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

A/D Channel D/A

Source Detector User
converter converter

 Digital waveforms over bandlimited baseband channels

 Band-limited channel and Inter-symbol interference
 Signal design for band-limited channels
 System design
 Channel equalization

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 Bandlimited Channel

5MHz, 10MHz
20MHz 15MHz, 20MHz

 Modeled as a linear filter with frequency response limited

to certain frequency range

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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.

s (t )
g 0 (t + 2TS )
g1 (t − 2TS )

∞
s (t ) = ∑ sn (t ) Ts
n = −∞
 g 0 (t − nTs ), with prob. P
sn (t ) = 
 g1 (t − nTs ), with prob. 1 − P
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 Power Spectral Density
 PSD of the baseband signal s(t) is

2
∞
1 1 m m m
S (= p (1 − p ) G0 ( f ) − G1 ( f ) + 2 ∑ pG0 ( ) + (1 − p )G1 ( ) δ ( f − )
2
f)
Ts Ts m = −∞ Ts Ts Ts

 1st term is the continuous freq. component

 2nd term is the discrete freq. component

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 For polar signalling with g 0 (t ) =
− g1 (t ) =
g (t ) and p=1/2

1
S ( f ) = G( f )
2

 For unipolar signalling with g 0 (t ) = 0 g1 (t ) = g (t ) and p=1/2,

and g(t) is a rectangular pulse
2
 sin π fT  T  sin π fT  1
G( f ) = T   =Sx ( f )   + δ( f )
 π fT  4  π fT  4

 For return-to-zero unipolar signalling τ = T/2

2
T  sin π fT / 2  1 1 1 m
=
Sx ( f )   + δ( f )+ ∑ δ( f − )
16  π fT / 2  16 4 odd m [ mπ ]2
T

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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

Source Σ

Input to tx filter =
xs (t ) ∑ A δ (t − iT )
i = −∞
i

∞
Output of tx filter =
xt (t ) ∑ A h (t − iT )
i = −∞
i T
Output of rx filter

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Source Σ

 Pulse shape at the receiver filter output

 Overall frequency response

 Receiving filter output

∞
v(t ) = ∑ A p(t − kT ) + n (t )
k = −∞
k o

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Source Σ

 Sample the rx filter output at (to detect Am)

Gaussian noise
Desired signal intersymbol interference (ISI)

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 Eye Diagram
 Distorted binary wave

 Eye pattern

15 Tb Meixia Tao @ SJTU

 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 Channel
Zero ISI
 Nyquist condition for Zero ISI for pulse shape

Echos made to be zero

at sampling points

or

 With the above condition, the receiver output simplifies to

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 Nyquist Condition: Ideal Solution
 Nyquist’s first method for eliminating ISI is to use

sin (π t / T ) t
p(t ) = = sinc 
π t /T T 

P(f) p(t) p(t-T)

“brick wall” filter
1

-1/2T 0 1/2T f

= Nyquist bandwdith,

The minimum transmission bandwidth for zero ISI. A channel with

bandwidth B0 can support a max. transmission rate of 2B0 symbols/sec
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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.

 P(f)
 1 0 ≤| f |< f1
 1
1   π (| f | − f1 )  
P( f ) =  1 + cos  f1 ≤| f |< 2 B0 − f1
2
   02 B − 2 f1 
2B0-f1
 0 f1 B0 2B0 f

0 | f |≥ 2 B0 − f1

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 Raised Cosine Spectrum

P(f)
α=0
Roll-off factor
1 α = 0.5 f1
α = 1−
0.5 α=1 B0
0 B0 1.5B0 2B0 f

 The sharpness of the filter is controlled by .

 Required bandwidth

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 Time-Domain Pulse Shape
 Taking inverse Fourier transform

cos(2πα B0t )
p(t ) = sinc(2 B0t )
1 − 16α 2 B02t 2
Ensures zero crossing at
desired sampling instants Decreases as 1/t2, such that the data
receiving is relatively insensitive to
sampling time error
0 Tb 2Tb
α=1
α=0.5
α=0

-2 -1 0 1 2 t/Tb

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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 ISI
Partial 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. b=
k ak + ak −1

Ideal LPF

 The resulting sequence {bk} is called duobinary signal

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Characteristics of Duobinary Signal
Frequency domain
T ( f ≤ 1/ 2T )
G ( f )= (1 + e − j 2π fT
)H L (f) HL ( f ) = 
0 (ot her wi se)

2Te − jπ fT cos π fT ( f ≤ 1/ 2T )
=
0 (ot her wi se)

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 Time domain Characteristics
sin π t / T sin π (t − T ) / T
g (t )= [δ (t ) + δ (t − T ) ]=
∗ hL (t ) +
πt /T π (t − T ) / T
t  t − T  T 2 sin π t / T
= sinc   + sinc  = ⋅
T   T  π t (T − t )
 g (t ) is called a duobinary
signal pulse
 It is observed that:
g (0)=g 0 = 1 (The current symbol)
g (T )=g1 = 1 (ISI to the next symbol)
g (iT =
)=gi 0 (i ≠ 0 ,1)
Decays as 1/t2, and spectrum within 1/2T
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 Decoding
 Without noise, the received signal is the same as the
transmitted signal
∞
yk = ∑ ai g k −i =ak + ak −1 =bk A 3-level sequence
i =0
 When {ak } is a polar sequence with values ＋1 or －1:
 2 (=
ak a=
k −1 1)

yk =
bk =
 0 (ak =
1, ak −1 =
−1 or ak =
−1, ak −1 =
1)
 −2 (ak =
ak −1 =
−1)

 When {ak } is a unipolar sequence with values 0 or 1
 0 (=
ak a=
k −1 0)

y=
k b=
k  1 (a=
k 0, ak −=1 1 or a=
k 1, ak −=
1 0)
 2 (=
ak a=
 k −1 1)

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 To recover the transmitted sequence, we can use

aˆk =bk − aˆk −1 =yk − aˆk −1

 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 {ak } so that c=
k ak ⊕ ck −1
 Then the output of the duobinary signal system is
b=
k ck + ck −1
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 Block Diagram of Precoded
Duobinary Signal

{ak } {ck } {bk } y (t )

+ + 0, 1, 2
Transformer HL ( f )
-2, 0, 2

T
{ck −1}

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 Modified Duobinary Signal
 Modified duobinary signal
b=
k ak − ak − 2
 After LPF
H L (f ), the overall response is

2Tje− j 2π fT sin 2π fT ( f ≤ 1/ 2T )
G ( f )= (1 − e − j 4π fT
)H L ( f ) =
0 otherwise

sin π t / T sin π (t − 2T ) / T 2T 2 sin π t / T

=
g (t ) − = −
πt /T π (t − 2T ) / T π t (t − 2T )

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Meixia Tao @ SJTU 32
 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
=
g (0) g=
0 1
g (T=
) g=
1 0
g (2T ) = g 2 = −1
g (iT=
) g=
i 0, i ≠ 0,1, 2
 g (t ) also delays as 1 / t 2 . But at t = T , the timing offset
may cause significant problem.

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 Decoding of modified duobinary signal

 To overcome error propagation, precoding is also needed.

c=
k ak ⊕ ck − 2
 The coded signal is

b=
k ck − ck − 2

+
+ +
-
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 (assume )

 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)
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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

By Parseval’s theorem

 Hence
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 Noise variance at the output of the receive filter is

Performance loss due to channel distortion

 Special case:
 This is the ideal case with “flat” fading
 No loss, same as the matched filter receiver for AWGN
channel
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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
1
response |Hc(f)| = ; |f| ≤ W where W= 4800 Hz
f
1 + ( )2
W
 The additive noise is zero-mean white Gaussian with
spectral density

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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

with

 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?

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 Monte Carlo Simulation
~ Vm

X
Threshold = γ

t m = mT

Let
1 error occurs
I ( x) = 
0 else

∴
1 L
Pe = ∑ I X ( l )
L l =1
( )
where X(1), X(2), ... , X(L) are i.i.d. (independent and
identically distributed) random samples

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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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 What is Equalizer

 A receiving filter with adjustable frequency response to

minimize/eliminate inter-symbol interference

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 Equalizer Configuration
t=mT
Transmitting Channel Equalizer
Filter HT(f) HC(f) HE(f)
v(t) vm

 Overall frequency response:

 Nyquist criterion for zero-ISI

 Ideal zero-ISI equalizer is an inverse channel filter with

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 Linear Transversal Filter
 Finite impulse response (FIR) filter

Unequalized
input
c-N ⊗ c ⊗
-N+1 cN-1 ⊗ cN ⊗
t=nT
∑
Output
(τ=T)
(2N+1)-tap FIR equalizer
 are the adjustable equalizer coefficients
 is sufficiently large to span the length of ISI

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 Zero-Forcing Equalizer
 : the received pulse from a channel to be equalized

Tx & Channel

 At sampling time

To suppress 2N adjacent interference terms

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 Zero-Forcing Equalizer

 Rearrange to matrix form

channel response matrix

or the middle-column of

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 Example
 Find the coefficients of
pc (t )
a five-tap FIR filter
equalizer to force two
1.0
zeros on each side of
the main pulse
response

− 4T + ∆t − 2T + ∆t T + ∆t 3T + ∆t

− 3T + ∆t − T + ∆t ∆t 2T + ∆t 4T + ∆t

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 Solution
 By inspection

 The channel response matrix is

 10 . 0.2 − 01 . 0.05 − 0.02
 − 01 . .
10 0.2 − 01 . 0.05 
 
[ Pc ] =  01 . − 01 . .
10 0.2 − 01. 
 
 − 0 . 05 01. − 01 . .
10 0.2 
 0.02 − 0.05 01 . − 01. . 
10

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 Solution
 The inverse of this matrix (e.g by MATLAB)
 0.966 − 0170
. 0117
. − 0.083 0.056 
 0118
. 0.945 − 0158
. 0112
. − 0.083
 
−1
[ Pc ] =  − 0.091 0133
. 0.937 − 0158
. 0117
. 
 
 0.028 − 0 .095 0133
. 0.945 − 0170
. 
− 0.002 0.028 − 0.091 0118
. 0.966 

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

 Equalized pulse response

 It can be verified that

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

peq (3) = (0117

. )(0.005) + ( −0158
. )(0.02) + (0.937)( −0.05)
+ (0133 . ) + ( −0.091)( −01
. )(01 .)
= − 0.027

peq ( −3) = (0117

. )(0.2) + ( −0158
. )( −01
. ) + (0.937)( −0.05)
+ (0133 . ) + ( −0.091)( −0.01)
. )(01
= 0.082

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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

Output from
the channel

N
z (t ) = ∑ cn y (t − nT )
n=− N

The output is sampled at :

Let Am = desired equalizer output

[ ]
MSE = E (z ( mT ) − Am ) = Minimum
2

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 MMSE Criterion

where
E is taken over and the
additive noise

 MMSE solution is obtained by

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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

Input Feedforward + Symbol Output

Filter -
Detection

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 when

a 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

31-taps Proakis & Salehi, 2nd

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Performance of DFE

Proakis & Salehi, 2nd

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 Maximum Likelihood Sequence
Estimation (MLSE)
t=mT
MLSE
Transmitting Channel Receiving
(Viterbi
Filter HT(f) HC(f) Filter HR(f) ym 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 Adaptive Blind

Equalizer Equalizer Equalizer

Linear Non-linear

ZF DFE
MMSE MLSE

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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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