On the SNR Exponent of Hybrid Digital Analog Space Time Coding

On the SNR Exponent ofHybrid Digital Analog Space Time Coding

Krishna R. NarayananTexas A&M University

http://ee.tamu.edu/~krn

Joint work with Prof. Giuseppe CaireUniversity of Southern California

Wireless Communications Lab, TAMU


What this talk is about

Transmit an analog source over a MIMO channel

Encoder

s1;s2; : : : ;sK s1; : : : ; sKReceiver

Quasi-static (Block) Fading

Performance criterion : End to End quadratic distortion

How to best use multiple antennas to minimize distortion?

T uses of the channel


Example 1: Streaming real-time video

Due to stringent delay constraints, we cannot code over many realizations of the channel

Channel changes from one block to another

Encoder


Quasi-static (Block) Fading


Example 2: Broadcasting an analog source

Empirical distribution of Hl converges to some ensemble distribution as l ! 1

Encoder

s1;s2; : : : ;sK

s1; : : : ; sKUser 1

s1; : : : ; sKUser 2

s1; : : : ; sKUser 3

H 1

H 2

H 3


Block Fading Channel Model

yt =p

½M H xt + wt; t = 1;: : : ;T

½denotes theSignal-to-Noise Ratio (SNR)

X = [x1; : : : ;xT ], is normalized such that tr(E[X HX ]) · M T

T is theduration (in channel uses)

M · N (the case M > N follows as a simple extension)

H 2 CN £ M , wherehi ;j » CN (0;1)


Diversity Multiplexing Tradeoff

Channel has zero capacity, we talk of outage capacity

For Pe() ! 0, we need SNR ! 1

Makes sense to look at the exponent of SNR

Diversity gain : Pe() / -d

Multiplexing gain: Use many antennas to transmit many data streams

Zheng and Tse: Both are simultaneously achievable, there is a fundamental trade off of how much of each we can achieve


Optimal D-M Tradeoff (Zheng and Tse)

Consider a family of space-time coding schemes fCr c(½)g

Rc = rc log½ rc is the muliplexing gain

Optimal exponent d¤(rc) = supd(rc)

whereas rate increases with ½as r = rc log½

For large ½, Poutage / ½¡ d(r c )


d*(r) for Rayleigh Fading Channels

d¤(r) = (M ¡ r) (N ¡ r)


Precise Problem Statement

SourceChannelEncoder


Fadingsi » N (0;1)

D(½) ¢= 1

KE[js ¡ esj2]

Family of source-channel coding schemes fSC (½)g

Rate of decay of MSE with SNR: D / ½¡ a(´)

SNR Exponent: a(´) = ¡ lim½! 1logD (½)

log½

a?(´) = supa(´)

Channel

SNR = ½

´ = 2KT


Exponent for Separation Scheme

Sp-TimeCode

Q MIMOChannel

Rs = rs log½bits Rc = rc log½bits

SinceRsK = RcT, wehaveRc = ´Rs

Equating exponents, we get the result in Theorem 1

Dsep(Rs) · DQ (Rs) +a P (e)

Dsep(Rs) · ½¡ 2´ rc +a ½¡ d¤ (r c )

Not in outage In outage



Main Results – Separation based approach

Comments – Separation is not optimal

Can we outperform the optimized separated scheme ? Yes, we will see that later

T heorem 1 [Exponent achievable by separation]

asep(´) = 2(j d? (j ¡ 1)¡ (j ¡ 1)d? (j ))2+´(d? (j ¡ 1)¡ d? (j )) ´ 2

h2(j ¡ 1)d? (j ¡ 1) ;

2jd? (j )

´for j = 1;:: : ;M ,

is achievable by a tandemsource-channel coding scheme ¤


Upper bound on the exponent

An upper bound is obtained by assuming that the channel is instantaneously known at the transmitter

Coding rate is chosen according to

Large SNR behavior can be analyzed using techniques similar to those in Zheng and Tse (Wishart distribution, Varadhan’s lemma)

Rc(H ) · logdet¡I + ½HH H

¢

D(½) ¸ Eh

1det(I +½H H H)2=´

i


Upper bound

T heorem 2 [Informed transmitter upper bound]

The optimal distortion SNR exponent a?(´) is upperbounded by

aub(´) =MX

i=1

min½

2´

;2i ¡ 1+ jM ¡ N j¾

(1

¤


The case for analog coding schemes – known SNR

n1; : : : ;nk

Channelcode

Q + Decoder

s1; : : : ;sk

OR

s1; : : : ;sk +MMSE

estimate

n1; : : : ;nk

rs = 12 log(1+ ½) D(½) = 1

1+½

D(½) = 11+½


Separation scheme with Fading

n1; : : : ;nk

Channelcode

Q + Decoder

s1; : : : ;sk

Separation based scheme is optimal only when ||hi||2 = 1

Can be quite bad otherwise

Cannot exploit higher instaneous channel gain

rs = 12 log(1+ ½)

hi


Analog scheme with Fading

s1; : : : ;sk +MMSE

estimate

n1; : : : ;nk

Analog scheme is simultaneously optimal for all SNRs

Graceful degradation of MSE with SNR

hi

D(½;hi ) = 11+jjhi jj2½


Why Hybrid then?

Alas! things are never that easy

The optimality is valid only for the SISO channel with T = K

If more bandwidth is available, it is difficult to take advantage

Same with lesser bandwidth also

Hence, we need to look for hybrid digital and analog solutions


HDA Solution for T > K (Bandwidth expansion)

Involves some math, but the exponent can be analyzed

Express MSE as a function of the Eigen values, use the Wishart distribution and Varadhan’s lemma

Quantizer

K rs log½bits

- Reconstruct

Space-TimeEncoder

X (d) 2 CM £ Td

SpatialMultiplexer

X (a) 2 CM £ K2M

s1; : : : ;sK

s1; : : : ; sK

e1; : : : ;eK


QAM


Exponent of Hybrid Digital Analog Coding Schemes

This is better than the separated exponent

T heorem 3 [H ybrid scheme lower bound]

For BW expansion, thehybrid digital-analog (HDA) space-timecoding achieves

ahybrid(´) = 1+µ

2´

¡1M

¶j d?(j ¡ 1) ¡ (j ¡ 1)d?(j ) ¡ 1

2´ ¡ 1

M + d?(j ¡ 1) ¡ d?(j )(1)


HDA Scheme for T < K (Bandwidth Compression)

QuantizerSpace-Time

Encoder

SpatialMultiplexer

+

X (D ) 2 CM £ T

X (a) 2 CM £ T

1 11,..., Ks s

1,...,K Ks s

1 logsK r bits

Main trick is in chosing ¯ to be dependent on SNR as ½¡ °

and optimizing °



Exponent of HDA Schemes

It is remarkable that this is equal to the upper bound

Hence it is optimal

T heorem 3 [Hybrid scheme lower bound]

Bandwidth Compression:

ahybrid(´) =2M´

; ´ ¸ 2M


Upper bound for the Parallel Channel

For M parallel channels

aub = min(M; M2´ )


Scalar channel M = N = 1


MIMO M = N = 2


Comments

SISO Channel with fading

Even if T > K, the best exponent is 1

More bandwidth does not buy us anything

It is the degrees of freedom, not the bandwidth that is important

For this case, the exponent for the entire region is fully known

Gunduz-Erkip - infinite layers of superposition coding is optimal

Our approach is much simpler


MIMO Case

However, in the MIMO case things are different

More bandwidth helps us buy diversity

Antennas can be used to compress

It is very easy to find practical schemes to get the correct exponent

For example, uniform scalar quantization is optimal at high SNRs!

We may require very large SNR for the asymptotics to kick in


Outlook

Our conjecture is that the upper bound is loose (it is not entirely clear) for the bandwidth expansion case

Better constructive schemes to improve the achievable part

In some sense, the key problem is to find a better exponent for the parallel channel

Documents

On the SNR Exponent of Hybrid Digital Analog Space Time Coding