Chapter 3 Baseband pulse and digital signaling...Chapter 3 Baseband pulse and digital signaling 2 Your site here 3.2 Pulse amplitude modulation 3 Your site here 3.2 Pulse amplitude

Chapter 3

Baseband pulse and digital signaling

22 Your site here

3.2 Pulse amplitude modulation

33 Your site here


Basis ---- The sampling theoremPAM. The amplitude of the carrier pulse varies with the analog baseband signal.Characteristics of PAM signal

Pulse-type signal, its amplitude denotes the analog information.Satisfied with sampling theoremPAM’s bandwidth is wider than that of analog waveform

Two classes of PAM signalsNatural Sampling (Gating)

Instantaneous Sampling

44 Your site here

3.2.1 Natural Sampling (Gating)

clock

S(t)

Analog bilateral switch

Ws(t)=w(t)s(t)W(t)

Generation of PAM with natural sampling


55 Your site here

PAM signal with natural sampling

)()()( tstwtws =

If w(t) is a analog waveform bandlimitedto B Hertz, The PAM signal that uses natural sampling (gating) is

where

∑ ∏∞

−∞=⎟⎠⎞

⎜⎝⎛ −

=k

skTttsτ

)(

66 Your site here

Spectrum of PAM waveform

[ ] )(sin)()( sn

ss nffWdn

dndtwFfW −== ∑∞

−∞= ππ

The spectrum for a naturally sampled PAM signal is

Where is the spectrum of the original unsampled waveform.

[ ])()( twFfW =

the spectrum of PAM signal with natural sampling is a function of the spectrum of the analog input waveform.

(3-3)

77 Your site here

3/1/ == sTd τ

Example.The case of an input waveform that has a rectangular spectrum, where the duty cycle of the switching waveform is

And the sampling rate is

Bfs 4=

Spectrum of PAM waveform

88 Your site here

Demodulation of PAM signal

Method 1 ---- Low-pass filter

Method 2 ---- Product detection

99 Your site here

3.2.2 Instantaneous sampling

W(t)

-τ/2 τ/2 t

τh(t)=∏(---)t

Ws(t)

Generation by using a sample-and-hold type of electronic circuit

δ(t-KTs)

1010 Your site here

Instantaneous sampled PAM

Characteristic

At t=kTs, the sampling values w(kTs) determine the amplitude of the flat-top rectangular pulses.

1111 Your site here

If w(t) is a analog waveform bandlimitedto B hertz, the instantaneous sampled PAM signal is given by

where h(t) denotes the sampling-pulse shape,

∑∞

−∞=

−=k

sss kTthkTwtw )()()(

⎪⎩

⎪⎨⎧

>

<=⎟

⎠⎞

⎜⎝⎛= ∏ 2/,0

2/,1)(

τ

τ

τ t

ttth

BffT sss 2/1 ≥=≤τwhere and

Instantaneous sampled PAM

1212 Your site here

Spectrum of Flat-top PAM

The spectrum for a Flat-top PAM signal is

[ ] ⎟⎟⎠

⎞⎜⎜⎝

⎛==

−= ∑∞

−∞=

ffthFfH

kffWfHT

fWk

ss

s

πτπττ sin)()(

)()(1)(

where

1313 Your site here

Spectrum of Flat-top PAM

1414 Your site here

Ws(f) Ws(f)´1/H(f) LPFWs(f)

Note;

1: Equalization filter ------ reduce the high frequency loss2: decreasing τ, The pulse width τ is called aperture.

Demodulation of flat-top PAM

Method 1 ---- Low-pass filter

1515 Your site here

Prefilter is needed before the multiplier to compensate for the spectral loss due to the aperture effect

Note;

Method 2 ---- Product detection

Ws(t)

PAM (flat-topsampling)

PrefilterH1(f)

Cos(nwst)

Oscillatorwo=nws

Low PassFilter ｜H(f)｜

Ws(t)^

H1(f)= Sin(πτf)πτf

fco-fco f

H(f)

Where B< fco<fs-B

Demodulation of flat-top PAM

1616 Your site here

Comparison -- in time domain

Fig3-1 PAM Signal with natural sampling Fig 3-5 PAM signal with flat-top sampling

waveform of PAM signal

1717 Your site here

Comparison -- in frequency domain

)(sin)( sn

s nffWdn

dndfW −= ∑∞

−∞= ππ

Spectrum of PAM signal

Fig3-3 The Spectrum of natural sampling PAM Fig 3-6 The Spectrum of Flat-top PAM

∑

∑∞

−∞=

∞

−∞=

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

−=

ns

ns

ss

nffWf

fd

nffWfHT

fW

)(sin

)()(1)(

πτπτ

1818 Your site here

The bandwith required for the transmission of PAM is much large than that of the original analog signal.

noise performance of the PAM system can never be better than that analog signal. ---- not very good for long-distance transmission.

Provide a means for converting an analog signal to a PCM signal. Time-division multiplexing etc.

Summary

1919 Your site here

3.3 Pulse code modulation

2020 Your site here

Coding & Streaming

1000 0101=-5

0001 0101=21

Pulse Code Modulation

Sampling

t1: -4.88 v

t2: +21.43 v

Analog waveform

Quantizing

-5 v

21 v

…, 0001 0101, 1000 0101, …

2121 Your site here

sampler quantizer encoder channel

Low-passfilter decoder

x(t)xn

x(t)

PCM transmitter (A/D conversion)

~

Three basic operationsSamplingQuantizingEncoding

xn~ s(t)


2222 Your site here

QuantizingApproximating the analog sample values by using finite number of levels

Uniform quantizing

Quantizing error

quantizing noise

PCM

2323 Your site here

EncodingThe quantized analog sample values are replaced by n-bit binary code

Quantized Sample Voltage Gray Code Word (PCM output)+7 110+5 111+3 101+1 100

-1 000-3 001-5 011-7 010

E.g. three-bit Gray code for M=8 levelsnM 2=


2424 Your site here

Bandwidth of PCM signals

The bit rate of PCM signal is

snfR =

Bit rate: R= fs (samples/s)×n(bits/sample)= 8 k sample/s ×8 bits/sample = 64 kbps

Example. Design of a PCM signal for telephone systemAssume: An analog audio voice-frequency (VF) telephone signal band: 300Hz ~ 3400 Hz The minimum sampling frequency is 2×3.4 = 6.8 ksample/sec. actually, using sampling frequency of 8 ksamole/sec.

2525 Your site here

For rectangular pulse, the first null bandwidth is

Example. the result for the case of the minimum sampling:

The bandwidth of (serial) binary PCM waveforms depends on:

The bit rateThe waveform pulse shape used to represent the data

sPCM nfRB ==

Number of quantizer levels, M

Length of the PCM, n(bit)

Bandwidth of PCM signal(the first null bandwidth)

2 1 2B

4 2 4B

8 3 6B

256 8 16B

……

Bandwidth of PCM signals

2626 Your site here

Effects of noise

Two main effects produce noise or distortion:

Bit errors in the recovered PCM signal . (channel noise, improper channel filtering, ISI etc. )

Quantizing noise that is caused by the M-step quantized at PCM transmitter

2727 Your site here

sampler quantizer encoder channel

Low-passfilter decoder

x(t)xn

x(t)

PCM transmitter (A/D conversion)

~

0101110……

0101010……

xn~

Effects of noise

2828 Your site here

Under certain assumptions, the ratio of the recover analog peak signal power to the total average noise power is:

The ratio of the average signal powerto the average noise power is

eoutpk PMM

NS

)1(413

2

2

−+=⎟

⎠⎞

⎜⎝⎛

eout PMM

NS

)1(41 2

2

−+=⎟

⎠⎞

⎜⎝⎛

Effects of noise

2929 Your site here

If Pe=0 (no ISI), the peak SNRresulting form only quantizing errors is

The average SNR due only to quantizing error is

23MNS

outpk

=⎟⎠⎞

⎜⎝⎛

2MNS

out

=⎟⎠⎞

⎜⎝⎛

α+=⎟⎠⎞

⎜⎝⎛ n

NS

dB

02.6

6-dB rule

α=4.77 for the peak SNR,α=0 for the average SNR.

Effects of noise

3030 Your site here

This equation points out the significant performance characteristic for PCM:

An additional 6-dB improvement in SNR is obtained for each bit added to the PCM word.

Assumptions:

① No bit errors② the input signal level is large enough to range over a significant number of quantizing levels

Effects of noise

3131 Your site here

Performance

3232 Your site here

Example SA3-2 (page 219)

In a communications-quality audio system, an analog voice-frequency (VF) signal with a bandwidth of 3200 Hz is converted into a PCM signal by sampling at 7000 samples/s and by using a uniform quantizer with 64 steps. The PCM binary data are transmitted over a noisy channel to a receiver that has a bit error rate (BER) of 10-4.

What is the null bandwidth of the PCM signal if a polar line code is used?What is the average SNR of the recovered analog signal at the receiving end?

PCM signal bandwidth and SNR

3333 Your site here

Nonuniform Quantizing

Characteristic voice analog signal

Nonuniform amplitude distributionThe granular quantizing noise will be a serious problem if uniform quantizing is used.

Solution: nonuiform quantizing

Nouniform Quantizing: a variable step size is used

3434 Your site here

Method:passing the analog signal through a compression (nonlinear) amplifier and then into a PCM circuit that uses uniform quantizer.

μ-Law and A-Low

AnalogSignal

A Compression(nonlinear)Amplifier

PCM(uniform quantized)

Nonuniform

Quantizing signal


3535 Your site here

μ-Law ( )( )μμ+

+=

1ln)(1ln

)( 12

twtw

Compressionquantizer

characteristic

Uniform quantizercharacteristic

(a) M=8 Quantizer Characteristic

00

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

0 0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

μ=0 μ=1μ=5

μ=100

μ=225

1)(0 1 ≤≤ tw


3636 Your site here

A-Law( )

( )⎪⎪

⎩

⎪⎪

⎨

⎧

≤≤+

+

≤≤+

=

1)(1,ln1

)(ln1

1)(0,1ln

)(

)(

11

11

2

twAA

twA

Atw

twA

twμ

0 0.2 0.4 0.6 0.8 1.0

0 0.2 0.4 0.6 0.8 1.0

0

0.2

0.4

0.6

0.8

1.0

A=1A=2

A=5

A=87.6A=100


3737 Your site here

In practice, the smooth nonlinear characteristics of μ-Law and A-Low are approximated by piecewise linear chords

Compression characteristics

3838 Your site here

Output SNR

Following 6-dB law:

α+=⎟⎠⎞

⎜⎝⎛ n

NS

dB

02.6

( )rmsxV /log2077.4 −=α

( )[ ]μα +−= 1lnlog2077.4

[ ]Aln1log2077.4 +−=α

whereUniform quantizing

μ-Law companding

A-Law companding

3939 Your site here

Comparison of output SNR

4040 Your site here

3.4 Digital signaling

4141 Your site here

Introduction

Baseband signalsthe signals involved in the analog-to-digital conversion

Bandpass signalsThe signals produced by using basebanddigital signals to modulate a carrier.

4242 Your site here

How do we mathematically represent the waveformfor a digital signal?

How do we estimate the bandwidth of the waveform?

In this section, we will answer the following questions:

Introduction

4343 Your site here

Vector representation

The voltage (or current) waveforms for digital signals can be expressed as an orthogonal series with a finite number of terms N.

• wk represents the digital data.Binary signaling , multilevel signaling.

• φk(t) are N orthogonal functions that give the waveform for a digital signal. • N is the number of dimensions required to describe the waveform.

∑=

=N

kkk twtw

1)()( ϕ

4444 Your site here

The orthogonal function representation of digital signaling corresponds to the orthogonal vector space represented by

A short-hand notation for the vector w is given by a row vector

∑=

=N

jjjww

1ϕ

W = ( w1, w2, …… wN )

Vector representation

4545 Your site here

baud rate and bit rate

Data rate:The baud (symbol rate)

The bit rate

For the case of the binary signal, n = N , R=D.For the case of the multilevel signal, n ≠ N, R ≠ D.

0/ TND =

0/TnR =

symbol / s

bits / s

Where N is the number of dimensions used in T0 s.

Where n is the number data bits sent in T0 s.

4646 Your site here

Bandwidth estimation

The lower bound for the bandwidth of the waveform w(t) is:

If theφk(t) are of the sin(x)⁄x type ,then

B = D⁄2 (Hz)otherwise

B＞D⁄2 (Hz)

DTNB

21

2 0

=≥

Equation (3-32) is useful for predicting the bandwidthof digital signals, especially when the exact bandwidth of the signal is difficult to calculate.

( 3-32 )(Hz)

4747 Your site here

Example 3.3.3 Bandwidth of PCM signals

The data rate of the binary encoded PCM signal is

The bandwidth of the binary encoded PCM waveformisbounded by

nBnfRB sPCM ==≥21

21

snfRD ==

The minimum bandwidth of is obtained only when (sinx)/x type pulse is used to generate the PCM waveform

snfR21

21

=

4848 Your site here

For rectangular pulse, the first null bandwidth is

Example. the result for the case of the minimum sampling:

3.3.3 Bandwidth of PCM signals

The bandwidth of (serial) binary PCM waveforms depends on:

The bit rateThe waveform pulse shape used to represent the data

sPCM nfRB ==

Number of quantizer levels, M Length of the PCM, n(bit) Bandwidth of PCM signal(the first null bandwidth)

2 1 2B

4 2 4B

8 3 6B

256 8 16B

……

4949 Your site here

3.4.3 Binary signaling

The orthogonal series cofficients wk take on binary values

Example 3-3A digital source can produce M=256 distinct messages. Each message is represented by n=8-bit binary words, assume that it takes T0=8 ms to transmit one message, a particular message corresponding to the code word 01001110 is to be transmitted.

w1=0, w2=1, w3=0, w4=0, w5=1, w6=1, w7=1, w8=0

Case 1. Rectangular pulse orthogonal functionCase 2. sin(x)/x pulse orthogonal function

5050 Your site here

1.5

1

0.5

0

-0.5 0 1 2 3 4 5 6 7 8t(ms)

1.5

1

0.5

0

-0.5 0 1 2 3 4 5 6 7 8

t(ms)

example

a. Rectangular pulse shape, Tb=1 ms

b. sin(x) / x pulse shape, Tb=1 ms

5151 Your site here

Binary signaling

When the rectangular pulse shape is used, the digital source information is transmitted via a binary waveform

When the sin(x)/x pulse shape is used, the digital source information is transmitted via an analog waveform.

Note:

5252 Your site here

Multilevel signaling

Question: the lower-bound bandwidth is B=N/(2T0) , can the bandwidth be made smaller ?

Thinking: if N could be reduced, the bandwidth could be made smaller.

N can be reduced by letting the wk takes on L>2 possible values.

When wk has L >2 possible values, the resulting waveform is called multilevel signal.

5353 Your site here

Example 3-4 L=4 multilevel signal

The source of example 3-3 will be encoded into L=4 multilevel signal, and the message will be sent in T0=8 ms.

ℓ bit D/A transferR bit/s D symbol/s=R/ℓ ,

R bit/s

Binary signal w1(t)

L-level waveform signalw2(t)

L=2ℓ

Multilevel signaling

5454 Your site here

One possible encoding scheme for an ℓ=2bit :

binary input(ℓ=2 bits)

11 +3

10 +1

00 -1

01 -3

output level(v)

For the binary code word 01001110 , the wk of Eq. (3-27) would be:

w1=-3, w2=-1, w3=+3, w4=+1

L=4 multilevel signal

5555 Your site here

Example 3-4

Note:The bit rate:R=n /T0 = /Ts=1 kbit/s

The baud rate:D=N /T0=1/Ts=0.5 kbaud

The bit rate and the baud rate are related by:


ℓ

R= ℓD

5656 Your site here

The null bandwidth of the rectangular-pulse multilevel waveform, fig. 3-14a:

B= 1/Ts = D = 500Hz

The absolute bandwidth of the sin(x)/x-pulse multilevel waveform, fig. 3-14b:

B= N/(2T0) = D/2 = 250Hz

In general, an L-level multilevel signal would have 1/ℓ the bandwidth of the corresponding binary signal, where ℓ =log2(L).


5757 Your site here

3.5 Line codes and spectra

5858 Your site here

Line codes and spectra

Self-synchronizationA spectrum that is suitable for the channelLow probability of bit errorTransmission bandwidth should be as small as possibleError detection capability……

Why line codes must be used?

Properties of a line codes

5959 Your site here

classification of the Line codes

Two major categories:

Binary line coding

• Return-to-zero (RZ)

• Nonreturn-to-zero (NRZ)

1 00101 1

1 00101 1

6060 Your site here

the line code may be further classified according to the rule used to assign voltage levels to represent the binary data.

Polar signaling

Bipolar (pseudoternary) signaling

Manchester signaling

Unipolar signaling1 100101

classification of the Line codes

6161 Your site here

Power spectra for binary line codes

The PSD can be evaluated byDeterministic techniqueStochastic technique

Deterministic technique

The stochastic approach will be used to obtain the PSD of the line codes, because the line codes have a random data sequence (instead of a particular data sequence)

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

∞→ TfW

fp T

Tw

2)(lim)(

∑∞=

−∞=

−=n

nn nffcfp )()( 0

2δFor a periodic waveform:

For a particular data sequence:

6262 Your site here

General expression for the PSD of a digital signal

∑∞

−∞=

=k

kfTj

ss

sekRTfF

fp π22

)()(

)(

∑=

+=I

iiiknn PaakR

1)()(

Where F(f) ↔ f(t), R(k) is the autocorrelation of the data:

∑∞

−∞=

−=n

sn nTtfats )()(

For a digital signal (or line code):


6363 Your site here

Unipolar NRZ

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛= )(11sin

4)(

22

fTfT

fTTAfpbb

bbNRZunipolar

δππ

Advantages:Disadvantages:

Easy to generate, only require one power supply.

Waste of power, dc coupled circuits are needed

1 100101


6464 Your site here

polar NRZ

22 sin)( ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

b

bb fT

fTTAfpNRZpolar π

π

Advantages:Disadvantages:

BER is superior to that of other signaling

Having a large PSD near dc; positive and negative power supply are required.

1 100101


6565 Your site here

unipolar RZ

⎥⎦

⎤⎢⎣

⎡−+⎟⎟

⎠

⎞⎜⎜⎝

⎛= ∑

∞

−∞=n bbb

bb

Tnf

TfTfTTAfp

RZunipolar)(11

2/)2/sin(

16)(

22

δππ

Advantages:

Disadvantages:

There is a discrete pulse at f=R, it can be used for recovery of the clock signal.

The first null bandwidth is twice that for NRZ signalRequires 3 dB more power than polar signal for the same PeThe spectrum is not negligible for frequency near dc

1 100101


6666 Your site here

bipolar RZ

( ))2cos(12/

)2/sin(8

)(22

bb

bb fTfT

fTTAfpRZbipolar

πππ

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

)(sin2/

)2/sin(4

)( 222

bb

bb fTfT

fTTAfpRZbipolar

πππ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1 100101


6767 Your site here

bipolar RZ

Advantages:

Disadvantages:

Has a spectral null at dc, so ac coupled circuit may be used.The clock signal can be easily be extracted for the bipolar

waveformHas single-error detection capabilities

The receiver has to distinguish between tree levels (+A, -A and 0)

Requires approximately 3 dB more power than polar signal for the same Pe. It has 3/2 the error of unipolar signal


6868 Your site here

Manchester NRZ

)2/(sin2/

)2/sin()( 22

2b

b

bb fT

fTfTTAfp

NRZManchesterπ

ππ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

0 0.5R R 1.5R 2R

0.5Tb

Pmanchester(f)

The PSD of line code(positive part of frequency)


6969 Your site here

Manchester NRZ

Advantages:

Disadvantages:

Has a zero dc levelA string of zeros will not cause a loss of the

clocking signal

The null bandwidth is twice that of the bipolar bandwidth


7070 Your site here

summery

The spectrum is a function of the bit pattern(via the bit autocorrelation) as well as the pulse shape.

The general result for the PSD, Eq. (3-36), is valid for multilevel as well as binary signaling.

7171 Your site here

3.5.3 Differential coding

7272 Your site here


Goal: to settle the question that the waveform is inverted.Method: differential coding is employed.

1−⊕= nnn ede

1−⊕= nnn eed ~~ ~

Encodeddifferential data are generated by

The received encoded data are decoded by

7373 Your site here

1−⊕= nnn edeEncoded sequence

Decoded sequence1−⊕= nnn eed

~~ ~


7474 Your site here

3.5.4 Eye patterns

The eye pattern provides the following information:

The timing errorallowed on the sampler at the receiver

The sensitivity to timing error

The noise margin

7575 Your site here

3.5.5 Regenerative repeaters

In long-distance digital communication systems, repeaters are necessary to amplify and “clean up” the signal periodicallyIn the analog system, linear amplifiers with appropriate filters can be used.In the digital system, Regenerative repeater can be used.

7676 Your site here

3.5.6 Bit synchronization

Digital communication usually need at least three types of synchronization signals:

Bit sync, to distinguish one bit interval from anotherFrame sync, to distinguish groups of dataCarrier sync. For bandpass signaling with coherent detection

Two kinds of ways to obtain synchronization signals:directly from the corrupted signalfrom a separate channel that is used only to transmit the sync. information

7777 Your site here

The complexity of the bit synchronizer circuit depends on the sync properties of the line code.

Example:The methods to obtain the bit sync clock signal for a unipolar RZ code:

To use narrowband bandpass filterTo use the phase-locked loop (PLL)

Example: the bit synchronizer for a polar NRZ line code.


7878 Your site here


7979 Your site here

Note: Unipolar, polar, and bipolar bit synchronizers will work only when there are a sufficient number of alternating 1’s and 0’s in the data.

Long strings of all 1’s or 0’s must be prevented.

To use bit interleaving ;To use a completely different type of line code that does not require alternating data for bit synchronization.


8080 Your site here

Multilevel signaling provides reduced bandwidth compared with binary signaling.

ℓ bit D/A transfer:

ℓ bit D/A transferR bit/s D symbol/s=R/ℓ ,

R bit/s

Single polar NRZBinary input

w1(t)

Multilevel polar NRZL-level waveform output

w2(t)

3.5.7 Power Spectra for multilevel polar NRZ signal

a binary signaling can be converted to a multilevel polar NRZ signal:

8181 Your site here

ℓ=3 , L= 2ℓ= 8

digital code output (an)i

000 +7001 +5010 +3011 +1

100 -1101 -3110 -5111 -7

Example Three-bit DAC code

8282 Your site here

Binary-to-multilevel polar NRZ signal conversion

The baud rateD = 1/Ts

= 1/(3Tb)= R/3

In generalD = R / ℓ

Tb

Ts

t0

w1(t)

w2(t)

Ts

t

76543

-2

21

-1

-3-4-5-6-7

output L=8

Figure 3-32 Binary-to-multilevel polar NRZ signal conversion

Bandwidth B ≥ D / 2

8383 Your site here

General expression for the PSD of a digital signal

∑∞

−∞=

=k

kfTj

ss

sekRTfF

fp π22

)()(

)(

∑=

+=I

iiiknn PaakR

1)()(

Where F(f) ↔ f(t), R(k) is the autocorrelation of the data:

Note:

The spectrum of the digital signal depend on two things:1. The pulse shape used2. Statistical properties of the data

Power Spectra for multilevel polar NRZ signal

8484 Your site here

For the where case ℓ=3 , Ts=3Tb, the rectangular pulse shape of width 3Tb:

2

33sin63)(

2 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

b

bb fT

fTTfpw π

π

the first null bandwidth for this multilevel polar NRZ signal is:

331 RT

Bb

null ==

----- one-third the bandwidth of the input binary signal


8585 Your site here

The null bandwidth is

2sin)(

2 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

b

b

fTfTKfp

w ππ

l

l

In general, for the case of L= 2ℓ levels, the PSD of a multilevel polar NRZ signal with rectangular pulse shape :

lRBnull =


8686 Your site here

Definition :

Where : R is the data rate B is the bandwidth

Spectral Efficiency

BR

=η

The spectral efficiency of a digital signal is given by the number of bits per second of data that can be supported by each hertz of bandwidth. That is, the spectral efficiency η is :

8787 Your site here

The spectral efficiency for multilevel polar NRZ signal

Where ℓ is the number of bits used in the DAC.

HzsbitRR

BR /)/(l

l===η

Note:ℓ Cannot be increased without limit to an infinite efficiency, because it is limited by the signal-to-noise ratio.

Spectral Efficiency

8888 Your site here

The maximum possible spectral efficiency is limited by the channel noise if the error is to be small

The maximum spectral efficiency ηmax is :

⎟⎠⎞

⎜⎝⎛ +==

NS

BC 1log2η

To approach this maximum spectral efficiency, practical systems usually incorporate error correction coding and multilevel signaling .

Spectral Efficiency

8989 Your site here

Code type First zero-pointBandwidth (Hz)

Frequency efficiencyη=R/B [(b/s)/Hz]

Unipolar NRZ R 1

polarity NRZ R 1

Unipolar RZ 2R 1/2

Bipolar RZ R 1

Manchester NRZ 2R 1/2

Multilevel polarity NRZ R/ ℓ ℓ

Spectral efficiency of Line codes

Chapter 3 Baseband pulse and digital signaling...Chapter 3 Baseband pulse and digital signaling 2 Your site here 3.2 Pulse amplitude modulation 3 Your site here 3.2 Pulse amplitude

Documents