alamouti

ISIT 2008, Toronto, Canada, July 6 - 11, 2008

Rotated and Scaled Alamouti Coding
Frans M.J. Willems
Philips Research Laboratories, High Tech Campus 37, 5656AE Eindhoven, The Netherlands

Abstract—Repetition-based retransmission is used in xk2 xk2
Alamouti-modulation [1998] for 2 × 2 MIMO systems. We £ £
+3
¢¡ ¢¡ +3
propose to use instead of ordinary repetition so-called ”scaled
repetition” together with rotation. It is shown that the rotated £ £
+1
¢¡ xk1
+1
¢ ¡ k1
x
and scaled Alamouti code has a hard-decision performance
which is only slightly worse than that of the Golden code [2005], -3
£-1 +1 +3
£-3 -1 +1 +3
¢¡ -1
¢¡ -1
the best known 2 × 2 space-time code. Decoding the Golden code
requires an exhaustive search over all the codewords (or sphere £ £
¢¡ -3 -3
¢¡
decoding for higher spectral efficiencies), while our rotated
and scaled Alamouti code can be decoded with an acceptable
Fig. 2. Two mappings from xk1 to xk2 . On the right the scaled-repetition
complexity. mapping, left the ordinary-repetition mapping.
I. S CALED - REPETITION FOR THE SISO C HANNEL
Fig. 3 shows the basic capacity C (black line) and repetition
First we consider transmission over a single-input single- capacity Cr (blue line) as a function of the signal-to-noise
output (SISO) additive white Gaussian noise (AWGN) channel ratio SNR which is defined as
(see Fig. 1), and introduce scaled-repetition retransmission.
Δ
It turns out that scaled-repetition improves upon ordinary- SNR = P/σ 2 . (5)
repetition retransmission.
It is easy to see that always Cr ≤ C. For large SNR we
A. Some information theory
may write Cr ≈ C/2 + 1/4, while for small SNR we obtain
n
Cr ≈ C.
x c y
transmitter E + E receiver
B. Ordinary and scaled repetition for 4-PAM

Fig. 1. The AWGN channel. When we use 4-PAM modulation, the channel inputs xk
The real-valued output yk for transmission k = 1, 2, · · · , K, assume values from A4-PAM = {−3, −1, +1, +3}, each with
see Fig. 1, satisfies probability 1/4. Ordinary repetition, see (3), leads to signal
points (x1 , x2 ) = (x, x) for x ∈ A4-PAM , see the left part
yk = xk + nk , (1) of Fig. 2. For this case the maximum transmission rate
I(X; Y1 , Y2 ) is shown in Fig. 3 with blue asterisks. Note that
where xk is the real-valued channel input for transmission
this maximum transmission rate is slightly smaller than the
k and nk is a real-valued Gaussian noise sample with mean
corresponding capacities Cr , mainly because uniform inputs
E[Nk ] = 0, variance E[Nk ] = σ 2 , which is uncorrelated with
2
are used instead of Gaussians.
all other noise samples. The transmitter power is limited, i.e.
2
we require that E[Xk ] ≤ P . It is well-known that an X which We can use Benelli’s [3] method to improve upon ordinary-
is Gaussian with mean 0 and variance P achieves capacity. repetition retransmission, i.e. by modulating the retransmitted
This basic capacity (in bit/transm.) equals symbol differently. We could e.g. take
1 P xk1 = xk , and xk2 = M2 (xk ) for xk ∈ A4-PAM , (6)
C= log2 (1 + 2 ). (2)
2 σ
When we retransmit (repeat) codewords , each symbol xk from where M2 (α) = 2α − 5 if α 0 and M2 (α) = 2α + 5 for
such a codeword (x1 , x2 , · · · , xK ) is actually transmitted and α 0. We call this method scaled repetition since we scale
received twice, i.e. xk1 = xk2 = xk , and a symbol by a factor (2 here) and then compensate (add -5
or +5) in order to obtain a symbol from A4-PAM . This results
yk1 = xk + nk1 , and yk2 = xk + nk2 . (3)
in the signal points (x, M2 (x)) for x ∈ A4-PAM , see Fig. 2,
yk1 +yk2
An optimal receiver can form zk = = xk + nk1 +nk2 .
2 2
right part. Also for the scaled-repetition case the maximum
Now the variance of the noise variable (Nk1 + Nk2 )/2 is transmission rate I(X; Y1 , Y2 ) is shown in figure 3, now with
σ 2 /2. Therefore the repetition capacity for a single repetition red asterisks. Note that this maximum transmission rate is only
in bit/transm. is slightly smaller than the basic capacity C. Ordinary repetition
1 2P is however definitively inferior to the basic transmission if the
Cr = log2 (1 + 2 ). (4) SNR is not very small.
4 σ

978-1-4244-2571-6/08/$25.00 ©2008 IEEE 1288

Authorized licensed use limited to: Eindhoven University of Technology. Downloaded on March 09,2010 at 04:35:07 EST from IEEE Xplore. Restrictions apply.


σ 2 (per two dimensions). Noise variable pairs in different
1.5

transmissions are independent.
We assume that the four channel coefficients H11 , H12 , H21 ,
1 and H22 are independent zero-mean circularly symmetric com-
plex Gaussians, each having variance 1 (per two dimensions).
The channel coefficients are chosen prior to a block of K
transmissions and remain constant over that block.
0.5
The complex transmitted symbols (Xk1 , Xk2 ) must satisfy
a power constraint, i.e.
∗ ∗
E[Xk1 Xk1 + Xk2 Xk2 ] ≤ P. (8)
0
−15 −10 −5 0 5 10 15

B. Telatar capacity
Fig. 3. Basic capacity C (black curve) and repetition capacity Cr (blue)
in bit/transm. as a function of SNR = P/σ 2 in dB (horizontally). Also
If the channel input variables are independent zero-mean
the maximum transmission rates achievable with 4-PAM in the ordinary- circularly symmetric complex Gaussians both having variance
repetition case (blue *’s). In red *’s the maximum rates achievable using P/2, then the resulting mutual information (called Telatar
scaled-repetition mapping.
capacity here, see [6]) is
P/2
CTelatar (H) = log2 det(I2 + HH † ), (9)
C. Demodulation complexity σ2
Scaled repetition outperforms ordinary repetition, but also h11 h12
where H = , i.e. the actual channel-coefficient
has a disadvantage. In an ordinary-repetition system the output h21 h22
yk = (yk1 + yk2 )/2 is simply sliced. In a system that uses matrix and I2 the 2 × 2 identity matrix. Also in the 2 × 2
scaled repetition we can only slice after having distinguished MIMO case we define the signal-to-niose ratio as
between two cases. More precisely note that xk2 = M2 (xk ) = Δ
SNR = P/σ 2 . (10)
2xk − D2 (xk ), where D2 (α) = 5 if α 0 and D2 (α) = −5
if α 0. Now we can use a slicer for yk1 + 2yk2 = xk + It can be shown (see e.g. Yao ([7], p. 36) that for fixed R
nk1 + 2(2xk − D2 (xk ) + nk2 ) = 5xk − 2D2 (xk ) + nk1 + 2nk2 . and SNR large enough Pr{CTelatar(H) R} ≈ γ · SNR−4 , for
Assuming that xk ∈ {−3, −1} we get that D2 (xk ) = −5 and some constant γ.
this implies that we should put a threshold at 0 to distinguish C. Worst-case error-probabilities
between −3 and −1. Similarly assuming that xk ∈ {+1, +3}
Consider M (one for each message) K × 2 code-matrices
we get D2 (xk ) = 5 and we must slice yk1 + 2yk2 again with
c1 , c2 , · · · , cM resulting in a unit average energy code. Then
a threshold at 0. Then the best overall candidate xk is found
ˆ
Tarokh, Seshadri and Calderbank [5] showed that for large
by minimizing (yk1 − xk )2 + (yk2 − M2 (xk ))2 over the two
ˆ ˆ
SNR
candidates.
Pr{c → c } ≈ γ (det((c − c)(c − c)† )−2 SNR−4 . (11)
II. F UNDAMENTAL P ROPERTIES FOR THE 2 × 2 MIMO
C HANNEL for some γ if the√ of the difference matrices c−c is 2, and
rank
we transmit x = P c. If this holds for all difference matrices
A. Model description
we say that the diversity order is 4. Therefore it makes sense
n1
x1 c
to maximize the minimum modulus of the determinant over
h11 E all code-matrix differences.
r ¨ +
rr 21
h ¨¨ 1
y
Tr. r¨¨ n2 Rec. III. A LAMOUTI : O RDINARY R EPETITION
¨r
¨ ¨ 12 rr 2
h c y Alamouti [1] proposed a modulation scheme (space-time
¨ r +
r E code) for the 2 × 2 MIMO cannel which allows for a very
x2 h22 simple detector. Two complex symbols s1 and s2 are trans-
mitted in the first transmission (an odd transmission) and in
Fig. 4. Model of a 2 × 2 MIMO channel.
the second transmission (the next even transmission) these
Next consider a 2 × 2 MIMO channel (see Fig. 4). Both symbols are more or less repeated. More precisely
the transmitter and the receiver use two antennas. The output x11 x12 s1 −s∗
2
vector (y1k , y2k ) at transmission k relates to the corresponding = . (12)
x21 x22 s2 s∗
1
input vector (x1k , x2k ) as given by
The received signal is now
y1k h11 h12 x1k n1k ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + (7) y11 h11 h12 n11
y2k h21 h22 x2k n2k ⎜ y21 ⎟ ⎜ h21 h22 ⎟ ⎜ n ⎟
⎜ ∗ ⎟=⎜ ∗ ⎟ s1
⎝ y12 ⎠ ⎝ h12 −h∗ ⎠ + ⎜ 21 ⎟ , (13)
⎝ n∗ ⎠
where (N1k , N2k ) is a pair of independent zero-mean cir- 11 s2 12
∗
cularly symmetric complex Gaussians, both having variance y22 h∗ −h∗
22 21 n∗
22

1289



Minimum Determinant, 16QAM−codesymbols, R=4 bits/transm.
or more compactly 8

7
y = s1 a + s2 b + n, with
∗ ∗
= (y11 , y21 , y12 , y22 )T ,
6
y
a = (h11 , h21 , h∗ , h∗ )T ,
12 22
5

b = (h12 , h22 , −h∗ , −h∗ )T , and
11 21
4

n = (n11 , n21 , n∗ , n∗ )T .
12 22 (14) 3

2
Since a and b are orthogonal the symbol estimates s1 andˆ
s2 can be determined by simply slicing (a† y)/(a† a) and
ˆ 1

(y † b)/(b† b) respectively. 0
0 0.5 1 1.5

Another advantage of the Alamouti method is that the
densities of a† a and b† b are (identical and) chi-square with Fig. 5. Minimum modulus of the determinant for rotated and scaled Alamouti
as a function of θ horizontally.
8 degrees of freedom. This results in a diversity order 4, i.e.

Pr{(S1 , S2 ) = (S1 , S2 )} ≈ γ · SNR−4 , (15)
B. Hard-decision Performance
for fixed rate and large enough SNR. We have compared the message-error-rate for several R = 4
A disadvantage of the Alamouti method is that only two space-time codes in Fig. 6. By message-error-rate we mean the
complex symbols are transmitted every two transmissions, but probability Pr{X = X}. Note that for each ”test” we generate
more-importantly that the symbols transmitted in the second a new message (8-bit) and a new channel matrix. The decoder
transmission are more or less repetitions of the symbols in is optimal for all codes, it performs M L-decoding (exhaustive
the first transmission. Section I however suggests that we can search). The methods that we have considered are:
improve upon ordinary repetition.
1) Uncoded, in green. We transmit
IV. T HE ROTATED AND SCALED A LAMOUTI METHOD x11 x12
X= , (19)
A. Method description x21 x22
Having seen in section I that scaled-repetition improves where x11 , x12 , x21 , and x22 are symbols from A4-QAM .
upon ordinary repetition in the SISO case, we use this concept 2) Alamouti, in blue, see (12), where s1 and s2 are
to improve upon the standard Alamouti scheme for MIMO symbols from A16-QAM .
transmission. Instead of just repeating the symbols in the 3) Tilted QAM, in cyan. Proposed by Yao and Wornell
second transmission we scale them. More precisely, when s1 [8]. Let sa , sb , sc , and sd symbols from A4-QAM . Then
Δ
and s2 are elements of A16-QAM = {a + jb|a ∈ A4-PAM , b ∈ we transmit
A4-PAM }, we could transmit for some value of θ the signals x11 cos(θ1 ) − sin(θ1 ) sa
= ,
x11 x12 s1 · exp(jθ) −s∗ x22 sin(θ1 ) cos(θ1 ) sb
2
= (16) cos(θ2 ) − sin(θ2 )
x21 x22 M2 (s2 ) M2 (s∗ )
1
x21
=
sc
,
(20)
s1 · exp(jθ) −s∗ 0 0 x12 sin(θ2 ) cos(θ2 ) sd
2
= − ,
2s2 2s∗1 D2 (s2 ) D2 (s∗ )
1 for θ1 = 1 arctan( 1 ) and θ2 = 1 arctan(2).
2 2 2
where M2 (α) = 2α − D2 (α) with D2 (α) = 5β when β is the 4) Rotated and scaled Alamouti, in red, see (16) for θ =
complex sign of α. 1.028, and with s1 and s2 from A16-QAM .
A first question is to determine a good value for θ. Therefore 5) Golden code, in magenta. Proposed by Belfiore et al.
we determine for 0 ≤ θ ≤ π/2 the minimum modulus of the [2]. Now
determinant mindet(θ) 1 α(z1 + z2 θ) α(z3 + z4 θ)
X= √ , (21)
5 j · α(z3 + z4 θ) α(z1 + z2 θ)
mindet(θ) = min | det(X(s1 , s2 , θ) − X(s1 , s2 , θ))|,
(s1 ,s2 ),(s1 ,s2 ) √ √
(17) with θ = 1+2 5 , θ = 1−2 5 , α = 1 + j − jθ, and α =
x11 x12 1 + j − jθ and where z1 , z2 , z3 , and z4 are A4-QAM -
where X = is the code matrix. The minimum
x21 x22 symbols.
modulus of the determinant as a function of θ can be found 6) Telatar, in black. This is the probability that the Telatar
in Fig. 5. The maximum value of the minimum determinant capacity of the channel is smaller than 4.
(i.e. 7.613) occurs for Clearly it follows from Fig. 6 that the winner is the Golden
θopt. = 1.028. (18) code. However rotated and scaled Alamouti is only slightly
worse, roughly 0.2 dB. Important is that Alamouti coding is
We will use this value for θ in what follows. roughly 2 dB worse than the Golden code.

1290



MER, 16QAM code−symbols, R=4 bits/transm., 1000 errors
we rewrite (16) and obtain
Telatar
uncoded
Alamouti x11 x12 −M2 (t1 )Θ M2 (t∗ )
2
= (24)
−1
10
Rot.Scal.Rep.
Tilted QAM
x21 x22 t2 t∗1
−2t1 Θ 2t∗ −D2 (t1 )Θ D2 (t∗ )
Golden Code
2 2
= − ,
t2 t∗
1 0 0
since t = M2 (s) implies that s = −M2 (t). Now
−2 ⎛ ⎞ ⎛ ⎞
−2h11 Θ h12
10
y11
⎜ y21 ⎟ ⎜ −2h21 Θ h22 ⎟ t1
⎜ ∗ ⎟=⎜ ⎟ (25)
⎝ y12 ⎠ ⎝ h∗
12 2h∗ ⎠ t2
11
∗
y22 h∗ 2h∗
⎛ ⎞22 21
⎛ ⎞ ⎛ ⎞
−3
10
−h11 Θ 0 n11
⎜ −h21 Θ ⎟ ⎜ ⎟ ⎜ n21 ⎟
− ⎜ ⎟ D2 (t1 ) − ⎜ 0 ⎟ D2 (t2 ) + ⎜ ⎟.
⎝ 0 ⎠ ⎝ h∗ ⎠
11
⎝ n∗ ⎠
12
10 11 12 13 14 15 16 17 18 19 20
0 h∗
21 n∗
22

Fig. 6. Message error rate for several R=4 space-time codes. We can write this as
y = t1 a + t2 b − D2 (t1 )c − D2 (t2 )d + n,
V. D ECODING COMPLEXITY a = (−2h11 Θ, −2h21 Θ, h∗ , h∗ )T ,
12 22
b = (h12 , h22 , 2h∗ , 2h∗ )T ,
11 21
Clearly the Golden code is better than rotated and scaled
Alamouti. However the Golden code requires the decoder to c = (−h11 Θ, −h21 Θ, 0, 0)T , and
check all 256 alternative codewords, since sphere-decoding d = (0, 0, h∗ , h∗ , 0, 0)T ,
11 21
is not a good alternative now. Here we will investigate the
complexity and performance of a suboptimal rotated and
scaled Alamouti decoder. Denote Θ = exp(jθopt. ). and for the ”cos(φ )” of the angle between a and b we can
A. In the rotated and scaled Alamouti case the received write
vector is |2(Θ − 1)(h11 h∗ + h21 h∗ )|
12 22
⎛ ⎞ ⎛ ⎞ cos(φ ) = . (26)
y11 h11 Θ 2h12 4|h11 |2 + 4|h21 |2 + |h12 |2 + |h22 |2
⎜ y21 ⎟ ⎜ h21 Θ 2h22 ⎟ s1 2 2
⎜ ∗ ⎟=⎜ ⎟ (22) C. It now follows from the inequality 2r1 r2 ≤ r1 + r2 (where
⎝ y12 ⎠ ⎝ 2h∗ 12 −h∗ ⎠ s2
11
∗ ∗ ∗
r1 and r2 are reals), that
y22 2h −h21
⎛ ⎞ 22 ⎛ ⎞ ⎛ ⎞ |h11 |2 + |h12 |2 + |h21 |2 + |h22 |2
0 h12 n11 cos(φ) ≤ |Θ − 1| · ,
⎜ 0 ⎟ ⎜ h ⎟ ⎜ n ⎟ |h11 |2 + |h21 |2 + 4|h12 |2 + 4|h22 |2
− ⎜ ∗ ⎟ D2 (s1 ) − ⎜ 22 ⎟ D2 (s2 ) + ⎜ 21 ⎟ .
⎝ h12 ⎠ ⎝ 0 ⎠ ⎝ n∗ ⎠ |h11 |2 + |h12 |2 + |h21 |2 + |h22 |2
∗
12 cos(φ ) ≤ |Θ − 1| · .
(27)
h22 0 n∗
22 4|h11 |2 + 4|h21 |2 + |h12 |2 + |h22 |2
We can write this as If
|h12 |2 + |h22 |2 ≥ |h11 |2 + |h21 |2 , (28)
y = s1 a + s2 b − D2 (s1 )c − D2 (s2 )d + n,
∗ ∗ then cos(φ) ≤ 2|Θ−1| = 0.393, else cos(φ ) ≤ 2|Θ−1| =
5 5
y = (y11 , y21 , y12 , y22 )T ,
0.393. Therefore it makes sense to decode (s1 , s2 ) when
a = (h11 Θ, h21 Θ, 2h∗ , 2h∗ )T ,
12 22 (28) holds and (t1 , t2 ) when (28) does not hold. Using zero-
b = (2h12 , 2h22 , −h∗ , −h∗ )T ,
11 21
forcing to decode, the noise enhancement is then at most
c = (0, 0, h∗ , h∗ )T , 1/(1 − 0.3932 ) = 1.183 which is 0.729 dB. We shall see
12 22
later that noise enhancement is un-noticeable in practise.
d = (h12 , h22 , 0, 0)T , and
D. The decoding procedure is straightforward. Focus on
n = (n11 , n21 , n∗ , n∗ )T .
12 22 the case where we decode (s1 , s2 ) for a moment. For all 16
alternatives of (D2 (s1 ), D2 (s2 )) the vector
For the ”cos(φ)” of the angle between a and b we can write
z = y + D2 (s1 )c + D2 (s2 )d = s1 a + s2 b + n (29)
|2(Θ − 1)(h11 h∗ + h21 h∗ )]
12 22
cos(φ) = . (23)
|h11 |2 + |h21 |2 + 4|h12 |2 + 4|h22 |2 and is determined. Then compute the sufﬁcient statistic
B. Instead of decoding (s1 , s2 ) we can also decode (t1 , t2 ) = a† z a† a a† b s1 a† n
= + . (30)
(M2 (s1 ), M2 (s2 )) which is equivalent to (s1 , s2 ). Therefore b† z b† a b† b s2 b† n

1291



MER, Rot.Scal.Alam., 16QAM code−symbols, R=4 bits/transm., 1000 errors Av.nr. slicings, Rot.Scal.Alam., 16QAM code−symbols, R=4 bits/transm., 1000 errors
8
full search method 1
method 1 method2
method2
−1
7
10

6

5

−2
10
4

3

2
−3
10

1

10 11 12 13 14 15 16 17 18 19 20 0
10 11 12 13 14 15 16 17 18 19 20

Fig. 7. Message error rate for three Rotated Scaled Alamouti decoders Fig. 8. Number of slicings for two Rotated Sclaed Alamouti decoders (R =
(R = 4), horizontally SNR. 4), horizontally SNR.

M3 (x)
b† b −a† b £
¢¡ +8
Use inverted matrix M = /D where D =
−b† a a† a £
s1
˜ a† z
+6
¢¡
(a† a)(b† b) − (b† a)(a† b) to obtain = M . £
s2
˜ b† z +4
¢¡
Next both s1 and s2 are sliced under the restriction that only
˜ ˜ £
¢¡ +2

alternatives that match the assumed values D2 (s1 ) and D2 (s2 )
£
¢¡
x
are possible outcomes. This is done for all 16 alternatives -8 -6 -4 -2 +2 +4 +6 +8

(D2 (s1 ), D2 (s2 )). The best result in terms of Euclidean dis- -2 £
¢¡
tance is now chosen. £
¢¡ -4
In considering all alternatives (D2 (s1 ), D2 (s2 )) we only
£
¢¡
need to slice when the length of z − s1 a − s2 b is smaller than
-6
˜ ˜
the closest distance we have observed so far. This reduces the -8 £
¢¡
number of slicing steps. We call this approach METHOD 1.
E. The number of slicing steps can even be further de- Fig. 9. The mapping M3 (·).
creased if we start slicing with the most promising alterna-
code, but can be decoded with an acceptable complexity. We
tive (D2 (s1 ), D2 (s2 )). This approach is called METHOD 2.
have obtained similar results for codes based on mapping
Therefore we note that the ”direct” s1 -signal-component in X
s1 Θ 0 M3 (·) for 9-PAM, see Fig. 9. Recently also Sezginer and Sari
is . Therefore we can slice (e† y)/(e† e1 ) [4] investigated complexity reducing methods for alternatives
0 −s∗ /2
1
1 1

in order to find a good guess for D2 (s1 ). Similarly we slice to the Golden code.
(e† y)/(e† e2 ) to find a good first guess for D2 (s2 ). Here
2 2 R EFERENCES
e1 = (h11 Θ, h21 Θ, −h∗ /2, −h∗ /2)T ,
12 22
[1] S.M. Alamouti, ”A simple transmit diversity technique for wireless
communications,” IEEE J. Sel. Areas. Comm. vol. 16, pp. 1451-1458,
e2 = (−h12 /2, −h22 /2, −h∗ , −h∗ )T .
11 21 (31) October 1998.
[2] J.-C. Belfiore, G. Rekaya, E. Viterbo, ”The golden code: A 2×2 full-rate
Then we consider the other 15 alternatives and only slice if space-time code with nonvanishin determinants,” IEEE Trans. Inform.
necessary. Similar methods apply if we want to decode (t1 , t2 ). Theory, vol. IT-51, No. 4, pp. 1432 - 1436, April 2005.
[3] G. Benelli, ”A new method for the integration of modulation and channel
F. We have carried out simulations, first to find out what the coding in an ARQ protocol,” IEEE Trans. Commun., vol. COM-40, pp.
degradation of the suboptimal decoders according to method 1594 - 1606, October 1992.
[4] S. Sezginer and H. Sari, ”Full-rate full-diversity 2 × 2 space-time codes
1 and method2 is relative to ML-decoding. The result is of reduced decoder complexity,” IEEE Comm. Letters, vol. 11, pp. 973
shown in Fig. 7. Conclusion is that the suboptimal decoders - 975, December 2007.
do not demonstrate a performance degradation. We have also [5] V. Tarokh, N. Seshadri, and A.R. Calderbank, ”Space-time codes for
high data rate wireless communication: performance criterion and code
considered the number of slicings for both method 1 and construction,” IEEE Trans. Inform. Theory, Vol. 44, pp. 744- 765, March
method 2. This is shown in Fig. 8. It can be observed that 1998.
method 1 leads to roughly 7 slicings (as opposed to 16). [6] I.E. Telatar, ”Capacity of multi-antenna Gaussian channels” European
Trans. Telecommunications, vol. 10, pp. 585-595, 1999. (Originally
Method 1 further decreases the number of slicing to roughly published as ATT Technical Memorandum, 1995).
3.5. [7] H. Yao, Efficient Signal, Code, and Receiver Designs for MIMO Com-
munication Systems,Ph.D. thesis, M.I.T., June 2003.
VI. C ONCLUSION [8] H. Yao and G.W. Wornell, ”Achieving the full MIMO diversity-
multiplexing frontier with rotation-based space-time codes,” in Proc.
Rotated and scaled Alamouti has a hard-decision perfor- Allerton Conf. Commun. Control, and Comput., Monticello, IL, Oct.
mance which is only slightly worse than that of the Golden 2003.

1292


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