topic iv: mimo communications › ~jkzhang › 4tm4_digital communication… · ee 4tm4 and 6tm4...

Topic IV: MIMO Communications

Instructor : Jian-Kang Zhang

Department of Electrical and Computer Engineering,McMaster University

Room: ITB A212, ext. 27599

Email: [email protected]

Website: http://www.ece.mcmaster.ca/∼jkzhang/

EE 4TM4 and 6TM4 Digital Communications II

Block by Block Transmission using Zero-Padding

Consider the discrete-time equivalent baseband intersymbol interferencechannel model:

yn =

L−1∑ℓ=0

hℓ xn−ℓ + ξn, (1)

Then, a block-by-block transmission for this channel with zero-paddingconsists of the following two basic steps:

• Zero-padding: Let s denote a length K data symbol column vector, i.e.,s = (s0, s1, · · · , sK−1)T . Then, L − 1 zeros are append to s to form xwhich is of length P = K + L− 1. That is, x = (x0, x1, · · · , xP−1)T =(s0, s1, · · · , sK−1, 0, 0, · · · , 0︸︷︷︸

L−1

)T .


• Block-by-block transmission: The elements of x are then seriallytransmitted through the ISI channel (1):

y0 = h0s0 + ξ0

y1 = h0s1 + h1s0 + ξ1

. . . . . . . . . . . . . . .

yk = h0sk + h1sk−1 + · · ·+ hL−1sk−L+1 + ξk. . . . . . . . . . . . . . .

yP−1 = hL−1sK−1 + ξP−1.

Collecting all these P equations leads to the following matrixrepresentation for block transmission:

y = Hs+ ξ,


where ξ is a P × 1 noise vector and H denotes a P ×K channel matrixdefined by

H =

h0 0 . . . 0h1 h0 . . . 0... h1 . . .

...hL−1 . . . . . . h00 . . . . . . h1... . . . . . . ...0 . . . 0 hL−1

P×K

.


Orthogonal Frequency Division Multiplexing (OFDM)

Consider the discrete-time equivalent baseband intersymbol interferencechannel model:

yn =

L−1∑ℓ=0

hℓ xn−ℓ + ξn, (2)

Then, the OFDM scheme captures the following four basic steps:

• Performing the inverse DFT: Let s denote a length N data symbolcolumn vector, i.e., s = (s0, s1, · · · , sN−1)T , and let W denote thenormalized N × N DFT matrix. Perform the inverse DFT on the data


vector s, i.e., S = WHs. In other words,

S[k] =1√N

N−1∑n=0

s[n]ej2πnkN (3)

for k = 0, 1, · · · , N − 1.

• Adding circular prefix: Add the last L− 1 elements of S to S as prefixand make a length K = N + L − 1 data symbol column vector, x =(x0, x1, · · · , xK−1)T = S[N−L+1], · · · , S[N−1], S[0], S[1], · · · , S[N−1])T for transmission over the ISI channel (2)

• Delete prefix: At the receiver side, during consecutive K time slots, we


have received K signals:

y0 = h0x0 + ξ0 = h0S[N − L + 1] + ξ0y1 = h0x1 + h1x0 + ξ1 = h0S[N − L + 2] + h1S[N − L + 1] + ξ1

. . . . . . . . . . . . . . .

yL−2 = h0xL−2 + · · · + hL−2x0 + ξL−2 = h0S[N − 2] + · · · + hL−2S[N − L + 1] + ξL−2yL−1 = h0xL−1 + · · · + hL−1x0 + ξL−1 = h0S[0] + · · · + hL−1S[N − L + 1] + ξL−1

yL = h0xL + · · · + hL−1x1 + ξL−1 = h0S[1] + · · · + hL−1S[N − L + 2] + ξL. . . . . . . . . . . . . . .

yK−1 = h0xK−1 + · · · + hL−1xK−L + ξK−1 = h0S[N − 1] + · · · + hL−1S[N − L] + ξK−1.

Deleting the first L − 1 received equations and then collecting all theremaining N equations lead to the following matrix representation:

y = CS+ ξ,


where y = (yL−1, yL, · · · , yK−1)T and ξ = (ξL−1, ξL, · · · , ξK−1)T is aP ×1 noise vector and C denotes a K×K circular channel matrix givenby

C =

h0 0 . . . h1h1 h0 . . . h2... ... ... ...

hL−1 hL−2 0 00 hL−1 0 0... ... ... ...0 0 . . . h0

• Perform DFT: Performing the DFT on the remaining received data


vector y results in

Y = Wy = WCS+Wξ

= WCWHs+ ξ̃

= Ds+ ξ̃,

where D = WCWH is an N × N diagonal matrix with the diagonalelements being D[k] =

∑L−1ℓ=0 hℓe

−j2πℓkN = H[k] for k = 0, 1, · · · , N − 1and ξ̃ = Wξ. Equation (4) tells us that the original ISI channel (2) isnow converted into the N parallel subchannels without ISI,

Y [k] = D[k]sk + ξ̃k

for k = 0, 1, · · · , N − 1.


Power Loading for OFDM Systems

It is known that the mutual information between the input signal andthe output signal is maximized when the input signal is Gaussian distributedand the maximum for the OFDM system is given by

I(s; Ȳ) =1

P

N−1∑k=0

log(1 +

pk|H[k]|2

2σ2

)where pk = E[|sk|2]. Hence, the channel capacity for the OFDM system isobtained by maximizing I(s; Ȳ) subject to a total power constraint, i.e.,

C = max I(s; Ȳ)

subject to∑N

k=1 pk = p. A solution to the above optimization problem is


called a water-filling solution, which is given by the following power-loadingalgorithm:

• Sort the magnitude of each subcarrier such that

|H[k1]| ≥ |H[k2]| ≥ · · · ≥ |H[kN̄ ]| > 0

and H[kN̄+1] = · · · = |H[kN ] = 0

• Determine a maximum positive integer r such that

p2σ2

+∑r

i=1 |H[ki]|−2

r> |H[ki]|−2

• Optimal power-loading: pki =p+2σ2

∑ri=1 |H[ki]|

−2

r − 2σ2|H[ki]|−2 for

i = 1, 2, · · · , r and pkr+1 = · · · = pkN = 0.


Multi-Antennas MIMO Channels

Consider a discrete-time baseband equivalent wireless communicationsystem with M transmitter antennas and N receiver antennas; i.e., multi-antennas MIMO system,

y = Hx+ ξ, (4)

where y is an N × 1 received signal vector, H denotes an N ×M channelmatrix with (n,m)-th entry hn,m being designated as the channel coefficientlinking the m-th transmitter antenna to the n-th receiver antenna , x is anM × 1 transmitted signal vector with the m-th entry xm denoting a signaltransmitted from the m-th transmitter antenna, and ξ is an N × 1 noisyvector with zero mean and covariance matrix 2σ2IN .


MIMO Channel Capacity

Let us now consider a discrete-time baseband equivalent MIMO system:

y = Hx+ ξ,

Here

• y is an Nr × 1 received signal vector,

• H denotes an Nr × Nt channel matrix and is assumed to be known atboth the transmitter and the receiver,

• x is an Nt × 1 transmitted signal vector with covariance matrix Σ, and

• ξ is an Nr × 1 complex circularly-symmetric Gaussian noisy vector withzero mean and covariance matrix 2σ2INr.


Water-Filling Solution

Under the above assumption, it is known that the mutual informationbetween the input signal and the output signal is maximized when the inputsignal is Gaussian distributed and the maximum for the MIMO system isgiven by

I(x;y) =1

Nculog det

(I+

HΣHH

2σ2

)where Ncu denotes the number of channel uses and Σ == E[xx

H]. Hence,the channel capacity for the MIMO system is attained by maximizing I(s; Ȳ)subject to a total power constraint, i.e.,

C = max I(x;y)


subject to Tr(Σ) = p. A solution to the above optimization problem is awater-filling solution, given by the following power-loading algorithm:

• Perform the eigenvalue decomposition (EVD) of Σ:

Σ = UPUH

where P = diag(p1, p2, · · · , pNt).

• Perform the eigenvalue decomposition (EVD) of HHH:

HHH = VΛVH

where Λ = diag(λ1, λ2, · · · , λR, 0, 0, · · · , 0) with λ1 ≥ λ2 ≥ · · · ≥ λR >0


• The optimal solution maximizing the channel capacity is that U = Vand the power-loading for each eign-subchannel follows the water-fillingprincipal:

1) Determine a maximum positive integer r such that

p2σ2

+∑r

i=1 λi−1

r> λ−1i

2) Optimal power-loading: pki =p+2σ2

∑ri=1 λ

−1i

r − 2σ2λ−1i for i =

1, 2, · · · , r and pkr+1 = · · · = pkN = 0.


Capacity for Multi-Antennas MIMO Sysytems

Now, let us consider a discrete-time baseband equivalent multi-antennasMIMO wireless communication system with M transmitter antennas and Nreceiver antennas:

y = Hx+ ξ,

where

• h = vec(HT ) is an MN×1 circularly symmetric Gaussian random vectorwith zero-mean and covariance matrix IMN ,

• E[xxH] = Σ, and

• ξ is an N × 1 noisy circularly symmetric Gaussian random vector withzero mean and covariance matrix 2σ2IN .


Under the above assumption, the average channel capacity is determined by

C = maxTr(Σ)=p

EH

[log det

(I+

HΣHH

2σ2

)]It can be proved that when Σ = pM I, the maximum is achieved and

C = EH

[log det

(I+

p

2σ2MHHH

)]= min{M,N} log SNR +O(1)

for large SNR, where

SNR =p

2σ2

is the common SNR at the each receiver antenna.


Orthogonal Space-Time Block Codes for MISO Channels

Consider a discrete-time baseband equivalent wireless communicationsystem with multiple transmitter antennas and a single receiver antenna;i.e., MISO system,

r = h1x1 + h2x2 · · ·+ hMxM + ξ, (5)

where M denotes the number of transmitter antennas, hm is the channelcoefficient linking the mth transmitter antenna to the receiver antenna andassumed to be known at the receiver, xm is a signal transmitted fromthe mth transmitter antenna through the channel hm, and ξ is a complexGaussian noise with zero mean and the real part and the imaginary partbeing independent and having the same variance σ2.


Let X(s) denote a T × 1 a space-time signal matrix:

X(s) =

K∑k=1

Aksk +

K∑k=1

Bks∗k. (6)

where Ak and Bk are T ×M matrices and sk are transmitted symbols froma certain scalar constellation A. Now, the transmission scheme is describedas follows:

1) During T consecutive time slots, the channel coefficients hk are fixed.

2) At the tth time slot, the tth row of matrix X(s) is selected fortransmission, i.e., by putting xm = xt,m in (5), we have

rt = xt,1h1 + xt,2h2 + · · ·+ xt,MhM + ξt (7)


for t = 1, 2, · · · , T . Stacking all T received signals yields the followingspace-time block coded channel model:

r = X(s)h+ ξ, (8)

where r is T × 1 received signal vector, h is an M × 1 channel vector andξ is T × 1 noise vector.

Definition 1. Let {Ai,Bi}Ki=1 be a sequence of T × M matrices withT ≥ M . It is said to constitute a complex orthogonal space-time blockcode if the following conditions are satisfied,

AHmAn +BHn Bm = δm,nIM (9)

AHmBn +AHn Bm = 0 (10)

for any 1 ≤ n,m ≤ K.


Example 1. The following Alamouti space-time block code is optimalin many senses for two transmitter antennas and a single receiver antennawireless communication systems,

X(s) =

(s1 s2−s∗2 s∗1

). (11)

It is clear that two symbols are independently transmitted during twoconsecutive time slots and thus the symbol rate of the Alamouti code is oneper channel use.

Example 2. Another typical example is a 4 × 4 orthogonal code with a


symbol rate 3/4,

X(s) =

s1 s2 s3 0−s∗2 s∗1 0 −s3−s∗3 0 s∗1 s20 s∗3 −s∗2 s1

. (12)

Notice that (6) can be rewritten as

r = Ha s+Hb s∗ + ξ, (13)

where

Ha = [A1h,A2h, · · · ,AKh] (14)Hb = [B1h,B2h, · · · ,BKh] . (15)


Taking the conjugate on the both sides of (13) leads to

r∗ = H∗bs+H⋆as

∗ + ξ∗. (16)

Equations (13) and (16) can be expressed in a compact matrix form as[rr∗

]=

[Ha HbH∗b H

∗a

] [ss⋆

]+

[ξξ∗

]= H

[ss⋆

]+

[ξξ∗

], (17)

where

H =[

Ha HbH∗b H

∗a

].


Properties of Orthogonal STBCs

Proposition 1. The following three statements are equivalent:

1. {Ai,Bi}Ki=1 constitutes a complex orthogonal space time block code;

2. XH(s)X(s) = ∥s∥2IK for any K × 1 complex vector s;

3. HHH = ∥h∥2I2K for any M × 1 complex channel vector h.

This property tells us that the orthogonal STBC is equivalent to the factthat the resulting virtual MIMO channel matrix H is orthonormal up to ascale for any channel coefficients.


Error Performance Analysis on Orthogonal STBCs

1. Assumptions: For the development of the analysis, the followingassumptions are needed:

• The transmitted symbols sk are independent and equally likely chosenfrom the square q-ary QAM constellation;

• The channel coefficients hm are samples of circularly symmetric zero-mean complex white Gaussian random variables with unit variancesand are fixed during one time frame (T time slots), but may changeindependently from one time frame to the next;

• ξ is circularly symmetric complex Gaussian noise with covariancematrix 2σ2I;

• Complete channel state information is available at the receiver andmaximum likelihood (ML) detection is employed.


2. Symbol error probability in terms of the given channel realization:

For the orthogonal STBCs, the channel matrix H is orthonormal up tothe scale ∥h∥2, i.e., HHH = ∥h∥2I2K. Therefore, the space-time blockcoded channel model (17) is simplified into the following K parallelindependent complex AWGN channel models for ML detection:

y = ∥h∥2s+ η,

where y = HHa r+HTb r

∗ and η = HHa ξ +HTb ξ

∗. In other words,

yk = ∥h∥2sk + ηk,

for k = 1, 2, · · · ,K. Using our assumptions and the property onthe orthogonal STBCs, we know that ηk for k = 1, 2, · · · ,K arejointly Gaussian and independent. Each has zero mean and variance


2σ2∥h∥2. Therefore, the symbol error probability for the square q-aryQAM constellation for given channel realization h is given by

Pe(h) = 4(1− 1√

q

)Q(∥h∥d

2σ

)− 4

(1− 1√

q

)2Q2

(∥h∥d2σ

)3. Average symbol error probability over random channel coefficients:

For analysis on the average symbol error probability over all channelrealizations, we need to take an average on Pe(h) over the randomchannel vector h, i.e., E

[Pe(h)

], where notation E[·] denote the

mathematical expectation operator. To do that, we use the followingtwo formulae

Q(t) =1

π

∫ π2

0

e− t

2

2 sin2 θdθ, Q2(t) =1

π

∫ π4

0

e− t

2

2 sin2 θdθ


for t > 0 to represent Pe(h) so that

Pe(h) =4

π×(1− 1√

q

)∫ π20

e− ∥h∥

2d2

8σ2 sin2 θ dθ − 4π

(1− 1√

q

)2 ∫ π40

e− ∥h∥

2d2

8σ2 sin2 θ dθ

=4

π×(1− 1√

q

) 1√q

∫ π4

0

e− ∥h∥

2d2

8σ2 sin2 θ dθ +4

π×

(1− 1√

q

)∫ π2π4

e− ∥h∥

2d2

8σ2 sin2 θ dθ

Hence, we have

E[Pe(h)

]=

4

π×

(1− 1√

q

) 1√q

∫ π4

0

E

[e− ∥h∥

2d2

8σ2 sin2 θ

]dθ

+4

π×(1− 1√

q

)∫ π2π4

E

[e− ∥h∥

2d2

8σ2 sin2 θ

]dθ.

Let hm = hre,m + jhim,m. According to our assumption on h, all hre,m


and him,m are Jointly Gaussian and independent. Each is Gaussiandistributed with zero mean and variance 1/2. As a result,

E

[e− ∥h∥

2d2

8σ2 sin2 θ

]= E

M∏m=1

e−

h2re,md2

8σ2 sin2 θ

M∏m=1

e−

h2im,md2

8σ2 sin2 θ

=

M∏m=1

E

e− h2re,md28σ2 sin2 θ M∏

m=1

E

e− h2im,md28σ2 sin2 θ

=

Ee− h2re,1d28σ2 sin2 θ

2M

Notice that E[e−

h2re,1d2

8σ2 sin2 θ

]= 1√

π

∫∞−∞ e

− x2d2

8σ2 sin2 θ e−x2dx =

(1 +

d2

8σ2 sin2 θ

)−12. Hence, E

[e− ∥h∥

2d2

8σ2 sin2 θ

]=

(1 + d

2

8σ2 sin2 θ

)−M. Substituting


this into (18) yields

E[Pe(h)

]=

4

π×(1− 1√

q

) 1√q

∫ π4

0

dθ(1 + d

2

8σ2 sin2 θ

)M+4

π×

(1− 1√

q

)∫ π2π4

dθ(1 + d

2

8σ2 sin2 θ

)M .Since 0 ≤ sin θ ≤ 1 for 0 ≤ θ ≤ π2 , we have

E[Pe(h)

]≤ q − 1

q

(1 +

d2

8σ2

)−M=

q − 1q

(1 +

3SNR

2(q − 1)

)−M<

(23

)M× (q − 1)

M+1

q× SNR−M


where we have used the fact that the average anergy for the square q-ar

QAM constellation per symbol is Es =(q−1)d2

6 and the signal to noise

ratio is defined by SNR = Es2σ2

. Therefore, the orthogonal STBCs enablefull diversity for the MISO systems with the ML detector.

topic iv: mimo communications › ~jkzhang › 4tm4_digital communication… · ee 4tm4 and 6tm4...

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