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: Cauchy-Schwarz inequality)
• 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,
• h is unknown to Tx (if known to Tx ⟹ same as AWGN channel!)
• 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

pdf of log(1+|h|2SNR) log(1+|h|2SNR)

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 at high SNR

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!

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pout(R) = Pr ( L X

l=1

log

• 1 + |hl|2SNR
• < LR

)

L

X

l=1

log

• 1 + |hl|2SNR
• bits/s/Hz

Fast ¡Fading ¡Channel ¡without ¡CSIT

• Block fading model:
• 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

• If 1 ≪ Tc ≪ N (comm. time)
• L parallel channels; L → ∞
• For finite L, capacity = 0 because
• is random
• But as L → ∞, C(h;SNR) →

58

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

l = 0 h[m] l = 1 l = 2 l = 3 m 

E ⇥ log

• 1 + |h|2SNR

⇤

C (h; SNR) := 1 L

L

X

l=1

log

• 1 + |hl|2SNR

Ergodic ¡Capacity

• What if Tc ≫ 1 does not hold?
• In particular, y[m] = h[m]x[m] + w[m], {h[m] | m ∈ [1:N]}: i.i.d.
• It turns out that the capacity of such channel is
• By LLN, we get
• In fact, for any fading process is stationary and ergodic,

the capacity of the fading channel is given by the above

• Stationary and ergodic ⟺ the long-term average (over time)

converges to the expectation (under the stationary distribution)

• Note: for slow fading channel, the fading process is not ergodic

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C = lim

N→∞

1 N

N

X

m=1

log

• 1 + |h[m]|2SNR
• C = E

⇥ log

• 1 + |h|2SNR

⇤

Ergodic ¡Capacity ¡vs. ¡AWGN ¡Capacity

60

CAWGN = log (1 + SNR) , CCSIR = E ⇥ log

• 1 + |h|2SNR

⇤

• In the fast fading channel, if channel state information is

available only at Rx, then CCSIR ≤ CAWGN

• Due to Jensen’s inequality and the concavity of log():
• Even if Tx can code over multiple coherence time

periods and average out the fluctuation of the fading channel, the capacity gets hit by fading

• With CSIT, the situation is changed

E ⇥ log

• 1 + |h|2SNR

⇤ ≤ log

• 1 + E

⇥ |h|2⇤ SNR

• = log (1 + SNR)

Transmitter ¡State ¡Information

• So far we assume that Tx does not know the realization
• f the fading coefficients (channel state information)
• Such assumption might be too conservative
• How to obtain CSI at Tx (CSIT)?
• TDD system: channel reciprocity
• FDD system: feedback from Rx
• How to use CSIT?
• Slow fading channel: Channel Inversion
• Control Tx power according to the current channel condition so that a fixed

data rate can be supported

• May not be feasible because many systems are also peak power constrained
• Fast fading channel: Water Filling

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Fast ¡Fading ¡Channel ¡with ¡CSIT ¡(1)

• Idea: with CSIT, Tx should allocate more power when the

channel is in good state! (Power ↑, Rate ↑)

• L parallel AWGN channels
• Power allocation problem
• Solution:
• The value of ν depends on all L channels {h1, … , hL}
• Not feasible because it requires the Tx to know the future channel

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max

P1,...,PL

1 L

L

X

l=1

log ✓ 1 + |hl|2Pl σ2 ◆ , subject to 1 L

L

X

l=1

Pl = P, Pl ≥ 0, l = 1, . . . , L P ∗

l =

✓ ν − σ2 |hl|2 ◆+ , ν satisfies 1 L

L

X

l=1

✓ ν − σ2 |hl|2 ◆+ = P

Fast ¡Fading ¡Channel ¡with ¡CSIT ¡(2)

• Final step: taking L → ∞
• Due to LLN, optimal solution becomes:
• Now the value of ν only depends on the distribution of h
• Ergodic capacity of fast fading channel with CSIT:

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C = E  log ✓ 1 + |h|2 P ∗(h) σ2 ◆

P ∗(h) = ✓ ν − σ2 |h|2 ◆+ , ν satisfies E "✓ ν − σ2 |h|2 ◆+# = P

Opportunistic ¡Transmission

• Fading creates peaks and nulls
• With CSIT, Tx is able to “ride the peaks”
• When channel is good, use a code with higher rate
• When channel is bad, use a code with lower rate
• Variable-rate (Full CSI) vs. Fixed-rate (CSIR)

64

Fixed-rate scheme Variable-rate scheme

1 5 10 h[m]2 Time m

Code over multiple channel states No need to code over multiple channel states; Simplify code design

Power ¡Allocation ¡at ¡High ¡SNR

65

σ2 |h[m]|2 m ν σ2 |h[m]|2 m

Optimal Uniform: Near-Optimal

Power ¡Allocation ¡at ¡Low ¡SNR

66

Optimal Best Only: Near-Optimal

σ2 |h[m]|2 m ν σ2 |h[m]|2 m

Performance ¡Comparison

67

CFull CSI = E  log ✓ 1 + |h|2 P ∗(h) σ2 ◆

P ∗(h) = ✓ ν − σ2 |h|2 ◆+ , ν satisfies E "✓ ν − σ2 |h|2 ◆+# = P

CAWGN = log ✓ 1 + P σ2 ◆ , CCSIR = E  log ✓ 1 + |h|2 P σ2 ◆ , E ⇥ |h|2⇤ = 1

–5 5 10 15 SNR (dB) 20

AWGN CSIR Full CSI

C (bits /s / Hz) –10 –15 –20 7 6 5 4 3 2 1

CFull CSI > CAWGN ≅ CCSIR CAWGN > CFull CSI ≅ CCSIR

Bene]it ¡of ¡Dynamic ¡Power ¡Allocation

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–10 –5 5 10 0.5 –15 –20 3 2.5 2 1.5 1

awgn

SNR (dB)

CSIR Full CSI

C CAWGN

Power gain due to dynamic power allocation For fading that is unbounded (like Rayleigh), such power gain is unbounded, but has to operate at very low SNR to see such gain

Performance ¡at ¡Low ¡and ¡High ¡SNR

• High SNR regime:
• A near-optimal power allocation is the uniform one
• Hence,
• Low SNR regime:
• A near-optimal power allocation is to uniformly allocate power

when |h|2 ≥ G for some large G

• Hence,

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CFull CSI ≈ CCSIR ≈ E  log ✓ |h|2 P σ2 ◆ = log SNR + E ⇥ log(|h|2) ⇤ ≈ CAWGN − 0.83 CCSIR ≈ E ⇥ |h|2SNR log2 e ⇤ = E ⇥ |h|2⇤ SNR log2 e = SNR log2 e ≈ CAWGN CFull CSI & Pr

• |h|2 > G

log 1 + GP/ Pr

• |h|2 > G

σ2 ! ≈ GSNR log2 e ≈ GCAWGN

Water ¡Filling ¡vs. ¡Channel ¡Inversion

• Water filling:
• Dynamic allocate the Tx power – riding peaks
• Power ↑ when channel ↑
• Large delay in getting the guaranteed performance
• Benefit: power is efficiently utilized
• Suitable for data application
• Channel inversion:
• Tight power control such that a fixed rate can be maintained
• Power ↑ when channel ↓
• Large power consumption to invert bad channel
• Benefit: delay is independent of the coherence time
• Suitable for voice application
• Which to use? Depends on the delay requirement

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Summary

• A slow fading channel is a source of unreliability
• Poor outage capacity
• Diversity is needed
• A fast fading channel with only CSIR:
• Ergodic capacity close to AWGN capacity
• Decoding delay is long compared to coherence time
• A fast fading channel with full CSI:
• Ergodic capacity can be greater than AWGN capacity
• Turn fading into a friend rather than an enemy
• Opportunistic communication
• Send more when the channel is good
• Even more powerful in multiuser situations

71