ECE 480

Wireless Systems

Lecture 13

Capacity of Wireless Channels

10 Apr 20101

To the optimist, the glass is half full

To the pessimist, the glass is half empty

To the engineer, the glass is twice as big as it needs to be

Thoughts to Live by

2

Channel Side Information at Receiver and Transmitter

• Transmitter can adapt its transmission relative to this CSI

• Transmitter will not send bits unless they can be coded correctly

• Assumptions

• Optimal power

• Optimal rate adaptation3

Shannon Capacity

• g [i] is known to both the transmitter and receiver

• Let s [i] be a stationary and ergodic stochastic process representing the channel state

• s [i] takes values on a finite set S of discrete memoryless channels

• C s = capacity of a particular channel s S

• p (s) = denote the probability (fraction of time) that the channel is in state s

ss S

C C p s

4

C B log bps 2 1Capacity of an AWGN channel with average received SNR

• Let p () = p ( [i] = ) be the distribution of the received SNR

C C p d B log p d

2

0 0

1

• Same as CSI at receiver only – no increase in capacity

• Must adapt power as well to increase capacity 5

Let the transmit power P () vary with subject to an average power constraint, P

P p d P

0

P

C max B log p dP

2

0

1

Fading channel capacity with average power constraint

P : P p d P

0 6

“Time diversity” system with multiplexed input and demultiplexed output

• Quantize the range of fading values to a finite set [ j: 1 j N]

• For each j we design an encoder – decoder pair for an AWGN channel with SNR j

7

• The input x j for encoder j has average power P ( j) and data rate R j = C j

• C j is the capacity of a time – invariant AWGN channel with received SNR

• These encoder – decoder pairs correspond to a set of input and output ports associated with each j

j jP

P

8

• When [i] j the corresponding pair of ports are connected through the channel

• The codewords associated with each j are multiplexed together for transmission and demultiplexed at the channel output

• Effectively, the system is reduces the time – varying channel to a set of time – invariant channels in parallel where the j th channel operates only with [i] j

9

P

C max B log p dP

2

0

1

P : P p d P

0

To optimize the power allocation P () form the Lagrangian

P

C B log p d P p dP

2

0 0

1

is a parameter that may set limitations 10

Take the derivative of the Lagrangian and set to zero

BJ P ln

pPP

P

20

1

Solve for P () with the constraint that () > 0

P

P

00

0

1 1

0

0 is a “cutoff” value below which no data is transmitted

The channel is used at time [i] only if 0 [i] < 11

C B log p d

0

20

• The time – varying data rate corresponding to the instantaneous data rate is

• Since 0 is constant, the data rate increases with

B log

20

12

• The optimal power allocation policy depends on the fading distribution only through 0

Pp d

P

0

1

p d

00

1 11

• This expression defines 0

• Depends only on p ()

• Must be solved numerically

13

P

P

0

1 1is called a “water – filling” formula

Optimum power allocation

• Shows power allocated to the channel vs. (t) =

• When conditions are good ( large) more power and a higher data rate are fed over the channel

Amount of power allocated for a given

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• For any power adaptation policy P ()

P

C B log p dP

2

0

1

the capacity can be achieved with arbitrarily small error probability

• Cannot exceed the case where power adaptation is optimized

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Example 4.4

Assume the same channel as in the previous example, with a bandwidth of 30 KHz and three possible received SNRs: 1 = 0.8333 with p () = 0.1, 2 = 83.33 with p () = 0.5, and 3 = 333.33 with p() = 0.4. Find the ergodic capacity of this channel assuming that both transmitter and receiver have instantaneous CSI.

Solution

• The optimal power allocation is water – filling

• Need to find 0 such that

i

ip

0 0

1 11

16

• Assume that all channel states are used to obtain 0

• Assume that 0 min i 0 and see if the resulting cutoff value is below that of the weakest channel

i i

i i

p p

3 3

1 10

1

ii

p

3

1 0 0

1

ii

p . . ..

. . .

3

10

1 0 1 0 5 0 41 1 1 1272

0 8333 83 33 333 33

..

0

10 8872

1 127217

0.8872 > 0.8333

Inconsistent result

ii

p. . ..

. .

3

20

0 9 0 5 0 41 1 1 0072

83 33 333 33

Assume that the weakest state is not used

ii

p . . .

3

2 0 0 0 0

0 5 0 4 0 9

. . . 0 1 0072 0 9 0 8936

1 0 2

Consistent result 18

iC B log p

. ., . log . log

. .

. kbps

3

22 0

2 2

83 33 333 3330 000 0 5 0 4

0 8936 0 8936

200 82

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Zero – Outage Capacity and Channel Inversion

• Suboptimal transmitter adaptation scheme

• Transmitter uses the CSI to maintain a constant received power (inverts the channel fading)

• Channel then appears to the encoder and decoder as a time – invariant AWGN channel

• Channel inversion:

P

P

= constant received SNR that can be maintained with the constraint

P p d P

0 20

satisfies the constraint

p d

E

1

1

1

With these definitions, fading channel capacity with channel inversion is the same as the capacity of an AWGN channel with SNR =

C B log B log

E

2 2

11 1

1

p d

E

1

1

1

C B log B log

E

2 2

11 1

121

• The transmission strategy uses a fixed – rate encoder and decoder designed for an AWGN channel with SNR

• Maintains a fixed data rate over the channel regardless of channel conditions

• (zero – outage capacity) – no channel outage

• Can exhibit a large data – rate reduction relative to Shannon capacity

• In Rayleigh fading, C = 0

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Example 4.5

Assume the same channel as in the previous example, with a bandwidth of 30 KHz and three possible received SNRs: 1 = 0.8333 with p () = 0.1, 2 = 83.33 with p () = 0.5, and 3 = 333.33 with p() = 0.4. Assuming transmitter and receiver CSI, find the zero – outage capacity of this channel,

Solution

E

1

1

C B log B log

E

2 2

11 1

1

23

. . .E .

. . .

1 0 1 0 5 0 40 1272

0 8333 83 33 333 33

C B log B log

E

2 2

11 1

1

C log . kbps.

2

130000 1 94 43

0 1272

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Outage Capacity and Truncated Channel Inversion

• Zero – outage capacity may be much smaller than Shannon capacity

• Requirement of maintaining a constant data rate in all fading states

• By suspending transmission in bad fading states (outage channel states) we can maintain a higher constant data rate in the other states

• Outage capacity: the maximum data rate that can be maintained in all non – outage channel states multiplied by the probability of non – outage

25

• Outage capacity is achieved with a truncated channel inversion policy for power adaptation that compensates for fading only above a certain cutoff fade depth, 0

P

P

0

00

0 is based on the outage probability

outP p , 0

• Channel is only used when > 0

E p d

0

0

1 1

26

The outage capacity associated with a given outage probability P out and corresponding cutoff 0 is

outC P B log p

E

0

2 0

11

1

The maximum outage capacity is obtained by maximizing outage capacity over the range of possible 0

outC P max B log p

E

0

2 0

11

10

27

• The maximum outage capacity will still be less than Shannon capacity

• Truncated channel inversion is a suboptimal transmission strategy

• The transmit and receive strategies may be easier to implement or have lower complexity

• Based on AWGN design

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Example 4.6

Assume the same channel as in the previous example, with a bandwidth of 30 KHz and three possible received SNRs: 1 = 0.8333 with p () = 0.1, 2 = 83.33 with p () = 0.5, and 3 = 333.33 with p() = 0.4. Find the outage capacity of this channel and associated outage probabilities for cutoff values 0 = 0.84 and 0 = 83.4. Which of these cutoff values yields a larger outage capacity?

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