Timing andsynch-

ronization

noiseh(t)channelimpulseresponse

gi(t) si(t)

z(T) r(t)

Digitalbaseband/bandpasswaveform

Digitalinput

Bitstream

ui

iuim

Digitaloutput

mi

Spreadspectrumdespread

Demod-ulate &sample

DetectDemul-tiplex

To other destinations

Channeldecode

DecryptVerify

Sourcedecode

FormatD/A

Information sink

Fig. 2. Block Diagram of a DCS

Informationsource

FormatA/D

XMT

Sourceencode Encrypt

Auth.Channelencode

Multi-plex

From other sources

Pulsemodu.

Band-passmodu.

Spreadspectrummodu.

Spreadcodegen.

OptionalEssential

RCV

Wave-formchan-nel(band-widthlimited)

Timing andsynch-

ronization

noise

gi(t) si(t)

y(T) r(t)

Digitalbaseband/bandpasswaveform

mi

Detect

To other destinations

FormatD/A

Information sink

Demod-ulate &sample

Block Diagram of a DCS in 411

Informationsource

FormatA/D

XMT

Multi-plex

From other sources

Pulsemodu.

Band-pass

modu.

RCV

h(t):Wave-formchan-nel

(band-width

limited)

ISI

Digitalinput

im

Digitaloutput

im

Digitaloutput

Demul-tiplex

h(t):channelimpulseresponse

Chapter 4. Baseband Transmission(Chapter 6 in the text)

Chapter 4. Baseband Transmission(Chapter 6 in the text)

1. Digital PAM Signals2. Power Spectra of Discrete PAM Signals3. Intersymbol Inteference4. Nyquist Criterion for Distortionless Signal

Baseband Transmission5. Correlative Coding and Equalization6. Remarks on Channel Bandwidth and Transmission

Rate

)(tx

)()( tthc δ≠

n(t)

)(tY

modulatorBandlimitedTransmission

channeldetector+

)(TY

decisiondevice

{ }1 ,0 },{ ∈nn aa

Figure 1.1 Baseband Transmission

pre-coderpulse shapefilter

nb{ }1 ,0sequencebinary

∈na)(thT

)( nTthb Tn −

We consider digital communications by means of PAM.

The modulator does the following tasks:1. The input binary data sequence is

subdivided into k-bit symbols and each symbol is mapped to a corresponding amplitude level .

2. The amplitude level modulates the output of the transmitting filter, the output of the modulator is the transmitted signal.

Thus, we can describe the modulator as a model with a pre-code which performs the task 1 and a pulse shape filter or the transmitting filter which performs the task 2.

{ }1 ,0sequencebinary

∈namodulator

)( nTthb Tn −

T = Tb: the symbol duration

modulator Pulse shape filterpre-coder=

{ } { }nn ba →

1. Digital PAM Signals

Pre-coder: transforming , desired form, which is a pre-coded signal format.

binarysource

pre-coder

)()( tthc δ≠

"0"or "1"

decisiondevice)(

detectorthd

+

n(t)

r(t) )(Ty

)(ty

x(t)

)(thT

{ }nb{ }na0or 1

bandlimitedchannel

∑ −=n

Tn nTthbtx )()(

pulse shape

: a pulse amplitude modulation (PAM) signal

Fig. 1.2. Block diagram of digital PAM system

{ }na

Objectives

(a) a good utilization of transmitted pulse energy

(b) a high bandwidth efficiency

(c) a high transmission reliability (intersymbol

interference (ISI) free transmission)

Non-return-to-zero (NRZ): a filter occupies the full duration of a signal.

Return-to-zero (RZ):a filter occupies a fraction (usually one-half) of the signal duration.

Two classes of digital PAM signals:

Consider {an}, a binary sequence.

Pre-coder: nn ba a

Pulse shaping filter: )( nTthbb Tnn −a

where T is the bit duration and is an impulse response of the filter.

)(thT

1). Unipolar (on-off) format (review):

0

1

⎩⎨⎧

==

=0 if 01 if

n

nn a

adb

na )( nTthb Tn −

T0

d

nb

2). Polar (antipodal) format (review):⎩⎨⎧

=−=

=0 if 1 if

n

nn ad

adb

0

1

na )( nTthb Tn −

−d

d

nb

)12( −= nn adbor equivalently,

3). Bipolar format:

⎩⎨⎧

=−+

=0 0

in s' 1 galternatin ,

n

nn a

addb

4). Manchester code:

0

1

na

−d

d

nb)( nTthb Tn −

0 1 1 0 1 0 0 0 1 1 Binary data

+1NRZ unipolar

+1

NRZ polar-1

+1

-1NRZ bipolar

+1Manchester

-1

PAM x(t) for different signing format

5). Polar quaternary signal (4-ary PAM):

Natural code Gray code

00 00

01 01

10 11

11 10

Level nb

−3

−1

1

3

na na

2T

)( nTthb Tn −

Polar quaternary format

+1

+3

-1-3

Natural-encoded

+1

+3

-1-3

Gray-encoded

0 1 1 0 1 0 0 0 1 1 Binary data

2. Power Spectra of Discrete PAM Signals2. Power Spectra of Discrete PAM Signals

∑ −=n

Tn nTthbtx )()(

The transmission signal is a discrete PAM:

where is a stationary random sequence, and depends on the different data formats and T is the symbol duration.

}{ nbB = nb

x(t) is a sample function of a random process X(t).

x(t)}{ nbB =)(thT

pulseshape

)(nRB

)( fHT

)(τXR

)( fSB )()()( 2 fSfHfS BTX = (1)

The power spectral of the random sequence is defined as

}{ nbB =

∑∞

−∞=

−=n

BB fnTjnRT

fS )2exp()(1)( π (2)

∑∞

−∞=

−=n

BTX fnTjnRfHT

fS )2exp()()(1)( 2 π

From (1) and (2), we obtain the psd of the PAM signal x(t) as follows

(3)

Remark. The results in (1) - (3) illustrate the dependence of the psd of the transmitted signal on(1) the spectral characteristics of of the pulse shape filter and(2) the spectral characteristics of of the pre-coded information sequence.

)( fSX)( fHT

)( fSB

Conclusion. Both and can be designed to control the shape and form of the psd of the transmitted signal.

)( fHT )( fSB

The formula given by (3) is the formula that we frequently use to determine the psd of x(t).

0-1/T-2/T-3/T 3/T1/T

2|)(| fHT

f2/T

Fig. 2.1. A rectangular pulse and its energy density spectrum 2|)(| fHT

)(thT

A

Tt

)(thT

0

Example 2.1. Determine the psd in (3) where is a rectangular pulse shown in Fig. 2.1.

)(thT

Solution. The Fourier transform of is as follows )(thT

)exp()(sin)( fTjfTcATfH T π−=

where x

xxcππ )sin()(sin =

Hence, we have )(sin|)(| 2222 fTcTAfH T =

Substituting it into (3), thus, we obtain that

∑∞

−∞=

−=n

BX fnTjnRfTcTAfS )2exp()()(sin)( 22 π (4)

)(thTif the pulse shape is the rectangular pulse then the psd of the PAM x(t) is given by

Definition. A discrete random sequence is said to be independent if for any k time instances ,

are independent.

}{ nX

kttt <<< L21

kttt XXX ,,,21L

Definition. A discrete random sequence is said to be mutually uncorrelated if any pair of , are uncorrelated, i.e., .

}{ nX

nk XX and nk ≠][][][ nknk XEXEXXE =

Property. If a discrete random sequence is independent and for each k, the random variable takes each value equally likely, then is an independent identical distributed (i.i.d.) random sequence.

}{ nX

kX}{ nX

][)0( 2kB bER =

Example 2.2. Assume that is an independent random sequence and each binary symbol occurs equally likely. Determine the psd of x(t) where the signal format is NRZ polar and the pulse shape is the rectangular pulse defined in Example 2.1.

)(thT

}{ nbB =

222 }{)(}{ ddbPddbPd kk =−=−+==

Solution. We will use (4) to determine the psd of x(t). First we need to compute the autocorrelation of the stationary random sequence . From the given condition, is an i.i.d. random sequence. Thus

)(nRB

}{ nb

21}{}{ =−=== dbPdbP kk

}{ nb

Thus 0)4/14/1()4/14/1()( 22 =+−+= ddnRB

For , . By independence and equally likely occurrence, the probability of is given by

0≠n [ ]nkkB bbEnR +=)(),( nkk bb +

),( nkk bb +),( nkk aa +

(d, d)(1, 1)

(d, −d)(1, 0)

(−d, d)(0, 1)

(−d, −d)(0, 0)},{

),( ybxbP

yxpnkk ===+

1/4

1/4

1/4

1/4

),( yxp

⎪⎩

⎪⎨⎧

≠=

=⇒0 00 )(

2

nndnRB

)(sin)( 222 fTcTdAfS X =⇒ (by (4))

@G. Gong 23

Summary on the psd of PAM signal:

∑ −=n

Tn nTthbtx )()(

• For a general pulse )(thT

∑∞

−∞=

−==n

BTBTX fnTjnRfHT

fSfHfS )2exp()()(1)()()( 22 π

• If is a rectangular pulse with magnitude A and duration T, then

)(thT

∑∞

−∞=

−=n

BX fnTjnRfTcTAfS )2exp()()(sin)( 22 π

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

fTb

SX(f

)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

fTb

SX(f

)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

fTb

SX(f

)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

fTb

SX(f

)

Figure 2.2 Power spectra of different binary data formats

NRZ polar

NRZ unipolar

NRZ bipolar Manchester

Considerations for Selection of Signaling Schemes:

Presence or absence of a DC levelPower spectral density, particularly its value at 0 HzSpectral occupancy (i.e., bandwidth)Bit error probability performance (i.e., relative immunity

from noise )Ease of clock signal recovery for symbol

synchronizationPresence or absence of inherent error detection properties

Example 2.3. Let be an uncorrelated binary valued (±1) random sequence, each having a zero mean and a unit variance. Let whose value is determined by

}{ na

}{ nb

1−+= nnn aab

Determine the psd of the transmitted signal.

Solution. The autocorrelation function of the sequence is }{ nb

[ ]nkkB bbEnR +=)( [ ]))(( 11 −++− ++= nknkkk aaaaE

⎪⎩

⎪⎨

⎧±=

==

otherwise 01 1

02nn

From (2), the psd of the input sequence is }{ nb

fTT

fTT

fSB ππ 2cos1)2cos1(21)( =+=

By substituting it into (1), the psd for the modulated signal is

fTfHT

fS TX π22 cos)(4)( =

Remark. This example explained that the transmitted signal spectrum can be shaped by having a correlated sequence as the input to the modulator.

}{ nb

decisiondevice

{ }na0or 1

)(Ty

)(ty{ }nb{ }na

0or 1∑ −=

nTn nTthbtx )()(

)(thT

PulseShape

Binarysequence

Pre-coder )()( tthc δ≠

bandlimitedchannel +

n(t)

r(t)

)(thd

Detector

AWGN:

3. Intersymbol Interference3. Intersymbol Interference

Fig. 3.1 Digital PAM Transmission through bandlimitedbaseband channel

• The transmission signal: ∑ −=n

Tn nTthbtx )()(

• The received signal is passed through an LTI with impulse response . If is matched to , then its output SNR is a maximum at the the sampling instant t = T.

)(thd )(thd )(th

• The channel output: which is the received signal at the demodulator

)()()( tnnTthbtrn

n +−=∑

is the impulse response of the channel, and n(t) represents AWGN.

where is the pulse of the cascade of the transmitting filter and the channel, i.e., )()()( ththth Tc ∗=

)(thc

)(th

Thus the sampler produces

)()()( kTnnTkTpbkTy on

n +−= ∑μ

kbμ= )(kTno+∑≠

−+kn

n nTkTpb )(μ

desired symbol ISI component

(1)

The output of the detect (receiving) filter

)()()( tnnTtpbty on

n +−= ∑μwhere )()()( ththtp d∗= )()()( ththth dcT ∗∗=

)()()( thtntn do ∗=which is normalized such that p(0) = 1 and

The first term on the RHS of (1) is the desired symbol , scaled by the gain parameter μ when the receiving filter is matched to H(f). The scale factor is given by

kb

dtth∫∞

∞−= )(2μ dffH∫

∞

∞−= )(2

dffHfH cT∫∞

∞−= 22 |)(||)(| hE≡

The second term on the RHS of (1) represents the effect of the other symbols at the sample instance t = kT, called intersymbolinterference (ISI).

The scale factor is given by

which is the energy of the pulse shape h(t).

Example 3.1. Illustrate the ISI effect when the input data is 011.

01

na

−dd

nb

+d

-d

0 1 1Binary data

T 3T

T2T

T ′ TT ′+3TT ′+2

TT ′+

0

1 1

T

)(thT

T ′

)(thc

T T ′ TT ′+

)()( thth cT ∗

*

Observation:

a) The effect of passing the transmitted pulse through a channel with memory is that the duration of the transmitted pulse is “stretched”.

b) The pulse stretching is referred to as a “time dispersion” and the channel is called “time dispersive channel”.

c) Time dispersion causes overlaps between adjacent symbol bits atthe output of the communication channel.

d) The distortion that arises from the overlapping between adjacentsymbols is called “intersymbol interference” (ISI).

)(thT)()( tthc δ≠

e) ISI is preserve only when the symbol is preserve. i.e. ISI is signal dependent of noise (distortion).

f) Unlike additive noise, ISI cannot be suppressed by simply increasing the signal energy.

Question: How to design such that the ISI term vanishes?

(t) and (t) dT hh

)()( kTnbkTy ok += μ

0)( =−∑≠kn

n nTkTpbIf (2)

then there is no ISI effect. In this case

The transmission performance is only degraded by AWGN.

)(tpμIf so, is said to be an effect channel without ISI.

)(thT )()( tthc δ≠ )(thd

{ }dd −,

decisiondevice

)(ty )(Ty ia

n(t)

+

A trivial solution: If , then the ISI can beremoved. However, it is not practical in practice.

)()( ttp δ=