b4_pulse_modulation1.pdf (6993k)

•

In the early 90’s, telecommunication networks is changing towards

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

microprocessor, several telecommunication components such as

transmission 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 Introduction

4.0 Introduction

-• Pulse Code Modulation (PCM)

• Differential PCM (DPCM)

• Adaptive Differential PCM (ADPCM) • Delta Modulation (DM)



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 communication system

◦

_{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 Digit

Voice : Analog : 4 kHz

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

BW_min  32 kHz

4.2 TRANSMISSION METHOD FOR ANALOG &

DIGITAL SIGNALS

Analog input

Analog channel Baseband

Analog output

Analog

input Modulator _modulatorDe

Analog output Analog

channel

Digital

input encoder

decoder Digital

channel

Digital output

Digital

input Modem Modem

Analog channel

Digital output

Analog input

ADC & encoder

Decoder & DAC

Analog output Digital

channel

Analog input

Analog output Analog

channel ADC &

encoder Modem

4.3 Pulse Modulation

4.3 Pulse Modulation

•PAM (Pulse Amplitude Modulation) => V_PAM V_m

•PWM (Pulse Width Modulation) => V_m

•PPM (Pulse Position Modulation) => _d (pulse delay) V_m

•PCM (Pulse Code Modulation)

Pulse Modulation consists of:

Easily effected by noise

Less susceptible to noise

m

s

f

f



2

Pemodulatan Digit

X

Digital signal

s(t)

m_s(t)

m(t)

m(t)

t

m_s(t)

t

s(t)

t

T

_s Nyquist theorem states that:







s s s s s s s

f

T

t

t

t

T

t

s













2

2

...

3

cos

2

2

cos

2

cos

2

1

1















where s s

f

T



1

 











2

cos

2

cos

2

2

cos

3

...



1













t

t

m

t

t

m

t

t

m

t

m

T

t

s

t

m

t

m

s s s s s







Fourier series for impulse train :

Pemodulatan Digit

s



 _s

m s 



m



 0 _m

m s    m s    m s     ) ( s M s T 1  m 

 0 _m 

) ( M 1 s 

 0 _s

) ( S s T  2  0 t ) (t m s T 6

 0 6T_s

) (t s t s T s T 6

 0 6T_s

) (t m_s t s T

Pemodulatan Digit

m



 0 m

 ) ( r M 1 m 

 0 m

 ) ( M 1 X Pulse signal

s(t)

m_s(t)

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

TX RX

Low pass filter

s



 _s

m s 



m



 0 _m

m s    m s    m s     ) ( s M s T 1  n 

 0 _n

 Sampling process shown previously uses an ideal pulse signal

 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

s n

s

A

A

nt

s t

c

T

T

T

n

T

c

n

T























Pemodulatan Digit

t

s(t)

T_s



A



- pulse width

T_s – pulse period

s

T

n



sinc

Pemodulatan Digit

Natural Sampling Flat-top Sampling

Information signal

Pulse signal

Sampled signal (PAM)

t m(t)

t s(t)

T_s



t m_s(t)

T_s



t m_s(t)

T_s

 In every sampling methods, the pulse amplitude is directly

proportional to the amplitude of the information signal

 Practically, an ideal sampling is difficult to generate

 However, by using an ideal and natural sampling, noise can be

eliminated, which is not the case for flat-top sampling

Pemodulatan Digit

Ideal Sampling Flat-top Sampling m_s(t)

Pemodulatan Digit

X

Pulse signal

s(t)

m_s(t)

m(t)

m(t)

t

m_s(t)

t

s(t)

t

Mathematical analysis:

s

n _s

s

s

T

nt

T

n

T

T

t

s

(

)



2



sinc



cos

2



1











Fourier series for pulse signal, s(t) :

)

(

)

(

)

(

t

m

t

s

t

m

_s

























 1

2

cos

2

).

(

)

(

n _s n s s s

T

nt

c

T

T

t

m

t

m









 





1

2

cos

2

)

(

)

(

)

(

n _s n s s s

T

nt

c

T

t

m

T

t

m

t

m







....

6

cos

2

)

(

4

cos

2

)

(

2

cos

2

)

(

)

(

)

(

3 2 1











s s s s s s s s

T

t

c

T

t

m

T

t

c

T

t

m

T

t

c

T

t

m

T

t

m

t

m















For n = 1, 2 , 3 …..

The above expression shows that the frequency components of the

....

6

cos

2

)

(

4

cos

2

)

(

2

cos

2

)

(

)

(

)

(

3 2 1











s s s s s s s s

T

t

c

T

t

m

T

t

c

T

t

m

T

t

c

T

t

m

T

t

m

t

m















f 3f

2_s 2f_s 3fs

f_s

0 fs-fm fs+fm 2fs-fm 2f_s+f_m 3fs-fm 3fs+fm

m_s(f)

Spectrum of the sampled signal

The choice of sampling frequency, f_s must follow the sampling theorem to overcome the problem of aliasing and loss of information

(a) Sampling frequency=> f_s1< 2f_{m (max)}

f

2f_s1 3f_s1 f_s1

f_m

Aliasing

m_s(f)

(b) Sampling frequency=> f_s2> 2f_{m (max)}

f

2f_s2 3f_s2 f_s2

f_m

m_s(f)

Shannon sampling theorem=> f_s2f_m

Nyquist frequency

 f_s= 2f_m= f_N

A bandlimited signal that has a maximum

4.4 Detection of Sampled Signal

4.4 Detection of Sampled Signal

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

LPF

m_s(t) m(t)

Cut-off frequency ,

f

_o

for LPF must be within the range:

f

_m



f

_o



f

_s

- f

_m

•

Eventhough the sampled signal can be detected easily at

f

_s

= 2f

_{m ,}

but usually

f

_s

> 2f

_m

. The main reason is to have a ‘guardband’ .

•

Therefore, the maximum frequency that can be processed by the sampled

data using sampling frequency,

f

_s

(without aliasing)

is:



 





1

2

cos

2

)

(

)

(

)

(

n s n s s s

T

nt

c

T

t

m

T

t

m

t

m







From the sampling process, the sampled signal:

s n

T

n

c



sinc



where : s

T





If:

1

sinc



s

T

n



therefore



 





1

2

cos

2

)

(

)

(

)

(

n _s s s s

T

nt

T

t

m

T

t

m

t

m







Therefore

 

t

t



m



1



cos



_m

Taking:

 

 





 





























_































































































































     

...

2

cos

2

cos

2

cos

2

cos

cos

cos

2

cos

1

...

cos

2

cos

2

2

cos

2

cos

cos

2

cos

2

cos

1

....

3

cos

2

2

cos

2

cos

2

1

cos

1

cos

2

1

cos

1

cos

2

cos

1

cos

1

cos

2

cos

1

cos

1

1 1 1

t

t

t

t

t

t

t

T

t

t

t

t

t

t

t

T

t

t

t

T

t

t

n

T

t

t

n

T

t

T

t

t

n

T

t

T

t

t

m

m s m s s m s m s s m s m s s m s s m s s s s s m n s s m n s s m s m n s s m s m s













































































replacing

m

 

t

t



m



cos

1







 





1

2

cos

2

)

(

)

(

)

(

n _s s s s

T

nt

T

t

m

T

t

m

t

m







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

s

T





that is, the pulse width



is much smaller compared to the pulse period T_s.

Voltage translator

v_m(t)

v_d(t)

v_PAM(t)

LPF

v_m(t) v_PAM(t)

(a) PAM generation

(b) PAM detection

4.5 Pulse Width Demodulation (PWD)

4.5 Pulse Width Demodulation (PWD)

)

(

t

v

_m





•



(pulse width) follows the instantaneous value of the information signal

v_m(t) :

t

m o

o













cos



_m

t



o









1



cos



_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 signal PWM

varies according to the

instantaneous value of the

information signal.

 





1



2

cos

t



2

cos

2

t



...



T

v

_{s s}

s

PWM _PWM







 

 





1



cos



1



2

cos

t



2

cos

2

t



...



T

t

v

_{s s}

s

m o

PWM _PWM









 



























...

cos

2

cos

2

cos

cos

2

cos

2

cos

2

cos

2

1

t

t

t

t

t

t

t

T

v

m s m s m s s s o

PWM





















































...

)

2

cos(

)

2

cos(

)

cos(

)

cos(

cos

2

cos

2

cos

2

1

t

t

t

t

t

t

t

T

v

m s m s m s m s m s s s o PWM

























Generation of PWM signal is by changing the value of sample signal of the PAM signal into a specific period

v

_PWM

(t)

v

_PAM

(t)

555 timer

(a) PWM generation using voltage to time converter

LPF

v

_PWM

(t)

v

m

(t)

f

_s

> 2f

_m

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

Analog signal

Anti aliasing

filter ADC : analog to digital converter PCM process:

4.6.1 Sampling

4.6.1 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

• Quantization Interval

Represent the voltage value for each quantized level

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

V

25

.

1

8

V

10

_





V

• Quantization value, V_k

The middle voltage for each quantized level

For example: for n = 3, quantized level, L = 8 and a sampled sinusoidal signal 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.

V

375

.

4

2

V

25

.

1

V

5

0











V

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 representation

Quantized value

Sampled signal

4.6.3 UNIFORM QUANTIZATION

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

Pemodulatan Digit

+m_p

-m_p

0

0 11

0 10

0 01

0 00

1 00

1 01

1 10

1 11

∆

value Sign but

t

000 001 011 011 011 010 001 100 110 111 111 110 100 001 010 010 010 000 Quantization error

Q_e

PCM code

t The same code representing several samples with different amplitudes

Pemodulatan Digit

May add to or substract from the actual signal

4.6.3.2 Quantization error

Input voltage range: –14 mV to +14 mV

Binary number

Input voltage range (mV)

1 11 10 to 14

1 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

Q

_n

= LSB voltage /2 =



/2



14 mV = 28 mV with 8 steps and 8 codes.

Therefore



= 28/8 = 3.5 mV.

Therefore : Q

_n

= 3.5 mV / 2 = 1.75 mV

SNR_q = [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:



_ln

 

₁



6

.

02

dB

3

log

10

₂

n

SNR

_q







PCM system

Example :

V_pp = 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 101101 ; +(13 x 0.492) = +6.4 V (e) Voltage value for 011001 ; –(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

4.6.4 Non uniform quantization

4.6.4 Non uniform quantization

nonuniform: to improve SNR (SQR)

 More levels is available for low level amplitudes compared to high amplitude

 Increase SNR for low level amplitude and decrease SNR for higher amplitudes

analog compression is done to the input signal before sampling and quantization at the transmitter

Expansion is done at the receiver

example : Non-Linear Quantization

Companding => Compress - Expanding

A method used to produce a uniform SNR for all input signal range is compression-expansion (Companding).

=> 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

compressor ADC

DAC Analog

expander

Analog signal (output)

PCM with analog compress-expand

To digital channel

analog signal

(input) ADC

DAC Digital

expander

Analog signal (output)

PCM with digital compress-expand

To digital channel Digital

Pemodulatan Digit

2 Popular companding system (standardized by ITU)

• EUROPE => A - Law

• USA/NORTH AMERICA => - Law

A

x

for

x

A

for

A

Ax

A

Ax

y

1

0

1

1

log

1

log

1

)

log(

1































Pemodulatan Digit

USA/NORTH AMERICA => - Law

Law is a standard compress-expand that is used in America and Japan. The value of used is 255 (8 bit).

 











1

log

)

1

log(

x

y

) (mak i

i

E

E

x



) (mak o

o

E

E

y



For both laws, the values of x and

Example 4.3 :

A compress-expand system using Law (= 255) is used for a signal with range 0 to 10V. Determine the output of the system if the input is 0 and 7.5V.

Solution :

Given = 255 and E_i(mak)= 10 V

For E_i = 0 V

) (mak i i

E

E

x



0

10

0

_



x

; Output :

 











1

log

)

1

log(

x

y

;



1

255



log

))

0

(

255

1

log(







y

0



y

For E_i = 7.5 V

) (mak i i

E

E

x



0

.

75

10

5

.

7

_



x

;

 











1

log

)

1

log(

x

y

Output :



1

255



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 SNR_q

dB

02

.

6

76

.

1

n

SNR

_q





dB

50

)

8

(

02

.

6

76

.

1







q

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

4.6.5 Bit rate for PCM transmission

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) : -Law

For 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/s

Needs Multiplexing

Needs Multiplexing

–

–

Process of transmitting two or

Process of transmitting two or

more signals simultaneously

Example : PCM

Example : PCM

-

-

TDM CEPT System

TDM CEPT System

Frame 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 s

3.9 s

125 s

125 s

2 ms

8 bits per time slot Bit duration

30 signal + 2 control = 32 channels = 1 frame

Signalling & synchronization

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

CEPT system – 32 channels (30 signals + 2 control)

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 s

MUX 1

MUX 2

MUX 3

MUX 4 30

Voice channels

. . . . . .

E1 line 2.048 Mbps

E2 line 8.448 Mbps E1

E1 E1

E2 E2 E2

E3 E3 E3

E3 line 34.368 Mbps

E4 line 139.264 Mbps

 There are 2 main components in the DM generator circuit, i.e

comparator and integrator.

Pemodulatan Digit



+Δ

-Δ

X

∑

-+

Pulse signal

s(t)

comparator

integrator

d(t)

x_DM(t)

m(t) e(t)

) ( ~ _t

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

 Output signal from comparator has the following function:

 The output from the comparator will be sampled with a pulse 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 signal will be

integrated first

 This signal will determine the error value

e

(

t

).

)

(

~

)

(

)

(

t

m

t

m

t

e









     













n s s n s DM

nT

t

nT

e

nT

t

t

e

t

x

)

(

)]

(

sgn[

)

(

)]

(

sgn[

)

(





Pemodulatan Digit





















sgn[

(

)]

)

(

t

e

t

d

0

)

(

0

)

(





t

e

t

e



  





n s

nT

e

t

Pemodulatan Digit

)

(

t

m

t

T_s

Δ

) ( ~ _t

m Effects 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 from happening, it has to follow the following condition:

dt

t

dm

mak

T

_s

)

(



Pemodulatan Digit • Binary 1 and 0 in PCM signal can be represented by several formats

known as line coding.

information

PCM Line

coder

channel

4.8.1 Line code format

4.8.1 Line code format

Digital Signal Encoding Formats

A. NRZ (Non Return to Zero)

- 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

Digital Signal Encoding Formats

B. RZ (Return to Zero)

• 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

Digital Signal Encoding Formats

C. Bi phase

• 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