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**Capacity of Wireless Channels**

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

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**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

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**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

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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

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**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

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

Fading channel capacity with average power constraint

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**“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

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**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

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**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

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** is a parameter that may set limitations**

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

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**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] <

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

Since 0 is constant, the data rate increases with

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**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

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**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

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**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

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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 = 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

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**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

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**Assume that the weakest state is not used**

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

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

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** satisfies the constraint**

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

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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 = 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

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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

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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

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**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

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**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 = 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?

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