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Principles of Communications

Chapter 5: Digital Transmission through Bandlimited Channels

Selected from Chapter 9.1-9.4 of Fundamentals of Communications Systems, Pearson Prentice Hall

2005, by Proakis & Salehi

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Topics to be Covered

Digital waveforms over bandlimited baseband channels Band-limited channel and Inter-symbol interference Signal design for band-limited channels System design Channel equalization

Source A/D converter

Channel Detector D/A converter

User

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5MHz, 10MHz15MHz, 20MHz

Bandlimited Channel

Modeled as a linear filter with frequency response limited to certain frequency range

20MHz

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Baseband Signalling Waveforms

To send the encoded digital data over a baseband channel, we require the use of format or waveform for representing the data

System requirement on digital waveforms Easy to synchronize High spectrum utilization efficiency Good noise immunity No dc component and little low frequency component Self-error-correction capability …

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Basic Waveforms

Many formats available. Some examples: On-off or unipolar signaling Polar signaling Return-to-zero signaling Bipolar signaling – useful because no dc Split-phase or Manchester code – no dc

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0 1 1 0 1 0 0 0 1 1

On-off (unipolar)

polar

Return to zero

bipolar

Manchester

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Spectra of Baseband Signals

Consider a random binary sequence g0(t) - 0，g1(t) - 1 The pulses g0(t）g1(t）occur independently with

probabilities given by p and 1-p，respectively. The duration of each pulse is given by Ts.

)(ts

Ts

0 ( 2 )Sg t T+1( 2 )Sg t T−

∑=∞

−∞=nn tsts )()(

0

1

( ), with prob. ( )

( ), with prob. 1s

ns

g t nT Ps t

g t nT P−

= − −7

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Power Spectral Density

PSD of the baseband signal s(t) is

1st term is the continuous freq. component 2nd term is the discrete freq. component

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0 1 0 12

1 1( ) (1 ) ( ) ( ) ( ) (1 ) ( ) ( )ms s s s s

m m mS f p p G f G f pG p G fT T T T T

δ∞

=−∞

= − − + + − −∑

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For polar signalling with and p=1/2

For unipolar signalling with and p=1/2, and g(t) is a rectangular pulse

For return-to-zero unipolar signalling

0 1( ) ( ) ( )g t g t g t= − =

21( ) ( )S f G fT

=

0 ( ) 0g t = 1( ) ( )g t g t=

sin( ) fTG f TfTπ

π

=

2sin 1( ) ( )

4 4xT fTS f f

fTπ δ

π

= +

/2Tτ =

[ ]

2

2odd

sin / 2 1 1 1( ) ( ) ( )16 / 2 16 4x

m

T fT mS f f ffT Tmπ δ δ

π π

= + + −

∑

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PSD of Basic Waveforms

unipolar

Return-to-zero polar

Return-to-zero unipolar

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Intersymbol Interference

The filtering effect of the bandlimited channel will cause a spreading of individual data symbols passing through

For consecutive symbols, this spreading causes part of the symbol energy to overlap with neighbouring symbols, causing intersymbol interference (ISI).

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Baseband Signaling through Bandlimited Channels

( ) ( )t i Ti

x t A h t iT∞

=−∞

= −∑Output of tx filter

Output of rx filter

Input to tx filter ( ) ( )s ii

x t A t iTδ∞

=−∞

= −∑

Source Σ

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Pulse shape at the receiver filter output

Overall frequency response

Receiving filter output( ) )()( tnkTtpAtv o

kk +−= ∑

∞

−∞=

Source Σ

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Sample the rx filter output at (to detect Am)

Desired signal intersymbol interference (ISI)Gaussian noise

Source Σ

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Eye Diagram

Distorted binary wave

Eye pattern

Tb15

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ISI Minimization

Choose transmitter and receiver filters which shape the received pulse function so as to eliminate or minimize interference between adjacent pulses, hence not to degrade the bit error rate performance of the link

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Signal Design for Bandlimited ChannelZero ISI

Nyquist condition for Zero ISI for pulse shape

With the above condition, the receiver output simplifies to

Echos made to be zero at sampling points

or

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Nyquist Condition: Ideal Solution

Nyquist’s first method for eliminating ISI is to use

f1/2T-1/2T 0

P(f)1

The minimum transmission bandwidth for zero ISI. A channel with bandwidth B0 can support a max. transmission rate of 2B0 symbols/sec

( )

==

Tt

TtTttp sinc

//sin)(

ππ

p(t) p(t-T)

= Nyquist bandwdith,

“brick wall” filter

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Achieving Nyquist Condition

Challenges of designing such or is physically unrealizable due to the abrupt transitions at ±B0

decays slowly for large t, resulting in little margin of error in sampling times in the receiver.

This demands accurate sample point timing - a major challenge in modem / data receiver design.

Inaccuracy in symbol timing is referred to as timing jitter.

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Practical Solution: Raised Cosine Spectrum

is made up of 3 parts: passband, stopband, and transition band. The transition band is shaped like a cosine wave.

( )

−≥

−<≤

−−

+

<≤

=

10

10110

1

1

2||0

2||22

||cos121

||0 1

)(

fBf

fBfffBff

ff

fP π

f2B00

P(f)

B0

1

f1

2B0-f1

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Raised Cosine Spectrum

The sharpness of the filter is controlled by .

Required bandwidth

f2B00

P(f)

B0

1

1.5B0

0.5

α = 0

α = 0.5

α = 10

11Bf

−=α

Roll-off factor

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Time-Domain Pulse Shape

Taking inverse Fourier transform

00 2 2 2

0

cos(2 )( ) sinc(2 )1 16

B tp t B tB t

παα

=−

Ensures zero crossing at desired sampling instants Decreases as 1/t2, such that the data

receiving is relatively insensitive to sampling time error

Tb0 2Tb

t/Tb0 1 2-1-2

α=1α=0.5α=0

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Choice of Roll-off Factor

Benefits of small Higher bandwidth efficiency

Benefits of large simpler filter with fewer stages hence easier to implement less sensitive to symbol timing accuracy

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Signal Design with Controlled ISIPartial Response Signals

Relax the condition of zero ISI and allow a controlled amount of ISI

Then we can achieve the max. symbol rate of 2W symbols/sec

The ISI we introduce is deterministic or controlled; hence it can be taken into account at the receiver

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Duobinary Signal

Let {ak} be the binary sequence to be transmitted. The pulse duration is T.

Two adjacent pulses are added together, i.e.

The resulting sequence {bk} is called duobinary signal

1k k kb a a −= +

Ideal LPF

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Characteristics of Duobinary SignalFrequency domain

( )2( ) 1 ( )j fTLG f e H fπ−= +

2 cos ( 1/ 2 )0 ( )

j fTTe fT f Tπ π− ≤= ot her wi se

( 1/ 2 )( )

0 ( )L

T f TH f

≤= ot her wi se

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[ ]( ) ( ) ( ) ( )Lg t t t T h tδ δ= + − ∗ sin / sin ( ) // ( ) /t T t T T

t T t T Tπ π

π π−

= +−

sinc sinct t TT T

− = +

is called a duobinary signal pulse

It is observed that:

( )g t

0(0)= 1g g =

1( )= 1g T g =

( )= 0 ( 0 1),ig iT g i= ≠

(The current symbol)

(ISI to the next symbol)

2 sin /( )

T t Tt T t

ππ

= ⋅−

Decays as 1/t2, and spectrum within 1/2T

Time domain Characteristics

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Decoding

Without noise, the received signal is the same as the transmitted signal

When is a polar sequence with values ＋1 or －1:

When is a unipolar sequence with values 0 or 1

0k i k i

iy a g

∞

−=

=∑ 1k k ka a b−= + =

1

1 1

1

2 ( 1)0 ( 1, 1 or 1, 1)2 ( 1)

k k

k k k k k k

k k

a ay b a a a a

a a

−

− −

−

= == = = = − = − = − = = −

{ }ka

{ }ka1

1 1

1

0 ( 0)1 ( 0, 1 or 1, 0)2 ( 1)

k k

k k k k k k

k k

a ay b a a a a

a a

−

− −

−

= == = = = = = = =

A 3-level sequence

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To recover the transmitted sequence, we can use

Main drawback: the detection of the current symbol relies on the detection of the previous symbol => error propagation will occur

How to solve the ambiguity problem and error propagation?

Precoding: Apply differential encoding on so that Then the output of the duobinary signal system is

1 1ˆ ˆ ˆk k k k ka b a y a− −= − = −

{ }ka 1k k kc a c −= ⊕

1k k kb c c −= +29

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Block Diagram of Precoded Duobinary Signal

+ Transformer

T

{ }ka+{ }kc { }kb

{ }1kc −

( )LH f ( )y t0, 1, 2

-2, 0, 2

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Modified Duobinary Signal

Modified duobinary signal

After LPF , the overall response is 2k k kb a a −= −

( )LH f

4( ) (1 ) ( )j fTLG f e H fπ−= −

22 sin 2 ( 1/ 2 )0

j fTTje fT f Tπ π− ≤= otherwise

sin / sin ( 2 ) /( )/ ( 2 ) /t T t T Tg t

t T t T Tπ π

π π−

= −−

22 sin /( 2 )

T t Tt t T

ππ

= −−

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Properties

The magnitude spectrum is a half-sin wave and hence easy to implement

No dc component and small low freq. component

At sampling interval T, the sampled values are

also delays as . But at , the timing offset may cause significant problem.

0

1

2

(0) 1( ) 0(2 ) 1( ) 0, 0,1,2i

g gg T gg T gg iT g i

= == == = −= = ≠

t T=( )g t 21 / t

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Decoding of modified duobinary signal

To overcome error propagation, precoding is also needed.

The coded signal is 2k k kc a c −= ⊕

2k k kb c c −= −

+ +2T

+

-

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Update

Now we have discussed: Pulse shapes of baseband

signal and their power spectrum ISI in bandlimited channels Signal design for zero ISI and

controlled ISI

We next discuss system design in the presence of channel distortion Optimal transmitting and receiving filters Channel equalizer

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Optimum Transmit/Receive Filter

Recall that when zero-ISI condition is satisfied by p(t) with raised cosine spectrum P(f), then the sampled output of the receiver filter is

Consider binary PAM transmission:

Variance of Nm =

with

Error Probability can be minimized through a proper choice of HR(f) and HT(f) so that is maximum (assuming HC(f) fixed and P(f) given)

(assume )

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Optimal Solution

Compensate the channel distortion equally between the transmitter and receiver filters

Then, the transmit signal energy is given by

Hence

By Parseval’s theorem

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Noise variance at the output of the receive filter is

Special case: This is the ideal case with “flat” fading No loss, same as the matched filter receiver for AWGN

channel

Performance loss due to channel distortion

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Exercise

Determine the optimum transmitting and receiving filters for a binary communications system that transmits data at a rate R=1/T = 4800 bps over a channel with a frequency response |Hc(f)| = ; |f| ≤ W where W= 4800 Hz

The additive noise is zero-mean white Gaussian with spectral density

2)(1

1

Wf

+

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Solution

Since W = 1/T = 4800, we use a signal pulse with a raised cosine spectrum and a roll-off factor = 1.

Thus,

Therefore

One can now use these filters to determine the amount of transmit energy required to achieve a specified error probability

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Performance with ISI

If zero-ISI condition is not met, then

Let

Then

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Often only 2M significant terms are considered. Hence

Finding the probability of error in this case is quite difficult. Various approximation can be used (Gaussian approximation, Chernoff bound, etc).

What is the solution?

with

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Monte Carlo Simulation

Let

I xerror occurselse

( ) =

10

( )∴ ==∑P

LI Xe

l

l

L11

( )

where X(1), X(2), ... , X(L) are i.i.d. (independent andidentically distributed) random samples

t mTm =

γ=ThresholdX

~ Vm

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If one wants Pe to be within 10% accuracy, how many independent simulation runs do we need?

If Pe ~ 10-9 (this is typically the case for optical communication systems), and assume each simulation run takes 1 msec, how long will the simulation take?

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We have shown that By properly designing the transmitting and receiving filters

one can guarantee zero ISI at sampling instants, thereby minimizing Pe.

Appropriate when the channel is precisely known and its characteristics do not change with time

In practice, the channel is unknown or time-varying

We now consider: channel equalizer

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A receiving filter with adjustable frequency response to minimize/eliminate inter-symbol interference

What is Equalizer

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Equalizer Configuration

Overall frequency response:

Nyquist criterion for zero-ISI

Ideal zero-ISI equalizer is an inverse channel filter with

TransmittingFilter HT(f)

Channel HC(f)

EqualizerHE(f)

t=mT

vmv(t)

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Linear Transversal Filter

Finite impulse response (FIR) filter

are the adjustable equalizer coefficients is sufficiently large to span the length of ISI

c-N⊗ ⊗ ⊗ ⊗c-N+1 cN-1 cN

∑

Unequalized input

(τ=T)

t=nT

(2N+1)-tap FIR equalizer

Output

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Zero-Forcing Equalizer

: the received pulse from a channel to be equalized

At sampling time

Tx & Channel

To suppress 2N adjacent interference terms

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Rearrange to matrix form

channel response matrix

Zero-Forcing Equalizer

or the middle-column of

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Example

Find the coefficients of a five-tap FIR filter equalizer to force twozeros on each side of the main pulse response

)(tpc

t∆

tT ∆+

tT ∆+2

tT ∆+3

tT ∆+4tT ∆+−

tT ∆+− 2

tT ∆+−3

tT ∆+− 4

1.0

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Solution By inspection

The channel response matrix is

[ ]

. . . . .. . . . .

. . . . .. . . . .

. . . . .

Pc =

− −− −

− −− −

− −

10 0 2 01 0 05 0 0201 10 0 2 01 0 05

01 01 10 0 2 010 05 01 01 10 0 2

0 02 0 05 01 01 10

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The inverse of this matrix (e.g by MATLAB)

Therefore,

Equalized pulse response

It can be verified that

[ ]

. . . . .

. . . . .. . . . .

. . . . .. . . . .

Pc− =

− −− −

− −− −

− −

1

0 966 0170 0117 0 083 0 0560118 0 945 0158 0112 0 0830 091 0133 0 937 0158 0117

0 028 0 095 0133 0 945 01700 002 0 028 0 091 0118 0 966

c1=0.117, c-1=-0.158, c0 = 0.937, c1 = 0.133, c2 = -0.091

Solution

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peq ( ) ( . )( . ) ( . )( . ) ( . )( . )

( . )( . ) ( . )( . ).

3 0117 0 005 0158 0 02 0 937 0 05

0133 01 0 091 010 027

= + − + −

+ + − −=−

peq ( ) ( . )( . ) ( . )( . ) ( . )( . )

( . )( . ) ( . )( . ).

− = + − − + −

+ + − −=

3 0117 0 2 0158 01 0 937 0 05

0133 01 0 091 0 010 082

SolutionNote that values of for or are not zero. For example:

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Minimum Mean-Square Error Equalizer

Drawback of ZF equalizer Ignores the additive noise

Alternatively, Relax zero ISI condition Minimize the combined power in the residual ISI and

additive noise at the output of the equalizer

MMSE equalizer: a channel equalizer that is optimized based on the minimum mean-

square error (MMSE) criterion

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MMSE Criterion

∑−=

−=N

Nnn nTtyctz )()(

( )[ ] Minimum)( 2 =−= mAmTzEMSE

The output is sampled at :

Let Am = desired equalizer output

Output from the channel

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MMSE solution is obtained by

E is taken over and the additive noise

where

MMSE Criterion

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MMSE Equalizer vs. ZF Equalizer

Both can be obtained by solving similar equations

ZF equalizer does not consider effects of noise

MMSE equalizer designed so that mean-square error (consisting of ISI terms and noise at the equalizer output) is minimized

Both equalizers are known as linear equalizers

Next: non-linear equalizer

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Decision Feedback Equalizer (DFE)

DFE is a nonlinear equalizer which attempts to subtract from the current symbol to be detected the ISI created by previously detected symbols

Feedforward Filter

Input Symbol Detection

Output

Feedback Filter

+

-

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Example of Channels with ISI

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Frequency Response

Channel B will tend to significantly enhance the noise whena linear equalizer is used (since this equalizer will have to introduce a large gain to compensate channel null).

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Performance of MMSE Equalizer

Proakis & Salehi, 2nd31-taps

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Performance of DFE

Proakis & Salehi, 2nd

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Maximum Likelihood Sequence Estimation (MLSE)

TransmittingFilter HT(f)

Channel HC(f)

Receiving Filter HR(f)

t=mT

ym

MLSE (Viterbi

Algorithm)

Let the transmitting filter have a square root raised cosine frequency response

The receiving filter is matched to the transmitter filter with

The sampled output from receiving filter is

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MLSE

Assume ISI affects finite number of symbols, with

Then, the channel is equivalent to a FIR discrete-time filter

T T T T

Finite-state machine

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Performance of MLSE

Proakis & Salehi, 2nd

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Equalizers

Preset Equalizer

AdaptiveEqualizer

BlindEqualizer

Linear Non-linear

ZF MMSE

DFEMLSE

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Suggested Reading

Chapter 9.1-9.4 of Fundamentals of Communications Systems, Pearson Prentice Hall 2005, by Proakis & Salehi

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