Lecture 4 Capacity of Wireless Channels I-Hsiang Wang - - PowerPoint PPT Presentation

Lecture ¡4 Capacity ¡of ¡Wireless ¡Channels

I-Hsiang Wang ihwang@ntu.edu.tw 3/20, 2014

What ¡we ¡have ¡learned

• So far: looked at specific schemes and techniques
• Lecture 2: point-to-point wireless channel
• Diversity: combat fading by exploiting inherent diversity
• Coding: combat noise, and further exploits degrees of freedom
• Lecture 3: cellular system
• Multiple access: TDMA, CDMA, OFDMA
• Interference management: orthogonalization (partial frequency

reuse), treat-interference-as-noise (interference averaging)

2

Information ¡Theory

• Is there a framework to …
• Compare all schemes and techniques fairly?
• Assert what is the fundamental limit on how much rate can be

reliably delivered over a wireless channel?

• Information theory!
• Provides a fundamental limit to (coded) performance
• Identifies the impact of channel resources on performance
• Suggest novel techniques to communicate over wireless channels
• Information theory provides the basis for the modern

development of wireless communication

3

Historical ¡Perspective

4

• First radio built 100+ years ago
• Great stride in technology
• But design was somewhat ad-hoc

1901 1948

• G. Marconi
• C. Shannon

Engineering ¡meets ¡science New ¡points ¡of ¡view ¡arise

• Information theory: every channel

has a capacity

• Provides a systematic view of all

communication problems

Modern ¡View ¡on ¡Multipath ¡Fading

• Classical view: fading channels are unreliable
• Diversity techniques: average out the variation
• Modern view: exploit fading to gain spectral efficiency
• Thanks to the study on fading channel through the lens of

information theory!

5

Time Channel quality

Plot

• Use a heuristic argument (geometric) to introduce the

capacity of the AWGN channel

• Discuss the two key resources in the AWGN channel:
• Power
• Bandwidth
• The AWGN channel capacity serves as a building block

towards fading channel capacity:

• Slow fading channel: outage capacity
• Fast fading channel: ergodic capacity

6

Outline

• AWGN Channel Capacity
• Resources of the AWGN Channel
• Capacity of some LTI Gaussian Channels
• Capacity of Fading Channels

7

8

AWGN ¡Channel ¡Capacity

9

Channel ¡Capacity

• Capacity := the highest data rate can be delivered

reliably over a channel

• Reliably ≡ Vanishing error probability
• Before Shannon, it was widely believed that:
• to communicate with error probability → 0
• ⟹ data rate must also → 0
• Repetition coding (with M-level PAM) over N time slots
• n AWGN channel:
• Error probability
• Data rate
• As long as M ≤ N⅓, the error probability → 0 as N → ∞
• But, the data rate
• still → 0 as N → ∞

∼ 2Q r 6N M 3 SNR ! = log2 M N = log2 N 3N

Channel ¡Coding ¡Theorem

• For every memoryless channel, there is a definite number C

that is computable such that:

• If the data rate R < C, then there exists a coding scheme that can

deliver rate R data over the channel with error probability → 0 as the coding block length N → ∞

• Conversely, if the data rate R > C, then no matter what coding

scheme is used, the error probability → 1 as N → ∞

• We shall focus on the additive white Gaussian noise

(AWGN) channel

• Give a heuristic argument to derive the AWGN channel capacity

10

AWGN ¡Channel

• We consider real-valued Gaussian channel
• As mentioned earlier, repetition coding yield zero rate if

the error probability is required to vanish as N → ∞

• Because all codewords are spread on a single dimension

in an N-dimensional space

• How to do better?

11

y[n] = x[n] + z[n] x[n] z[n] ∼ N(0, σ2)

Power constraint:

N

X

n=1

|x[n]|2 ≤ NP

Sphere ¡Packing ¡Interpretation

12

y = x + z RN

• By the law of large numbers,

as N → ∞, most y will lie inside the N-dimensional sphere of radius

p N(P + σ2)

• Also by the LLN, as N → ∞,

y will lie near the surface of the N-dimensional sphere centered at x with radius √

Nσ2

How many non-overlapping spheres can be packed into the large sphere?

• Vanishing error probability

⟹ non-overlapping spheres

p N(P + σ2) √ Nσ2

Why ¡Repetition ¡Coding ¡is ¡Bad

13

y = x + z RN

It only uses one dimension

• ut of N !

p N(P + σ2)

Capacity ¡Upper ¡Bound

14

y = x + z RN

p N(P + σ2) √ Nσ2

Maximum # of non-overlapping spheres = Maximum # of codewords that can be reliably delivered = ⇒ R ≤ 1 N log p N(P + σ2)

N

√ Nσ2N ! = 1 2 log ✓ 1 + P σ2 ◆ 2NR ≤ p N(P + σ2)

N

√ Nσ2N This is hence an upper bound of the capacity C. How to achieve it?

Achieving ¡Capacity ¡(1/3)

• (random) Encoding: randomly generate 2NR codewords

{x1, x2, ...} lying inside the “x-sphere” of radius

• Decoding:
• Performance analysis: WLOG let x1 is sent
• By the LLN,
• As long as x1 lies inside the uncertainty sphere centered at αy

with radius

• , decoding will be correct
• Pairwise error probability (see next slide) =

15

√ NP y − → MMSE − → αy − → Nearest Neighbor − → b x

α := P P + σ2

||αy − x1||2 = ||αw + (α − 1)x1||2 ≈ α2Nσ2 + (α − 1)2NP = N Pσ2 P + σ2 r N Pσ2 P + σ2 ✓ σ2 P + σ2 ◆N/2

Achieving ¡Capacity ¡(2/3)

16

√ NP

x-sphere

r N Pσ2 P + σ2 x1

When does an error occur? Ans: when another codeword x2 falls inside the uncertainty sphere of αy What is that probability (pairwise error probability)? Ans: the ratio of the volume of the two spheres

Pr {x1 → x2} = p NPσ2/(P + σ2)

N

√ NP

N

= ✓ σ2 P + σ2 ◆N/2

Union bound: Total error probability ≤ 2NR

✓ σ2 P + σ2 ◆N/2 αy x2

Achieving ¡Capacity ¡(3/3)

• Total error probability (by union bound)
• As long as the following holds,
• Hence, indeed the capacity is

17

Pr {E} → 0 as N → ∞ R < 1 2 log ✓ 1 + P σ2 ◆ Pr {E} ≤ 2NR ✓ σ2 P + σ2 ◆N/2 = 2

N R+ 1

2 log 1 1+ P σ2

!!

C = 1 2 log ✓ 1 + P σ2 ◆ bits per symbol time

18

Resources ¡of ¡ AWGN ¡Channel

Continuous-­‑Time ¡AWGN ¡Channel

• System parameters:
• Power constraint: P watts; Bandwidth: W Hz
• Spectral density of the white Gaussian noise: N0/2
• Equivalent discrete-time baseband channel (complex)
• 1 complex symbol = 2 real symbols
• Capacity:

19

Power constraint:

N

X

n=1

|x[n]|2 ≤ NP

y[n] = x[n] + z[n] x[n] z[n] ∼ CN(0, N0W) CAWGN(P, W) = 2 × 1 2 log ✓ 1 + P/2 N0W/2 ◆ bits per symbol time = W log ✓ 1 + P N0W ◆ bits/s = log (1 + SNR) bits/s/Hz

SNR := P/N0W SNR per complex symbol

Complex ¡AWGN ¡Channel ¡Capacity ¡

• The capacity formula provides a high-level way of

thinking about how the performance fundamentally depends on the basic resources available in the channel

• No need to go into details of specific coding and

modulation schemes

• Basic resources: power P and bandwidth W

20

CAWGN(P, W) = W log ✓ 1 + P N0W ◆ bits/s = log (1 + SNR) bits/s/Hz Spectral Efficiency

Power

• High SNR:
• Logarithmic growth with power
• Low SNR:
• Linear growth with power

21 3 4 5 6 7 20 40 60 80 100 1 2 SNR log (1 + SNR)

C = log(1 + SNR) ≈ log SNR C = log(1 + SNR) ≈ SNR log2 e

SNR = P N0W

Fix W:

Bandwidth

22

30 5 Bandwidth W (MHz) Capacity Limit for W → ∞ Power limited region 0.2 1 Bandwidth limited region (Mbps) C(W ) 0.4 25 20 15 10 1.6 1.4 1.2 0.8 0.6 P N0log2 e

C(W) = W log ✓ 1 + P N0W ◆ ≈ W P N0W log2 e = P N0 log2 e

Fix P

Bandwidth-­‑limited ¡vs. ¡Power-­‑limited

• When SNR ≪ 1: (Power-limited regime)
• Linear in power; Insensitive to bandwidth
• When SNR ≫ 1: (Bandwidth-limited regime)
• Logarithmic in power; Approximately linear in bandwidth

23

CAWGN(P, W) = W log ✓ 1 + P N0W ◆ bits/s SNR = P N0W CAWGN(P, W) ≈ W ✓ P N0W ◆ log2 e = P N0 log2 e CAWGN(P, W) ≈ W log ✓ P N0W ◆

24

Capacity ¡of ¡Some LTI ¡Gaussian ¡Channels

25

SIMO ¡Channel

• MRC is a lossless operation: we can generate y from :
• Hence the SIMO channel capacity is equal to the

capacity of the equivalent AWGN channel, which is

h x y e y = ||h||x + e w, e w ∼ CN

• 0, σ2

↓ MRC, h∗ ||h|| ↓ y = hx + w ∈ CL

Power constraint: P w ∼ CN

• 0, σ2IL
• e

y y = e y (h/||h||)

CSIMO = log ✓ 1 + ||h||2P σ2 ◆ Power gain due to Rx beamforming

MISO ¡Channel

• Goal: maximize the received power ||h*x||2
• The answer is ||h||2P ! (check. Hint: SVD)
• Achieved by Tx beamforming
• Send a scalar symbol x on the direction of h
• Power constraint on x : still P
• Capacity:

26

h x y y = h∗x + w ∈ C, x, h ∈ CL h∗ = ⇥h1 h2 ⇤ y = x||h|| + w

↓ Tx Beamforming x = xh/||h|| ↓ Power constraint:

N

X

n=1

||x||2 ≤ NP

CMISO = log ✓ 1 + ||h||2P σ2 ◆

Frequency-­‑Selective ¡Channel

• Key idea 1: use OFDM to convert the channel with ISI

into a bunch of parallel AWGN channels

• But there is loss/overhead due to cyclic prefix
• Key idea 2: CP overhead → 0 as Nc → ∞
• First focus on finding the capacity of parallel AWGN

channels of any finite Nc

• Then take Nc → ∞ to find the capacity of the frequency-

selective channel

27

y[m] =

L−1

X

l=0

hlx[m − l] + w[m]

Recap: ¡OFDM

28 d[N–1] ˜ y0 x [N + L – 1] = d[N – 1] Cyclic prefix y [N + L – 1] ˜ dN–1 IDFT DFT Remove prefix ˜ yN–1 y[L] y[N + L – 1] y[1] y[L – 1] y[L] x [L – 1] = d[N – 1] x [L] = d[0] x [1] = d[N – L + 1] ˜ d0 d[0] Channel

y := y[L : Nc + L − 1], w := w[L : Nc + L − 1], h := ⇥h0 h1 · · · hL−1 · · · 0⇤T

e yn = e hn e dn + e wn, n = 0, 1, . . . , Nc − 1

e yn := DFT (y)n , e dn := DFT (d)n , e wn := DFT (w)n , e hn := p NcDFT (h)n

Nc parallel AWGN channels

Parallel ¡AWGN ¡Channels

29

e d0[m] e h0 e y0[m] e w0[m]

. . .

e d1[m] e h1 e y1[m] e w1[m] e dNc−1[m] e hNc−1 e wNc−1[m] e yNc−1[m]

Parallel Channels e yn = e hn e dn + e wn, n ∈ [0 : 1 : Nc − 1] m = 1, 2, . . . , M (M channel uses) Power Constraint

M

X

n=1

||e d[n]||2 ≤ MNcP

Due to Parseval theorem of DFT

Equivalent Vector Channel e y = e He d + e w e H = diag ⇣ e h0, . . . ,e hNc−1 ⌘ e w ∼ CN

• 0, σ2I

Independent ¡Uses ¡of ¡Parallel ¡Channels

• One way to code over such parallel channels (a special

case of a vector channel): treat each channel separately

• It turns out that coding across parallel channels does not help!
• Power allocation:
• Each of the Nc channels get a portion of the total power
• Channel n gets power Pn , which must satisfy
• For a given power allocation {Pn}, the following rate can

be achieved:

30

Nc−1

X

n=0

Pn ≤ NcP R =

Nc−1

X

n=0

log 1 + |e hn|2Pn σ2 !

Optimal ¡Power ¡Allocation

• Power allocation problem:
• It can be solved explicitly by Lagrangian methods
• Final solution: let (x)+ := max(x , 0)

31

max

P0,...,PNc−1 Nc−1

X

n=0

log 1 + |e hn|2Pn σ2 ! , subject to

Nc−1

X

n=0

Pn = NcP, Pn ≥ 0, n = 0, . . . , Nc − 1 P ∗

n =

ν − σ2 |e hn|2 !+ , ν satisfies

Nc−1

X

n=0

ν − σ2 |e hn|2 !+ = NcP

Water\illing

32

σ2 |e hn|2 ν

Note: e hn = Hb ✓nW Nc ◆ Baseband frequency response at f = nW/Nc

P1 = 0 Subcarrier n P2 P3 * * * 1 λ

Frequency-­‑Selective ¡Channel ¡Capacity

• Final step: making Nc → ∞
• Replace all
• by Hb(f), summation over [0 : Nc – 1]

becomes integration from 0 to W

• Power allocation problem becomes
• Optimal solution becomes

33

e hn = Hb ✓nW Nc ◆

max

P (f)

Z W log ✓ 1 + |H(f)|2P(f) σ2 ◆ d f, subject to Z W P(f) = P, P(f) ≥ 0, f ∈ [0, W] P(f)∗ = ✓ ν − σ2 |H(f)|2 ◆+ , ν satisfies Z W ✓ ν − σ2 |H(f)|2 ◆+ d f = P

Water\illing ¡over ¡the ¡Frequecy ¡Spectrum

34

P ( f ) Frequency ( f ) 0.4W 0.2W – 0.2W – 0.4W 4 3.5 3 2.5 2 1.5 1 0.5

|2

* λ

ν σ2 |H(f)|2

35

Capacity ¡of ¡Fading ¡ Channels

36

Flat ¡vs. ¡Frequency ¡Selective ¡Fading

• Frequency selectivity induces ISI
• Frequency selective underspread channel can be

converted to parallel flat fading channels

• We focus on flat fading channels (single tap):

x[m] h[m] w[m] y[m] Channel Input Channel Output Channel Gain Noise

y[m] = h[m]x[m] + w[m]

E ⇥ |h[m]|2⇤ = 1, ∀ m Power constraint:

N

X

n=1

|x[n]|2 ≤ NP

Slow ¡vs. ¡Fast ¡Fading

• Slow fading (quasi-static)
• h[m] = h for all m
• h is random, unknown to Tx
• h is known to the Rx (if not, it is the non-coherent setting)
• Fast fading
• h[m] = hl for all m within the l-th coherence time (Tc) period
• hl : i.i.d. over l, that is, i.i.d. over different coherence time periods
• hl is random, could be known or unknown to Tx
• hl is known to the Rx (if not, it is the non-coherent setting)

37

x[m] h[m] w[m] y[m]

Slow ¡Fading ¡Channel

• h[m] = h for all m and h is random
• Conditional on h, channel capacity = log(1+|h|2SNR)
• Tx does not know the channel h
• Suppose Tx send at rate R bits/s/Hz:
• If R < log(1+|h|2SNR), Shannon: “Pe → 0 as N → ∞”
• If R > log(1+|h|2SNR), Shannon: “Pe → 1 as N → ∞”
• Total Error probability:

38

P (N)

e

= Pr

• R < log
• 1 + |h|2SNR
• Pr
• E | R < log
• 1 + |h|2SNR
• + Pr
• R > log
• 1 + |h|2SNR
• Pr
• E | R > log
• 1 + |h|2SNR
• → Pr
• R < log
• 1 + |h|2SNR
• 0 + Pr
• R > log
• 1 + |h|2SNR
• 1

= Pr

• R > log
• 1 + |h|2SNR
• as N → ∞

Shannon ¡Capacity ¡= ¡0

• Error probability → Pr{R > log(1+|h|2SNR)} as N → ∞
• Vanishing error probability: impossible due to deep fade!
• According to Shannon’s definition, capacity = 0 …
• Outage probability pout(R) := Pr{R > C(h ; SNR)}
• Determines the reliability level of sending at a particular rate R
• C(h ; SNR) = log(1+|h|2SNR) for the point-to-point channel

39

Outage ¡Probability: ¡Computation

40

pout(R) := Pr

• R > log
• 1 + |h|2SNR
• = Pr
• |h|2 < (2R − 1)SNR−1

for h ∼ CN(0, 1): Rayleigh fading

0.15 0.2 0.25 0.3 0.35 0.4 0.45 1 2 3 4 5 0.05 0.1 R Area = pout (R)

= 1 − e

−(2R−1) SNR

Outage ¡Capacity

• Shannon capacity of a slow fading channel is 0
• Because we insist that “Pe → 0 as N → ∞”!
• More realistic capacity measure:
• Set a reliability level ϵ
• Find the maximum rate R* such that Pout(R) ≤ ϵ
• ϵ-outage capacity:
• It’s just the inverse function of outage probability! (why?)

41

C✏ := max {R | pout (R) ≤ ✏} C✏ = log

• 1 + F −1 (1 − ✏) SNR
• F (x) := Pr
• |h|2 > x

: complementary CDF of |h|2

Outage ¡Capacity: ¡Computation

42

⇐ ⇒ Pr

• 2C✏ − 1
• SNR−1 > |h|2

= ✏ ⇐ ⇒ 1 − F

• 2C✏ − 1
• SNR−1

= ✏ ⇐ ⇒ 2C✏ − 1 = F −1 (1 − ✏) SNR ⇐ ⇒ C✏ = log

• 1 + F −1 (1 − ✏) SNR
• pout (C✏) = ✏

⇐ ⇒ Pr

• C✏ > log
• 1 + |h|2SNR
• = ✏

For Rayleigh fading: h ∼ CN(0, 1) : y = F(x) = e−x ∈ [0, 1], F −1(y) = − ln y ≥ 0 F −1(1 ✏) = ln(1 ✏) ⇡ ✏, for ✏ ⌧ 1 C✏ ⇡ log (1 + ✏SNR) when ✏ ⌧ 1

Fade ¡Margin

• AWGN vs. ϵ-outage capacity:
• Under a reliability level ϵ, to achieve the same capacity

as the AWGN channel, in slow fading channel the Tx has to use an extra 10log10(1/F –1(1-ϵ)) dB of power

• Fade margin: the extra amount of Tx power to improve

the system when channel is in deep fade

43

C✏ = log

• 1 + F −1 (1 − ✏) SNR
• CAWGN = log (1 + SNR)

Typically less than 1 because ϵ ≪ 10% ⟹ Extra Tx power improve deep fade

Impact ¡of ¡Fading

• High SNR:
• Additive loss
• Smaller when SNR → ∞
• Low SNR:
• Multiplicative loss
• For Rayleigh,
• Significant loss when SNR is small
• Impact of fading is also significant at low SNR

44

C✏ ≈ log

• F −1(1 − ✏)SNR
• ≈ CAWGN − log

✓ 1 F −1(1 − ✏) ◆ C✏ ≈ F −1(1 − ✏)SNR log2 e ≈ F −1(1 − ✏)CAWGN F −1(1 ✏) = ln(1 ✏) ⇡ ✏, for ✏ ⌧ 1

AWGN ¡Capactiy ¡vs ¡Outage ¡Capacity

45

1 –10 –5 5 10 15 20 25 30 0.6 0.4 0.2 0.8

= 0.1 = 0.01

awgn

SNR (dB) 35 40

∋ ∋

C✏ CAWGN

Diversity ¡Order

• Recall:
• Error probability → Outage probability pout(R) as N → ∞
• Recall from Lecture 2:
• For uncoded transmission and some coding scheme, we see that

Error probability ~ SNR–1 for some fixed N

• Is this true for the optimal coding scheme and arbitrarily large N ?
• Outage probability at high SNR:
• Even for optimal coding scheme and large N, error prob. ~ SNR–1
• Optimal diversity order is 1

46

pout(R) = 1 e

−(2R−1) SNR

⇡ 2R 1 SNR when SNR 1

Optimal ¡Diversity ¡Order ¡and ¡pout(R)

• Hence we are able to define the optimal diversity order

for point-to-point slow fading channels:

• Note: in taking the limit, we assume that R is a constant
• This view will be modified in later lectures when we

discuss the diversity-multiplexing tradeoff

47

d := lim

SNR→∞

− log pout(R) log SNR where pout(R) := Pr {C (h; SNR) < R}

48

Receive ¡Diversity

• Outage probability
• Outage capacity

h x y y = hx + w ∈ CL

Power constraint: P w ∼ CN

• 0, σ2IL
• SNR := P

σ2

pout(R) := Pr

• log
• 1 + ||h||2SNR
• < R

= Pr ⇢ ||h||2 < 2R − 1 SNR

• ≈ (2R − 1)L

L! SNR−L for Rayleigh faded h’s

Lecture 2 Slide #25 ✏ = pout (C✏)

C✏ ≈ log ⇣ 1 + (L!)

1 L (✏) 1 L SNR

⌘

E ⇥ ||h||2⇤ = L

C (h; SNR) = log

• 1 + ||h||2SNR

SIMO ¡Outage ¡Capacity

49

5 10 15 20 25 30 35 40 –10 1 0.8 0.6 0.4 0.2 –5

awgn

L = 2 L = 4 L = 5 L = 3 L = 1 SNR (dB)

✏ = 1% C✏ CSIMO CSIMO = log

• 1 + ||h||2SNR
• = log (1 + LSNR)

Transmit ¡Diversity

• Tx beamforming is impossible since Tx does not know h
• For SIMO, Tx does not need to know h to achieve C(h ; SNR)
• How to find the optimal outage probability?
• For i.i.d. Rayleigh fading, it can be shown (cf. Appendix B.8 and

Exercise 5.15, 5.16) that the optimal outage probability

50

h x y y = h∗x + w ∈ C, x, h ∈ CL h∗ = ⇥h1 h2 ⇤

Power constraint:

N

X

n=1

||x||2 ≤ NP SNR := P σ2 E ⇥ ||h||2⇤ = L

y = x||h|| + w

↓ Tx Beamforming x = xh/||h|| ↓

pout(R) = Pr ⇢ log ✓ 1 + ||h||2 SNR L ◆ < R

Impact ¡of ¡CSIT: ¡Loss ¡in ¡Power ¡Gain

• Comparison of outage probability
• The same diversity order L
• SIMO has L-fold power gain over MISO
• Lack of channel state information at the transmitter

(CSIT) ⟹ Loss in power gain

51

pTx

• ut(R) = Pr

⇢ log ✓ 1 + ||h||2 SNR L ◆ < R

• pRx
• ut(R) = Pr
• log
• 1 + ||h||2SNR
• < R

Repetition ¡Coding

52

Time 1 Time 2 Equivalent Channel: h1 h2 h1 h2

u u

y1 y2

• =

h1 h2

• u +

w1 w2

• Projection

− − − − − − → e y = h∗ ||h||y = ||h||u + e w Supports rates up to 1 2 log

• 1 + ||h||2SNR
• Outage probability:

pRepetition

• ut

(R) = Pr ⇢1 2 log

• 1 + ||h||2SNR
• < R
• 2 blocks, but just 1 scalar channel

Alamouti ¡Scheme

53

Time 1 Time 2

u1 u2 −u∗

2

u∗

1

X = u1 −u∗

2

u2 u∗

1

• u1, u2 ∈ C

space-time codeword Equivalent Channel: y1 y∗

2

• =

h1 h2 h∗

2

−h∗

1

u1 u2

• +

w1 w2

• = u1

h1 h∗

2

• + u2

 h2 −h∗

1

• +

w1 w2

• h1

h2 h1 h2 Projection onto the two column vectors respectively, we can get two clean channels for u1 and u2! e h1 e h2 e h1 ⊥ e h2

Performance ¡of ¡Alamouti ¡Scheme

• Projection onto two orthogonal directions
• Power allocation: Lack of CSIT ⟹ No idea which

channel is better ⟹ Uniform power allocation (P/2 each)

• Each channel supports rates up to
• Outage probability: achieves optimal

54

y1 y∗

2

• = u1

h1 h∗

2

• + u2

 h2 −h∗

1

• +

w1 w2

• e

h1 ⊥ e h2 e y = u1e h1 + u2e h2 + e w Two parallel channels, each for one symbol!

e y1 := e h∗

1

||e h1|| e y = u1||e h1|| + e w1 = u1||h|| + e w1 e y2 := e h∗

2

||e h2|| e y = u2||e h2|| + e w2 = u2||h|| + e w2

log ✓ 1 + ||h||2 SNR 2 ◆ pAlamouti

• ut

(R) = Pr ⇢ log ✓ 1 + ||h||2 SNR 2 ◆ < R

Repetition ¡Coding ¡vs. ¡Alamouti

55

Repetition Alamouti C(h;SNR) pout(R) when SNR ≫ 1 Cϵ for ϵ ≪ 1 Diversity Order 2 2

1 2 log

• 1 + ||h||2SNR
• log

✓ 1 + ||h||2 SNR 2 ◆ (22R − 1)2 2 SNR−2 1 2 log ⇣ 1 + √ 2✏SNR ⌘ 2(2R − 1)2SNR−2 log ✓ 1 + r ✏ 2SNR ◆

≤ ≥ ≤ =

Time ¡and ¡Frequency ¡Diversity

• Recall from Lecture 2:
• Time diversity is obtained by coding + interleaving across multiple

(L) coherence time

• Frequency diversity is obtained by coding + hopping across

multiple (L) coherence bandwidth

• Hence, time and frequency diversity techniques are

equivalent to coding over L parallel channels:

• Channel l has power constraint Pl , P1+P2+…+PL ≤ LP,

where P is the power constraint of the original channel

• No CSIT ⟹ cannot do water-filling

56

yl[m] = hlxl[m] + wl[m], l = 1, 2, . . . , L

Outage ¡Probability

• Instead, first use uniform power allocation Pl = P, ∀l
• Given h and SNR, L parallel channels can support up to
• Subtlety: due to lack of CSIT, coding across parallel

channel is necessary

• Because Tx does not know for each of the L channels, how high

the rate should be!

• Outage probability:
• For i.i.d. Rayleigh fading, it can be shown (cf. Exercise 5.17)

that uniform power allocation is optimal!

57

pout(R) = Pr ( L X

l=1

log

• 1 + |hl|2SNR
• < LR

)

L

X

l=1

log

• 1 + |hl|2SNR
• bits/s/Hz