E6712: Midterm Exam - Solutions
(Total 100 points)
Given:
March 22, 2005
1. Consider the following vector channel detection problem. There are three possible signals:
√ √
√ √
√
√
√
√
√
√
s1 = { 1 − α, 0, α} s2 = {− 1 − α/2, 3 1 − α/2, α} s3 = {− 1 − α/2, − 3 1 − α/2, α}
The received signal is given by
r = si + n, i ∈ {1, 2, 3},
where n is a 3-dimensional noise vector whose components are i.i.d. Gaussian RVs with zero mean
and variance σ2 . Assume that the prior probabilities of the three possible messages are equal.
(a) (25 pts) Consider the case when α = 0.
i. Derive an optimum receiver decision rule and draw a block diagram of the optimum receiver.
ii. Draw the signal constellation and decision regions.
iii. Derive an expression for this receiver's error probability, and its Union-Cherno bound.
(b) (25 pts) Consider the case when α = 1/3.
i. Derive an optimum receiver decision rule and draw a block diagram of the optimum receiver.
ii. Draw the signal constellation. Hint: Compute the angles between the signal vectors; consider a change of coordinates.
iii. Derive an expression for this receiver's error probability, and its Union-Cherno bound.
(c) (10 pts) Consider now transmitting the signals over a uniformly random-phase channel
r = ejθ si + n, i ∈ {1, 2, 3}.
i. Draw a block diagram of the optimum receiver.
ii. Would you choose the constellation with α = 0 or with α = 1/3 for this channel? Why?
Solution
(a) α = 0:
i. The optimal detector nds (see Lec. 3)
arg min
3
hX
i
(rj − smj )2 ,
m∈{1,2,3}
or, equivalently,
arg max
m
3
X
j=1
1
j=1
smj rj = arg maxhsm , ri.
m
A block diagram of the optimum receiver:
r
- hr, s1 i
-
- hr, s2 i
-
- hr, s3 i
-
choose
largest
-m
^
ii. Signal constellation with decision regions:
s2
s1
s3
iii. By the genie, union, and Cherno bound, the probability of error can
s
Q

s
d2
min  < P < 2Q 
e
4σ2

µ 2 ¶
min  < exp − dmin ,
4σ2
8σ2
d2
where d2min = 3.
(b) α = 1/3:
i. See (a)-i.
ii. Computing angles shows that the constellation consists of three pairwise orthogonal signals.
iii. Same as (a)-iii, but d2min = 2.
(c) Random phase channel:
i. The optimal detector nds (see Lec. 5)
arg max
|hsm , ri|2 .
m
2
A block diagram of the optimum receiver:
r
- kr, s k2
1
-
- kr, s k2
2
-
- kr, s k2
3
-
choose
largest
-m
^
ii. HW4
2. Consider the following partial response channel:
vk = Ik + Ik−1 + ηk ,
where {ηk } is discrete AWGN with zero mean and spectral density N0 /2. The user data Dk ∈ {0, 1}
is binary, and the channel can take only bipolar signal values Ik ∈ {−1, 1}.
(a) (10 pts) Suppose that user data is directly mapped into the channel sequence as
Ik = 2Dk − 1
Suppose that {Ik } is the sequence 1, −1, 1, −1, 1, . . . , and the MLSE decodes a sequence {Ibk } with
an error event starting at time n0 and ending at time nl−1 . What is the most likely incorrect
ck }?
sequence {D
(b) (10 pts) Derive a lower bound on the sequence error probability for the MLSE.
(c) (10 pts) Suppose that you have to use a symbol-by-symbol detector rather than the MLSE.
How would you organize transmission and detection for this channel?
(d) (10 pts) Derive a lower bound on the error probability for your receiver.
Solution
(a) The minimum (merged) distance on the bipolar channel with transfer function (1 − D) is given
by
d2<> = minε(D)6=0 k(1 + D)ε(D)k2 ,
where the squared norm of a polynomial is dened as the sum of its squared coecients, and
ε(D) is the polynomial corresponding to a nite length normalized error sequence:
ε = {εn0 , ε1 , . . . , εnl −1 }
of length l < ∞, εi ∈ {−1, 0, 1}, εn0 , εnl −1 6= 0
Since, the normalized error sequence corresponding to {In } and {bIn } is
1
εn = (In − bIn ),
2
3
we have εi ∈ {0, 1}, εn0 = 1, and thus
d2<> =
=
=
min k(1 + D)
ε(D)6=0
where the equality holds i
εni Di k2
i=0
min kεn0
ε(D)6=0
min (1 +
ε(D)6=0
=2+
l−1
X
min
l−1
X
+
(εni + εni−1 )Di + εnl−1 Dl k2
i=1
l−1
X
(εni + εni−1 )2 + 1)
i=1
l−1
X
(εni + εni−1 )2 ≥ 2,
ε(D)6=0
i=1
l−1
X
ε(D) = 1 +
(−1)i Di .
i=1
ck } is the sequence 0, 1, 0, 1, . . .
Therefore bIn = −In , and thus {D
(b) A lower bound to the probability of error (see Lec. 8):
s
Pe ≥ Q 

2d2
N0
Ãs
d2<>  = Q
4
N0
!
(c) We can use precoding for the duobinary channel as follows (see Lec. 7):
Data sequence {Dn }
Precoded sequence Pn = Dn + Pn−1 mod 2
Transmitted sequence In = 2Pn − 1
Received sequence Bn = In + In−1
Decoded sequence: Dn = 12 Bn + 1 mod 2
(d) Note that Bn can take 3 possible values: {−2, 0, 2}. Therefore d2min = 4, and thus
s

d2
min  = Q
2N 0
Pe ≥ Q 
4
Ãs
2
N0
!