EE6604 Personal & Mobile Communications. Week 9. Co-Channel Interference

Transcription

1 EE6604 Personal & Mobile Communications Week 9 Co-Channel Interference 1

2 Co-channel interference on the forward channel d 1 d 6 d 2 mobile subscriber d 0 d 5 d 3 d 4 The mobile station is being served by the center base station. 2

3 At a particular location, let d = (d 0, d 1,, d NI ) be the vector of distances between a mobile station and the serving base station BS 0 and N I co-channel base stations BS k,k = 1,...,N I. The received signal power power at distance d, Ω p (dbm) (d), is a Gaussian random variable that depends on the distance d through the path loss model, i.e., µ Ωp (dbm) (d) = E[Ω p (dbm) (d)] = µ Ωp (dbm) (d o ) 10βlog 10 (d/d o ) Experiments have verified that co-channel interferers add noncoherently (power addition) rather than coherently (amplitude addition). The C/I a function of the vector d is or in decibel units Λ(d) = Ω p(d 0 ) N I k=1ω p (d k ) Λ(d) (db) = Ω p (dbm) (d 0 ) 10log 10 The outage probability given vector d is N I k=1 O(d) = P r ( Λ(d)(dB) < Λ th(db) ) Ω p (d k ) Although the Ω p (d k ) are log-normal random variables, the sum N I k=1ω p (d k ) is not a log-normal random variable. 3

4 Multiple Log-normal Interferers Consider the sum of N I log-normal random variables I = N I k=1 Ω k = N I 10 Ω k(dbm) /10 k=1 where the Ω k (dbm) are independent Gaussian random variables with mean µ Ωk (dbm) and variance σ 2 Ω k. The sum I is commonly approximated by another log-normal random variable with appropriately chosen parameters, i.e., I = N I k=1 10 Ω k(dbm) /10 10 Z (dbm) /10 = Î where Z (dbm) is a Gaussian random variable with mean µ Z (dbm) and variance σ 2 Z. The task is to find µ Z (dbm) and σ 2 Z. 4

5 Fenton-Wilkinson Method The mean µ Z (dbm) and variance σ 2 Z of Z (dbm) are obtained by matching the first two moments of I and Î. Switching from base 10 to base e: Ω k = 10 Ω k (dbm) /10 = e ξω k (dbm) = eˆω k where ˆΩ k = ξω k (dbm) and ξ = (ln10)/10 = Note that µˆωk = ξµ Ωk (dbm) σ 2ˆΩk = ξ 2 σ 2 Ω k The nth moment of the log-normal random variable Ω k can be obtained from the moment generating function of the Gaussian random variable ˆΩ k as E[Ω n k] = E[e nˆω k ] = e nµˆωk +(1/2)n 2 σ 2ˆΩ Here we have assumed identical shadow variances, σ 2ˆΩk = σ 2ˆΩ, which is a reasonable assumption. 5

6 Suppose that ˆΩ 1,..., ˆΩNI are independent with means µˆω1,..., µˆωni and identical variances σ 2ˆΩ. The appropriate moments of the log-normal approximation are obtained by equating the means on both sides of where Ẑ = ξz (dbm). This gives µ I = E[I] = N I N I k=1 k=1 e µˆωk Also equate the variances on both sides of E[eˆΩ k ] E[eẐ] = E[Î] = µ Î e (1/2)σ2ˆΩ = e µ Ẑ +(1/2)σ2 Ẑ (1) σ 2 I = E[I 2 ] µ 2 I E[Î2 ] µ 2 Î = σ2 Î This gives N I k=1 e 2µˆΩk e σ2ˆω(e σ2ˆω 1) = e 2µ Ẑe σ2 Ẑ(e σ2 Ẑ 1) (2) 6

7 To obtain µẑ and σ 2 Ẑ 1. Square Eq. (1) and divide by Eq. (2) to obtain σ 2 Ẑ. 2. Obtain µẑ from Eq. (1) The above procedure yields σ 2 Ẑ = ln µẑ = σ2ˆω σ 2 Ẑ 2 (eσ2ˆω 1) ( NI +ln N I k=1e 2µˆΩk k=1e µˆωk ) 2 +1 N I k=1 e µˆωk Given the means µˆω1,..., µˆωni and variance σ 2ˆΩ, µẑ and σ 2 Ẑ are easily obtained. Finally, we convert back to base 10 by scaling, such that µ Z (dbm) = ξ 1 µẑ σz 2 = ξ 2 σ 2 Ẑ where ξ =

8 Fenton s method breaks down in the prediction of the first and second moments for σ Ω > 4 db. Schwartz and Yeh s method yields the exact first and second moments. However, Fenton s method accurately predicts the tails of the complementary distribution function cdfc F c I (x) = P r(i x) and the cdf F I (x) = 1 F c I (x) = P r (I < x). We are interested in the accuracy of the approximations FI(x) c Q lnx µẑ F I (x) 1 Q when x is large and small, respectively. σẑ lnx µẑ σẑ The cdfc is more important than the cdf for outage calculations and predictions, since outages typically occur when the interference is large. 8

9 Wilkinson Schwartz and Yeh Farley Simulation P(I<x) N I =2 N I = x Comparison of the cdf for the sum of two and six log-normal random variables with various approximations; σ Ω = 6 db. 9

10 P(I>x) Wilkinson Schwartz and Yeh Farley Simulation x Comparison of the cdfc for the sum of two log-normal random variables with various approximations; σ Ω = 6 db. 10

11 P(I>x) Wilkinson Schwartz and Yeh Farley Simulation x Comparison of the cdfc for the sum of six log-normal random variables with various approximations; σ Ω = 6 db. 11

12 P(I>x) Wilkinson Schwartz and Yeh Farley Simulation x Comparison of the cdfc for the sum of six log-normal random variables with various approximations; σ Ω = 12 db. 12

13 Outage with Multiple Interferers 1. First obtain the mean and variance µ Z = µẑ/ξ σ 2 Z = σ 2 Ẑ /ξ2 ξ = Treat the average CIR as Gaussian distributed with mean and variance µ Λ(d) = µ Ω(d0 ) µ Z (dbm) σ 2 Λ(d) = σ 2 Ω +σ2 Z. 3. Compute the outage for a given location, described by d O(d) = Q µ Ω(d0 ) µ Z Λ th(db) σ 2 Ω +σ 2 Z 4. Average over all locations d by Monte Carlo integration O = R N O(d)p d (d)dd 13

14 Single Co-channel Interferer For a single co-channel interferer where p Λ(d)(dB) (x) = The outage for a given d is 1 exp 4πσΩ (x µ Λ(d) (db) ) 2 4σ 2 Ω µ Λ(d)(dB) = µ Ω(d0 ) (db) µ Ω(d1 ) (db) O(d) = Pr(Λ(d) (db) < Λ th(db) ) = Λ th(db) = Q 1 4πσΩ exp µ Λ(d)(dB) Λ th(db) 2σΩ (x µ Λ(d) (db) ) 2 4σ 2 Ω dx 14

15 R D-R serving MS co-channel BS BS Worst case interference from a single co-channel base-station. In this case d = (R,D R). The worst case outage due to a single co-channel interferer is O(R) = Q µ Ω(R)(dB) µ Ω(D R)(dB) Λ th (db) 2σΩ 15

16 Using a simple inverse-β path loss characteristic µ Ω(dB) = Ω (db) (d o ) 10βlog 10 (d/d o ) gives O(R) = Q 10log 10 ( D R 1) β Λth (db) 2σΩ The minimum CIR margin on the cell fringe is M Λ = 10log 10 D R 1 For an ideal hexagonal layout D R = 3N, so that N = β Λ th (db) M Λ +Λ 2 th (db) 10β +1 A small cluster size is achieved by making the margin M Λ and receiver threshold Λ th small. 16

17 Rician/Multiple Rayleigh Interferers Sometimes propagation conditions exist such that the received signals experience fading, but not shadowing. In this section, we calculate the outage probability for the case of fading only. The received signal may consist of a direct line of sight (LoS) component, or perhaps a specular component, accompanied by a diffuse component. The envelope of the received desired signal experiences Ricean fading. The interfering signals are often assumed to be Rayleigh faded, because a direct LoS is unlikely to exist due to the larger physical distances between the co-channel interferers and the receiver. Let the instantaneous power in the desired signal and the N I interfering signals be denoted by s 0 and s k, k = 1,, N I, respectively. Note that s i = α 2 i, where α2 i is the squared-envelope. The carrier-to-interference ratio is defined as λ = s 0 / N I k=1s k, and for a specified receiver threshold λ th, the outage probability is O I = P(λ < λ th ). 17

18 Single Interferer For the case of a single interferer, the outage probability reduces to the simple closed form O I = λ th exp λ th +A KA 1 1 λ th +A 1, where K is the Rice factor of the desired signal, A 1 = Ω 0 /(K + 1)Ω 1, and Ω k = E[s k ]. If the desired signal is Rayleigh faded, then the outage probability can be obtained by setting K = 0. 18

19 Multiple Interferers For the case of multiple interferers, each with mean power Ω k, the outage probability has the closed form O I = 1 N I k=1 1 λ th exp λ th +A KA k k λ th +A k N I j=1 j k A j A j A k, where A k = Ω 0 /(K +1)Ω k. This expression is only valid if Ω i Ω j when i j, i.e., the different interferers have different mean power. If all the interferers have the same mean power, then the outage probability can be derived as λ th O I = exp λ th +A KA 1 1 λ th +A 1 N I 1 k=0 A 1 (λ th +A 1 k k m=0 k m 1 m! Kλ th λ th +A 1 If the desired signal is Rayleigh faded, then the probability of outage with multiple Rayleigh faded interferers can be obtained by setting K = 0. m. 19

20 Probability of Outage, O I K = 0, N I = 1 K = 7, N I = 1 K = 7, N I = 2 K = 7, N I = Λ (db) Probability of outage with multiple interferers. The desired signal is Ricean faded with various Rice factors, while the interfering signals are Rayleigh faded and of equal power; λ th = 10.0 db. Λ = Ω 0 N I Ω 1 20