b4 pulse modulation ver1 -comm theorem

8/14/2019 b4 Pulse Modulation Ver1 -comm theorem

1/52

SZE 3533SZE 3533

Topic IV Pulse ModulationTopic IV Pulse Modulation


2/52

4.0 Digital Communication System4.0 Digital Communication System

Pemodulatan Digi


3/52

In the early 90s, telecommunication networks is changing towards

digital world. With the rapid advancement in the fields of VLSI and

microprocessor, several telecommunication components such astransmission line and switching has been using digital signals in

their operation.

Therefore, information signals must be changed to digital form so

that it can be transmitted through this network.

Several techniques requiring full coding of the original signal will

be used:

4.0 Introduction4.0 Introduction

-

Pulse Code Modulation (PCM)

Differential PCM (DPCM)

Adaptive Differential PCM (ADPCM)

Delta Modulation (DM)

Adaptive Delta Modulation (ADM)


4/52

4.1 Digital Modulation4.1 Digital Modulation Advantages :

Immunity to noise

Easy storage and processing:

Regeneration

Easy to measure

Enables encryption

Data from several sources can be integrated and

transmitted using the same digital communicationsystem

Error correction detection can be utilized

Disadvantages :

Requires a bigger bandwidth Analog signal need to be changed to digital first

Not compatible to analog system

Need synchronization

Pemodulatan Digi

Voice : Analog : 4 kHz

Digit : 2 x 4 kHz x 8 bit = 64 kb/s

BWmin 32 kHz

MP, DSP, RAM, ROM, Computer


5/52

4.2 TRANSMISSION METHOD FOR ANALOG &

DIGITAL SIGNALS

Analog

input

Analog channel

Baseband

Analog

output

Analog

inputModulator De

modulator

Analog

output

Analog

channel

Digital

input

encoder decoderDigital

channel

Digital

output

Digital

inputModem Modem

Analog

channel

Digital

output

Analoginput ADC &encoderDecoder& DAC

Analogoutput

Digitalchannel

Analog

input

Analog

outputAnalog

channel

ADC &

encoderModem

ADC &

decoderModem


6/52

4.3 Pulse Modulation4.3 Pulse Modulation

PAM (Pulse Amplitude Modulation) => VPAM Vm PWM (Pulse Width Modulation) => Vm PPM (Pulse Position Modulation) => d (pulse delay) Vm PCM (Pulse Code Modulation)

Pulse Modulation consists of:

Easily effected by

noise

Less susceptible to

noise

Less susceptible to

noise compared to

PAM


7/52

4.3 Sampling Theorem4.3 Sampling Theorem

ms ff 2

Pemodulatan Digi

X

Digital signal

s(t)

ms(t)m(t)

m(t)

t

ms(t)

t

s(t)

t

Ts

Nyquist theorem

states that:

( ) [ ]

s

s

s

sss

s

fT

tttT

ts

22

.....3cos22cos2cos211

==

++++=

where

s

sf

T1

=

( ) ( ) ( )

( ) ( ) ( ) ( )[ ].....3cos22cos2cos21

++++=

=

ttmttmttmtmT

tstmtm

sss

s

s

Fourier series for impulse train :

Therefore :


8/52Pemodulatan Digi

s s

ms +

m m0

ms ms +

ms

)(sM

sT

1

m m0

)(M

1

s s0

)(S

sT

2

0t

)(tm

sT6 sT60

)(ts

t

sT

sT6 sT60

)(tms

tsT

Time domain Frequency domain


9/52

ReconstructionReconstruction

Pemodulatan Digi

m m0

)(rM

1

m m0

)(M

1

X

Pulse signal

s(t)

ms(t)

m(t) h(t) mr(t)

TX RX

Low pass filter

s s

ms +

m m0

ms ms +

ms

)(sM

sT

1

n n0

)(H

sT


10/52

Sampling process shown previously uses an ideal pulsesignal

However, it is quite difficult to generate an ideal pulse

signal practically The usual pulse signal generated is as shown below:

1

2 2( ) kos

di mana

sin

n

n s s s

sn

s

A A nt s t c

T T T

n

Tc

n

T

=

= +

=

Pemodulatan Digi

t

s(t)

Ts

A

-pulse widthTs pulse period

sT

nsinc=


11/52Pemodulatan Digi

Natural Sampling Flat-top Sampling

Information signal

Pulse signal

Sampled signal (PAM)

t

m(t)

t

s(t)

Ts

t

ms(t)

Ts

t

ms(t)

Ts

( )sT


12/52


13/52


14/52

+=

=1

2cos

2).()(

n s

n

ss

sT

ntc

TTtmtm

=

+=1

2cos

2)()()(

n s

n

ss

sT

ntc

T

tm

T

tmtm

....6

cos2)(

4cos

2)(2cos

2)()()(

3

21

+

+++=

ss

sssss

s

T

tc

T

tm

T

tc

T

tm

T

tc

T

tm

T

tmtm

For n = 1, 2 , 3 ..

The above expression shows that the frequency components of the

sampled signal is at fs, 2f

sand 3f

s. Components 2f

sand 3f

sis a replica of

the spectrum of the sampled signal.


15/52

....6cos2)(

4cos

2)(2cos

2)()()(

3

21

+

+++=

ss

sssss

s

Ttc

Ttm

T

tc

T

tm

T

tc

T

tm

T

tmtm

f

3f2s 2fs3fsfs0 fs-fm

fs+fm 2fs+fm 3fs+fm3fs-fm2fs-fm

ms(f)

Spectrum of the sampled

signal

The spectrum of the sampled signal has sidebands fs f

m, 2f

s f

m, 3f

s

fm

and so on.


16/52

The choice of sampling frequency, fs

must follow the sampling theorem

to overcome the problem of aliasing and loss of information

(a) Sampling frequency=> fs1 < 2fm(max)

f2fs1 3fs1fs1fm

Aliasingms(f)

(b) Sampling frequency=> fs2 > 2fm(max)

f

2fs2 3fs2fs2fm

ms(f)

Shannon sampling

theorem=> fs 2fmNyquist frequency

fs = 2fm= fN

A bandlimited signal thathas a maximum

frequency, fmax can be

regenerated from the

sampled signal if it is

sampled at a rate of at

least 2fmax.


17/52

4.4 Detection of Sampled Signal4.4 Detection of Sampled Signal

By using LPF to the sampled signal, ms(t)

LPFms(t) m(t)

Cut-off frequency , fo for LPF must be within the range: fm fo fs - fm Eventhough the sampled signal can be detected easily atfs = 2fm, but usually

fs > 2fm . The main reason is to have a guardband .

Therefore, the maximum frequency that can be processed by the sampleddata using sampling frequency, fs (without aliasing) is:

=> fm = fs/ 2 = 1 / 2Ts


18/52

=+=

1

2cos2)()()(n s

n

ss

sTntc

Ttm

Ttmtm

From the sampling process, the sampled signal:

s

nT

nc

sinc=where :

sT


19/52

( ) ( ) ( )

( ) ( )

( )

( )

( ) ( )

( ) ( )

+++++

+++++=

+++

+++=

+++++=

++=

+++=

+++=

=

=

=

.........2cos2cos2cos2

coscoscos2cos1

...cos2cos22cos2

coscos2cos2cos1

....3cos22cos2cos21cos1

cos21cos1

cos2cos1cos1

cos2

cos1cos1

1

1

1

ttt

tttt

T

ttt

tttt

T

tttT

t

tnT

t

tnT

tT

t

tnT

tT

ttm

msmss

msmssm

s

mss

mssm

s

sss

s

m

n

s

s

m

n

s

s

m

s

m

n

s

s

m

s

ms

replacing ( ) ( )ttm mcos1+=

=

+=1

2cos

2)()()(

n sss

sT

nt

T

tm

T

tmtm

inside


20/52

It can be shown that the output sampled signal is the same as the output

PAM signal when :

sT


21/52

4.5 Pulse Width Modulation (PWM)4.5 Pulse Width Modulation (PWM)

)(tvm

(pulse width) follows the instantaneous value of the informationsignal vm(t) :

tmoo cos+=

( )tmo cos1+=

o

represents the width that is

fixed according to the minimum

value of the information signal

The equation shows that the pulse

width, of the output signalPWM varies according to the

instantaneous value of the

information signal.

( ) { }...2cos2cos21 +++== ttT

v sss

PWMPWM

Replacing inside the general equation of the sampled signal:


22/52

( )

( ){ }...2cos2cos21

cos1+++

+== tt

T

tv ss

s

moPWM PWM

++

++++= ...cos2cos2

coscos2cos2cos2cos21

tt

ttttt

Tv ms

msmss

s

oPWM

+++++

++++=

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

)cos(cos2cos2cos21

ttt

tttt

Tv

msmsms

msmss

s

oPWM

Generation of PWM signal is by changing the value of sample signal of the

PAM signal into a specific period

vPWM (t)vPAM (t)555 timer

(a) PWM generation using voltage to time converter

LPFvPWM (t) vm(t)

(b) PWM detection using LPF


23/52

fs

> 2fm

fs = 2fm fs < 2fm


24/52

4.6 Pulse Code Modulation (PCM)4.6 Pulse Code Modulation (PCM)

Sampling Quantization Coding

A method used to represent an analog signal in terms of digital word

Constitutes 3 process:

1. Sampling the analog signal

2. Quantization of the amplitude of the sampled signal

3. Coding of the quantized sample into digital signal

LPF S/H ADC PCM

S/H : Sample and hold

circuit

Analogsignal

Anti aliasing

filter ADC : analog to digital converter

PCM process:

fs


25/52

4.6.1 Sampling4.6.1 Sampling

An analog signal must be sampled at Nyquist rate to avoid

aliasing

4.6.2 Quantization & Coding

Process of estimating the sampled amplitude into a value suitable for

coding (ADC).

A fixed number of levels including the maximum and minimum value of

the analog signal

Number of levels is determined by the number of bits used for coding

3 t th t l d i th ti ti


26/52

Quantization Interval

Represent the voltage value for each quantized level

For example: For a sampled signal that has 5V amplitude, Vpp

= 10 V

divide by the quantized level,L = 8 level,

Therefore, quantized interval ,

Quantization level,L= 2n

Quantization level depends on the number of binary bits, n used to

represent each sample.

For example:For= 3; Quantization level,L = 23 = 8 level.

In this example, first level (level 0) is represented by 000, whereas bit

111 represents the eigth level

V25.18

V10==V

3 terms that are commonly used in the quantization

process:


27/52

Quantization value, Vk

The middle voltage for each quantized level

For example: forn = 3, quantized level,L = 8 and a sampled sinusoidalsignal with +5 V ,

The middle quantized value for level 0,

In this example, for a sample that is in level 0 segment will be

represented by bit 000 with a voltage value of 4.375 V. The difference

between the sampled value and the quantized value results in

quantization noise.

V375.4

2

V25.1V50 =+=V

For voice communication 256 levels

are commonly used (i.e n = 8)

4 6 3 UNIFORM QUANTIZATION


28/52

t

Level 0 : 000

Level 1 : 001

Level 2 : 010

Level 3 : 011

Level 4 : 100

Level 5 : 101

Level 6 : 110

Leve l 7 : 111

1.9V

+5.0V

-5.0V

4.375V

3.125V

1.875V

0.625V

-0.625V

-1.875V

-3.125V

-4.375V

4.3V

1.9V

-3.2V

-4.5V

Quantization level &

binary representationQuantized

value

Sampled signal

4.6.3 UNIFORM QUANTIZATION

Uniform quantization is a quantization process with a uniform (fixed)

quantization interval.

Example : n = 3 ,L = 8 , signal +5 V ; => Vk = 1.25 V . Bit rate: sb nff =


29/52

4.6.3.1 Uniform Quantization using Folded Binary4.6.3.1 Uniform Quantization using Folded Binary

Code (sign bit)Code (sign bit)

Pemodulatan Digi

+mp

-mp

0

0 11

0 10

0 01

0 00

1 001 01

1 10

1 11

valueSign bit

t

000 001 011 011 011 010 001100 110 111 111 110 100 001 010 010 010 000

Quantization error

Qe

PCM code

t

The same code representing several

samples with different amplitudes

Step size


30/52

4.6.3.2 Quantization error4.6.3.2 Quantization error

Pemodulatan Digi

May add to or substract from the

actual signal

Quantization error (Qe) is also called Quantization noise (Qn) . And its

maximum magnitude is one half of the voltage of the minimum step

size .


31/52

4.6.3.2 Quantization error4.6.3.2 Quantization error


32/52

Input voltage range: 14 mV

to +14 mV

Binary

number

Input voltage

range (mV)

1 11

10 to 141 10 6 to 10

1 01 2 to 6

1 00 0 to 2

0 00 -2 to 0

0 01 -6 to -2

0 10 -10 to -6

0 11 -14 to -10

Example : Uniform Quantization error

Qn= LSB voltage /2 = /2 14 mV = 28 mV with 8 steps and 8 codes.Therefore = 28/8 = 3.5 mV.Therefore : Qn= 3.5 mV / 2 = 1.75 mV

SNRq = [1.76 + 6.02n] dB : (for details, refer

to monograph page 122)

Noise from quantization error can be

reduced by increasing the quantization

level i.e increase n.

Nonuniform quantization

using Law: ( )[ ]dB02.6

1ln

3log10

2nSNRq +

+=


33/52

PCM

system

Example :

Vpp = 31.5 V

6 bit code (5 bits for

magnitude and 1 bit

for sign

(a) No of levels: 26 = 64

(b) LSB voltage, : 31.5/64 = 0.492 V(c) Maximum quantization level, /2 = 0.25 V(d) Voltage value for 001101 ; +(13 x 0.492) = +6.4 V

(e) Voltage value for 111001 ; (25 x 0.492) = -12.3 V

(f) Code for input +13.62 V

= 13.62/0.492 = 27.68 28 => 111100(g)Code for input 9.37 V

= 9.37/0.492 = 19.04 19 => 010011


34/52


35/52

4.6.4 Non Uniform Quantization4.6.4 Non Uniform Quantization

example : Non-Linear

Quantization

Pemodulatan Digi

Companding => Compress - Expanding


36/52

Companding Compress Expanding

A method used to produce a uniform SNR for all input signal range is

compression-expansion (Companding).

Input signal is compressed at the transmitter and expanded at the

receiver


37/52

=> Analog Compression process is done on the input signal

before sampling and coding

=> Digital compression process is done after the signal is

sampled

Companding => Compress - Expanding

analog signal

(input)analog

compressorADC

DACAnalog

expander

Analog signal

(output)

PCM with analog compress-expand

To digital channel

analog signal

(input)ADC

DACDigital

expander

Analog signal

(output)

PCM with digital compress-expand

To digital channelDigital

compressor


38/52

Pemodulatan Digi

2 Popular companding system (standardized by ITU)

EUROPE => A - Law

USA/NORTH AMERICA => - Law

Axfor

xA

for

A

AxA

Ax

y1

0

11

log1

log1

)log(1

+

+

+

=

A - compressor paramater. Usually

the value ofA is 87.6.


39/52

Pemodulatan Digi

USA/NORTH AMERICA => - Law

Law is a standard compress-expand that is used in America

and Japan. The value of usedis 255 (8 bit).

( )

++=1log

)1log( xy

)(maki

i

E

E

x = )(makoo

E

E

y =

For both laws, the values ofxand

y refers to the equation below:


40/52

Example 4.3 :

A compress-expand system using Law ( = 255) is used for a signalwith range 0 to 10V. Determine the output of the system if the input is 0 and

7.5V.Solution :

Given = 255 and Ei(mak)

= 10 V

ForEi= 0 V

)(maki

i

E

Ex = 0

10

0==x;

Output :

( )

++

= 1log)1log( x

y ; ( )2551log))0(2551log(

+

+=y

0=y

ForEi= 7.5 V

)(maki

i

E

E

x= 75.0

10

5.7==

x;

( )

+

+=

1log

)1log( xy

Output :

( )2551log))75.0(2551log(

+

+=y

948.0=y


41/52

Example 4.4 :

A random signal has gone through a 256 level quantization process.

Determine the quantization signal to noise ratio for this system.

Solution :

From the above statement, the number of sampling bits is not known.

But, given L=256

L = 2n

therefore, n = 8

Given SNRq

dB02.676.1 nSNR q +=

dB50)8(02.676.1 +=qSNR

4.6.5 Bit rate for PCM transmission


42/52

Europe bit rate(Mb/s)

2.048

8.448

34.368

139.264

565.148

Telephone

channel

30

120

480

1920

7680

SDH 2.5Gb/s

Telephone

channel

North America bit

rate(Mb/s)

24 1.544

48 3.152

96 6.321

672 44.736

4032 274.176

European standard : A-Law

30 + 2 control channel = 32

Bit rate= 32 x 8 bit/sample x 8000 sample/s

= 2.048 Mb/s

North American standard (NAS) : -LawFor every 24 sample, 1 bit is added for

synchronization

For 24 sampel => 24 x 8 bit/sample+ 1 bit = 193 bits

Bit rate= 193 x 8000 = 1.544 Mb/sNeeds Multiplexing Process of transmitting two orNeeds Multiplexing Process of transmitting two or

more signals simultaneouslymore signals simultaneously

Example : PCM TDM CEPT SystemExample : PCM-TDM CEPT System


43/52

Example : PCM-TDM CEPT SystemExample : PCM-TDM CEPT SystemFrame structure and Timing : European standard PCM system : E Line

(a) bits per time slot (b) time slots per frame (c) frames per multiframe

488 ns

3.9 s3.9 s

125 s125 s

2 ms

8 bits pertime slot

Bit duration

30 signal + 2 control = 32 channels = 1 frame

Signalling & synchronization

16 frames = 1 multiframeDuration of multiframe

CEPT system 32 channels (30 signals + 2 control)


44/52

Frame structure and timing

Number of channel = 32

Number of bits in one time slot = 8

32 channels = 1 frame

Number of bits in a frame = 32 x 8 = 256 bits

This frame must be transmitted within the sampling period

and thus 8 x 103 frames are transmitted per second.

Therefore :

Transmission rate = 8 x 103 x 256 = 2.048 Mb/s

Bit duration = 1 / 2.048 x 106 = 488 ns

Duration of a time slot = 8 x 488 ns = 3.9 sDuration of a frame = 32 x 3.9 s = 125 s => (= 1 / 8 kHz = 125 s)Duration of a multi frame = 16 x 125 s = 2 ms

CEPT telephone system hierarchy


45/52

MUX

1

MUX

2

MUX3

MUX4

30

Voice

channels

.

.

.

.

.

.

E1 line

2.048 Mbps

E2 line

8.448 MbpsE1

E1

E1

E2

E2

E2

E3

E3

E3

E3 line

34.368 Mbps

E4 line

139.264 Mbps


46/52

4.7 Delta Modulation (DM)4.7 Delta Modulation (DM)

There are 2 main components in the DM generator circuit,i.e comparator and integrator.

Pemodulatan Digi

+

-

X

-+

Pulse signal

s(t)

comparator

integrator

d(t)xDM(t)m(t)

e(t)

)(~ tm

Comparator will compare the error signal e(t), where)(~)()( ttt


47/52

Output signal from comparator has the following function:

The output from the comparator will be sampled with apulse signal at a rate of 1/T

s.

Next, DM signal will be generated with the equation below:

The DM signal will be feed back, but before that this signalwill be integrated first

This signal will determine the error value e(t).

)()()( tmtmte =

=

=

=

=

n

ss

n

sDM

nTtnTe

nTttetx

)()](sgn[

)()](sgn[)(

Pemodulatan Digi

+== )](sgn[)( tetd

0)(

0)(

te

te

=

=n

snTetm )](sgn[)(~


48/52

4.7.1 Delta Modulation (DM) signal4.7.1 Delta Modulation (DM) signal

Pemodulatan Digi

)(tm

t

Ts

)(~ tmEffects of steep

slope

0001010111111101100010000000

If e(t) < 0 or - , it will be coded as 0

If e(t) > 0 or +, it will be coded as 1

A steep slope results in noise in DM signal. To avoid this fromhappening, it has to follow the following condition:

dt

tdmmak

Ts

)(>


49/52

4.8 Line Coding4.8 Line Coding

Pemodulatan Digi

Binary 1 and 0 in PCM signal can be represented by several formatsknown as line coding.

informationPCM

Line

coder

channel

Reasons for line coding:

1. Synchronization

2. Error detection

3. Error correction

4 8 1 Line code format4.8.1 Line code format A. NRZ (Non Return to Zero)


50/52

4.8.1 Line code format4.8.1 Line code format

Digital Signal Encoding Formats

- Popular method

- easy

- Data does not return to 0 in one

clock interval

- No synchronization. Can use start

bit for synchronization purposes

1. NRZ-L (NRZ-Level)

1 => High level

0 => Low level

2. NRZ-M (NRZ-Mark)

1 => transition at the starting interval

0 => no transition

3. NRZ-S (NRZ-Space)

1 => no transition

0 => transition at the starting interval

B. RZ (Return to Zero)


51/52


Return to 0 at the half bit interval

The same

advantages/disadvantages with

NRZ

Overcome by using bipolar signal

and alternating pulse for

synchronization

4. RZ (Unipolar)

1 => High level

0 => Low level

5. RZ (Bipolar)

1 => Alternately +ve

0 => Alternately ve

6. RZ (AMI Alternately Mark Inversion)

1 => Alternately +ve and -ve

0 => Low level

C. Bi phase


52/52


Used in optical communication

system, satellite and video

recorder

Self synchronizing

7. Bi phase M

1 => transition at the middle of the

interval

0 => no transition at the middle of the

interval

8. Bi phase L (Manchester Coding)

1 => transition from HI to LO at the

middle of the interval

0 => transition from LO to HI at the

middle of the interval

used in Ethernet IEEE 802.3 standard in

LAN

9. Bi phase S inverse of Bi phase M

1 => no transition in the middle of the

interval

b4 pulse modulation ver1 -comm theorem

Documents