1Wireless Communication
Lecture Note 2
Yingbo Hua
I. OPTIMAL DETECTOR AND ITS PERFORMANCE
Consider two hypotheses H0 and H1 with the prior probabilities p0 and p1, respectively. Let f(y|H0)
and f(y|H1) be the probability density functions (pdf) of the observation y conditioned upon H0 and
H1, respectively. The optimal (minimum error) detector D(y) of H0 and H1 based on y is known to be
D(y) =
 H1, f(y|H1)p1 > f(y|H0)p0H0, f(y|H1)p1 < f(y|H0)p0 (1)
The probability of decision error is
pe = p0 +
∫
S0
[f(y|H1)p1 − f(y|H0)p0] dy (2)
where S0 = {y|f(y|H1)p1 < f(y|H0)p0}.
A. Binary phase shift keying (BPSK)
A BPSK signal has the form
x(t) =
 p(t) cos(2pifct), symbol 0−p(t) cos(2pifct), symbol 1 (3)
where p(t) has the (effective) duration −T/2 < t < T/2, T = 1/W , and p(0) = a. Its discrete-time
baseband form is simply
x(n) =
 a, symbol 0−a, symbol 1 (4)
Consider a flat fading channel where h(n) = h0δ(n). The discrete-time baseband form of the received
signal is
y(n) = h0x(n) + w(n) (5)
where w(n) is the noise. It is a common situation that both x(n) and w(n) are temporally independent.
Then, the optimal detector of x(n) only depends on y(n). For convenience, we will write
y = hx+ w (6)
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2Assume that w is circular complex Gaussian CN(0, σ2w). Also assume that h is known at the receiver.
Then, it follows that f(y|x = a) ∼ CN(ha, σ2w) and f(y|x = −a) ∼ CN(−ha, σ2w). The optimal
detector can be simplified as
D(y) =
 a, Re{h∗y} > 0−a, Re{h∗y} < 0 (7)
The probability of error can be shown to be
pe = Q
(√
2SNRh
)
(8)
where SNRh =
|h|2a2
σ2w
and Q(x) =
∫∞
x
1√
2pi
exp(−τ2/2)dτ .
We see that SNRh depends on |h|. A meaningful measurement of the performance of BPSK is given
by the average of pe over the distribution of |h|, i.e., Pe .= E{pe}. Using f|h|2(x) = 1σ2h exp(−x/σ
2
h) or
equivalently fSNRh(x) =
1
SNR exp(−x/SNR) with SNR = σ
2
ha
2
σ2w
, it can be shown that
Pe =
1
2
(
1−
√
SNR
1 + SNR
)
≈ 1
4SNR
+O
(
1
SNR2
)
(9)
Although pe decreases exponentially as SNRh increases, Pe decreases linearly (on log-log scale) as
SNR increases.
It can be shown that
Pe > Q(
√
2)Prob{SNRh < 1} (10)
where Prob{SNRh < 1} ≈ 1SNR which is known as “deep fade” probability. We see that Pe has the
same order as the deep fade probability.
II. TIME DIVERSITY
Assume that the same symbol is transmitted repeatedly using L time slots. The corresponding received
symbols are
yi = hix+ wi (11)
where i = 1, · · · , L. The above is equivalent to the vector form y = hx+w. Assuming w ∼ CN(0, σ2w),
the optimal detector of a binary symbol x based on y can be shown to be
D(y) =
 a, Re{hHy} > 0−a, Re{hHy} < 0 (12)
Note that Re{hHy} is, up to a positive scalar, the projection coefficient of y onto the direction of ha
(where a is known to be positive real). The probability of error of this detector can be shown to be
pe = Q
(
d
2
)
(13)
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3where d is the distance between E{y|x = a} and E{y|x = −a} normalized by the noise deviation along
the direction from E{y|x = a} to E{y|x = −a}. The equation (13) holds for any (optimal) binary
detector under the white Gaussian noise assumption. In the current case, d = 2‖h‖a
σw/
√
2
. Hence,
pe = Q
(√
2SNRh
)
(14)
with SNRh =
a2‖h‖2
σ2w
.
If all L time slots are in different coherence periods, we can assume h ∼ CN(0, I). Then, 2‖h‖2 is
equivalent to a standard Chi-square random variable with degree 2L. The pdf of ‖h‖2 is known to be
f‖h‖2(x) =
1
(L− 1)!x
L−1e−x (15)
It can be shown that
Pe = E{pe} =
(
1− µ
2
)L L−1∑
l=0
 L− 1 + l
l
(1 + µ
2
)l
(16)
with µ =
√
SNR0
1+SNR0
and SNR0 = a
2
σ2w
. The averaged SNR is SNR = E{SNRh} = L × SNR0. At
high SNR, (1 + µ)/2 ≈ 1 and
1− µ
2
≈ 1
4SNR0
(17)
Hence,
Pe ≈ 1
4LSNRL0
L−1∑
l=0
 L− 1 + l
l
 = 1
4LSNRL0
 2L− 1
L
 (18)
where L is called the diversity gain.
The deep fade probability at high SNR is
Prob{SNRh < 1} = Prob
{
‖h‖2 < 1
SNR0
}
≈
∫ 1/SNR0
0
1
(L− 1)!x
L−1dx =
1
L!
1
SNRL0
(19)
A. Diversity gain by rotation in time
The diversity gain achieved by repetition in time is at the expense of spectral efficiency. To solve this
problem, we now consider a rotation method using two time slots. To preserve the spectral efficiency of
one symbol per time slot, we must transmit effectively two independent symbols during two time slots.
However, the actual symbol transmitted in slot one must be correlated to that in slot two, or otherwise
there would be no diversity gain. Let u1 and u2 be two independent binary symbols, and x1 and x2 are
two symbols transmitted in slot one and slot two, respectively. These two pairs of symbols are related
to each other via the following rotation: x1
x2
 =
 cos θ sin θ
− sin θ cos θ
 u1
u2
 (20)
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4or equivalently x = Qu. The channel model over the two time slots is y1
y2
 =
 h1 0
0 h2
 x1
x2
+
 w1
w2
 (21)
or equivalently y = Hx+w. Therefore y1
y2
 =
 h1 cos θ h1 sin θ
−h2 sin θ h2 cos θ
 u1
u2
+
 w1
w2
 (22)
or equivalently y = Tu+w where T = HQ. There are four possible choices for u, i.e., u0 = [a, a]T ,
u1 = [−a, a]T , u2 = [−a,−a]T , u3 = [a,−a]T . Associated with the decisions 0, 1, 2 and 3, there are
corresponding four decision regions Y0, Y1, Y2 and Y3 which comprise the space of y. The partition of
the regions is governed by a minimum distance principle. Namely, y ∈ Yi if and only if y is the closest
to E{y|i} among all E{y|j}, j = 0, 1, 2, 3, where E{y|j} .= E{y|u = ui} = Tuj .
The probability of detection error is
pe =
1
4
3∑
i=0
∑
j 6=i
Prob{y ∈ Yj |i}
<
1
4
3∑
i=0
∑
j 6=i
Q
(
di,j
2
)
<
1
4
3∑
i=0
∑
j 6=i
exp
(
−d
2
i,j
8
)
(23)
where di,j is the normalized distance between E{y|i} and E{y|j}. The first inequality is due to the fact
that Prob{y ∈ Yj |i} < Prob{i→ j} where Prob{i→ j} is the probability of the binary decision error
of “mistaking i as j”. The binary decision between i and j is based on whether y is closer to E{y|i}
or E{y|j}.
It can be shown that
d0,1 =
‖T(u0 − u1)‖
σw/
√
2
=
√
8[|h1|2 cos2 θ + |h2|2 sin2 θ]SNR0 (24)
and, assuming h ∼ CN(0, I),
E
{
exp
(
−d
2
0,1
8
)}
= E
{
exp
(−SNR0[|h1|2 cos2 θ + |h2|2 sin2 θ])}
=
(
1
1 + SNR0 cos2 θ
)(
1
1 + SNR0 sin
2 θ
)
<
1
SNR20 cos
2 θ sin2 θ
=
4
SNR20 sin
2 2θ
(25)
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5Similarly, we can show that
E
{
exp
(
−d
2
0,1
8
)}
= E
{
exp
(
−d
2
0,3
8
)}
(26)
E
{
exp
(
−d
2
0,2
8
)}
<
1
SNR20 cos
2 2θ
(27)
It follows that
Pe = E{pe}
= E

3∑
j=1
Prob{y ∈ Yj |0}

< E

3∑
j=1
exp
(
−d
2
0,j
8
)
<
12
min
(
sin2 2θ, 4 cos2 2θ
) × 1
SNR20
(28)
where the first equality is due to the symmetry among the four cases: i = 0, 1, 2, 3. We see that the
diversity gain is two.
The constant before 1SNR20 is also important. We want this constant to be as small as possible. If
min
(
sin2 2θ, 4 cos2 2θ
)
is maximized, the coding gain is said to be maximized. This is achieved when
sin2 2θ = 4 cos2 2θ, i.e., θopt = 12 tan
−1 2.
B. Diversity gain through a generalized code matrix
In the previous coding scheme, only two time slots were used, and x is a column vector from the 2×4
code matrix X = Q[uA,uB,uC ,uD]. To generalize this idea, we now consider a coding scheme over L
time slots, for which we have a L ×M code matrix X = [x0, · · · ,xM−1]. The data rate is log2ML bits
per slot. The optimal detector is the minimum distance detector under the channel model y = Hx+w
where w ∼ CN(0, σ2wI).
The probability of detection error is
Pe < E
 1M
M−1∑
i=0
∑
j 6=i
Prob{i→ j}

= E
 1M
M−1∑
i=0
∑
j 6=i
Q
(
di,j
2
)
< E
 1M
M−1∑
i=0
∑
j 6=i
exp
(
−d
2
i,j
8
) (29)
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6Since the L elements in y are taken from L time slots, the matrix H is diagonal with the diagonal
elements h1, · · · , hL. We will assume that h1, · · · , hL are i.i.d. Gaussian random variables with zero
mean and unit variance. It follows that d2i,j =
2
σ2w
‖H(xi − xj)‖2 = 2SNR0
∑L
l=1 |hl|2|xi,l − xj,l|2 and
Pe <
1
M
M−1∑
i=0
∑
j 6=i
L∏
l=1
1
1 + SNR0|xi,l − xj,l|2/4 (30)
If xi,l − xj,l 6= 0 for all i 6= j and all l, then
Pe <
1
M
M−1∑
i=0
∑
j 6=i
L∏
l=1
1
SNR0|xi,l − xj,l|2/4
=
4L
M
M−1∑
i=0
∑
j 6=i
1∏L
l=1 |xi,l − xj,l|2
1
SNRL0
≤ 4L M − 1
minj 6=i
∏L
l=1 |xi,l − xj,l|2
1
SNRL0
(31)
where the diversity gain L is achieved.
To maximize the coding gain, we should maximize minj 6=i
∏L
l=1 |xi,l−xj,l|2 = minj 6=i det[∆Xi,j∆XHi,j ]
where ∆Xi,j = diag(xi)−diag(xj) subject to ‖xi‖ = 1 for i = 0, · · · ,M −1. There are many possible
choices of the structure of X. One possible choice is such that x = Qs where Q is a complex unitary
matrix and s is a vector of independent symbols.
For mobile communications, the coherence time of a channel response varies. It is difficult (if not
impossible) to predetermine L time slots that are incoherent. Furthermore, subject to a delay requirement,
the time diversity order may be limited.
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