1. Capacity of Wireless Channels
ECE 480
Wireless Systems
Lecture 13
Capacity of Wireless Channels
10 Apr 2010

2. Thoughts to Live by 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

3. 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 adaptation

4. 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

5. Capacity of an AWGN channel with average received SNR 
Let p () = p ( [i] = ) be the distribution of the received SNR
Same as CSI at receiver only – no increase in capacity
Must adapt power as well to increase capacity

6. Let the transmit power P () vary with  subject to an average power constraint,
Fading channel capacity with average power constraint

7. “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

8. 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

9. 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

10.  is a parameter that may set limitations
To optimize the power allocation P () form the Lagrangian
 is a parameter that may set limitations

11. Take the derivative of the Lagrangian and set to zero
Solve for P () with the constraint that () > 0
0 is a “cutoff” value below which no data is transmitted
The channel is used at time [i] only if  0   [i] < 

12. The time – varying data rate corresponding to the instantaneous data rate  is
Since  0 is constant, the data rate increases with 

13. The optimal power allocation policy depends on the fading distribution only through  0
This expression defines  0
Depends only on p ()
Must be solved numerically

14. is called a “water – filling” formula
Optimum power allocation
Amount of power allocated for a given 
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

15. For any power adaptation policy P ()
the capacity can be achieved with arbitrarily small error probability
Cannot exceed the case where power adaptation is optimized

16. Example 4.4
Assume the same channel as in the previous example, with a bandwidth of 30 KHz and three possible received SNRs:  1 = with p () = 0.1,  2 = with p () = 0.5, and  3 = with p() = 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

17. 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

18. Assume that the weakest state is not used
>
Inconsistent result
Assume that the weakest state is not used
Consistent result

20. 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:
 = constant received SNR that can be maintained with the constraint

21.  satisfies the constraint
With these definitions, fading channel capacity with channel inversion is the same as the capacity of an AWGN channel with SNR = 

22. 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

23. Example 4.5
Assume the same channel as in the previous example, with a bandwidth of 30 KHz and three possible received SNRs:  1 = with p () = 0.1,  2 = with p () = 0.5, and  3 = with p() = Assuming transmitter and receiver CSI, find the zero – outage capacity of this channel,
Solution

25. 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

26. 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
 0 is based on the outage probability
Channel is only used when  >  0

27. The outage capacity associated with a given outage probability P out and corresponding cutoff  0 is
The maximum outage capacity is obtained by maximizing outage capacity over the range of possible  0

28. 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

29. Example 4.6
Assume the same channel as in the previous example, with a bandwidth of 30 KHz and three possible received SNRs:  1 = with p () = 0.1,  2 = with p () = 0.5, and  3 = with p() = Find the outage capacity of this channel and associated outage probabilities for cutoff values  0 = 0.84 and  0 = Which of these cutoff values yields a larger outage capacity?