Wireless Communications Lecture 10

Transcription

1 Wireless Communications Lecture 1 [SNR per symbol and SNR per bit] SNR = P R N B = E s N BT s = E b N BT b For BPSK: T b = T s, E b = E s, and T s = 1/B. Raised cosine pulse shaper for other pulses. T s 1 B The SNR becomes SNR = E s or ( E b BPSK) N N The SNR per symbol γ s and the SNR per bit γ b are defined as γ s = E s N γ b = E b N [Symbol error rate and bit error rate] One approximation: P s one bit in error = P b1 (1 P b2 )(1 P b3 )... + P b2 (1 P b1 )(1 P b3 )... + = MP b1, M : the number of bits

2 2 Lecture 1 [QPSK] P s = P s 1 transmitted 1 ( ) 4 4 = 1 Q si1 Q σ = 1 (1 Q( SNR b )) 2, SNR symbol = 2a2 σ 2 2Q( SNR b ) ( si2 σ ) P b P s /2 [General case] For MPSK

3 Wireless Communications 3 = r = s i + n need angle distribution. In-phase and quadrature Gaussian bivariate distribution of r and θ Integrate for θ distributiion. β P s = 1 P (θ)dθ α Closed form does not exist. The nearest neighbor approximation P s 2Q( 2 SNR b ) sin(π/m) [Average symbol error probability(general)] Many systems have : P s α M Q( β M γ s ) Finding the average probability of symbol error P s = α M Q( β M γ)λe λγ dγ, [ ] = α M.5β M γ s β M γ s α M 2β M γ s λ = 1 γ s Show P b for AWGN and fading channel (figure 6.1, 6.2 Goldsmith) [Alternative form of Q-function by Craig] Q(z) = 1 π e z 2 2 sin 2 ϕ dϕ, z >

4 4 Lecture 1 [Moment Generating Function] M γ (s) = p γ (γ)e sγ dγ Setting α = α M and g =.5β M in P s = α M Q( β M γ s ) P s = α M Q( β M γ s ) α=αm,g=.5β M P s = αq( 2gγ s ) = α π Using the alternative Q-function e gγs sin 2 ψ dϕ The average error probability in fading for modulation with P s = αq( 2gγ s ) is P s = α π = α π = α π [ The MGF M γs (s) for Rayleigh distribution is e gγ sin 2 ϕ dϕp γs (γ)dγ ] e gγ sin 2 ϕ p γs (γ)dγ dϕ M γs ( g sin 2 ϕ ) dϕ M γs (s) = (1 s γ s ) 1, s = g sin 2 ϕ [Example] Find outage probability of BPSK at P b = 1 3 for Rayleigh fading with selection diversity for M = 2. Equal branch SNR s of γ = 15 db. P b = 1 3 γ b = 1 db γ = 1 P out = [1 e γ / γ ] 2 = [1 e 1/11.5 ] 2 [Average probability of symbol error(mrc)] The distribution of γ Σ for MRC p γσ (γ) = γm 1 e γ/ γ γ M (M 1)!, γ.

5 Wireless Communications 5 The average probability of symbol error becomes P s = αq( βγ)p γσ (γ)dγ Capacity of fading channels Capacity of AWGN channel: C = B log 2 (1 + γ) C: bits/sec, B:Bandwidth, γ = P N ob (received SNR) Meaning of capacity Fading channels: SNR is a random variable. Finding capacity if only channel distribution is and RX is hard. Channel capacity with RX CSI (Channel Side Information) RX knows the channel. TX sends a fixed rate Two types of capacity: Shannon and Outage Shannon capacity: ergodic capacity, max rate with which you can still decode with zero error probability. Capacity with outage: you allow some probability of error. [Shannon (Ergodic) Capacity] For RX CSI, we will have C = B log 2 (1 + γ)p(r)dr log is a concave function applying Jenson s Inequality. For concave f(x), E(f(x)) f(e(x)) C = B log 2 (1 + γ)p(γ)dγ B log 2 (1 + γ) It is worse than AWGN with γ.

6 6 Lecture 1 [Capacity with outage] RX CSI, outage case: RX assumes γ min for received SNR & sends C = B log 2 (1 + γ min ) if γ < γ min bits cannot be decoded without error P out = Probability of not decoding correctly = P (γ < γ min ) C out = number of received correctly = (1 P out )B log 2 (1 + γ min ) γ min C but γ min Choose γ min that gives the optimum. { 1 + γmin 1 P out Channel side information at Transmitter and Receiver [Shannon capacity, TX & RX CSI:] Adapt the rate to SNR. C = max ( B log P (γ)γ ) p(γ)dγ P P (γ)p(γ)dγ P Optimum P (γ): Form Lagrangian : ( J = B log P (γ)γ ) ( ) p(γ)dγ λ P (γ)p(γ)dγ P P differentiate with respect to P (γ), J P (γ) (B/ ln 2)(γ/P ) = p(γ) λp(γ) = 1 + P (γ)γ/ P then, P (γ) = (B/ ln 2)(γ/P ) λ γ(λ/p ) γ > P λ B ln 2 = γ P (γ) = { P γ P γ γ γ else >

7 Wireless Communications 7 Finding γ : C = γ γ B log 2 (γ/γ )p(γ)dγ P (γ) P p(γ)dγ = 1 ( 1 1 ) p(γ)dγ = 1 γ γ [Wideband Channels:] want to distribute total TX power over the frequency bins: C = ( B log H j 2 ) P j N o B Pj P Similar Lagrangian method: P j P = { 1 γ 1 γ j γ j γ else Find γ j 1 γ 1 γ j = 1 where γ j = H j 2 P j N ob Water-filling (Multicarrier Modulation) : DSL