122/272

time

)sin( 0 tMkt mpc ++=

tMkdt

dmmpc

cos+==

mpMk= mpMfkf =

Frequency deviation depends on both: amplitude and also frequency of the modulatingsignal !!

e chage hae, e hae a :

The a ada feec fdf:

The unmodulated frequency (M=0) is c, whereas the modulated frequency deviatesfrom this value in both the positive and negative directions by a maximum value of:

122

Frequency modulation

Frequency modulation occurs when the frequency of a carrier signal changes in accordance

to a modulating signal.

F l i i h f d i i i li i d d h

Introduction


123/272

tcosMkff mfc +=

Mkf f=

m

f

m

f

f

Mk

f

fm =

=Index of modulation:

Bandwidth: (Carsons rule): )(2 mffBW +=

For almost communication system, the frequency deviation is linearity depend on to the

amplitude of the modulating signal.

The expression for frequency of an FM signal with sinusoidal modulation is:

The feec dea f he FM ga

he a chage feec ad:

Frequency Deviation depends on only the amplitude of modulating signal:

The de f da f a edaed FM ga defed a he afeec dea f he aef dded b he da feec:

where kfis constant andMis magnitude of the

modulating signal applied to the frequency

modulator.

123

Introduction


124/272

Dg each cce f he dag ga, , he feec f e f fc fa,he bac fc, he fhe bac fc. The agde f ea ca.

The feec dea:

F FM, a chage dag feec , f, de chage f a d PM. A ceae dag ga ade, M, ceae fa, deceae f, ad hee b

ceae f.

F a ca fc f 100.1 MH ceca FM, he a feec dea, ed b

he FCC, fa= 75 H. Th, fa= 100,175 H ad f= 100,025 H. The afeec chage e ha 0.1%.

minmax fffff cc ==

124

2. Phase modulation

)tsinMktcos(Cvmpcout

++=0

mp = kpMmax

The phase varies linearly with the applied modulating signal. The maximum phase shift is

given by the index of phase modulation, which is:


125/272

p p max

Using 2 quadrature sinusoids of the same frequency create a resultant sinusoidwith phase shift

)sin( += tDv c

tAv csin1 =

tBv ccos2 =

22 BAD +=

ABactg=

ttmVv cmc cos)]cos1([2 +=

ttkMVv cmc cos)]cos1([2 +=

the phase will be modulated in accordance withM cos mt.The constant kis afunction of the amplitude modulator and kM equals m, the index of modulation.

Phase modulators based on amplitude modulators

g y p

125

Phasor representation

Phase modulator


126/272

of signals

With the index of modulation for AM signal 100% then results in a phase

deviation only of 7

A

tkMVactg mc

)cos1(

+=

The phase modulator must exhibit a linear relationship between the phase angle

and the modulating signal. It is adapted when actg

A

tkMV mc )cos1( +

126

Phase modulator

Phase modulator


127/272

90

A

1

2

out

- Limitation of the output amplitude

- The operating range is narrow: with the index of modulation for AM

signal 100% then results in a phase deviation only of 7

127

Circuit for large-deviation phase modulator

Phase modulator


128/272

Circuit used a varistor (silicon-carbide voltage variable resistor). The modulator

allows linear phase shifts of 50or greater and does not require an amplitude

limiter.

128

The frequency modulator should produce a frequency deviation

that is a function of the amplitude of the modulating signal, but is

3. Frequency modulation


129/272

p g g

independent of modulation frequency.

The phase modulator produce an output with a frequency that isdependent on both modulation signal amplitude and frequency.

If a modulating signal is integrated and then applied to a phase

modulator, a frequency-modulated signal results, with a frequency

deviation that is not a function of modulation frequency.

129

The indirect method of frequency modulation

Using the phase modulator with the frequency deviation do not depend on

f f th d l ti i l

Frequency modulator


130/272

frequency of the modulating signal

mpMk =

tfMkff mmpc cos+=

Phase

modulation

tMkff mfc cos+=Frequencymodulation

Mkff

=

tM

mm

sin1

11 Mk

Mk pm

m

p ==

130

== tMktdtMke mm

ImII

sincos

Frequency modulator


131/272

If a modulating signal is integrated and then applied to a phase modulator, a frequency-

modulated signal results, with a frequency deviation that is not a function of modulation

frequency

131

Frequency modulator


132/272

The Armstrong or Indirect method of FM waveform generation

132

The direct method of FM generation

Frequency modulator


133/272

Frequency reference signal is produced from the crystal oscillator

- dc error signal is a result of the frequency comparison between VCO and reference

133

1. Convert the frequency variation of the signal into an

amplitude variation and then applies an AM detector.

T a:

3.5. FM detection


134/272

2. Use Phase-Locked Loop for FM detection if an

accurate frequency reference is available

134

Foster-Seeley discriminator

FM Detection


135/272

Reace

Off eace

135

FM detector using the phase-locked loop PLL

FM Detection


136/272

PLL used as an FM detector

136

2 kind of Phase Comparator:

- Multiplier (balanced modulator) then low-pass filter

- Change to square-waveforms then XOR gate then LF or Integrator

FM Detection


137/272

137

FM Receiver

FM Detection


138/272

(max)][2 mffBW +=)(2 mffBW +=

138

FM Detection


139/272

Spectrum of baseband signal applied to FM modulator

An FM stereo transmitter

139

FM Detection


140/272

140

FM Detection


141/272

141


142/272

Htruyn thng s


143/272

Definition: the purpose of digital modulation is to convert an information bearingdiscrete-time symbol into a continuous-time waveform.

Why Digital ?

Increase System Capacity (compression, more efficient modulation)

Error control coding, equalizers, etc. possible to combat noise and interference =>lower power needed

Reduce cost and simplify designs

Improve security (encryption possible)

Basic concepts

Data rate : rate at which data can be communicated . In general it is the baud rate(the number of symbolsper second). In binary transmission, data rate is bit rate(the number of bit per second).

Each symbolrepresents nbits, and hasMsignal states, whereM = 2n. This is calledM ary signaling

Cc thng scbn


144/272

M-ary signaling.

Channel Capacity (C): The maximum rate at which data can be transmitted over agiven communication path, or channel, under given conditions the maximum rate ofinformation transfer through a baseband channel is given by:

Capacity fb= 2 B log2M bits per second,

where B = bandwidth of modulating baseband signal.

Bandwidth (B) : The bandwidth of the transmitted signal as constrained by thetransmitter and the nature of the transmission medium (Hertz)

Noise (N) : Impairments on the communications path

Bit error rate (BER) - rate at which errors occur : Error = transmit 1 and receive 0;transmit 0 and receive 1.

Nyquist Bandwidth :- For binary signals (two voltage levels) : C = 2B;

- Formultilevel signaling (M-ary signaling) : C = 2B log2M

Shannon Boundfor AWGN non-fading channel : C = Blog2(1+ S/N)

SNR l tsgia cng sut tn hiu v cng sut tp m. Khi nim ny dng chung cho

cthng tin sln analog. i vi thng tin sth SNR c thhin cthqua ts

Eb/No.

Tstn hiu trn tp trong thng tin sEb/No


145/272

b o

SNR (digital com.) = Ps/Pnvi Psv Pnln lt l cng sut tn hiu hu ch v cng suttp m.Ps= Es/Tsvi Esl nng lng ca mt symbol cn Tsl rng thi gian ca

symbol.

Khi ghp bt thnh mt symbol, Es= .Eb, vi Ebl nng lng mt bt. Do

Ps= .Eb/Ts.Cng sut tp m Pn= Ws.No, vi Wsl rng bng tn hiu truyn bng cc symbol cn

Nol mt phcng sut mt pha ca tp m cng trng chun.

Tuy nhin Ws= 1/Ts, do vy ta c: Ps/Pn = [.Eb/Ts]/[Ws.No] = . Eb/No.

phthuc vo siu chtn hiu nhiu mc, trnh phthuc vo kiu

iu ch, tstn/tp (SNR) trong thng tin sc thhin thng qua tsEb/No.

Given any modulation scheme, it is possible to obtain its signal constellation.

Represent each possible signal as a vector in a Euclidean space spanned by an

orthonormal basis.

If we know the signal constellation, we can estimate the performance in terms of theprobability of symbol error or probability of bit error given the noise parameters

Signal Constellation


146/272

probability of symbol error or probability of bit error given the noise parameters.

Probability of error depends on the minimum distance between the constellation points.

The receiver implementation can affect the performance.

Coherent detection

Receiver will exploit the exact knowledge of the phase of the carrier to detectthe signal better.

Non-coherent detection

Involves making some approximations to the phase information that results in

a loss in performance. However, it simplifies the circuitry.

In symbol detection decode incoming signal as closest symbol in the signalconstellation space

D

Constellation Diagram used to represents possible symbols that may be selected by a

given modulation scheme as points in 2-D plane X-axis is related to in-phase carrier: cos(ct)

The projection of the point on the X-axis defines the peak amplitude of the in-

phase component


147/272

p p

Y-axis is related to quadrature carrier: sin(ct)

The projection of the point on the Y-axis defines the peak amplitude of the

quadrature component

The length of line that connects the point to the origin is the peak amplitude of thesignal element (combination of X & Y components)

The angle the line makes with the X-axis is the phase of the signal element

Sng mang analog vi tn sthch hp c thtryn i xa trong mi trng truyn dn

(nhdy ng, cp ng trc, hay khong khng).

Cc kthut iu chsda trn bin i sng mang analog c phn loi cbn nhsau:

Phn loi kthut iu chs


148/272

iu chng bgm:

- ng bnhphn: ASK (t c dng), PSK, FSK

- ng bhng M: ASK hng M, PSK hng M, FSK hng M.

iu chkhng ng bgm:

- Khng ng bnhphn- Khng ng bhng M

One binary digit represented by presence of carrier, at constant amplitude.Other binary digit represented by absence of carrier. Commonly, one of theamplitudes is zero.

Amplitude Shift Keying (ASK)


149/272

Pulse shaping can be employed to remove spectral spreading.

ASK demonstrates poor performance, as it is heavily affected by noise andinterference.

d(t)

ASK(t)

A

C

Phtn sca tn hiu ASK


150/272

Mch iu chASK

DataModulated Signal


151/272

Sinusoidal Carrier

g

Mch gii iu chASK

LPF

Frequency Shift Keying (FSK)

Trong kthut ny i lng mang thng tin 1, 0 l tn sf1v f2ca sng mang. Cp

sng sin biu din c m tl:


152/272

f1v f2 lch so vi tn ssng mang fcmt lng bng nhau nhng ngc du.

B = 2([f2 f1]/2 + fb), trong fb = tc bit.

d(t)

FSK(t)

F

Hiu 2 tn ssng mang c tnh l : f2-f1=1/Tb= tn sbit.

Tn hiu FSK m ty l tn hiu pha lin tc (khi chuyn bit ttn sny sang tn skhc,

khng c snhy pha v chu kbit lun l bi ca chu ksng mang).Tp hm cssl:


153/272

Do 2 tn sl trc giao vi nhau (c thkim tra bng php ly tch phn tch 2 hm nytrong khong thi gian bit sbng zero) v cc hssij tng ng l:

HFSK c trng bng khng gian tn hiu 2 chiu v 2 im symbol (N = 2, M = 2).

Ch khong cch Euclid gia 2 vectl

Dliu s: vd(t) c iu chvi 1

vd(t) = 1 vd(t) c iu chvi 2.

Tn hiu c iu ch:

vFSK(t) = cos1t.vd(t) + cos2t(1-vd(t)) =

Phtn sca FSK


154/272

FSK c pha gin on:

Trong iu chFSK, tn stc thi ca sng mang c c thay i ty thuc dliu sbng gc.


155/272

FSK c pha lin tc:

u im: FSK t chu nh hng sai lch hn ASK do my thu nhn nhng tn sring


156/272

nn t chu nh hng ca nhiu kim. Nhc im: FSK c phrng gp i phASK.

ng dng: over voice lines, in high-freq. radio transmission, etc.

Mch iu chFSK


157/272

Ch l trong s 1 hoc 0 nhnh trn th qui tcngc li nhnhdi ().

Mch gii iu ch FSK

Vectquan st c (sau khi tn hiu qua knh) c 2 thnh phn l:


158/272

Khng gian quan st c chia thnh 2 vng (hnh v) c x1>x2v vng x2>x1


159/272

a vo mt bin mi l l=x1-x2khi :

V x1v x2l cc bin c lp thng k (do gn vi 2 hm trc giao) c phng sai = N0/2 nn

var[l] = var[x1] + var[x2] = N0.

Gis0 c truyn, hm khnng sau knh sl:

V x1>x2 tng ng l>0, nn:


160/272

:

:

Cui cng khi xt thm Pe1mt cch tng tta c

Dliu nhphn c biu din bi 2 tn hiu sng mang vi pha khc nhau trong BPSK. in

hnh hai pha ny l 0 v . 0t


161/272

vi m l snguyn, fbl tc bit dliu (data bit rate) th pha ban u ca mi bit schnh xcl 0 hay (hnh a). Nu khng th biu thc ca s1v s2skhng chnh xc vpha nhtrn (hnh

b). iu kin ny l cn thit m bo xc xut li bit l cc tiu. Tuy nhin nu fc>> fbth

iu kin ny c thc bqua.

Trong biu chm sao, cc tn hiu ny c biu din trong hta 2 chiu vi cc hm:

vi :


162/272

Ta ca 2 symbol sl:

u tin, lung dliu a(t) c to bi cc dliu nhphn:

trong ak{+1, -1} v p(t) l xung chnht c bin n vtrong khong [0,T].

Mch iu chBPSK


163/272

g k { } p( ) g g g

Sau a(t) c nhn vi sng mang hnh sin Acos 2fct.

Mch gii iu ch BPSK


164/272

Tn hiu chun phi c ng bvi tn hiu thu c vpha v tn s. Tn hiu nyc pht ra bi bhi phc sng mang CR (carrier recover).Khi khng c n (A = 1), li ra ca btch phn l:

Nu fc = mRb, shng thhai bng 0, nhvy tn hiu a(t) c hi phc hon ho.Nu fcmRb, shng thhai khc 0, tuy vy, nu fc >> Rbth shng ny c thbquaso vi shng thnht.

Tnh xc sut li bitchia khng gian thnh 2 vng: 1) vng gn v 2) vng gn


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T tnh c xc sut li loi 1 (pht 0 li quyt nh l 1 ti ni thu),

Vng quyt nh k hiu l 1 (tn hiu s1(t)) l Z1 vi 0 < X1< vi

y, x(t) l tn hiu thu c sau knh.

Hm xc sut iu kin l:


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Do :

i bin tch phn:

Ta c:

Tng tc thtnh c xc sut li pht 1 m thu c 0 c gi trcng nhvy.

Mt phcng sut PSD (power spectral density) ca tn hiu PSK:

Bit rng PSD ca cc sng skhng tng quan thng tng ng vi mt phnnglng ca xung chia cho di k hiu (symbol).

Vi xung cbn hnh chnht:


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Chuyn i F ca n l:

Nhvy PSD ca tn hiu bng gc PSK l:

Tkt qucho thy bng thng ca tn hiu l:

B = 2 /T = 2 fb


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So snh cc gi trPbca cc tn hiu PSK v FSK.

The error function (also called the Gauss error function) is a special function (non-elementary) of sigmoid shape which occurs in probability, statistics and partialdifferential equations. It is defined as:

The complementary error function, denoted erfc, is defined as:


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which also defines erfcx, the scaled complementary error function(which can be usedinstead of erfc to avoid arithmetic underflow.

Differential PSK (DPSK)

Gii iu chPSK yu cu phc hi tn hiu dliu my thu da vo sng mang chun c

pha tuyt i bit. iu ny i hi my pht phi gi mt tn hiu my thu tham khopha phc hi sng mang.

My thu PSK vi phn (DPSK) khng cn tn hiu sng mang chun. N dng s thay i

ca dliu iu chsng mang chkhng phi chnh dliu. thc hin vic ny, so

snh dliu hin hnh vi dliu vo trc , nu hai tn hiu ny ging nhau ta c mt


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y g g

pha ca sng mang v nu chng khc nhau ta c mt pha ngc li. Ni thu v pht phi

tha thun vi nhau vbit tham kho u tin trc khi pht dliu tn hiu c phc

hi ng nh pht i.

Mch iu chDPSK

Mch gii iu chDPSK


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Tn hiu ra mch tch phn l:

Khi khng c n v suy knh:

vi skv sk-1l cc k hiu hin ti v trc .

Li ra btch phn l dng nu tn hiu vo ging trc , ngc li l m. C ngha l

bgii m lm mt quyt nh da trn ssai khc gia 2 tn hiu. Do dliu thngtin phi c m ha nhl skhc bit gia cc tn hiu ln cn, l chnh xc nhng

g m bm ha khc bit thc hin.

Nu bt 1 c chn lm chun th quy tc m ha vi sai l:

Ngc li, bin i tdksang ak:

Nu dkv dk-1nhnhau, chng i din cho 1 dy ak, ngc li l 0. So snh cc chui

{dk} v {ak} trong bng. Dy {dk} c iu chvo sng mang vi pha 0 hoc . Li ra

l dy bit ca bn tin.


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Tlli bit Pbca dy li ra bgii m phthuc vo tc bit Pb,d ca dy m ha

trong gii iu ch l:

Thay gi trPb,d vo ta c:


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Pbca DPSK v PSK

iu chPSK hng M (M-ary PSK)

Trong MPSK, mt symbol c biu din bi (n = log2M ) bit dliu. Trong M l

hng ca iu ch.

Tp tn hiu M-ary PSK c nh ngha nhsau:

si(t) = A cos (ct+i) vi 0 t T , i = 1, 2, , M

trong :


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Tn ssng mang c chn l bi ca tc k hiu (symbol rate).

Thng M c chn l smca 2 (v d: M = 2n, n = log2M). Do vy dy sliu

nhphn c chia thnh cc nhm c n bit. Mi nhm c biu din bi 1 k hiu

(symbol) c pha ban u xc nh.

Vit li biu thc si:

trong l cc hm trc giao cs, v:

trong A = A2T l nng lng k hiu

Biu chm sao do c 2 chiu. Mi tn hiu si(t) c biu din bi mt im (si1, si2)

trong ta

Ta cc ca tn hiu l vi cc im tn hiu nm trn vng trn bn knhc tm ti gc ta .

M Grray thng c dng trong biu din tn hiu ca MPSK bi l 2 tn hiu lin k

trong trng hp ny chkhc nhau 1 bit. Khi mt k hiu bli, c nhiu khnng xy ra

t hi l t h d h t t bt l b li


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tn hiu ln cn trn chm sao,, do chmt trong cc bt vo l bli.

Hnh di l chm sao 8-PSK

dng m Gray. Lu rng PSK(BPSK) v QPSK l MPSK vi M

= 2 v M = 4.

Trn ton trc thi gian, tn hiu MPSK l:

trong :


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y, kl mt trong M pha c xc nh bi nhm n bit vo v p(t) l xung chnht n

vtrong khong[0,T]. Biu thc ca s(t) chng trng tn ssng mang l bi ca thi

gian k hiu pha ban u ca tn hiu trong bt kchu kk hiu no l k.

V rng tn hiu MPSK l 2 chiu nn vi cc M 4, cc biu chc thc hin kiutrc giao. Vi cc M khc nhau, chc kt cu ca bphn tch bit l thay i. Mi nhm n

bit li vo iu khin bphn tch cp cho knh I (ng pha) v Q (vung pha) cc tn hiu

v mc cho cc ta ngang v dc ca thchm sao. i vi trng hp QPSK, b

phn tch n gin chl bbin i ni tip song song.

Mch iu chMPSK.


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Hiu sut bng thng c nh ngha l tsvn tc bit (bps) trn bng thng yu cu (Hz).

Thng thng khi vn tc bit tng th bng thng tng, tuy nhin trong cc cch iu chkhc

nhau tshai i lng ny c thkhc nhau, do ngi ta dng hiu sut bng thng

nh gi cht lng ca hthng (hiu sut cao ng ngha vi tn dng c bng thng).

Nguyn nhn ca vic dng iu chM hng MPSK l n cho php tng hiu sut bng thng

Hiu sut bng thng:


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ca tn hiu ln n ln. V dvi QPSK (n = 2) bng thng knh truyn chcn bng so vi

PSK.

V d, vi ASK c vn tc bit 2400bps, tn ssng

mang l 1200Hz, bng thng sl 2400 Hz vy :

Vi QPSK 2400bps, iu chvn tc ca knh I v Q

sl 1200 bps, tn smang l 600Hz, bng thng cn

thit chl 1200 Hz vy:

V tp tn hiu MPSK chc 2 hm csnn my thu ng bMPSK c thdng 2 bnhn nhhnh sau:

Mch gii iu chMPSK.


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Psca MPSK (ng lin nt) v ca DMPSK(ng t nt).

4-PSK (QPSK) l loi mch MPSK thng c dng nht do n khng chu nh hng ca ts

li bit BER khi tng hiu sut sdng bng thng.V QPSK l trng hp c bit ca MPSK nn:

trong : Nhvy, c 4 pha ban u l:

Trong ta chm sao c th biu din 4 tn hiu ny bi 4 vector (im):

iu ch4-PSK (QPSK)


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Trong ta chm sao, c thbiu din 4 tn hiu ny bi 4 vector (im):

vi i = 1, 2, 3, 4.

Mch iu chQPSK


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- Btch bit (bit splitter) : chuyn dng dliu vo theo hai ng I v Q.

- Nhng bit vo I siu chsng mang c pha ban u v nhng bit vo Q siu ch

sng mang c lm lch pha 90 .

- V cc dliu vo c thl bit 1 hoc 0, nn tn hiu li ra mch nhn I c thl

sinct hoc - sinct v li ra Q c thl cosct hoc -cosct, cc tn hiu ny c

tng hp mch tng cho ra 1 trong 4 tn hiu c bin bng pha ban u

khc nhau.

cosctsinct + cosct-sinct + cosct

(11)(01)


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-cosct

sinct-sinct

-sinct - cosct sinct - cosct

(10)(00)

Dng sng QPSK


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Gii iu chQPSK


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Mch phc hi sng mang cho li sng sinct ttn hiu nhn c, tn hiu ny c cho

thng vo mch nhn knh I v c lm lch pha 90 trc khi vo mch nhn knh Q.

Tn hiu ra cc mch nhn c a vo mch lc thng thp loi bthnh phn tn

scao. Cc thnh phn DC sc tng hp mch tng cho li dng dliu. Gis

tn hiu vo l tn hiu nhn c, th d: r(t) ~ (cosct - sinct )

Tn hiu ra mch nhn knh I l:

sinct ( cosct - sinct) = 1/2sin2ct - 1/2(1-cos2ct)

Tn hiu ra sau mch lc l in thdc m, tng ng bit 0

Tn hiu ra mch nhn knh Q l:cosct ( cosct - sinct) = -1/2sin2ct + 1/2(1+cos2ct)Tn hiu ra sau mch lc l in thdc dng, tng ng bit 1

Mch thp bit scho li dliu nh pht : 01 (vit theo thtab)

Xc sut li bit trung bnh cho mi knh l:

V li ra ca bgii m l li ra dn knh I v Q nn tc li bt bng sbng gi tr

ny trn mi knh. Mi k hiu biu din cho 2 bit trn mi knh nn mt li k hiu

xy ra nu bt kknh ny c li/ Do xc sut li k hiul:


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Vi trng hp m Gray v c SNR ln:

S ph thuc Pb (Eb/No) caQPSK so vi PSK c ch ra

trn hnh.

Tc truyn thng thng ca

QPSK l 2400 bps v vy mch iu ch tc ca knh I


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mch iu chtc ca knh I

v Q l 1200 bps. Tc bin

i ln nht ca tn hiu tng

ng vi chui lin tip cc bit 1

v 0, chui ny c biu dinbi tn hiu hnh vung tn s

600 Hz, tn hiu hnh vung bao

gm tn scbn v cc ha tn

bc l.

iu chvi sai QPSK (DEQPSK)

Trong DEQPSK, cc nhm bit thng tin (dbit) c biu din bi cc lch pha i, tk

hiu ny ti k hiu khc. C cc php gn pha khc nhau gia v cc nhm bit. Mttrong cc khnng nhtrong bng:


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Quy tc lp m nhsau:

trong : Ik(0,1) v Qk (0,1) l cc bit thng tin tng ng lv chn

uk(0,1) v vk(0,1) l cc bit knh I v Q c m ha.Cp (IK,Qk) v (uk-1,vk-1) c dng sinh ra cp (uk,vk) dng iu khin pha tuyt i

ca sng mang.

Biu chDEQPSK vcbn ging nhtrong QPSK c thm blp m vi sai trn mi

knh cho tn hiu trc khi i vo bnhn vi sng mang.Bng m ha vi sai cho DEQPSK:


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Gii m DEQPSK

Vcbn nhgii m QPSK nhng phi thm vo bgii m vi sai sau khi c

gii iu ch. Quy tc gii m l:

V dkt qumt bng gii m DEQPSK:


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Sbgii m ng bDEQPSK ng b:


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Xc sut li bit c tnh xp x:

Offset QPSK (OQPSK) l cch iu chda trn nguyn tc ca QPSK nhng to s

lch pha ca hai tn hiu trn hai knh I v Q bng cch cho mt tn hiu trmt bit so

vi tn hiu kia.

Vic lm ny khin cho schuyn trng thi ca tn hiu knh ny (th dknh I)

lun lun xy ra ngay im gia ca tn hiu ca knh kia (knh Q), nhvy trong

mt cp bit IQ bt kchc sthay i ca mt bit duy nht v iu ny a n ktqu l cc tn hiu ng ra tng hp ch lch pha 0 hoc 90 ch khng phi 180

iu chOQPSK (Offset QPSK)


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qul cc tn hiu ng ra tng hp chlch pha 0 hoc 90 chkhng phi 180

nhQPSK.

im thun li ca OQPSK l gii hn c slch pha ca tn hiu ra v trnh c

cc xung t bin khi phc hi tn hiu nhphn. Tn hiu OQPSK c thc biu din nhsau:

V OQPSK chkhc QPSK mt chu ktrnn mt phcng sut v cc chtiu v

sai scng nhca QPSK.

c thso snh cc tn hiu cc li ra, xt chui tn hiu vo nhhnh (a) v chui

tn hiu ca 2 knh I v Q trong hai trng hp QPSK hnh (b) v OQPSK hnh (c).


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V tn hiu tng hp tng ng (a) v (b)


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C () () :

- Nu 2 bit trn 2 knh I v Q khc nhau hon ton th cc tn hiu tng ttng ng khc

nhau 180o

- Nu 2 bit trn 2 knh I v Q chkhc nhau mt bit th cc tn hiu tng ttng ng khc

nhau +90o hoc -90o.im bt li ca phng php OQPSK l sthay i pha ca tn hiu ra xy ra trong tng

khong thi gian T (chkhng phi 2T), do vn tc iu ch(baud rate) v bng thng ti

thiu ca knh truyn tng gp i so vi phng php QPSK .

iu ch/4- PSK

/4-PSK l loi m vi phn DQPSK, nhng khc vi DQPSK quy tc m ha vi sai:

trong uk l bin ca u(t) trong k hiu thk, v.v Ik v Qk ly gi tr(-1,1). Nu

ban u u0 = 1 v v0 = 0 th uk v vk c bin l 1,0 v 1/ .

Tn hiu ra b iu ch s l:


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Tn hiu ra biu chsl:

y: phthuc vo dliu c m ha,v :

Mi quan hvpha gia 2 k hiu lin tip bng:

trong l lch pha c xc nh bi

dliu vo.

Thay vo phng trnh trn ta c:

Bng sau cho kt quxc nh k theo dliu vo:

Thy rng:

- cc thay i vpha l bi slca /4 (khng c pha 90cng nh180)


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- Thng tin vsbin i pha k chkhng phi pha tuyt i k.

- Trong biu chm sao, gc ca mt vector i

vi hng dng ca trc u l pha ca k hiu

k.

- Mt k hiu c biu din bi () chc thtr

thnh cc k hiu c biu din bi (x) vngc li.

Bgii m ng b/4-QPSK nhhnh, trong d bgii m vi sai c thc hin trn

cc pha tn hiu vo. Bgii m DEQPSK khng thch hp y do nhm bit (dbit) c

gn khc nhau.


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GisAk= 1, tn hiu gii iu chl hai mc (1/ ) tt cthi gian ly mu. Giacc khong l 3 mc (0, 1). Trn biu , cc tn hiu () ng vi 2 mc v (x) ng

vi 3 mc. Khi tn hiu l 2 mc, chuyn mch vtr A v vic tch sng ging nh

QPSK. Khi tn hiu l 3 mc, chuyn mch vtr B, tn hiu c bin i thnh 2 mc

theo cng thc:

Dthy kt qubin i l squay vector (x1k, y1k)

i +/4 v khuch i bin ln Ni cch khc

quay vector (x) ti vector () bn cnh v nhn binln , tc l tng cng sut tn hiu ln gp i.

Tuy vy, cng sut n cng tng gp di do n trn 2

knh l khng tng quan nhau. Do BER cng vn

nhtrong gii m ng bQPSK.


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Tn hiu v c gii m bi mt bgii m DEQPSK, sau c a qua b

bin i song song ni tip . Xung nhp nhn c tnhp k hiu chia hai. Pha ca

nhp ng hny c ng bvi chuyn mch chn tn hiu ly mu.

Trong iu chQAM cbin v pha ca sng mang u thay i

Mch iu chQAM 8 pha

iu chQAM (iu chbin trc giao - Quadrature Amplitude Modulation)


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Trong mch iu chny a,b xc nh cc tnh ca tn hiu ra mch bin i, ring bit c

uc a thng vo hai mch bin i m khng qua mch o nhPSK 8 pha, nu c =1

chai li ra c bin cao v nu c = 0 chai li ra c bin thp. Nhvy, vi QAM 8pha, cc tn hiu cc li ra ca mch bin i lun c cng bin , gin vtr cc

im c trng cc tribit cho hnh sau.


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Cc tn hiu ra ca QAM 8 pha c 2 bin v 4 pha khc nhau.

So snh cc cch iu chQAM v PSK ngi ta thy QAM tt hn vmt ts

tn hiu nhiu. Th dvi hthng QAM 16 pha xc sut li l 10-8 trong lc PSK

16 pha xc sut ny l 10-4. Do trong cc hthng truyn vi vn tc cao ngi

ta thng dng cch iu chQAM hn.

Mch iu chQAM 16 pha


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Trong s, mch chia bit chia thp 4 bit theo hai knh vo hai mch bin i 2 ra 4mc, cc bit a,b xc nh cc tnh tn hiu ra v cc bit c,d xc nh bin

a,b = 0, tn hiu ra m c,d = 0 bin = 0,22 V

a,b = 1 tn hiu ra dng c,d = 1 bin = 0,821 V

Mi li ra ca mch bin i c thc 1 trong 4 tn hiu 0,22 hoc 0,821. Mch LPF

loi bcc ha tn cao. Cc tn hiu sau vo mch iu chcn bng nhtrong ccphn trc v li ra ta c 1 trong 16 tn hiu, cc tn hiu ny nhn 3 gi trbin v

12 gc pha khc nhau, khong cch cc gc pha l 30 .

Vi cch iu chQAM 16 pha, mi 4 bit tng ng mt tn hiu ra nn vn tc bit bng 4

ln vn tc baud. Nu chn vn tc baud l 2400 baud/s tha bng thng ca knh thoi th

vn tc bit l 9600 bps v hiu sut bng thng l 4 bps/Hz. Trong trng hp ny bng thngtn hiu trong khong t500 Hz (1700 Hz - 1200 Hz) n 2900 Hz (1700 Hz + 1200 Hz)

Trong gin trn gc A xc nh bi:


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Th dvi thp cc bit li vo nhtrong hnh l

1001, ta c cc kt qusau:Li ra knh I : +0,22 V

Li ra knh Q : -0,821 V

Li ra mch iu chknh I : +0,22 cosct

Li ra mch iu chknh Q : -0,821sinct

Li ra mch lc di thng : 0,22 cosct -0,821sinctTn hiu ra tng ng c xc nh trn gin bi

du X.

Phc hi sng mang

Vi kthut iu chFSK vic phc hi sng mang khng cn thit.

Tuy nhin, iu chPSK hay QAM tng tvi kthut iu chtrit sng mang, do

cn thit phi c mch phc hi sng mang my thu. Hn na, sng mang c phc hiphi c tn sv pha ging nhmy pht mch gii iu chmy thu hot ng hu

hiu.

Skhi mt mch phc hi sng mang cho trng hp iu chBPSK cho nhhnh.


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Tn hiu nhn c my thu l +cosct hoc -cosct, sau khi qua mch lc di thng (

hn chdi tn) squa mch bnh phng cho li ra cos2ct. Dng bin i lng gc

ta c:

cos2ct =(1/2)(1+cos2ct)

Tn hiu ny li qua mch lc loi bthnh phn mt chiu, cn li tn hiu tn s2c ,

tn hiu ny li qua mch chia tn c sng mang. Vng kha pha trong mch c tc

dng gipha ca tn hiu ra khng blch so vi tn hiu vo.i vi cc tn hiu iu chPSK bc cao hn (4-PSK, 8-PSK, 16-QAM . . .) th mch

gii iu chsnng tn hiu vo ln theo cc ly tha bc cao hn. Dnhin mch sphc

tp hn.


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i mt sng cha tin analog thnh tn hiu ri rc, trc thi gian phi c ri rcho. Si trc thi gian lin tc thnh mt trc ri rc c thc hin nhphngphp ly mu.

nh l ly mu ( nh l Shannon) chng trng: Nu bin i F ca mt hm thigian l zero vi |f| > fm v nhng trgi ca hm thi gian c bit vi t = n TS(vimi trnguyn ca n ) th hm thi gian c bit mt cch chnh xc cho mi trca t.

iu kin hn chl TS< 1/2fm.

Ly mu tn hiu

Gii thiu v iu chxung


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S m

Ni cch khc, s(t) c thc xc nh tnhng trgi ca n ti mt lot nhng thiim cch u nhau.

Tn sly mu, k hiu l fS

= 1/TS

, fS > 2fmNhvy, tn sly mu t nht phi 2 ln cao hn tn sca tn hiu c ly mu.Nhp ly mu ti thiu, 2 fm, c gi l nhp ly mu Nyquist. Th d, nu mtting ni c tn smax 4 kHz, n phi c ly mu t nht 8.000 ln/sec. Ta thy rngkhong cch gia nhng thi im ly mu th tlnghch vi tn scao nht ca tnhiu ( fm).

nh l ly mu gi ra mt kthut i mt tn hiu analog s(t) thnh mt tn

hiu ri rc. Ta chcn ly mu tn hiu lin tc ti nhng thi im ri rc, th d

mt danh sch cc sc ly mu s(0), s(T), s(2T)... Trong T< 1/2fm. truyn tn hiu ri rc mu ho , danh sch cc ssc c trn mt

telephone hoc c vit trn mt mnh giy gi FAX.

C thiu chvi thng sca mt sng mang ty vo danh sch cc s. Tn

hiu c iu chsau c truyn trn dy hoc trong khng gian (nu bngtn n chim cho php ).

V thng tin c dng ri rc nn ch cn dng tn hiu mang sng ri rc (thay v


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V thng tin c dng ri rc, nn chcn dng tn hiu mang sng ri rc (thay v

dng sng sin lin tc nhtrc).

Nu chn mt chui xung tun hon lm sng mang. Cc thng sc thlm thayi l bin , brng v vtr ca mi xung. Slm thay i mt trong ba thng

sy sa n 3 kiu iu ch::

- PAM ( Pulse Amlitude Modulation): iu chbin xung .

- PWM ( Pulse Width Modulation): iu chrng xung.

- PPM ( Pulse Position Mod: iu chvtr xung.

Hnh di vmt sng mang sc(t) mt tn hiu cha tin s(t) v tn hiu PAM sm(t).

ta thy chc bin ca xung sng mang bthay i, cn dng xung vn gi

khng i.

Nhl sm(t) khng phi l tch ca s(t) vi sC(t).

Ta gi sm(t) trong trng hp ny l PAM nh phng ( flat top PAM ) hoc PAM ly

mu tc thi (Instantanous Sampling PAM )

iu chbin xung PAM


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Nu ly tch ca sC(t) v s(t), ta c kt qul sng PAM vnhhnh . , chiucao cc xung khng phi l hng m thay i theo ng cong ca s(t). Trng hpny, ta gi l PAM ly mu tnhin (Natural Sampling ).

Phtn sca PAM


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By gita ly bin i Fourier ca PAM xc nh knh sng cn thit. Trc htl xem trng hp ca PAM ly mu tnhin. Da vo nh l ly mu. Khai trin

sC(t) thnh chui F. Ri nhn vi s(t). Kt quthu c l 1 tng gm nhiu sngAM vi cc tn ssng mang l tn scn bn v cc hotn sC(t).

Bin i F ca PAM ly mu tnhin

Xc nh bin i F ca PAM nh phng th kh hn. n gin ta xem hthng ly

mu s (t) bng mt chui xung lc l tng. Ri nh dng mi xung lc thnh dngxung nh mun, trong trng hp ny l mt xung vung nh phng.


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Bin i F ca tn hiu ly mu ca lc c tm tnh l ly mu.

Chui F ca chui xung lc c nhng trCn bng nhau vi mi n.

Bin i F ca sng c ly mu xung lc vhnh

Hm truyn mch lc nhhnh sau:


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Bin i F ca output ca mch lc l tch ca bin i trn y vi hm ca mchlc. Nhrng phn tn sthp ca n khng phi l mt phin bn bmo ca S(f).


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Bin i F ca PAM nh phng

Th d1: Mt tn hiu cha tin c dng: s(t) = sint/tc truyn bng cch dng PAM. Sng mang l chui xung tam gic tun hon nhhnh di. Tm bin i F ca sng bin iu.

ca mch ly mu bng xung lc l tng c bin i F.


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Trong S(f) l bin i F ca sin/t. Bin i ny l mt xung nhhnh v.

Mch lc phi thay i mi xung lc thnh mt xung tam gic. p ng xung lc cachng l mt xung tam gic m bin i ca n l:

Cui cng, bin i F ca sng PAM c cho bi tch ca S(f).H(f)


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ng PAMchim tt cnhng tn stzero n v hn. Nhvy, n bxem lkhng thtruyn c hiu qutrong khng kh cng nhMultiplexing.

V phn c ngha nht ca bin i F ca sng PAM nm xung quanh tn szero,

ta thng dng AM hoc FM gi sng PAM. l, taxem sng PAM nhl tn

hiu cha tin v n bin iu mt sng mang hnh sin. Nhng ti sao ta phi thchin mt bin iu kp, m khng truyn tn hiu gc bng AM hoc FM ? Hy nh

l tn hiu gc khng c dng nalog lin tc m l tn hiu ri rc.


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t u gc g c d g a og tc t u c.

Sau khi bin iu AM hoc FM vi sng PAM, khbng trnn rt rng. V l do

ny bin iu xung c kt hp vi AM hoc FM thng khng c truyn theocng cch thc nhtn hiu bin iu khc. N thng truyn trn cp ng trc,

vn c khnng truyn mt khong rng ca tn s. i khi n cng c truyn

qua khng kh ti tn smicrowave. Tn sny cao khbng rng khng b

xem nhl squ cng sut ( over powering ) i vi sng mang.

Nhng mch cng dng bin iu AM u c thdng PAM ly mu

tnhin. Chcn loi blc dy thng tkhi (nh a). Hnh b chkhi

dng cu diode.


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Khi iu chcho PAM phng nh th n gin hn cho PAM ly mu tnhin. Ta

chcn dng mt mch ly mu v gi( Sample and Hold ). Mch ny c l tng

ho nhhnh sau.

S1ng tun hon ti nhng thi im ly mu T C np in n tr mu mi khi S1


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S1 ng tun hon ti nhng thi im ly mu. TC np in n trmu mi khi S1ng v ri ngt. V tkhng c ng phng in, nn sgitrgi mu v to nn

ng phng ca nh sng PAM. Khi S2ng, tsphng in n zero.Cn tnh trc tv mch in tsao cho thi gian np tht nhanh v ta cng chn

mch c tng trra tht nhhng sthi gian phng in ngn.

Tch sng PAM ly mu tnhin da trc tip vo nh l ly mu. Shi phc tn

hiu analog gc tphin bn mu ho ca n cn mt LPF.

Tch sng PAM nh phng cn thm mt svic. Dng mch S/H phc hi mt

dng sng hnh bc thang xp xvi dng sng tn hiu gc. t thi gian gibng

chu kly mu. Kt quvhnh di. Hm bc thang c thc lc bi mt LPF

dng sng c trn phng, gn ging vi dng sng gc.


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Do c c bin i F ca sng PAM nh phng bng cch nhn bin i F ca tn hiu

mu ho cho H(f) (l hm truyn ca mch lc). Phn bng gc ca bin i F c dng

S(f)H(f). Vy s(t) c thc hi phc tsm(t) bng cch dng mt mch lc LPF mhm truyn ca n th vhnh di. Mch lc vi hm truyn 1/H(f) c xem nhl

mt mch cn bng v n trit nhng hiu quca sto dng xung.


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Nhtrng hp ca PAM, cng c sng mang l mt chui xung tun hon. Hnh di ch

mt sng mang cha bin iu, mt tn hiu cha tin s(t) v sng bin iu PWM. rng

ca mi xung bin iu thay i tutheo trmu tc thi ca s(t). Trmu ln hn slm rng xung bin iu rng hn. V rng xung thay i, nn nng lng ca sng cng thay

i. Vy khi bin tn hiu tng, cng sut truyn cng tng.

iu chrng xung PWM (Pluse Width Modulation)


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Cng nhtrong trng hp FM, PWM l mt php phi tuyn. Xem mt th d

n gin minh chng iu . Gistn hiu cha tin l mt hng, s(t) = 1. Sng

PWM sgn nhng xung c rng bng nhau, v mi trmu th bng vi mi trmukhc. By ginu ta truyn s(t) = 2 theo PWM, th ta li c mt chui xung c rng

bng nhau, nhng rng ca chng ln hn khi truyn s(t) = 1. Nguyn l tuyn tnh s

cho kt qul rng xung ca trng hp sau gp i trng hp trc. Nhng y

khng phi nhvy, nhhnh .Nu ta gistn hiu s(t) bin i chm ( ly mu

vi nhp nhanh hn so vi nhp Nyquist ) th cc xung ln cn sc rng hu nhbngnhau. Vi githit ny, c thphn gii xp xcho sng bin iu, theo chui Fourier.

Mi shng ca chui l mt sng FM, thay v l mt sng sin thun tu.


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Ta strnh by mt dng ca khi bin iu v mt dng ca khi hon iu cho PWM.Trong chai, ta u dng sng rng ca chuyn i gia thi gian v bin . iu nytng tnhcch thc cho FM, ta thy rng cch dnht bin iu mt tn hiu ltrc tin i n thnh AM. Tn hiu rng ca c dng vhnh.

Mch iu chPWM


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c thc hin bng cch tch phn sng PWM trong mi khongthi gian. V chiu cao ca xung th khng i, tch phn tlvi rng xung. Nuoutput ca tch phn c ly mu v giti trgi cui ca n, kt qusl mt sngPAM.

Trc tin tn hiu s(t) c ly mu vgic s1(t).

Tn hiu rng ca bdi xung 1 n vto nn s2(t). Tng ca s1(t) v s2(t) tonn s3(t) v vo mch so snh. Nhngkhong thi gian m s3(t) dng lnhng khong m rng tlvi trgi mu gc. Output ca mch

so snh l 1 khi s3

(t) dng v l 0 khis3(t) m. Kt qul s4(t), l mt sngPWM. rng xung c thc hiuchnh bng cch tng gim s(t). Tronghnh v, ta gisrng bnh thng s(t)


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, g g g ( )nm gia 0 v 1.

Shon iu c thc hin bng cchtch phn sng PWM trong mi khongthi gian. V chiu cao ca xung thkhng i, tch phn tlvi rngxung. Nu output ca tch phn cly mu v giti trgi cui ca n,kt qusl mt sng PAM.

PPM c li hn PWM vmt trit nhiu v cng khng c vn cng sut thay itheo bin tn hiu.

Mt tn hiu cha tin s(t) v sng PPM tng ng vhnh.

iu chvtr xung PPM (Pulse Position Modulation)


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Nu trgi mu ln hn sc

xung tng ng di xa hn (sovi vtr khng bin iu ca

n).

Mt sng PPM c thc

suy tmt sng PWM. Slin

hgia chng l, trong khi vtr ca xung thay i trong

PPM th sn ca xung thay

i t PWM Gi t d


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i trong PWM. Gista d

mi sn iu khin ca

PWM, (ly o hm v xem

nhng xung m).

By ginu ta t mt xung c

rng khng i ti mi

im ny, kt qul sngPPM. c vra nhhnh.

R rng, cPWM v PPM u rt phc tp so vi PAM. Chng phc tp hn v cn

c nhng tnh cht khc. Trong cc hphn knh theo thi gian TDM, ta phi bo

m rng cc xung mu ln cn khng c phnhau. Nu cc xung di tdo hoc

rng hn (nhtrong PPM v PWM), ta khng thchen vo mt cch n gin cc

xung khc trong khng gian m tin chc rng khng c stc ng qua li sxy ra.

Khong cch cn thit phi c gic thtruyn cc trmu ln nht. iu

ny lm gim sknh khi Multiplex.


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Bc thnht chuyn i mt tn hiu analog lin tc thnh dng digital l i tn hiuthnh mt danh mc cc s. ( iu ny c thc hin bng cch ly mu hm thi gian).

Danh mc cc skt qubiu din cho nhng trlin tc. l mc d mt mu no c

thtrng ra nhl mt slm trn, nhng thc tn sc tip tc nhmt sthp phn

v hn.

Danh mc cc sanalog sau phi c m ho thnh cc Code Words ri rc. Bin php

trc nht hon tt vic l lm trn mi strong danh mc. Th d, nu cc mu nm

trong khong t0 n 10V, mi mu sc lm trn n snguyn gn nht. Vy cc tm

( d d ) t t 11 ( t 0 10 )

Lng tha tn hiu Bin i ADC


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( code words ) srt ra t11 snguyn ( t0 n 10 ).

Schuyn i A/ D c xem nhl slng tho ( quantizing ). Trong slng thou n, cc trlin tc ca hm thi gian c chia thnh nhng vng u n, v mt m s

nguyn c kt hp cho mi vng. Nhvy, tt ccc trca hm trong mt vng no

u c m ho thnh mt snhphn ging nhau.

ng tho 3 bit theo hai cch khc nhau.

Hnh a, chkhong cc trca hm c chia lm 8 vng nhau. Mi vng kt hp vi mt

snhphn 3 bit. Chn 8 vng v 8 l lutha ca 2 ( = 23

). Tt cthp 3 bit u c dng,lm hiu quln hn.

Hnh b chslng tho bng cch dng slin hca input v output. Trong khi input th

lin tc, output chly nhng trri rc. Brng ca mi bc khng i. V slng tho th

u n.


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A/D

Cc mch bin i A/D

1. Lng tho m, m ln lt ng vi snhn c thng qua mi mc lng

t.

2. Lng tho ni tip, to ra mt tm, tng bit mt. l, chng bt u vi bit

c trng sln nht ( MSB ) v lm vic n bit c trng snhnht ( LSB ).

3. Lng tho song song, to ra cng lc tt ccc bit ca mt tm hon chnh.


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