Create Presentation
Download Presentation

Download

Download Presentation

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

145 Views
Download Presentation

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

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**ECE 480**Wireless Systems Lecture 13 Capacity of Wireless Channels 10 Apr 2010**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**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**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**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**Let the transmit power P () vary with subject to an**average power constraint, Fading channel capacity with average power constraint**“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**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**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**To optimize the power allocation P () form the Lagrangian** is a parameter that may set limitations**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] < **The time – varying data rate corresponding to the**instantaneous data rate is • Since 0 is constant, the data rate increases with **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**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**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**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**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**0.8872 > 0.8333**Inconsistent result Assume that the weakest state is not used Consistent result**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** satisfies the constraint**With these definitions, fading channel capacity with channel inversion is the same as the capacity of an AWGN channel with SNR = **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**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**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**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**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**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**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?