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 Wireless Channel Wired Channel X X y [ m ] = h l x [ m − l ] + w [ m ] y [ m ] = h l [ m ] x [ m − l ] + w [ m ] l l • Deterministic channel gains • Random channel gains • Main issue: combat noise • Main issue: combat fading • Key technique: coding to • Key technique: coding to exploit degrees of freedom exploit diversity and increase and increase data rate reliability (coding gain) (diversity gain) • Remark : In wireless channel, there is still additive noise, and hence the techniques developed in wire communication are still useful. 2

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 0 , σ 2 � � w ∼ CN y = x + w, BPSK: x = ± a Transmitted constellation is real, it suffices to consider the real part: 0 , σ 2 / 2 � � < { y } = x + < { w } , < { w } ⇠ N ( if | < { y } � a | < | < { y } � ( � a ) | a, ML rule: x = b otherwise � a, Probability of error: ! ⇣ p ⇢ < { w } > a � ( � a ) � a ⌘ Pr {E} = Pr = Q = Q 2 SNR 2 p σ 2 / 2 a 2 SNR := average received signal energy per (complex) symbol time noise energy per (complex) symbol time σ 2 6

Gaussian ¡Scalar ¡Detection � { y | x = u A } � { y | x = u B } If y < ( u A + u B ) / 2 If y > ( u A + u B ) / 2 choose u A choose u B y ( u A +u B ) u A u B 2 7

Gaussian ¡Vector ¡Detection 0 , σ 2 I � � w ∼ CN y = x + w , • Sufficient statistic for detection: y 2 u A − u B projection on to v := || u A − u B || y ✓ ◆ y − u A + u B y := v ∗ e = e x + e w 2 u B ✓ ◆ x − u A + u B x := v ∗ e y 1 2 ( || u A − u B || y ˜ x = u A , 2 u A = − || u A − u B || U B x = u B , 2 • Since w is circular symmetric, U A � 0 , σ 2 � w ∼ CN ⇒ e = 8

Binary ¡Detection ¡in ¡Gaussian ¡Noise 0 , σ 2 I � � w ∼ CN y = x + w , Binary signaling: x = u A , u B ( u A − u B ) It suffices to consider the projection onto � 0 , σ 2 � x = ± 1 y = x || u A − u B || + e w ∼ CN e e w, 2 , Probability of error: ! ⇢ � < { w } > || u A � u B || || u A � u B || = Q Pr p 2 σ 2 / 2 2 9

Rayleigh ¡Fading ¡Channel 0 , σ 2 � � y = hx + w, h ∼ CN (0 , 1) , w ∼ CN ⇥ | h | 2 ⇤ a 2 = a 2 SNR = E BPSK: x = ± a σ 2 σ 2 • 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 ! a | h | ⇣p ⌘ 2 | h | 2 SNR Pr {E | h } = Q = Q p σ 2 / 2 • Probability of error: Check! ! r = 1 SNR h ⇣p ⌘i Hint: exchange Pr {E} = E 2 | h | 2 SNR Q 1 − the order in the 1 + SNR 2 double integral 10

Non-­‑coherent ¡Detection 0 , σ 2 � � y = hx + w, h ∼ CN (0 , 1) , w ∼ CN • If Rx does not know the realization of h : - Scalar BPSK ( � � ) completely fails x = ± a - Because the phase of h is uniform over [0, 2 π ] • Orthogonal modulation: m = 1 - Use two time slots m = 0,1 x B  y [0] �  x [0] �  w [0] � = ⇒ y := = h + y [1] x [1] w [1] y | y [1] | x A := h x + w m = 0 - | y [0] | Modulation:  a �  0 � x A = or x B = 0 a 11

Non-­‑coherent ¡Detection 0 , σ 2 I 2 � � y = h x + w , h ∼ CN (0 , 1) , w ∼ CN  a �  0 � SNR = a 2 Orthogonal modulation: x A = or x B = 0 2 σ 2 a • ML rule:  a 2 + σ 2 ✓ �◆ 0 - ⇒ y ∼ CN Given x = x A = 0 , σ 2 0 ✓  σ 2 �◆ 0 ⇒ y ∼ CN x = x B = 0 , a 2 + σ 2 0 - LLR: a 2 Λ ( y ) := ln f ( y | x A ) | y [0] | 2 − | y [1] | 2 � � f ( y | x B ) = ( a 2 + σ 2 ) σ 2 σ 2 + a 2 � | y [0] | 2 + σ 2 | y (1) | 2 σ 2 | y (0) | 2 + σ 2 + a 2 � �� � � | y [0] | 2 − ( a 2 + σ 2 ) σ 2 • Energy detector: x = x A ⇐ b ⇒ | y [0] | > | y [1] | x = x B ⇐ b ⇒ | y [0] | < | y [1] | 12

Non-­‑coherent ¡Detection 0 , σ 2 I 2 � � y = h x + w , h ∼ CN (0 , 1) , w ∼ CN  a �  0 � SNR = a 2 Orthogonal modulation: x A = or x B = 0 2 σ 2 a • Probability of error:  a 2 + σ 2 ✓ �◆ 0 - ⇒ y ∼ CN Given x = x A = 0 , σ 2 0 ( a 2 + σ 2 ) − 1 � ⇒ | y [0] | 2 ∼ Exp , | y [1] | 2 ∼ Exp ( σ 2 ) − 1 � � � = | y [0] | 2 and | y [1] | 2 are independent - ( a 2 + σ 2 ) − 1 � Hence ( σ 2 ) − 1 � � � � Pr {E} = Pr Exp > Exp ( a 2 + σ 2 ) − 1 1 Check! = ( σ 2 ) − 1 + ( a 2 + σ 2 ) − 1 = 2 + a 2 / σ 2 1 = 2(1 + SNR ) 13

Comparison: ¡AWGN ¡vs. ¡Rayleigh • AWGN: Error probability decays faster than e -SNR ⇣ √ 1 ⌘ e − SNR Pr {E} = Q 2 SNR at high SNR √ ≈ √ 2 SNR 2 π 1 2 π e − x 2 / 2 Q ( x ) := Pr {N (0 , 1) > a } ⇡ when x � 1 p x • Rayleigh fading: Error probability decays as SNR -1 - Coherent detection: ! r Pr {E} = 1 SNR ≈ (4 SNR ) − 1 1 − at high SNR 2 1 + SNR ◆ 1 / 2 r ✓ 1 2(1 + x ) ⇡ 1 � 1 1 x 1 + x = 1 � ⇡ 1 � when x � 1 1 + x 2 x - Non-coherent detection: 1 2(1 + SNR ) ≈ (2 SNR ) − 1 Pr {E} = at high SNR 14

Comparison: ¡AWGN ¡vs. ¡Rayleigh 1 3 dB 10 –2 10 –4 15 dB 10– 6 Pr {E} 10 –8 BPSK over AWGN 10 –10 Non-coherent orthogonal 10 –12 Coherent BPSK 10 –14 10 –16 –20 –10 0 10 20 30 40 SNR (dB) 15

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 • QPSK: Im x ∈ { b (1 + j ) , b (1 − j ) , b ( − 1 + j ) , b ( − 1 − j ) } SNR = 2 b 2 Double of BPSK! σ 2 b • (Bit) Probability of error Re - – b b Simply a product of two BPSK – b - Analysis is the same - Simply replace SNR by SNR/2 ! r SNR Pr {E} Rayleigh = 1 ⇣ √ ⌘ Pr {E} AWGN = Q SNR 1 − 2 + SNR 2 ≈ (2 SNR ) − 1 at high SNR 16

Degrees ¡of ¡Freedom QPSK 4-PAM QPSK Im Im r ! 2 b 2 symbol error probability ≈ 2 Q σ 2 b Re 4-PAM Re – b b –3 b – b b 3 b r ! 2 b 2 symbol error probability = 3 – b 2 Q σ 2 • A complex scalar channel has 2 degrees of freedom - BPSK only uses 1 but QPSK uses 2 ⟹ QPSK rate is doubled • QPSK is 2.5 × more energy efficient than 4-PAM - QPSK avg. Tx energy = 2 b 2 - 4-PAM avg. Tx energy = 5 b 2 17

Typical ¡Error ¡Event: ¡Deep ¡Fade • In Rayleigh fading channel, regardless of constellation size and detection method (coherent/non-coherent), 1 Pr {E} ∼ SNR ⇣p ⌘ • For BPSK, Pr {E | h } = Q 2 | h | 2 SNR - | h | 2 � SNR − 1 = If the conditional probability is very small ) - | h | 2 < SNR − 1 = If the conditional probability is very large ⇒ - | h | 2 > SNR − 1 E | | h | 2 > SNR − 1 � � Hence, Pr {E} = Pr Pr | h | 2 < SNR − 1 E | | h | 2 < SNR − 1 � � + Pr Pr probability of deep fade | h | 2 < SNR − 1 = 1 − e SNR − 1 � / Pr ≈ SNR − 1 18

Diversity 0 , σ 2 � � y = hx + w, h ∼ CN (0 , 1) , w ∼ CN | h | 2 < SNR − 1 � Deep Fade Event: • Reception only relies on a single “path” h • If h is in deep fade ⟹ trouble (low reliability) • Increase the number of “paths” ⟺ Increase diversity - If one path is in deep fade, other paths can compensate! • Diversity over time, space, and frequency 19

Outlook • Time diversity - Coding + Interleaving: obtain diversity over time - Repetition coding - Rotation coding: utilize degrees of freedom better • Space (Antenna) diversity - Receive diversity: multiple Rx antennas - Transmit diversity: multiple Tx antennas - Space-time codes • Frequency diversity - ISI mitigation - Time-domain equalization - Direct-sequence spread spectrum - OFDM 20

Time ¡Diversity

Repetition ¡Coding ¡+ ¡Interleaving • A simple idea: Repetition Coding - Repeat the symbol over L time slots (note: L is NOT the # of taps) ⇥ b b ⇤ Info. Symbol b → ENC → Codeword x := b · · · y l = h l x l + w l , l = 1 , 2 , . . . , L - As long as the channels { h l | l = 1, 2, ... L } are not ALL in deep fade, there is a good probability that we can decode the symbol • Interleaving: - Channels within coherence time are highly correlated - Diversity is obtained if we interleave the codeword across multiple coherence time periods 22

Coding ¡+ ¡Interleaving ¡Increases ¡Diverirsit | h l | L = 4 l All are bad h 1 h 2 h 3 h 4 No interleaving Codeword Codeword Codeword Codeword x 0 x 1 x 2 x 3 Only one is bad h 1 h 2 h 3 h 4 Interleaving 23