Fading Channels: Capacity, BER and Diversity

Home , Rayleigh fading

Master Universitario en Ingenier´ıade Telecomunicaci´on

I. Santamar´ıa Universidad de Cantabria Introduction Capacity BER Diversity Conclusions

Contents

Introduction

Capacity

BER

Diversity

Conclusions

Fading Channels: Capacity, BER and Diversity 0/48 Introduction Capacity BER Diversity Conclusions

Introduction

I We have seen that the randomness of signal attenuation (fading) is the main challenge of wireless communication systems I In this lecture, we will discuss how fading aﬀects 1. The capacity of the channel 2. The Bit Error Rate (BER)

I We will also study how this channel randomness can be used or exploited to improve performance diversity →

Fading Channels: Capacity, BER and Diversity 1/48 Introduction Capacity BER Diversity Conclusions

General communication system model

sˆ[n] bˆ[n] b[n] s[n] Channel Channel Source Modulator Channel ⊕ Demod. Encoder Decoder

Noise

I The channel encoder (FEC, convolutional, turbo, LDPC ...) adds redundancy to protect the source against errors introduced by the channel

I The capacity depends on the fading model of the channel (constant channel, ergodic/block fading), as well as on the channel state information (CSI) available at the Tx/Rx Let us start reviewing the Additive White Gaussian Noise (AWGN) channel: no fading

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AWGN Channel

I Let us consider a discrete-time AWGN channel

y[n] = hs[n] + r[n]

where r[n] is the additive white Gaussian noise, s[n] is the transmitted signal and h = h ejθ is the complex channel | | I The channel is assumed constant during the reception of the whole transmitted sequence

I The channel is known to the Rx (coherent detector)

I The noise is white and Gaussian with power spectral density N0/2 (with units W/Hz or dBm/Hz, for instance) I We will mainly consider passband modulations (BPSK, QPSK, M-PSK, M-QAM), for which if the bandwith of the lowpass signal is W the bandwidth of the passband signal is 2W

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Signal-to-Noise-Ratio

I Transmitted signal power: P 2 I Received signal power: P h | | I Total noise power: N = 2WN0/2 = WN0 I The received SNR is P h 2 SNR = γ = | | WN0

I In terms of the energy per symbol P = Es /Ts I We assume Nyquist pulses with β = 1 (roll-oﬀ factor) so W = 1/Ts I Under these conditions

2 2 Es Ts h Es h SNR = | | = | | Ts N0 N0

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I For M-ary modulations, the energy per bit is

2 Es Eb h Eb = SNR = | | log(M) ⇒ N0 log(M)

I To model the noise we generate zero-mean circular complex 2 Gaussian random variables with σ = N0

2 r[n] CN(0, σ ) or r[n] CN(0, N0) ∼ ∼

I The real and imaginary parts have variance

σ2 N σ2 = σ2 = = 0 I Q 2 2

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Average SNR and Instantaneous SNR Sometimes, we will ﬁnd useful to distinguish between the average and the instantaneous signal-to-noise ratio

P Average SNR =γ ¯ = ⇒ WN0 P Instantaneous SNR =γ ¯ h 2 = h 2 ⇒ | | WN0 | |

I AWGN channel: N0 and h are assumed to be known (coherent detection), therefore we typically take (w.l.o.g) h = 1 SNR =γ ¯ = P , and the SNR at the receiver is WN0 constant⇒

I Fading channels: The instantaneous signal-to-noise-ratio SNR = γ =γ ¯ h 2 is a random variable | |

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Capacity AWGN Channel

I Let us consider a discrete-time AWGN channel y[n] = hs[n] + r[n] where r[n] is the additive white Gaussian noise, s[n] is the transmitted symbol and h = h ejθ is the complex channel | | I The channel is assumed constant during the reception of the whole transmitted sequence I The channel is known to the Rx (coherent detector) The capacity (in bits/seg or bps) is given by the well-known Shannon’s formula

C = W log (1 + SNR) = W log (1 + γ)

2 where W is the channel bandwidth, and SNR = P|h| ; with P WN0 2 being the transmit power, h the power channel gain and N0/2 the power spectral density| (PSD)| of the noise

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I Sometimes, we will ﬁnd useful to express C in bps/Hz (or bits/channel use)

C = log (1 + SNR)

A few things to recall 1. Shannon’s coding theorem proves that a code exists that achieves data rates arbitrarily close to capacity with vanishingly small probability of bit error

I The codewords might be very long (delay) I Practical (delay-constrained) codes only approach capacity 2. Shannon’s coding theorem assumes Gaussian codewords, but digital communication systems use discrete modulations (PSK,16-QAM)

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10 Shaping gain (1.53 dB)

8 Gaussian inputs log(1+SNR) 64-QAM 6

16-QAM 4

Capacity (bps/Hz) Capacity 4-QAM (QPSK) 2

Pe (symbol) = 10-6 (uncoded) 0 5 10 15 20 25 30 SNR dB For discrete modulations, with increasing SNR we should increase the constellation size M (or use a less powerful channel encoder) to increase the rate and hence approach capacity: Adaptive modulation and coding (more on this later)

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Capacity fading channels 1 I Let us consider the following example Fading channel

AWGN C = 1 bps/Hz Channel Random Encoder Switch AWGN C = 2 bps/Hz

I The channel encoder is ﬁxed I The switch takes both positions with equal probability I We wish to transmit a codeword formed by a long sequence of coded bits, which are then mapped to symbols I What is the capacity for this channel? 1Taken from E. Biglieri, Coding for Wireless Communications, Springer, 2005 Fading Channels: Capacity, BER and Diversity 10/48 Introduction Capacity BER Diversity Conclusions

The answer depends on the rate of change of the fading 1. If the switch changes position every symbol period, the codeword experiences both channels with equal probabilities so we could use a ﬁxed channel encoder, and transmit at a maximum rate 1 1 C = C + C = 1.5 bps/Hz 2 1 2 2

2. If the switch remains ﬁxed at the same (unknown) position during the transmission of the whole codeword, then we should use a ﬁxed channel encoder, and transmit at a maximum rate

C = C1 = 1 bps/Hz

or otherwise half of the codewords would be lost

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What would be the capacity if the switch chooses from a 2 continuum of AWGN channels whose SNR, γ = P|h| , 0 γ < , WN0 follows an exponential distribution (Rayleigh channel) ≤ ∞ 

C1 = log(1+ γ 1)

Channel Encoder C2 = log(1+ γ 2 ) Random Switch

C3 = log(1+ γ 3 ) 

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Fast fading (ergodic) channel

I If the switch changes position every symbol period, and the codeword is long enough so that the transmitted symbols experience all states of the channel (fast fading or ergodic channel) Z ∞ C = E [log(1 + γ)] = log(1 + γ)f (γ)dγ (1) 0

1 −γ/γ where f (γ) = γ e , with γ being the average SNR I Eq. (1) is the ergodic capacity of the fading channel I For Rayleigh channels the integral in (1) is given by 1 1 1 C = exp E 2 ln(2) γ 1 γ

R ∞ e−t 2 where E1(x) = x t dt is the Exponential integral function 2y=expint(x) in Matlab Fading Channels: Capacity, BER and Diversity 13/48 Introduction Capacity BER Diversity Conclusions

I The ergodic capacity is in general hard to compute in closed form (we can always resort to numerical integration)

I We can apply Jensen’s inequality to gain some qualitative insight into the eﬀect of fast fading on capacity Jensen’s inequality If f (x) is a convex function and X is a random variable

E [f (X )] f (E [X ]) ≥ If f (x) is a concave function and X is a random variable

E [f (X )] f (E [X ]) ≤ I log( ) is a concave function, therefore · C = E [log(1 + γ)] log (1 + E[γ]) = log (1 + γ) ≤

Fading Channels: Capacity, BER and Diversity 14/48 ' $ IntroductionSISO FadingCapacity Channel:BER ErgodicDiversity Capacity Conclusions

The capacity of a fading channel2 with2 receiver CSI is2 less than the Assume iid Rayleigh fading with σ = E[ h ] = 1, and σz = 1. capacity of an AWGN channelh with| the| same average SNR

5 AWGN Channel Capacity Fading Channel Ergodic Capacity 4

3 Capacity (bps/Hz)

0 −10 −5 0 5 10 15 20 SNR (dB)

17 & Fading Channels: Capacity, BER and Diversity 15/48 % Introduction Capacity BER Diversity Conclusions

Block fading channel

I If the switch remains ﬁxed at the same (unknown) position during the transmission of the whole codeword, there is no nonzero rate at which long codewords can be transmitted with vanishingly small probability or error I Strictly, the capacity for this channel model would be zero I In this situation, it is more useful to deﬁne the outage capacity

Cout = r Pr(log(1 + γ) < r) = Pr(C(h) < r) = Pout ⇒ Suppose we transmit at a rate Cout with probability of outage Pout = 0.01 this means that with probability 0.99 the instantaneous capacity of the channel (a realization of a r.v.) will be larger than rate and the transmission will be successful; whereas with probability 0.01 the instantaneous capacity will be lower than the rate and the all bits in the codeword will be decoded incorrectly

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I Pout is the error probability since data is only correctly received on 1 Pout transmissions − I What is the outage probability of a block fading Rayleigh channel with average SNR = γ for a transmission rate r?

Pout = Pr(log(1 + γ) < r)

I The rate is a monotonic increasing function with the SNR = γ; therefore, there is a γmin needed to achieve the rate

r log(1 + γmin) = r γmin = 2 1 ⇒ −

and Pout can be obtained as

Z γmin Z γmin 1 −γ/γ Pout =Pr(γ < γmin) = f (γ)dγ = e dγ = 0 0 γ r − γmin − 2 −1 =1 e γ = 1 e γ − −

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Pout vs Cout (rate) for γ = 100 (10 dB)

0.8

0.6

0.4 Rate (bps/Hz) Rate 0.2

0 10-4 10-3 10-2 P out

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Summary

AWGN L L L h h h 1 1  1 C = log(1+ γ )

Ergodic (fast fading) L L

h h hL h h h 1 2  1 2  L C = E[log(1+ γ )]

Block fading L L L h h h C 1 2  n outage

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Final note

I All capacity results seen so far correspond to the case of perfect CSI at the receiver side (CSIR)

I If the transmitter also knows the channel, we can perform power adaptation and the results are diﬀerent

I We’ll see more on this later Now, let’s move to analyze the impact of fading on the Bit Error Rate (BER)

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BER analysis for the AWGN channel (a brief reminder)

I BPSK s[n] 1, +1 ∈ {− } r ! 2Es Ps = Pb = Q = Q √2SNR N0

∞ where Q(x) = R √1 exp (x2/2) 1 exp (x2/2) x 2π − ≤ 2 − −1−j −1+j 1−j 1+j I QPSK s[n] √ , √ , √ , √ ∈ { 2 2 2 2 } 2 Ps = 1 1 Q √SNR − −

I Nearest neighbor approximation Ps 2Q √SNR ≈ I With Gray encoding Pe = Ps /2

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I M-PAM constellation Ai = (2i 1 M)d/2, i = 1,..., M − − I Distance between neighbors: d I Average symbol energy

2 2 2 1 2 d (M 1)d Es = (M 1) = − 3 − 2 12

I Symbol Error Rate

M 2 d 2 d Ps = − 2Q + Q M 2σr M 2σr ! 2(M 1) r 6 SNR = − Q M M2 1 −

I With Gray encoding Pe Ps /M ≈

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I M-QAM: the constellations for the I and Q branches are Ai = (2i 1 M)d/2, i = 1,..., M − − I At the I and Q branches we have orthogonal √M PAM signals, each with half the SNR −

I Symbol Error Rate

!!2 2(√M 1) r3 SNR Ps = 1 1 − Q − − √M M 1 −

I Nearest neighbor approximation (4 nearest neighbors) ! r3 SNR Ps 4Q ≈ M 1 −

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The impact of fading on the BER

I Let us consider for simplicity a SISO channel with a BPSK source signal x[n] = hs[n] + r[n] s[n] +1, 1 , r[n] CN(0, σ2) is the noise (additive, white∈ and { Gaussian)− } ∼

I An AWGN channel or a fading channel behave also diﬀerently in terms of BER AWGN channel: h is constant Rayleigh fading channel: h ~ CN(0,1)

h h

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I The receiver knows perfectly the channel coherent det. → I Since we are sending a BPSK signal over a complex channel the optimal coherent detector is h∗x[n] +1 Re ≷ 0 (2) h −1 | | I Assuming that the transmitted power is normalized to unity E[ s[n] 2] = 1, the average SNR is | | 1 SNR = σ2 and the instantaneous Signal-to-Noise-Ratio is SNR h 2 2 | | I For an AWGN channel, SNR h is a constant value and the BER is | | ! √ q 2 h 2 Pe = Q | | = Q 2 h SNR σ | |

∞ where Q(x) = R √1 exp (x2/2) 1 exp (x2/2) x 2π − ≤ 2 − Fading Channels: Capacity, BER and Diversity 25/48 Introduction Capacity BER Diversity Conclusions

2 I For a fading channel SNR h is now a random variable | | p 2 I We can therefore think of Pe = Q 2 h SNR as another | | random variable

I The BER for the fading channel would be the average over the channel distribution E[Pe ] I This applies to a fast fading channel model BER for a Rayleigh channel h is now a Rayleigh random variable (z = h 2 is exponential with pdf| | f (z) = exp( z), the BER is given by | | − Z ∞ r ! −z 1 SNR 1 Pe = Q √2zSNR e dz = 1 0 2 − 1 + SNR ≈ 4SNR

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0 10

−2 10

−4 10 AWGN Rayleigh

−6 BER 10

−8 10

−10 10

0 2 4 6 8 10 12 14 16 SNR (dB) −3 At an error probability of Pe = 10 , a Rayleigh fading channel needs 17 dB more than an AWGN (constant) channel, even though we have perfect channel knowledge in both cases !!

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I The main reason of the poor performance is that there is a signiﬁcant probability that the channel is in a deep fade, and not because of noise 2 I The instantaneous signal-to-noise-ratio is h SNR, and we may consider that the channel is in a deep| fade| when h 2SNR < 1 | | I In this event, the separation between constellation points will be of the order of magnitude of σ (noise standard deviation) and the probability of error will be high I For a Rayleigh channel, the probability of being in a deep fade is

Z 1/SNR Pr( h 2SNR < 1) = e−z dz = 1 e−1/SNR = | | 0 − ! 1 1 2 1 = 1 1 + ... − − SNR SNR − ≈ SNR

which is of the same order as Pe Fading Channels: Capacity, BER and Diversity 28/48 Introduction Capacity BER Diversity Conclusions

Remarks

I Although our example considered a BPSK modulation, the same result is obtained for other modulations On the Rayleigh channel, all modulation schemes are equally bad and for all them the BER decreases as SNR−1

I The gap between the AWGN and the Rayleigh channel in terms of BER is much higher than in terms of capacity

This suggests that coding can be very beneﬁcial to compensate fading

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Diversity

I The negative slope (at high SNR) of the BER curve with respect to the SNR is called the diversity gain or diversity order log(P (SNR)) d = lim e − SNR→∞ log(SNR)

I In other words, in a system with diversity order d, the Pe at high SNR varies as −d Pe = SNR

I For a SISO Rayleigh channel the diversity order is 1 −1 (remember that Pe 1/(4SNR) SNR !! ≈ ∝

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SIMO system Consider now that the Rx has N antennas and that the resulting SIMO (single-input multiple-output) channel is spatially uncorrelated r1[n] h1  h1s[n] r1[n] h r2[n] s[n] 2  h2s[n] r2[n]

 

rN [n] hN hN s[n] rN [n] If the channels hi are faded independently, the probability that the SIMO channel is in a deep fade will be approximately

N 2 2 2 1 Pr((|h1| SNR < 1)& ... &(|hN | SNR < 1)) = Pr((|h| SNR < 1) ≈ SNRN

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I In consequence, the diversity order of a SIMO system with N uncorrelated receive antennas is N

I Notice the impact of antenna correlation: if all antennas are very close to each other so that the channels become fully correlated, then h1 h2 hN , and the diversity reduces to that of a SISO channel≈ ≈

I There are diﬀerent ways to process the receive vector and extract all spatial diversity of the SIMO channel T I Assuming that the channel h = (h1, h2,..., hN ) is known at the Rx, the optimal scheme called maximum ratio combining (MRC)

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Maximum Ratio Combining The MRC schemes linearly combines (with optimal weights wi ) the outputs of the signals received at each antenna

r1[n] h1s[n] r1[n] w1 h1  

r2[n] h2 h2s[n] r2[n] w2    z[n]   r [n] N hN s[n] rN [n] wN hN

The signal z[n] can be written in matrix form as     h1 r1[n] h2  r2[n] z[n] = w w ... w   s[n] +   = wT (hs[n] + r[n]) 1 2 N  .   .   .   .  hN rN [n]

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I As long as ||w|| = 1 the noise distribution after combining does not change: i.e., wT r[n] ∼ CN(0, σ2) I It is easy to show that the optimal weights are given by h∗ w = , ||h|| which is just the matched ﬁlter for this problem! I These are the optimal weights because they maximize the signal-to-noise-ratio at the output of the combiner:

||h||2 SNR = = ||h||2SNR MRC σ2 I For a BPSK transmitted signal, the optimal detector is

hH x[n] +1 Re(z[n]) = Re ≷ 0 ||h|| −1 where (·)H denotes Hermitian (complex conjugate and transpose). Notice the similarity with the optimal coherent detector for the SISO case, which was given by (2)

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I Similarly to the SISO case, the Pe can be derived exactly p 2 Pe = Q 2||h|| SNR ,

which is a random variable because h is random (fading channel)

I Under Rayleigh fading, each hi is i.i.d CN(0, 1) and

N 2 X 2 x = ||h|| = |hi | i=1 follows a Chi-square distribution with 2N degrees of freedom 1 f (x) = x N−1 exp(−x), x ≥ 0 (N − 1)!

I Note that the exponential distribution (SISO case with N = 1) is a Chi-square with 2 degrees of freedom I The average error probability can be explicitly computed, here we only provide a high-SNR approximation ! Z ∞ √ 2N − 1 1 Pe = Q 2xSNR f (x)dx ≈ N 0 N (4SNR)

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The probability of being in a deep fade with MRC is

Z 1/SNR 1 Pr( h 2SNR < 1) = xN−1 exp( x)dx || || (N 1)! − ≈ 0 − Z 1/SNR 1 1 (for x small) xN−1dx = ≈ (N 1)! N!SNRN 0 −

Conclusions

I The Pe is again dominated by the probability of being in a deep fade

I Both (the Pe and the probability of being in a deep fade) decrease with the number of antennas roughly as (SNR)−N

I MRC extracts all spatial diversity of the SIMO channel !! However, MRC is not the only multiantenna technique that extracts all spatial diversity of a SIMO channel

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Antenna selection

r1 h1 

r2 h2 x[n]  hmax s[n] r[n]  RF chain ADC



rN hN Selection of the best link

I In antenna selection the path with the highest SNR is selected and processed: x[n] = hmax s[n] + r[n], where

hmax = max (|h1|, |h2|,..., |hN |)

I In comparison to MRC, only a single RF chain is needed !!

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I Intuitively, the probability that the best channel is in a deep fade is

N 2 Y 2 1 Pr( hmax SNR < 1) = Pr(( hi SNR < 1) | | | | ≈ SNRN i=1 and, therefore, the diversity of antenna selection is also N

I The same conclusion can be reached by studying the Pe (we 2 would need the distribution of hmax ) | | I However, the output SNR of MRC is higher than that of antenna selection (AS)

2 2 h1 hN SNR = | | + ... + | | MRC σ2 σ2 2 hmax SNR = | | AS σ2

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Array gain

I The SNR increase in a multiantenna system with respect to that of a SISO system is called array gain

I Whereas the diversity gain is reﬂected in slope of the BER vs. SNR, the array gain provokes a shift to the left in the curve 0 10

−1 10 N=1

−2 10 BER −3 10 N=2

−4 Array Gain 10

−5 10 0 5 10 15 20 25 SNR (dB)

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I For a receiver with N antennas, MRC is optimal because it achieves maximum spatial diversity and maximum array gain

I MRC achieves the maximum array gain by coherently combining all signal paths

I On average, the output SNR of MRC is N times that of a SISO system, so its array gain is 10 log10 N (dBs) I For instance, using 2 Rx antennas, MRC provides 3 extra dBs in comparison to a SISO system and in addition to the spatial diversity gain

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0 10

Antenna Selection −1 10 MRC

−2 10 N=2 SER −3 N=3 10

−4 10

0 5 10 15 20 25 SNR (dB)

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Maximum Ratio Transmission Similar concepts apply for a a MISO (multiple-input single-input) channel: the optimal scheme is now called Maximum Ratio Transmission (MRT) w 1 h  1 w 2 h r[n] s[n]  2 T  z[n]  w h s[n] r[n] hN wN

Again, the optimal transmit beamformer that achieves full spatial diversity (N) and full array gain (10 log10 N) is h∗ w = , h || || The main diﬀerence is that now the channel must be known at the Tx (through a feedback channel)

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Spatial diversity of a MIMO channel

An nR nT MIMO channel with i.i.d. entries has a spatial × diversity nR nT , which is the number of independent paths oﬀered to the source signal for going from the Tx to the Rx

I Some schemes extract the full spatial diversity of the MIMO channel

I Optimal combining at both sides (MRT+MRC) I Antenna selection at both sides I Antenna selection at one side and optimal combining at the other side I Repetition coding at the Tx side (same symbol transmitted through all Tx antennas) + optimal combining at the Rx side

I But others might not: think of a system that transmits independent data streams over diﬀerent transmit antennas (more on this later)

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Time and frequency diversity

I We have mainly focused the discussion on the spatial diversity concept, but the same idea can be applied to the time and frequency domains I Time diversity:

1. The same symbol is transmitted over diﬀerent time instants, t1 and t2 2. For the channel to fade more or less independently: 1 t1 t2 > Tc = | − | Ds I Frequency diversity: 1. The same symbol is transmitted over diﬀerent frequencies (subcarriers in OFDM), f1 and f2 2. For the channel to fade more or less independently: 1 f1 f2 > Bc = | − | τrms I To achieve time or frequency diversity, interleaving is typically used

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Interleaving

I The symbols or coded bits are dispersed over diﬀerent coherence intervals (in time or frequency)

I A typical interleaver consists of an P Q matrix: the input signals are written rowwise into the matrix× and then read columnwise and transmitted (after modulation)

I Example: A 4 8 interleaving matrix ×

from source to channel s1 s2 s3 s4 s5 s6 s7 s8 s1s2s3s4s5  s28s29s30s31s32 s1s9s17s25s2  s31s8s16s24s32 s9 s10 s11 s12 s13 s14 s15 s16

s17 s18 s19 s20 s21 s22 s23 s24

s25 s26 s27 s28 s29 s30 s31 s32

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s1 s2 s3 s4 s5 s6 s7 s8

s9 s10 s11 s12 s13 s14 s15 s16

s17 s18 s19 s20 s21 s22 s23 s24 s s s s s s s s burst of errors 25 26 27 28 29 30 31 32

h h 1 2 h3 h4 s s s s s s s s s s s s Without interleaving 1 2 3 4 5 6 7 8 9 10 11 12 s13 s14 s15 s16

h h h 1 2 h3 4 s s s s With interleaving s1 s9 s17 s25 s2 s10 s18 s26 s3 s11 s19 s27 4 12 20 28

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Remarks:

I Doppler spreads in typical systems range from 1 to 100 Hz, corresponding roughly to coherence times from 0.01 to 1 sec. 4 6 I If transmissions rates range from 2.10 to 2.10 bps, this would imply that blocks of length L ranging from L = 2.104 0.01 = 200 bits to L = 2.106 1 = 2.106 bits would be aﬀected× by approximately the same× fading gain

I Deep interleaving might be needed in some cases

I But notice that interleaving involves a delay proportional to the size of the interleaving matrix

I In delay-constrained systems (transmission of real-time speech) this might be diﬃcult or even unfeasible

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Conclusions

I We have analyzed the eﬀect of fading from the point of view of capacity (only CSIR) and BER

I Depending of the channel model (ergodic or block fading) we use ergodic capacity or outage capacity

I With CSIR only, the capacity of a fading channel is always less than the capacity of an AWGN with the same average SNR

I The eﬀect of fading is more signiﬁcant in terms of BER than in terms of capacity (power of coding gain + interleaving)

I Diversity: slope of the BER wrt SNR at high SNRs

I Diversity techniques with multiantenna systems: MRT, MRC, antenna selection, etc

I We can extract the time and frequency diversity of the channel through interleaving

Fading Channels: Capacity, BER and Diversity 48/48