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Capacity of MIMO channels
Nghi H. Tran
Supervisor: Prof. Ha H. Nguyen
Department of Electrical Engineering
University of Saskatchewan
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Outline
Channel Model
Mathematical Preniminaries
Channel with fixed transfer function
Channel with Rayleigh fading
Example with Orthogonal Design
Non-ergodic channel
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System Model
We deal with a linear model with the received vector y Cr andthe transmitted vector x Ct:
y = Hx + n (1)
with H is a r t complex matrix and n is complex Gaussian
noise with independent components, E[nn] = Ir.
The power constraint: E[xx] P, or equivalently
tr(E[xx]) P. So does for x E(x) interested inzero-mean x.
Three cases ofH:
Deterministic
H is random, chosen according to a pdf, and each use of the
channel corresponds to an independent realization.H is random, but is fixed once it is chosen.
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Mathematical Preliminaries
A complex vector x Cn is Gaussian ifx =
Re(x)
Im(x)
is Gaussian.
To specify the distribution ofx, it is necessary to specify E[x] andE
(x E[x])(x E[x])
A complex Gaussian x is circularly symmetric if:
E
(x E[x])(x E[x])
=1
2
Re(Q) Im(Q)Im(Q) Re(Q)
(2)for some Hermitain non-negative definite Q.
For circularly symmetric x, the first and the second statistics can be specified by
E[x] and E
(x E[x])(x E[x])
In Tse book: Circular symmetric: x and exp(j)x has the same distribution. Itleads to the mean = 0
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Mathematical Preliminaries
The density function of a circularly symmetric complex Gaussian
x with mean and covariance Q is given as:
,Q(x) = det(Q)1 exp (x )Q1(x ) (3)
The differential entropy ofx is then computed as:
H(Q) = EQ [ log Q(x)]
= log det(Q) + (log(e))E[xQ1x]
= log det(Q) + (log(e))tr(E[xx]Q1
)
= log det(eQ) (4)
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Mathematical Preliminaries
The importance of the circularly symmetric complex Gaussian is
due to the fact that, given Q, x maximize the entropy.
How to prove: Using following inequality:
Cnp(x)logp(x)dx Cnp(x)logpN(x)dx (5)where p(x) and pN(x) are arbitrary pdfs. After that, apply it by
using pN(x) = Q(x).
Ifx Cn is a circularly symmetric complex Gaussian then so is
y = Ax for any A Cmn
Ifx and y are independent circularly symmetric complexGaussian, then so is z = x + y.
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Channel With Fixed H
Using singular value decomposition theorem:
H = UDV (6)
where U Crr and V Ctt are unitary, and D Rrt is non-negative and
diagonal.
In fact, the diagonal entries ofD are the singular values ofH, the columns ofU
and U are the eigenvectors ofHH and HH, respectively.
Then we have:
y = UDVx+ n (7)
The fact thatUn and n has the same distribution, we have an equivalent channel:
y = Dx+ n (8)
where n = Un, y = Uy and x = Vx
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Channel With Fixed H
Since rank(H) is at most min{r, t}, at most min{r, t} of its singular values are
non-zero, denoted by 1/2i
Therefore, we have parallel Gaussian channels:
yi = 1/2i xi + ni, 1 i min{r, t} (9)
Clearly, yi, i > min{r, t} is just noise component and xi does not play any role.
The well-known result: {xi i min{r, t}} to be independent Gaussians.
The variances ofxi are chosen via water filling" as:
E[Re(xi)2
] = E[Im(xi)2
] =
1
2 ( 1
i )+
(10)
where is chosen to meet the power constraint P() =
i( 1i )
+
The maximal mutual information:
C() =i
(ln(i))+
(11)
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Alternative Derivation
The mutual information I(x,y) is given:
I(x,y) = H(y) H(n) (12)
The covariance ofy: E[yy] = HQH + Ir
To maximize H(y) with given Q, x should be circularly
symmetric Gaussian.Hence, we have:
I(x,y) = log det(Ir + HQH) = log det(It + QH
H) (13)
(Using determinant identity det(Ia + AB) = det(Ib + BA))
Now we need to choose Q subject to tr(Q) P and Q is
non-negative definite.
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Alternative Derivation
Diagonalize HH as follows (it is Hermitian):
HH = UU (14)
with unitary U and = diag(1, . . . , t)
It then follows:det(It +QH
H) = det(It +QUU) = det(It +UQU
)
= det(It + 1/2UQU1/2) (15)
Observe that the properties ofQ = UQU is the same withQ Can maximizeover Q.
Furthermore, for non-negative definite matrixA, detA i Aii. Then we have:det(It + 1/2Q1/2) i
(1 + Qiii) (16)
The equality holds when Q is diagonal. Thus we see that the maximizing Q is
diagonal and the optimal diagonal one can be obtained using water filling.
Results: Qii = ( 1i ) where
i Qii = P.
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The Gaussian Channel with Rayleigh
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The Gaussian Channel with Rayleigh
Fading
Consider H is random, independent realization for each block, each entry is
zero-mean, complex Gaussian, variance 1/2, independent component.
Lemma: For any unitary U Crr
and V Ctt
,UHV
,UH,HV and Hhas the same distribution.
The mutual information is given as:
I(x,y) = EH[I(x,y)|H] (17)
For a given covariance Q ofx, I(x,y)|H is maximized when x is circularly
symmetric Gaussian.
Therefore, now, we need to maximize
(Q) = E
log det(Ir +HQH
(18)
over the choice of non-negative definiteQ subject tr(Q) P
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Capacity
Since Q is non-negative definite,Q = UDU with unitaryU and diagonal
non-negative D
Then we have:
(Q) = E
log det(Ir + (HU)D(HU)
(19)
Since H and HU have the same distribution, (Q) = (D) restrict attention
to non-negative diagonalQ
Considering t! permutation matrix corresponding to Q = Q. It is easy to
see (Q) = (Q)
Observe: log det is concave on the set of positive definite matrices. It then
follows that is concave.
(aQ1 + (1 a)Q2) a(Q1) + (1 a)(Q2) (20)
where 0 a 1.Nghi H. Tran, UoS Capacity of MIMO channels 12
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Evaluation of Capacity
The capacity of channel with power constraint P is:
0
log(1+P/t)m1
k=0
k!
(k + n m)! Lnmk ()2
nm exp()d
(23)where m = min{r, t}, n = max{r, t}, Lij: associated Laguerre
polynomials.
At high SNR, we have:
C(SN R) = m log(SN R) + O(1) (24)
where SN R is average signal to noise ratio at each receiveantenna. In this model, SN R = P
Clearly, C increases with m log(SN R), in contrast to log(SN R)
for single antenna. MIMO can be viewed as m parallel spatialchannels.
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Evaluation of Capacity: Low P
Rewrite C based on singular values ofH 1/2i :
C(P) = Em
i=1 log1 +P
ti (25)
At low P, using log2(1 + x) x log2 e:
C
mi=1
P
t E[i]log2 e =
P
t E[tr(HH
)]log2 e
=P
tE
i,j
|hi,j |2 = P.r. log2 e (26)
where using the fact: sum of eigenvalues equal to trace.
Clearly, with low P, multiple transmitter antennas are not veryuseful.
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Example with Alamouti Scheme
For the convenience, to get SNR at each receive antenna is :
y =
tHx+ n (27)
The capacity:
C(,t,r) = E[log det(Ir +
tHH) = E[log det(It +
tHH) (28)
With r = 1, Alamouti Scheme is equivalent to :
y1
y2
=
2
h1 h2
h2 h1
x1
x1
+
n1
n2
(29)
The capacity of this equivalent channel is:
Corth() =1
2
log detI2 +
2
(|h1|2 + |h2|
2)I2 = C(, 2, 1) (30)Clearly, there is no loss in terms of capacity with r = 1.
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Example with Alamouti Scheme
For r = 2, an equivalent channel:
y11
y21
y12
y22
=
2
h11 h21
h21 h11
h12 h22
h22 h12
H
x1x2
+
v11
v21
v12
v22
(31)
With HH =
i,j |hi,j |2I2, we have:
Corth() =1
2E
log det
I2 +
2(|h11|
2 + |h12|2 + |h21|
2 + |h22|2)I2
= Elog1 + 2
4(|h11|2 + |h12|2 + |h21|2 + |h22|2)
= C(2, t = 4, r = 1) < C(, t = 2, n = 2) (32)
It says that Alamouti scheme achieves the capacity oft = 4, r = 1 and at twiceSNR. General result: C(r, 2r, 1) < C(, 2, r).
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2 2
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Capacity 2 2 Channel
0 5 10 15 20 25 300
2
4
6
8
10
12
14
16
18
SNR(dB)
C(bits
)
r=2, t=2
Actual 2x2 channelAlamouti scheme
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N di Ch l
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Non-ergodic Channels
H is chosen randomly at the beginning and is held fixed for all the
uses of the channel.
In this case, given data rate R, when log det(Ir +HQH) < R, noreliable communication System is in outage.
Outage probability given R and P:
Pout(R, P) = P(log det(Ir + HQH) < R) (33)
Given R, if system is not in outage, there exists a universal codingstrategy that achieves reliable communication
A transmit strategy can be chosen (parameterized by the
covariance) to minimize the probability of outage event.
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