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✩
EE4601
Communication Systems
Week 12
Partial Response Signals
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0
c
2015,
Georgia Institute of Technology (lect12 1)
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Objective
Objective: Signals with a baud rate of 2W symbols/sec in a bandwidth of W Hz
with realizable filters.
{a k}
g(t)
c(t)
a k ε {−1,+1}
+
h(t)
kT
w(t)
Assume c(t) =
Then h(t) =
p(t) =
P (f ) =
✫
0
c
2011,
Georgia Institute of Technology (lect11 2)
δ(t) (ideal channel)
g(T − t)
g(t) ∗ g(−t)
|G(f )|2
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Duobinary Signaling
Assume that P (f ) has the following form
a(t)= Σ a δ(t-nT)
n
n
{a k}
{c k}
c(t)
H (f)
+
N
T
T
1
f
HN (f ) =
rect
2W
2W
!


=
1 |f | < W
0 else
where W = 1/2T i.e., the baud rate is R = 1/T = 2W symbols/sec
P (f ) = (1 + e−j2πf T )HN (f )


jπf T
−jπf T
e
+
e
 H (f )
= 2e−jπf T 
N
2
= 2 cos(πf T )e−jπf T HN (f )
!
f
−jπf T 1
= 2 cos(πf T )e
rect
2W
2W
✫
0
c
2013,
Georgia Institute of Technology (lect11 3)
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Duobinary


P (f ) = 
2T cos(πf T )e−jπf T |f | < W
0
else
|P(f)|
arg(P(f))
π/2
2
f
-W=-1/2T
f
W=1/2T
−π /2
To get p(t) we write
!
!)
(
n
o
t−T
t
j2πf T
+ sinc
P (f ) = HN (f ) + HN (f )e
↔ p(t) = sinc
T
T
p(t)
0
✫
0
−3T
−2T
−T
c
2011,
Georgia Institute of Technology (lect11 4)
T
2T
3T
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Duobinary
a(t)= Σ a δ(t-nT)
n
n
p(t)
T
{a k}
T
c(t) =
ck = c(kT ) =
X
n
X
n

an p(t − nT )
an p((k − n)T ) =
1 j = 0, 1
 0 j 6= 0, 1
= ak + ak−1
But pj = p(jT ) =
Therefore, ck
{c k}
c(t)
X
n
an pk−n
Since ak ∈ {−1, +1} ck ∈ {−2, 0, 2} (3-level)
We can recover {ak } from {ck } by ak = ck − ak−1 assuming an initial value, e.g.
a0 = −1 or + 1. This is called decision feedback detection.
Problem : Errors due to noise propagate, i.e., aˆk = ck − ak−1
ˆ .
If ak−1
ˆ is in error then aˆk is likely to be in error.
✫
0
c
2011,
Georgia Institute of Technology (lect11 5)
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Precoding
k
{a k}
dk ε {0,1}
{b }
+
k
bk ε {0,1}
D
a(t)
Impulse
generator
a = 2d -1
k
0 - -1
1 - +1
logic
D-flip flop
Same as before
{c k}
c(t)
a(t)
+
H (f)
N
T
T
P(f)
dk = bk ⊕ dk−1 = bk + dk−1(mod 2)
Example:
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0
{bk }
0 0 1 1 1 0 1 0 0 0
{dk } 1 1 1 0 1 0 0 1 1 1 1
{ak } +1 +1 +1 −1 +1 −1 −1 +1 +1 +1 +1
{ck }
+2 +2 0 0 0 −2 0 +2 +2 +2
c
2011,
Georgia Institute of Technology (lect11 6)
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Modified Duobinary


±2 if bk = 0
0 if bk = 1
Therefore {bk } can be recovered from {ak } by using symbol by symbol detection.
Note that ck = 
Modified Duobinary
{bk }
{a k}
dk ε {0,1}
+
k
2D
a(t)
Impulse
generator
a = 2d -1
k
0 - -1
1 - +1
Same as before
{c }
c(t)
a(t)
+
k
H (f)
N
kT
2T
P(f)
a(t) =
P
n an δ(t
− nT ), dk = bk ⊕ dk−2, ck = ak − ak−2 .
✫
0
c
2011,
Georgia Institute of Technology (lect11 7)
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Modified Duobinary
P (f ) = (1 − e−j4πf T )HN (f )


j2πf T
−j2πf T
e
−
e
 H (f )
= j2e−j2πf T 
N
j2
= j2HN (f ) sin 2πf T e−j2πf T

 2T sin 2πf T ej(π/2−2πf T ) |f | < 1/2T
= 
0
|f | > 1/2T
|P(f)|
arg(P(f))
π/2
f
-W=-1/2T
f
W=1/2T
−π /2
✫
0
c
2011,
Georgia Institute of Technology (lect11 8)
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Modified Duobinary Pulse
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P (f ) = HN (f ) − e−j4πf T HN (f )
p(t) = sinc(t/T ) − sinc((t − 2T )/T )
−3T
−2T
−T
0
✫
0
c
2011,
Georgia Institute of Technology (lect11 9)
T
2T
3T
4T
5T
6T
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Modified Duobinary with Precoding
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Example:
{bk }
0 0 1 1 1 0 1 0 0 0
{dk } 1 1 1 1 0 0 1 0 0 0 0 0
{ak } +1 +1 +1 +1 −1 −1 +1 −1 −1 −1 −1 −1
{ck }
0 0 −2 −2 2 0 −2 0 0 0


Note: ck = 
±2 if bk = 1
0 if bk = 0
✫
0
c
2011,
Georgia Institute of Technology (lect11 10)
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Duobinary Error Probability
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Here we consider the error probability of precoded duobinary signaling with
symbol-by-symbol detection.
The transmit and receiver filters are implemented as the root-duobinary pulse,
such that
q
|G(f )| = |H(f )| = P (f )
We note that the noise process at the output of the receiver filter has power
spectral density
No
No
Φnn(f ) =
|H(f )|2 =
P (f )
2
2
Since p(t) is not a Nyquist pulse, i.e., pk = p(kT ) 6= δk0 , the noise samples are
correlated, and, hence the symbol-by-symbol detector is suboptimal. This loss
can be recovered by using a sequence detector, but we will not discuss here.
The Gaussian noise samples at the output of the receiver matched filter H(f )
are zero mean and have variance
σn2 =
No Z 1/2T
2No
No Z ∞
P (f )df =
2 cos(πf T )df =
2 −∞
2 −1/2T
π
✫
0
c
2013,
Georgia Institute of Technology (lect11 10)
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✬
Duobinary Error Probability
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Note that the sampled outputs of the matched filter have the Gaussian density
function

 N (±2, 2N /π) , x = 0
o
k
yk ∼ 
N (0, 2No/π) , xk = 1
where the means ±2 each occur with probability 1/4 and the mean 0 occurs with
probability 1/2.
Assuming that the receiver makes decisions according to


b̂k = 
1,
0,
|yk | < 1
|yk | > 1
we have the probability of error
Pb =
where
1
1
1
3
· Q + · Q + · 2Q = Q
4
4
2
2
!
✫
0
v

u
1
u π

Q=Q
= Q t
σ
2No
c
2013,
Georgia Institute of Technology (lect11 10)
✪
✬
Duobinary Error Probability
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The energy per bit is
Eb =
Z ∞
2
−∞
|G(f )| df =
Z ∞
−∞
P (f )df =
4
π
Hence, π4 Eb = 1 and
Q=
and
v
u
u

Q t
v
u

u π
π π 
t

· Eb = Q 
2No 4
4
Pb =
v
u
3 u
Q t
2
π
4
!2
!2

2Eb 

No

2Eb 

No
When compared to binary antipodal signaling with
Pb =
v

u
u 2Eb

Q t
No
the loss in Eb/No performance is −10log10(π/4)2 = 2.1 dB.
✫
0
c
2013,
Georgia Institute of Technology (lect11 10)
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