3F4 Line Coding
Dr. I. J. Wassell
Introduction
• Line coding is the procedure used to convert an
incoming bit stream, bk to symbols ak which are
then sent as pulses akhT(t-kT) on to the channel
• The line coding technique affect the properties of
the transmitted signal
• Desirable properties are
– Self Synchronisation. There should be sufficient
information in the transitions and zero-crossings to
permit symbol timing clock regeneration
Introduction
– Spectrum suited to the channel. The PSD of the
transmitted signal should be compatible with the
channel frequency response Hc(w), eg,
• Many channels cannot pass dc (zero frequency) owing to ac
coupling
• Lowpass response limits the ability to carry high frequencies
Line Codes for Copper Cables
• Following a qualitative introduction to some
practical line codes, we will look at a framework
for their quantitative analysis
• Unipolar Binary
–
–
–
–
binary symbols transmitted
pulse weightings, ak=0 or A Volts, eg, ‘0’ and ‘1’
no guarantee of transitions (for timing regeneration)
transmitted signal has non-zero mean, so the channel
must be able to pass dc. If not, errors will occur with
long runs of ‘1’s in the data
Unipolar Binary
1
A
0
0
1
1
0
0
0
0
1
1
0
1
Line Codes for Copper Cables
• Polar Binary (do not confuse with bipolar)
–
–
–
–
binary symbols transmitted
pulse weightings, ak=-A or A Volts, eg, ‘0’ and ‘1’
no guarantee of transitions (for timing regeneration)
transmitted signal is zero mean (provided data is
equiprobable), but there is still a high PSD at low
frequencies, therefore still a problem with channels
which do not pass dc
Polar Binary
1
A
0
-A
0
1
1
0
0
0
0
1
1
0
1
Line Codes for Copper Cables
• Bipolar (or Alternate Mark Inversion (AMI))
– ternary (3 level) symbols transmitted
– pulse weightings, ak=0 or either -A or +A Volts, eg, ‘0’
or ‘1’, that is,
• ‘0’ sent as ak=0
• ‘1’ sent as ak=-A or +A alternately
– no guarantee of transitions (for timing regeneration)
– signal mean is always zero (independent of data)
– Spectral null at dc means there is no longer a problem
if channel does not pass dc
– 1 binary symbol sent as 1 ternary symbol, 1B1T code
Bipolar (AMI)
1
A
0
-A
0
1
1
0
0
0
0
1
1
0
1
Line Codes for Copper Cables
• High Density Bipolar (HDBn)
– modified bipolar codes which guarantee transitions
despite runs of zeros.
– thus substitute runs of more than n zeros
– an alternative name is BnZS
• in this case substitute runs of n or more zeros
– HDB3 is a popular code
• note HDB3 is equivalent to B4ZS
HDB3
• Any run of 4 zeros is replaced by the special
pattern: B00D
– D is sent as +/- A, such that successive D’s have
alternating polarity (1st D is arbitrary)
– B is sent as 0 or +/- A. Select B such that the next D
violates the AMI alternating polarity result, ie send B
as 0 if D violates the polarity rule. Otherwise send
B=D to force a violation of the alternating polarity rule
• Equivalently,
HDB3
– B is sent as 0 if there has been an odd number of input
‘1’s since the last special sequence B00D
– B is sent as +/- A if there has been an even number of
input ‘1’s since the last special sequence B00D
• All other symbols obey the AMI rules
• The scheme allows the unique detection of the
special sequences, since polarity violations
correspond with the D symbols.
• An overall mean of zero is achieved
HDB3
1
A
0
-A
0
1
1
0
0
0
0
1
1
0
1
HDB3
• There are never more than 3 consecutive zeros, so
plenty of edges for timing regeneration
• The channel is not required to pass dc
• The transmit power requirement is a little greater
than AMI (about 10%)
Power Spectra for Line Codes
• We can use the earlier results concerning power
spectra to derive the PSD for PAM schemes
• Initially we will consider the case where the
symbols are transmitted as weighted impulses.
• We will then generalise to arbitrary pulse shapes
Signal TX as impulse train
• A PAM signal sent as a weighted impulse train is,
x(t ) 

 a  (t  nT )
n  
n
s
Where Ts is the symbol period. We will show that the PSD of
x(t) is given by,
and
1
S x (w ) 
Ts

 R(m)e
m
jmwTs

1

  R(0)  2 R(m) cos(wmTs ) 
Ts 
m1

R(m)  E[an anm ]
Which is the discrete Autocorrelation function (ACF). Also note
that R(m) = R(-m) for real valued an
Proof
• Using direct method, ie,
lim E[ X T (w ) ]
S x (w ) 
T 
T
2
• The truncated signal xT(t) is,
xT (t ) 
N
 a  (t  nT ),
n N
n
s
T  (2 N  1)Ts
• Now,

X T (w )   xT (t )e jwt dt

And substituting for xT(t) gives,
Proof
X T (w ) 


Now,

N
  a  (t  nT )e
n   N
n
N

n N

 jwt
s
dt
 jwt
a

(
t

nT
)
e
dt
 n
s
N
 jnwTs
a
e
 n
n N
X T (w )  X T (w ) X (w )
2
*
T
Proof
| X T (w ) |2 

• Now,

N
 jnwTs
a
e
 n
n N
N

N
jkwTs
a
e
 k
k  N
N
j ( k  n )wTs
a
a
e
 nk
n N k  N

N
N

j ( k  n )wTs 
2
E | X T (w ) |  E    an ak e

n   N k   N


N

N
j ( k  n )wTs
E
[
a
a
]
e
 n k
n N k  N
Proof
k  nm
• Let,
E | X
(w ) |     E[a a
N
N n
2
T
m  k n
and so
n
n N m N n

N

nm
]e
j ( n  m  n )wTs
N n
jmwTs
R
(
m
)
e

n N m N n
• Replace the outer sum over index n by 2N+1,


E | X T (w ) |2  (2 N  1)
N n
jmwTs
R
(
m
)
e

m N n
Proof
• Now,
lim E[ X T (w ) ]
S x (w ) 
T 
T
2
Where T=(2N+1)Ts, so
N n
1
S x (w ) 
(2 N  1)  R(m)e jmwTs
N   (2 N  1)Ts
m N n
lim
1

Ts

jmwTs
R
(
m
)
e

m  
Proof
1

1
jmwTs
jmwTs 
  R(0)   R(m)e
  R ( m) e

Ts 
m  
m 1



1
 jmwTs
jmwTs 
  R(0)   R(m)e
  R ( m) e

Ts 
m 1
m 1




1
jmwTs
 jmwTs 
  R(0)   R(m) e
e

Ts 
m 1


1

  R(0)  2 R(m) cos(wm Ts ) 
Ts 
m 1

Remembering that R(m) = R(-m) for real valued an
Proof
• Note that for R(m)=0 for all m except zero,
the PSD reduces to
R(0)
S x (w ) 
Ts
e.g., for polar binary line coding.
For Arbitrary Pulse Shapes
• Recall that a PAM signal with a desired pulse
shape h(t) may be generated by filtering (actually
convolving) the weighted impulse train with a
filter whose impulse response is h(t)
• From the power spectra results we obtain,
S y (w )  H (w ) S x (w )
2

1

 H (w )
 R(0)  2 R(m) cos(wmTs ) 
Ts 
m1

2
PSD of Specific Schemes
• Given a particular line coding scheme which
generates symbols ak, the transmitted PSD is
calculated as follows:
– Determine the discrete ACF
M
R(m)  E[ak ak  m ]   Ri pi
i 1
Where,
•M is the number of possible values that akak+m can take on
•Ri is the ith value of akak+m
•pi is the probability that Ri occurs
PSD of Specific Schemes
• Evaluate impulse train PSD
– Substitute R(m) into the PSD formula,

1

S x (w )   R(0)  2 R(m) cos(wmTs ) 
Ts 
m1

• Evaluate PSD with pulse shaping
– Multiply Sx(w) by |H(w)|2,
S y (w )  H (w ) S x (w )
2
Example- Polar Binary
bk : 0,1
ak :  1,1
m
0
1
bk ....bk m ak ....ak m Ri  ak ak m
M
pi
R(m)   Ri pi
i 1
0...0
 1... 1
R1  1
p1  0.5
1...1
1...1
R2  1
p2  0.5
0...0
0...1
 1... 1
R1  1
1...0
1...1
1... 1
1...1
 1...1
R(0)  (1 0.5)  (1 0.5)
R(0)  1
p1  0.25 R( 1)  (1 0.25) 
R2  1 p2  0.25 (1 0.25)  (1 0.25)
R3  1 p3  0.25  (1 0.25)
R4  1 p4  0.25 R( 1)  0
Example- Polar Binary
• So,

1

S x (w )   R(0)  2 R(m) cos(wmTs ) 
Ts 
m1

1

Ts
Example- Polar Binary
• Now extend to pulses with a rectangular
shape and duration Ts. To do this we
convolve the impulse stream with a filter
h(t) with a rectangular impulse response,
i.e., from the E and I Data Book,
h(t)
-b/2
a
0
FT
b/2
t
 wb 
H w   ab sinc 

 2 
sin  x 
where,
sinc x  
x
Example- Polar Binary
• In our example, b=Ts and a=1, so the filter
frequency response is,
 wTs 
H w   Ts sinc 

 2 
This response has an amplitude of Ts at w = 0 and zero
crossings at multiples of 2p/Ts rad/s or 1/Ts Hz.
Example- Polar Binary
• Now the PSD at the output of the filter H(w)
is,
S y (w )  S x (w ) H (w )
2
1
 wTs 
 Ts sinc 

Ts
 2 
2
2
 wTs 
 Ts sinc 

 2 
This response has an amplitude of Ts at w = 0 and zero
crossings at multiples of 2p/Ts rad/s or 1/Ts Hz. This is
consistent with the PSD plot for polar binary signalling with
rectangular pulses.
Example- Polar Binary
PSD*Ts
PSD*Ts
f *Ts
Impulse Train PSD
f *Ts
Rectangular Pulse PSD
bk : 0,1
Example- Bipolar (AMI)
ak : 0,1,1
m
0
1
2
2
bk ....bk m ak ....ak m Ri  ak ak m
M
pi
R(m)   Ri pi
i 1
Example- Bipolar (AMI)
• So,

1

S x (w )   R(0)  2 R(m) cos(wmTs ) 
Ts 
m1


1  1  1
   2  cos(wTs ) 
Ts  2  4 

1
1  cos(wTs ) 

2Ts
1 2  wTs 
 sin 

Ts
 2 
Example- Bipolar (AMI)
• Note that if rectangular pulses are employed
as in the previous Polar example, the
resulting PSD is given by,
S y (w )  S x (w ) H (w )
Where,
2
 wTs 
H w   Ts sinc 

 2 
Note the zero crossings at multiples of 2p/Ts rad/s or 1/Ts Hz
and the lowering of the frequency sidelobes evident in the
PSD plot.
Example- Bipolar (AMI)
PSD*Ts
PSD*Ts
f *Ts
Impulse Train PSD
f *Ts
Rectangular Pulse PSD
ACF With Non-Zero Mean Data
• Sometimes the line coded data will have a
non-zero mean value.
• This could be due to
– The line coding scheme, e.g., unipolar
– The probability distribution of the data
– Incorrect signalling voltages owing to faults
• The result is that the R(m) will have finite
values for m in the range +/- infinity
ACF With Non-Zero Mean Data
• The result for Sx(w) is valid but is hard to
interpret physically
• The solution is to express R(m) as the sum
of two parts
– The first part has a constant value of R over the
range of m from + to - infinity
– The second part has non-zero values of R over a
finite range of m
ACF With Non-Zero Mean Data
• The advantage of this approach is that the
first term can be represented in a format that
is consistent with physical observations.
• We will now show that this representation
comprises a sequence of spikes (impulses)
in the frequency domain, occurring at
multiples of the bit rate.
ACF With Non-Zero Mean Data
• Suppose that R(m)=R for all m, then we can
express the PSD as follows
1
S x (w ) 
Ts
R

Ts

jmwTs
R
(
m
)
e

m  

jmwTs
e

m  
• Note the similarity in form to the Fourier series
representation of an impulse train,

1
 (t  kT0 ) 

T0
k  

e
k  
jk
2p
t
T0
Note in time
domain
ACF With Non-Zero Mean Data
• Reminder, the Fourier Series (FS) for a
periodic function
is,

x(t ) 

ck e jkwot
where, w0  2p
T0
k 
i.e., a weighted (complex) sum of phasors.
• We now wish to find the FS of the
rectangular pulse train x(t),
t
x(t)
A
0
To
t
ACF With Non-Zero Mean Data
• Now the coefficients are given by,
t
1
ck 
T0
2
 jkw 0t
Ae
dt

t
2
A  kpt 

 sin 
pk  T0 
For the case where t goes to zero and At = 1 (i.e., a unit
impulse train),
 kpt 

sin 
lim 1
T0  1

ck 

t  0 T0  kpt 
T0


 T0 
ACF With Non-Zero Mean Data
• So we know,

x(t ) 

ck e jkwot
k 
Remembering that ck=1/T0 when x(t) is a sequence of
unit impulses we can write,


1
 (t  kT0 ) 
T0
k 
1

T0


e jkwot
k 

e
k  
jk
2p
t
T0
ACF With Non-Zero Mean Data
• If we substitute, w=t, m=k and T0=2p/Ts, to get the
equivalent relation in the frequency domain,
Ts  jmTsw
2p
 (w  m ) 
e


Ts
2p m
m 


2
p
2p
jmTsw
e

 (w  m )


Ts m
Ts
m 

And hence we may express Sx(w) as a series of impulses in the
frequency domain,

R
jmwTs
S x (w ) 
e

Ts m
2pR 
2p
 2   (w  m )
Ts m
Ts
ACF With Non-Zero Mean Data
• The PSD due to a constant R for all m consists of
spikes, known as line spectral components, at
multiples of wo=2p/Ts
• This result enables simplified calculation of PSDs
when R(i) can be written in the form,
R(i)  C (i)  R
with C (i)  0 for | i | L
First calculate the PSD for R(i)=C(i), then add on the line
spectrum for R(i)=R
Example- Unipolar Binary
bk : 0,1
ak : 0,1
m
0
1
bk ....bk m ak ....ak m Ri  ak ak m
M
pi
R(m)   Ri pi
i 1
0...0
0...0
R1  0
p1  0.5
1...1
1...1
R2  1
p2  0.5
0...0
0...1
0...0
R1  0
1...0
1...1
1...0
1...1
0...1
R(0)  (0  0.5)  (1 0.5)
R(0)  0.5
p1  0.25 R( 1)  (0  0.25) 
R2  0 p2  0.25 (0  0.25)  (0  0.25)
R3  0 p3  0.25  (1 0.25)
R4  1 p4  0.25 R( 1)  0.25
Example- Unipolar Binary
• So,
1 
jmwTs
S x (w ) 
R
(
m
)
e

Ts m
1  1 1  jmwTs 
   e

Ts  4 4 m  


1 
jmwTs 

1   e

4Ts  m  

1

4Ts
 2p
1 
 T
s

1
p

 2
4Ts 2Ts

2p
  w  m

Ts
m   


2p
  w  m

Ts
m   


 





The second term is known as a line spectrum
Example- Unipolar Binary
PSD*Ts
PSD*Ts
f *Ts
Impulse Train PSD
f *Ts
Rectangular Pulse PSD
Scrambling
• Some components of a transmission system work
better if the bit sequence bk is random, ie,
independent and uncorrelated
– Timing regeneration not possible with some line codes
if long sequences of ‘0’s or ‘1’s occur
– Equalisers (see next section) rely on random bit
sequences for successful operation
• A common solution is to randomise (or scramble)
the input bit sequences (in a known manner) prior
to line coding
Scrambling
• The received signals then appear as if they
come from a random source
• Unscrambling after detection restores the
correct bit sequence
• A simple way to generate scramblers is to
XOR the data with the output of a feedback
shift register arrangement.
Scrambling
• For example we may utilise the so called
frame-synchronised scrambler (also called a
cryptographic scrambler
S
XOR
PRBS
T
T
XOR
R
PRBS
AB
00
01
10
11
XOR
0
1
1
0
PRBS- Pseudo Random Binary Sequence generator. Usually
a Maximal Length (ML) linear feedback shift register
arrangement.
PRBS at TX and RX must be synchronised
Scrambling
• An alternative is the Self-Synchronised
scrambler. Is susceptible to error propagation.
• Similar arrangements used for error rate testing.
5
5
4
4
XOR
S
XOR
3
2
3
2
1
1
T
XOR
T
R
XOR
Summary
• In this section we have
– Examined the need for line codes and looked at
some practical examples
– Seen how to calculate the power spectra for
various line codes
– Briefly looked at scrambling to randomise data