EE 379A
Digital Communication: Signal Processing
Handout #9
Thursday, January 31, 2008
Homework #2 (52 + 1 pts)
Solutions
1. (1.8) Optimum Receiver (From Wozencraft and Jacobs)
(a) (2 pts) The data signals xi (t) being equiprobable, the optimum decision rule is the
Maximum Likelihood (ML) rule, given by, (in vector form) mini ||y − x i ||2 From the
invariance of the inner product, the ML rule is equivalent to,
min
i
Z
T
0
||y(t) − xi (t)||2 dt
The integral is then written as a sum of three integrals,
Z
0
T
Z
||y(t) − xi (t)||2 dt =
t1
0
Z
||y(t) − xi (t)||2 dt +
t2
t1
||y(t) − xi (t)||2 dt +
Z
T
t2
||y(t) − xi (t)||2 dt
The second integral over the interval [t1 , t2 ] being constant as a function of i, the optimum
decision rule reduces to,
min
i
Z
0
t1
2
||y(t) − xi (t)|| dt +
Z
T
t2
||y(t) − xi (t)||2 dt
And therefore, the optimum receiver may ignore the interval [t1 , t2 ].
(b) (2 pts) In an appropriate orthonormal basis of dimension N ≤ M , the vectors x i and
y are given by,
xTi
=
yT
=
h
h
xi1 xi2 . . . xiN
y1 y2 . . . yN
i
i
Assume that xim = x1m for all i, the optimum decision rule becomes,
min
i
M
X
j=1
||yj − xij ||2 ⇔ min
i
M
X
j=1,j6=m
||yj − xij ||2 + ||ym − xim ||2
Since ||ym −xim ||2 = ||ym −x1m ||2 is independent of i, the optimum decision rule becomes,
min
i
M
X
j=1,j6=m
||yj − xij ||2
Therefore, the projection xim might be ignored by the optimum receiver.
1
(c) (2 pts) The result does not hold true if the noise is colored Gaussian noise. This is due
to the fact that the noise along one component is correlated with other components and
hence might not be irrelevant. In such a case, all components turn out to be relevant.
Equivalently, by duality, the same result holds in the time domain.
2. (1.9) Receiver Noise with MATLAB
We need to compute the Q = orth(A) matrix. From matlab, we have,





Q=



0.3803 0.2260 −0.0120
0.7606 0.4520 −0.0239
0.0245 −0.2395 −0.0876
0.0464 0.1157
0.3710
0.2129 −0.3713 0.8501
0.4783 −0.7321 −0.3624









(a) (2 pts) At the output of the basis detector, we have,
y = (x + n)
The signal power is given by,
2
E[kxk ] =
P7
2
i=0 kxi k
8
=
trace(AT A)
= 181
8
and hence
ε̄x =
181
= 30.17
6
The noise power per dimension is easily seen to be
σ2 =
E[knk2 ]
= 1.
6
Hence,
SN R = 30.17 = 14.8dB
(b) (4 pts) The basis detector is now using φ̂(t) = φ(t)Q, where Q=orth(A) as in problem
1.4. The new matrix  is the representation of our original constellation in terms of the
new basis vectors. As in 1.4, this is given by
 = Q∗ A
Using Q from above we have
 = Q∗ A =

−4.9564 −7.3362 −9.7160 −12.0958 −12.5841 −14.9639 −17.3437 −19.7235


0.9914
1.3894
1.7873
2.1853 
 −3.8839 −3.4860 −3.0881 −2.6901
0.5912
0.1691 −0.2530 −0.6751
0.8108
0.3887
−0.0335 −0.4556

2
Now, obviously, E[kx k2 ] remains the same since our new basis still includes in its span
all the original constellation input vectors,
E[kxk2 ] = 181
However, since the number of system dimensions changed,
ε̄x =
E[kxk2 ]
181
=
= 60.33 = 17.8dB
3
3
As for the noise, some of its energy now lies outside the new basis vectors. One way to
calculate the noise energy that resides in the new basis is as follows. Note that in the
original basis the total energy of the noise is the sum of the elements of the diagonal of
the noise autocorrelation matrix and so,
1
σ 2 = trace(E[nn∗ ]) = 1
6
After projecting the noise vector n on the new basis we will have a new noise vector,
n̂ = Q∗ n
Once again, the energy of this new noise vector will be the sum of the diagonal elements
of the new autocorrleation matrix, and so the new variance per dimension will be,
σ2 =
=
=
=
=
=
1
trace(E[n̂n̂∗ ])
3
1
trace(E[Q∗ nn∗ Q])
3
1
trace(Q∗ E[nn∗ ]Q)
3
1
trace(Q∗ I6×6 Q)
3
1
trace(I3×3 )
3
1
Therefore, we have,
SN R = ε̄x = 60.33 = 17.8dB
(c) (1 pts) The error performance of these two detectors will be identical because the extra
noise admitted by the first detector is orthogonal to the space spanned by the signal set.
As such, the detector from part a will disregard it.
(d) (2 pts)
We have for the system from part a:
b̄ =
log2 (8)
6
ε̄x =
181
6
=
1
2
= 30.17 = 14.8dB
3
For system from part b:
b̄ =
log2 (8)
3
ε̄x =
181
3
=1
= 60.33 = 17.8dB
There are a few ways to compare the two systems, resulting in different answers to this
question. The systems could be viewed as equivalent since in reality they have the same
transmission characteristics and all we did was simply change the basis representation of
the constellation. But one could also argue that the systems have the exact same performance yet the system in part a possesses more resources than system in part b so that
system from part b is superior. So for this problem any reasonable answer was accepted.
(e) (4 pts)
Notice that the 8 new symbols we added are actually independent of the original 3dimensional constellation from part b. So the 8 additional messages we now transmit will
be decoded independently of the original constellation and this will add 3 more bits (per
symbol) to our data rate. Therefore, we have,
b̄ =
log2 (8)+log2 (8)
6
ε̄x =
2∗181
6
=1
= 60.33 = 17.8dB
So the new system has the exact same b̄ and ε̄x as system from part b. Therefore, it
compares with a in these two parameters just like the comparison in part d.
3. (1.10) Tilt
(a) (1 pt) Pe is invariant under rotation or translation of signal constellation. Therefore, Pe
does not depend on L or θ .
(b) (2 pts) First, we get the average number of nearest neighbors:
Ne =
M
−1
X
Ni px (i)
i=0
=
=
Pe ≤
=
=
4+4·2+4·3
9
8
.
3 8
d
Q
3
2σ
8
2
Q √
3
2 0.1
2.09 × 10−3 .
(c) (5 pts) By part (a), the constellation shown in the figure gives the same Pe :
4
phi2
1
2
4
3
5
6
phi1
7
8
9
Figure 1: Equivalent Signal Constellation
Pc|i=5
Pc|i=2,4,6,8
Pc|i=1,3,5,9
2
d
=
1 − 2Q
2σ
d
d
=
1 − 2Q
1−Q
2σ
2σ
2
d
=
1−Q
.
2σ
Hence,
i
1 h
1 − 4Q + 4Q2 + 4 1 − 3Q + 2Q2 + 4 1 − 2Q + Q2 ,
9
= 1 − Pc
8
16
=
Q − Q2 ,
3
9
Pc =
Pe
where Q = Q
d
2σ
= Q(3.162) = 7.827 × 10−4 . So,
Pe = 2.09 × 10−3 − 1.09 × 10−6
= 2.09 × 10−3 .
In this case, NNUB was off by
16 2
9 Q
= 1.09 × 10−6 , a small quantity compared to Pe .
(d) (2 pts) To get a constellation with minimum energy, we subtract from the constellation its
mean. So, a possible choice is the constellation of part c). The energy of the original constellation would change with θ , whereas the energy of the minimum-energy constellation
would be independent of θ .
4. (1.12) Disk storage channel
5
2.5
2
1.5
1
0.5
0
−1
−0.5
0
0.5
1
1.5
2
Figure 2: Plot of output probability density functions.
(a) (2 pts) The plot is shown in Figure 2.
(b) (3 pts) The inputs being equally likely, the optimum decision rule is the ML rule. To get
the boundary points, solve,
1
q
−
2πσ02
e
v2
2σ 2
0
1
−
=q
e
2πσ12
(v−1)2
2σ 2
1
After simplification, the equation is,
(1 −
By letting α =
σ0
σ1
σ02 2
σ02
σ0
σ02
2
)v
+
2
v
+
2σ
log(
)
−
=0
0
σ1
σ12
σ12
σ12
and solving the equation,
v± =
q
−α2 + α4 − (1 − α2 )(2σ02 log α − α2 )
1 − α2
The optimal receiver is then,
X̂ = 0 if v− ≤ v ≤ v+
= 1 otherwise.
(c) (1 pts) The SNR is given by,
1
Ex
1
= 1 22 2 =
En
32σ02
2 (σ0 + σ1 )
1
SN R = 15 dB = 10 log10
⇒ σ02 = 9.88 × 10−4
32σ02
SN R =
and σ12 = 3.06 × 10−2
6
(d) (3 pts) For the values of σ02 and σ12 obtained above, the roots of the quadratic equation
are, v− = −.2281 and v+ = .1615. The optimum receiver chooses X̂ = 0 when v− ≤ v ≤
v+ and X̂ = 1 otherwise. The corresponding probability of error is then,
.1615
.2281
) + Q(
)
σ0
σ0
= Q(5.14) + Q(7.25)
Pe|x=0 = Q(
Pe|x=1
⇒ Pe
= 1.37 × 10−7
1 − .1615
1 + .2281
= Q(
) − Q(
)
σ1
σ1
= Q(4.79) − Q(7.02)
= 8.34 × 10−7
1
1
Pe|x=0 + Pe|x=1 = 4.86 × 10−7
=
2
2
(e) (1 pts) As α → 0, v → 0. Thus, we choose X̂ = 1 unless v = 0 exactly.
5. (1.18) Pattern Recognition
(a) (1pt) The total number of possible patterns is
4!
2!2!
= 6.
(b) (3pts) The optimum decision rule is the ML rule since the signals are equiprobable,
maxhy, xi i
i
The signal detector is shown in the figure below.
y1
y2
y3
y4
<y,x0>
<y,x1>
Matrix Multiply
[x0 x1 x2 x3 x4 x5]
<y,x2>
<y,x3>
<y,x4>
Max
+
Decode
m
<y,x5>
Figure 3: Signal Detector
(c) (2 pts) The union bound for the probability of error is,
√ !
2
Pe ≤ 5Q
σ
since d2min = 22 + 22 = 8 (xi must differ by at least 2 squares).
(d) (2 pts) The number of nearest neighbors at minimum distance is 4, and so,
√ !
2
Pe ≤ 4Q
σ
7
6. (1.21) Comparing Bounds
(a) (2 pts) The signal constellation is shown in the figure below.
From the signal constellation, we get dmin = 2. Since we have 5 signals, the Union Bound
is given by,
dmin
≤ 4Q
2σ
1
= 4Q
.
σ
Pe
x4
3
x2
x3
1
-1
1
-1
x0
x1
Figure 4: Signal Constellation
(b) (2 pts) The number of Nearest Neighborhood Ne = 1/5(2+2+3+3+2) = 2.4. Therefore,
the Nearest Neighborhood Union Bound is given by,
Pe = 2.4 Q
1
.
σ
(c) (1 pt) Since Ex = 15 (2 × 4 + 32 ) = 3.4,
σ=
s
E¯x
=
SN R
s
1
2 Ex
SN R
=
r
1.7
101.4
Therefore, we get
s
Pe (NNUB) = 2.4 × Q 
8

101.4 
= 1.45 × 10−4 .
1.7