advance communication system lectures part 4

8/2/2019 Advance Communication System Lectures Part 4

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Learning Outcomes

29 April 20121

Interpret various type of PulseModulation (PAM, PWM,PPM,PCM).

Analyze Pulse Modulation anddiscuss the process of Sampling,

Quantization and Coding, whichforms the fundamental for digitaltransmission of any signal


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Pulse Modulation

29 April 20122

Pulse Modulation is a process of sampling analogsignal and then converting them into discretepulses and transporting the pulses from a sourceto a destination over a transmission medium.

A device to perform this is called ADC (Analog-to-Digital Converter) & DAC (Digital-to-AnalogConverter).


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Pulse Modulation

29 April 20123

1. PAM (Pulse Amplitude Modulation)2. PWM (Pulse Width Modulation)

3. PPM (Pulse Position Modulation)

4. PCM (Pulse Code Modulation)


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PAM (Pulse Amplitude Modulation)

29 April 20124

It is used to describe the conversion ofanalog signal to pulse-type signal in whichthe amplitude of the pulse denotes the

analog information.

In addition, it is a series of pulses in whichthe amplitude of each pulse represents the

amplitude of the information signal at agiven time.


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Pulse Modulation

29 April 20125

PWM (Pulse Width Modulation)

It is a pulse duration modulation (PDM) or pulselength modulation. The width of pulse is varied

proportional to the Amplitude of the analog signal atthe time signal is sampled.

PPM (Pulse Position Modulation)

It is a series of pulses in which the timing of each

pulse represents the amplitude of the informationsignal at a given time.


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PCM (Pulse Code Modulation)

29 April 20126

It is a series of pulse in which the amplitude of theinformation signal at a given time is coded as abinary number. The pulses are of fixed length and

fixed amplitude.

PCM is generated by 3 processes; Sampling,Quantization & Encoding.

An Integrated circuit that perform PCM encodingand decoding function is called CODER ORDECODER.


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29 April 20127

Pulse Modulation

Analog signal

Sample pulse

Pulse widthmodulation

Pulse position modulation

Pulse amplitudemodulation

Pulse codemodulation

8 bit

ts


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Pulse Modulation

Pulse Modulation consists of:

Easily effected bynoise

Less susceptible tonoise

Less susceptible tonoise compared toPAM


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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


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Sampling An analog signal must be sampled at Nyquist rate to avoid aliasing

Quantization & Coding

Quantization : Process ofMapping samples of a continuousamplitude waveform to a finite set of amplitudes.

A fixed number of levels including the maximum and

minimum value of the analog signal

Number oflevels is determined by the number of bits usedfor coding

Coding : translate the quantized sample into a code

number.


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29 April 201211


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Quantization Interval (V)

Represent the voltage value for each quantized level

For example: For a sampled signal(mp) 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 torepresent 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

V102

L

mV

p

quantization


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13

M = L


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+mp

-mp

0

1 11

1 10

1 01

1 00

0 000 01

0 10

0 11

t

V

We assume that the amplitude of

m(t) is confined to the range (-mp,

mp)

L

mV

p2

nL 2


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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 of4.375 V.

The difference between the sampled value and the quantized value

results in quantization noise.

V375.4

2

V25.1V50 V

V


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+mp

-mp

0

1 11

1 10

1 01

1 00

0 000 01

0 10

0 11

valueSign bit

t

100 101 111 111 111 110 101 000 010 011 011 010 000 001 110 110 110 100

Quantization error

Qe

PCM code

t

The same code representing several

samples with different amplitudes

Step size


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The binary codes used for PCM are n-bit codes (sign-magnitudecode) where the MSB bit is the sign bit. If PCM is 3-bit codes,

then the sign and magnitude are shown below:

In terms of Voltage, the maximum signal voltages are 3 V or -3 V

and the minimum signal voltages are 1 V or -1 V.

Sign Magnitude Decimal value Quantization range

(V)

1 1 1 +3 +2.5 to +3.5

1 10 +2 +1.5 to +2.5

1 01 +1 +0.5 to 1.5

1 00 +0 0 to +0.5

0 00 -0 0 to -0.5

0 01 -1 -0.5 to -1.50 10 -2 -1.5 to -2.5

0 11 -3 -2.5 to -3.5

Folded binary code

0 V codes each

have an input

range equal to

only one half aquantum


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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

UNIFORM QUANTIZATION

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

interval/step size.

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


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Input analog signal

Sampling pulse

PCM code

Quantization

PAM signal

What is the PCM code for +2.6 V??

+2.6

V


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Input analog signal

Sampling pulse

PCM code

PAM signal

Question: What is the quantized interval and PCM code for +1.75 V??


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Quantization Error/ Noise

Folded PCM code = sample voltage

resolutionFor input at 2.6 V, the PCM code is therefore:

2.6/1 = 2.6

But since there is no code for +2.6, the magnitude is

rounded off to the nearest valid code, which is 111(+3V)

Thus there is difference of 0.4

QUANTIZATION ERROR (Qe)

or also known as quantization noise (Qn)

Qe =sample voltage - original analog signal


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Maximummagnitude Qe isequal to one-half

a quantum

Low Resolution

, more accuratethe quantizedsignal willresemble the

original analogsample

2229 April 2012

resolution

2eQ


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Linear input-output transfer curve

Linear

Quantization

Error


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Dynamic Range

max max

min2 1resolution

nV V

DR V

DR = dynamic range (unitless)

Vmin= the quantum value

Vmax= the maximum voltage magnitude of the DACs

n = number of bits in a PCM code (excl. sign bit)

20log 2 1n

dBDR

Ratio of the largest possible magnitude to the smallest

(other than 0) magnitude that can be decoded by the

digital-to-analog converter (DAC) in the receiver


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Number of bits used for a PCM code depends onthe dynamic range

DR = 2n-1

Thus 2n= DR + 1

And therefore, The minimum number of bit used:n = log ( DR + 1 )

log 2

2 1 2n nDR

For n > 4

20log 2 1 20 log2 6n

dBDR n n

D i R


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No of Bits No of Levels DR (dB)

1 2 6.02

2 4 12

3 6 18.1

4 16 24.1

5 32 30.1

6 62 36.1

7 128 42.1

8 256 48.2

9 512 54.2

10 1024 60.2

11 2048 66.2

12 4096 72.2

13 8192 78.3

14 16348 84.3

15 32768 90.3

16 65536 96.3

Dynamic Range


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Dynamic Range

Total dynamic range in (dB) = 6 x (number of bits)


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Coding Efficiency

minimum number of bitscoding efficiency= 100

actual number of bits

Coding efficiency is a numerical indication of how

efficiently a PCM code is utilized


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EXAMPLEA PCM systems has the following specification:

Maximum Analog Input Frequency = 4 kHz

Maximum decoded voltage at the receiver =2.55 V

The dynamic range = 46 dB

Determine the following :

(a) Minimum Sampling Rate

(b) Minimum number of bits used in PCM code

(c) Resolution

(d) Quantization Error(e) Coding Efficiency


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Solution

(a) The minimum sampling rate:

fs

= 2fa

= 2(4 kHz) = 8 kHz

(b) Calculate the Dynamic range :

46 = 20log(Vmax/ Vmin)

Vmax/ Vmin = antilog (46/20) = 199.5Thus, the minimum number of bit

used:

n = log (199.5 + 1) / Log 2 = 7.63

(c) Resolution is defined as:

Vmax / 2n - 1 = 0.01 V

(d) Quantization Error :Q = resolution / 2 = 0.01 V / 2 =

(e) Coding Efficiency

Coding efficiency= (8.63/9)(100)= 95.89%


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PCM

system

Example :

Vpp= 31.5 V

6 bit code (5 bits for

magnitude and 1 bit for

sign

(a) No of levels:

(b) LSB voltage, V :(c) Voltage value for 101101 ;

(d) Voltage value for 011001 ;

(e) PCM Code for input +13.62 V

(g)PCM Code for input9.37 V


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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, V : 31.5/64 = 0.492 V(c) Voltage value for 101101 ; +(13 x 0.492) = +6.4 V

(d) Voltage value for 011001 ; (25 x 0.492) = -12.3 V

(e) Code for input +13.62 V

= 13.62/0.492 = 27.68 28 => 111100(g)Code for input9.37 V

= 9.37/0.492 = 19.04 19 => 010011

advance communication system lectures part 4

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