Lecture 2 Point-to-Point Communications 1 I-Hsiang Wang - - PowerPoint PPT Presentation

Lecture ¡2 Point-­‑to-­‑Point ¡Communications ¡1

I-Hsiang Wang ihwang@ntu.edu.tw 2/27, 2014

Wire ¡vs. ¡Wireless ¡Communication

2

y[m] = X

l

hlx[m − l] + w[m] y[m] = X

l

hl[m]x[m − l] + w[m]

Wired Channel Wireless Channel

• Deterministic channel gains
• Main issue: combat noise
• Key technique: coding to

exploit degrees of freedom and increase data rate (coding gain)

• Random channel gains
• Main issue: combat fading
• Key technique: coding to

exploit diversity and increase reliability (diversity gain)

• Remark: In wireless channel, there is still additive noise,

and hence the techniques developed in wire communication are still useful.

Plot

• Study detection in flat fading channel to learn
• Communication over flat fading channel has poor performance

due to significant probability that the channel is in deep fade

• How the performance scale with SNR
• Investigate various techniques to provide diversity across
• Time
• Frequency
• Space
• Key: how to exploit additional diversity efficiently

3

Outline

• Detection in Rayleigh fading channel vs. static AWGN

channel

• Code design and degrees of freedom
• Time diversity
• Antenna (space) diversity
• Frequency diversity

4

Detection ¡in ¡Rayleigh ¡ Fading ¡Channel

Baseline: ¡AWGN ¡Channel

6

y = x + w, w ∼ CN

• 0, σ2

BPSK: x = ±a

Transmitted constellation is real, it suffices to consider the real part:

ML rule: Probability of error:

Pr {E} = Pr ⇢ <{w} > a (a) 2

• = Q

a p σ2/2 ! = Q ⇣p 2SNR ⌘ b x = ( a, if |<{y} a| < |<{y} (a)| a,

• therwise

SNR := average received signal energy per (complex) symbol time noise energy per (complex) symbol time a2 σ2 <{y} = x + <{w}, <{w} ⇠ N

• 0, σ2/2

Gaussian ¡Scalar ¡Detection

7

y If y < (uA + uB) / 2 choose uA If y > (uA + uB) / 2 choose uB uA

2

uB (uA+uB)

{y | x = uA} {y | x = uB}

• Sufficient statistic for detection:

projection on to

• Since w is circular symmetric,

Gaussian ¡Vector ¡Detection

8

y ˜ y

uA uB UA UB y2 y1

v := uA − uB ||uA − uB|| e y := v∗ ✓ y − uA + uB 2 ◆ = e x + e w e x := v∗ ✓ x − uA + uB 2 ◆ = ( ||uA−uB||

2

, x = uA − ||uA−uB||

2

, x = uB

y = x + w, w ∼ CN

• 0, σ2I
• =

⇒ e w ∼ CN

• 0, σ2

Binary ¡Detection ¡in ¡Gaussian ¡Noise

9

Binary signaling: It suffices to consider the projection onto Probability of error: y = x + w, w ∼ CN

• 0, σ2I
• x = uA, uB

(uA − uB) Pr ⇢ <{w} > ||uA uB|| 2

• = Q

||uA uB|| 2 p σ2/2 ! e y = x||uA − uB|| + e w, x = ±1 2, e w ∼ CN

• 0, σ2

Rayleigh ¡Fading ¡Channel

• Note: |h| is an exponential random variable with mean 1
• Fair comparison with the AWGN case (same avg. signal power)
• Coherent detection:
• The receiver knows h perfectly (channel estimation through pilots)
• For a given realization of h, the error probability is
• Probability of error:

10

y = hx + w, h ∼ CN (0, 1) , w ∼ CN

• 0, σ2

Pr {E | h} = Q a|h| p σ2/2 ! = Q ⇣p 2|h|2SNR ⌘

Check!

Hint: exchange the order in the double integral

Pr {E} = E h Q ⇣p 2|h|2SNR ⌘i = 1 2 1 − r SNR 1 + SNR !

BPSK: x = ±a

SNR = E ⇥ |h|2⇤ a2 σ2 = a2 σ2

Non-­‑coherent ¡Detection

• If Rx does not know the realization of h:
• Scalar BPSK (

) completely fails

• Because the phase of h is uniform over [0, 2π]
• Orthogonal modulation:
• Use two time slots m = 0,1
• Modulation:

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y = hx + w, h ∼ CN (0, 1) , w ∼ CN

• 0, σ2

x = ±a xA = a

• r

xB = 0 a

• m = 1

m = 0 y xB |y[1]| |y[0]| xA

= ⇒ y := y[0] y[1]

• = h

x[0] x[1]

• +

w[0] w[1]

• := hx + w

Non-­‑coherent ¡Detection

• ML rule:
• Given
• LLR:
• Energy detector:

12

Orthogonal modulation: xA =

a

• r

xB = 0 a

• y = hx + w,

h ∼ CN (0, 1) , w ∼ CN

• 0, σ2I2
• x = xA =

⇒ y ∼ CN ✓ 0, a2 + σ2 σ2 ◆ x = xB = ⇒ y ∼ CN ✓ 0, σ2 a2 + σ2 ◆ Λ(y) := ln f(y | xA) f(y | xB) = a2 (a2 + σ2)σ2

• |y[0]|2 − |y[1]|2
• σ2 + a2

|y[0]|2 + σ2|y(1)|2 −

• σ2|y(0)|2 +
• σ2 + a2

|y[0]|2 (a2 + σ2)σ2

b x = xA ⇐ ⇒ |y[0]| > |y[1]| b x = xB ⇐ ⇒ |y[0]| < |y[1]| SNR = a2 2σ2

Non-­‑coherent ¡Detection

• Probability of error:
• Given
• Hence

13

Orthogonal modulation: xA =

a

• r

xB = 0 a

• y = hx + w,

h ∼ CN (0, 1) , w ∼ CN

• 0, σ2I2
• x = xA =

⇒ y ∼ CN ✓ 0, a2 + σ2 σ2 ◆ = ⇒ |y[0]|2 ∼ Exp

• (a2 + σ2)−1

, |y[1]|2 ∼ Exp

• (σ2)−1

|y[0]|2 and |y[1]|2 are independent

Check!

Pr {E} = Pr

• Exp
• (σ2)−1

> Exp

• (a2 + σ2)−1

= (a2 + σ2)−1 (σ2)−1 + (a2 + σ2)−1 = 1 2 + a2/σ2 = 1 2(1 + SNR) SNR = a2 2σ2

Comparison: ¡AWGN ¡vs. ¡Rayleigh

• AWGN: Error probability decays faster than e-SNR
• Rayleigh fading: Error probability decays as SNR-1
• Coherent detection:
• Non-coherent detection:

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Pr {E} = Q ⇣√ 2SNR ⌘ ≈ 1 √ 2SNR √ 2π e−SNR at high SNR

Q (x) := Pr {N(0, 1) > a} ⇡ 1 x p 2π e−x2/2 when x 1 r x 1 + x = ✓ 1 1 1 + x ◆1/2 ⇡ 1 1 2(1 + x) ⇡ 1 1 2x when x 1

Pr {E} = 1 2 1 − r SNR 1 + SNR ! ≈ (4SNR)−1 at high SNR Pr {E} = 1 2(1 + SNR) ≈ (2SNR)−1 at high SNR

Comparison: ¡AWGN ¡vs. ¡Rayleigh

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10 20 30 40 Non-coherent

• rthogonal

Coherent BPSK BPSK over AWGN

SNR (dB)

10–8 –10 –20 1 10–2 10–4 10–6 10–10 10–12 10–14 10–16

Pr {E}

15 dB 3 dB

Coherent ¡Detection ¡under ¡QPSK

• BPSK only makes use of the real dimension (I channel)
• Rate can be increased if an additional bit is sent on the

imaginary dimension (Q channel)

• QPSK:
• (Bit) Probability of error
• Simply a product of two BPSK
• Analysis is the same
• Simply replace SNR by SNR/2

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b –b b –b QPSK Im Re

Pr {E}AWGN = Q ⇣√ SNR ⌘ Pr {E}Rayleigh = 1 2 1 − r SNR 2 + SNR !

Double of BPSK! SNR = 2b2 σ2 x ∈ {b(1 + j), b(1 − j), b(−1 + j), b(−1 − j)}

≈ (2SNR)−1 at high SNR

Typical ¡Error ¡Event: ¡Deep ¡Fade

• In Rayleigh fading channel, regardless of constellation

size and detection method (coherent/non-coherent),

• For BPSK,
• If
• If
• Hence,

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Pr {E} ∼ 1 SNR Pr {E | h} = Q ⇣p 2|h|2SNR ⌘

probability of deep fade |h|2 SNR−1 = ) the conditional probability is very small |h|2 < SNR−1 = ⇒ the conditional probability is very large / Pr

• |h|2 < SNR−1

= 1 − eSNR−1 ≈ SNR−1 Pr {E} = Pr

• |h|2 > SNR−1

Pr

• E | |h|2 > SNR−1

+ Pr

• |h|2 < SNR−1

Pr

• E | |h|2 < SNR−1