B32SF3 Signals and Information e 1

8/7/2019 B32SF3 Signals and Information e 1

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Pulse Code Modulation

The advantages of digital communication systems(cf. analogue communication) Easier to store as a pattern of 1's and 0's

Increased Immunity

non-linearities

Easier to process in computers and digital signalprocessors

Can be coded for security and error correction purposes Several digital signals can easily be interleaved

(multiplexed) and transmitted on one channel

Noisy digital signals can be regenerated more effectivelythan analogue signals can be amplified.


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0000

1111

1110

1101

1100

1011

1010

1001

0001

0010

0011

0100

0101

0110

0111

A brief aside about ADCs

0000 0110 0111 0011 1100 1001 1011

Numbers passed from ADC to computer to represent analogue voltage

ADCs are used to convert an analogue input voltage into a number that can

be interpreted as a physical parameter by a computer.

Resolution=1 part in 2n


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Sampling The input signal is sampled prior to digitisation and an

approximation to the input is reconstructed by the digital-to-

analogue converter:

Sampling Digitisation code, modulate

Transmission

Wire/optical fibre

Aerial/free-space

input

FilteringDigital-to-analogue

conversionDemodulate, Decode

output


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Sampling an analogue signal Prior to digitisation, signals must be sampled

With a frequency fs=2B=1/T

ADC converts the height of each pulse into binary representation

Sampling involves the multiplication of the signal by a train of samplingpulses

vi(t) vo(t)

T TT

time time

1

0

vi(t)

Input Signal

WaveformSampled Signal

Waveform

(i)(ii) Sampling (iii)

...011101010011010011...

Samples are coded for transmission

(iv)

Analogue-to-digital

Converter

Digital-to-analogue

Converter

Sampled signal is recovered

(v)


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Sampling as multiplication by a sampling

waveform:

vs(t)

time

T

X1

0

Vi(f) V

o(f)

vs(t)

T TT

timetime

vi(t)

time

" ample-

a d-Hold" Sampling pulse is

short enough so thatcan normallyconsidered have zero

duration DAC, however

produces pulseslength T

Multiplication = Amplitude modulation

Amplitude modulation produces sidebands


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Sidebands produced by multiplication with acarrier That is, amplitude modulation

[c

[c+ [

s[

c-[

s[

s0

s

1/2cA

s

LSB USBBaseba d


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Sidebands at each harmonic of the sampling pulse

Digital-to-analogue conversion involves recovery of the baseband How?

What is the minimum value of fs for which there is no overlap of the Harmonics withthe baseband?

vs(t)

Vi(f) V

o(f)

freq

Vi(f)

B

fs= 1/T 2f

sfreq

Vo(f)

fs/2

B

armonics of vs(t)

(iv)

(v)

(vi)

time

T

X1

0

vs(t)


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fs= 1/T 2f

sfreq

Vo(f)

fs/2

B

GuardBand

B fs-B

H(f)1

0freq

Reconstructionfilter

Originalsignal

armonics of vs(t)

(vi)

freq

Vi(f)

If the sidebands do not overlap the signalcan be recovered


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Practical sampling the "Sample-and-hold" system:

This is Nyquists theorem For a signal of bandwidth B Hz, the minimum sampling

rate is 2B samples/s

T TTtime

Sample-

and- old


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Effect of sampling rate

sampling at more than the Nyquist Rate

Sample

sidebands

(i) Oversample - F s 4B

FsB

Fs- B Fs+ B

0freq

spectr m

Original signal

can easily be recovered


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Sampling at the Nyquist Rate

cannot build an ideal filter -

(ii) Nyquistsam ling- s = 2B

FsB

Fs-B Fs+B

0freq

Original signal can just be recoveredwith ideal filter


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Undersampling

produces aliasing distortion!

(iii) Undersam ling-F s = 1.5B

FsB

Fs-BFs+B

0freq

Original signal

cannot be recovered

Alias

signal


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Aliasing-time domain

Oversampled signal

Reconstructed signal

Undersampled signal

Reconstructed signal

Sampling:aliasing & Nyquist:time domain


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The Anti-alias

(Pre-sampling)

filter

ensures that

sampling

obeys the

Nyquisttheorem

fs= 1/T freq

Vo(f)

B

Aliasing orFoldback

distortion

Originals ectrum

freq

fs/2

fs/2

fs= 1/T freq

Vo(f)

B

Filtereds ectrum

fs/2

(ii) Anti-aliasfilter

bandwidth less than fs/2

high frequency components removed

(iii) Guardband

created

1st pair of

sampling sidebands

fs-B

H(f)

1

0

(i) Signalbandwidth > fs/2


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Examples

For the compact disc (Audio CD) themaximum signal frequency is 20 kHz and thesampling rate is 44.1 kHz.

The Nyquist Sampling Rate is 40 kHz

Hence the guard band is 4.1 kHz wide.

In the telephone system (see Section 5.8),

the speech signal has a bandwidth up to 3.4kHz and a sampling rate of 8 kHz,

The Nyquist Sampling Rate is 6.8 kHz

Hence the guard band is 1.2 kHz wide.


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Regeneration vamplification:

Gain of amplifiers equals loss in transmission lines

SNR analog: S/kN

SNR digital: S/N In practice finite S/N means there will be a low level of bit errors

Some accumulation of bit-error noise with repeaters, but much lower levelthan with analogue amplification

1 2 k

Atte atio A i eac sectio , Gai A i eac amplifier

xA xAxAo rce est

N

S/N

N

A(S+N)

S/2N

N N

S/kN

A(S+2N)

S/(k-1)N

O tp t SNR

(i) Analog e Channel

1 2 k

Atte atio A i eac sectio , Gai A i eac repeater

xA xAxASo rce est n

N

S/N

N

AS

S/N

N N

S/NAS

S/N

O tp t SNR

(ii) Digital Channel


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A Pulse-Code Modulation communicationsystem

"PCM"

Anti-alias

filter

Sampleand

HoldADC

The channel

codi g, mod latio , repeaters,

demod latio , decodi g

DACReconstr ction

Filter

Sig al

I

Sig al

O t


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A digital communication system - "PCM" Anti-alias Filter*

Digitiser/Sample-and-Hold circuit* Analogue-to-Digital Converter*

Coding- Source codingfor data compression,

Line codingfor signalling efficiency

Errorcoding to reduce the effect oferrors

Modulator

Physical Channel (with repeaters if necessary)* Copper cables

Fibre Optic cables

Radio

Sonar

Recording medium

Demodulator

Decoder (Source-, Line- and Error-)

Digital-to-Analogue Converter*

Reconstruction Filter*


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Time-division Multiplexing "TDM" Allocate interleaved time-slots to each signal

Assemble the binary coded samples into Frames: 2-channel time-division multiplexing scheme:

Frame n Frame n+1

Slot1 Slot2 Slot1 Slot2

Channel1Sample1

Channel2Sample1

Channel1Sample2

Channel2Sample2

Frame n Frame n+1

slot 1 slot 2 slot 1 slot 2

Sampling and

A/D conversionDigitised samples

Frame length 1/fs

Slot length =No. of channels

Frame length

Analogue

inputs

Channel1

Channel2

10111001

Sample 2

10111010

Sample 1

00110000

Sample 2

00101111

Sample 1

Sample rate f s 10111010001011111011100100110001

T

wo channels share a single physical channel Cost?


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The 32-channel PCM Transmission system

30speech signals plus two control channels

for signalling and synchronising:

Signal bandwidth 3.4 kHz

Sampling rate 8 kHz

Hence frame length?

Sample size 8 bits/sample

Hence bit rate from each signal 64 kbit/s

32 channels Hence each time slot3.906Qs

1/(8000*32)

Overall data rate 2.048 Mbit/s

8000*32*8

125Qs


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1

2

3

4 1

2

3

4

13

14

15

1661

62

63

64

1

2

3

4

2.048 Mbit/s

8.448 Mbit/s

34.368 Mbit/s

139.264 Mbit/s

x16 x4

(32 channels x 64

= 2048 channels)


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A number of frames can be time-division multiplexed

together in a TDM heirachy. 4 frames of 32 channels

= 128 basic PCM channels,

Has data rate of 4 x 2.048 Mbit/s = 8.192 Mbit/s

8.448Mbit/s including extra signalling bits

4 x 128 = 512 channels

Has data rate = 4 x8.192 Mbit/s (+ signalling bits) = 34.368 Mbit/s

etc

Up to a multiplex of32768 channels with an overall data

rate of2.48832G

bit/

s.


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1

2

3

4 1

2

3

4

13

14

15

1661

62

63

64

1

2

3

4

2.048 Mbit/s

8.448 Mbit/s

34.368 Mbit/s

139.264 Mbit/s

x16 x4

(32 channels x 64

= 2048 channels)


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Spectrum of a train of pulses:

Samplingp lsespectr mfrequency

Harmonic spacing 1/T

N llspacing 1/ X

Pulse

idt X

Sampling pulse

period T

time

Samplingp lse waveform

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