Spring 2016 Network Science Solution of Homework Assignment #5

Transcription

1 Spring 216 Network Science Solution of Homework Assignment #5 Problem 1 [PGFL and Its Application] (4%) In a wireless cellular network, suppose users associate with a base station (BS) by the nearest BS association policy. Namely, they associate with the BS that is nearest to them. Due to the limited resource at each BS, the probability that a base station is able to accept a new connection request from users is ρ. (a) If all BSs form a homogeneous PPP of intensity λ, use the probability generating functional (PGFL) of a PPP to show the following: E [R1] ɛ = Γ(1 + ɛ ) 2, (1) (πρλ) ɛ 2 where ɛ >, R 1 is the distance from a user to its associated accessible (nearest) BS and Γ(x) = t x 1 e t dt is the Gamma function. Sol: Since the event that a base station accepts a new connection happens indepently over the entire network, the base stations that can accept a new connection request form a thinning PPP of intensity ρλ. Let Φ c be the set of base stations that can accept a new connection and the associated BS for the user located at the origin can be written as X = arg min { X i }. X i Φ c Hence, the CDF of R1 ɛ can be written as [ ] P[R1 ɛ x] = 1 P[R 1 x 1/ɛ ] = 1 P min { X i } x 1/ɛ X i Φ c { [ ]} = 1 E P min X i x 1/ɛ X i Φ c Φ c { } = 1 E. X i Φ c P [ X i x 1/ɛ ] Now let v(x) = P [ X i x 1/ɛ] and it follows that [ ] [ ] [ G(v) = E v(x i ) = E v(x i ) = E P [ X i x ] ] 1/ɛ X i Φ c X i Φ c X i Φ c ( ) ( ) x 1/ɛ = exp 2πρλ P[r x 1/ɛ ]rdr = exp 2πρλ rdr = exp ( πρλx 2/ɛ). Thus, P[R ɛ 1 x] = 1 exp ( πρλx 2/ɛ) 1

2 and E[R ɛ 1] is given by E[R ɛ 1] = = = P[R ɛ 1 x]dx = ɛ 2 (πρλ) Γ(1 + ɛ/2). (πρλ) ɛ 2 ɛ 2 u ɛ exp ( πρλx 2/ɛ) dx 2 1 e u du = ɛ Γ(ɛ/2) 2 (πρλ) ɛ 2 (b) (Computer Simulation) Run a computer simulation to simulate E[ R 1 ] and verify whether the simulation results coincide with the theoretical result in (1) for ρ =.5,.75. Sol: See the Simulation results in Fig. 1 and Matlab code in below. Mean of the ɛ-moment of R Simulated Result (ρ =.5) Theoretical Result (ρ =.5) Simulated Result (ρ =.75) Theoretical Result (ρ =.75) x 1 3 Intensity (Nodes/m 2 ) Figure 1: E[R ɛ 1] v.s. Intensity for ρ =.5,.75. Matlab Code for Problem 1 clear all lambda = 4e -4:1 e -4:2 e -3; % TX intensity L =5; A =(2* L )^2; % Network area Nsamp =8; % Number of Samples rho =.75; 2

3 epsilon =.5; EpsMomDis = zeros (1, length ( lambda )); Dmin = zeros (1, Nsamp ); for i =1: length ( lambda ) N= poissrnd ( lambda (i)*a,[ Nsamp,1]); for k =1: Nsamp V= binornd (1,rho,[1,N(k )]); X= unifrnd (-L,L,sum (V),2); DisVec = sqrt (X (:,1).^2+ X (:,2).^2); Dmin (k)= min ( DisVec ); EpsMomDis (i)= mean ( Dmin.^( epsilon )); TheEpsMomDis = gamma (1+ epsilon /2)./(( pi*rho * lambda ).^( epsilon /2)); plot ( lambda, EpsMomDis, b.-o, lambda, TheEpsMomDis, k.-s ); xlabel ( Intensity ( Nodes /m$ ^2$) ); ylabel ( Mean of the epsilon - moment of $R_1$ ); grid on; Problem 2 [Simulation of Outage Probability] (6%) Consider a large-scale interference-limited wireless ad hoc network in which all transmitters (TXs) form a homogeneous PPP Φ of intensity λ and each of them only has a unique inted receiver (RX). Namely, the network consists of many TX-RX pairs and the transmission distance in each TX-RX pair is a fixed constant d (A schematic illustration of this network model can be referred to page 43 of Lecture 7). For successful decoding, the SIR of the typical RX Y o located at the origin must be no less than the threshold θ, i.e., SIR o = P oh o d α I o θ, where H o is the channel power gain from the TX X o of typical RX Y o to the typical RX, P o is the transmit power used by the TX, α > 2 is the pathloss exponent and I o is the interference given by I o = P i Hi X i α, X i Φ\X o where P i is the transmit power used by TX X i and H i is the interference channel gain from TX X i to typical RX Y o. All channels in the network undergo the same Rayleigh fading, i.e., all channel gains (H i s and H i s) are i.i.d. exponential random variables with unit mean and variance. (a) Suppose every TX tosses a coin to determine if they should transmit at current time slot. Every TX will transmit if they get Head after tossing their coin. If the probability of having a Head after tossing a coin is q, then the outage probability p out can be found as [ Po H o p out (λ) P d α I o ] < θ = 1 exp ( 2π2 λqd 2 θ 2 α α sin(2π/α) if all transmit powers are the same (i.e., P o = P 1 = P 2 = = P ). Run a computer simulation to verify whether this theoretical outage probability is correct or not. Draw 3 )

4 a figure of p out (λ) vs. λ to show the simulated and theoretical results for transmit distance d = 2, threshold θ = 1, q =.3679, and α = 4, suggested range of λ: Check whether your simulated outage probability perfectly coincides with the theoretical outage probability derived in above. Sol: See the Simulation results in Fig. 2 and Matlab code in below Simulation Result Theortical Result.45 Outage Probability Intensity (Nodes/m 2 ) Figure 2: Simulation result of outage probability when each TX transmit with probability q. Matlab Code for Problem 2 (a) %% Parameter Setting lambda = 1e -4:1e -4:1e -3; SimRad =5; SimArea = SimRad ^2* pi; NumSam =1 e1; NumDrops =1 e5; D =2; alpha =4; theta =1; thin =.3679; % Thinning Parameter q lambda_thin = lambda * thin ; q= zeros (1, length ( lambda )); qth =q; 4

5 %% Theoretical Result num =2* pi ^2* lambda * thin *D ^2* theta ^(2/ alpha ); den = alpha * sin (2* pi/ alpha ); qth =1- exp (- num./ den ); %% Simulation code for m =1: length ( lambda_thin ) MeanTX = round ( lambda_thin (m)* SimArea ); NumTX = poissrnd ( MeanTX,1, NumDrops ); temp =; for k =1: NumDrops % Generate node position Rad = sqrt ( rand (1, NumTX (k )))* SimRad ; Ang = rand (1, NumTX (k ))*2* pi; Xt=Rad.* cos ( Ang )+j* Rad.* sin ( Ang ); % Received signal power S= exprnd (1,1, NumSam ).* D^(- alpha ); % Channel fading between each node % to the reference point Ht= exprnd (1, NumTX (k), NumSam ); % Interference power I=abs (Xt ).^( - alpha )* Ht; SIR =S./ I; temp = temp +( length ( find (SIR < theta ))/ NumSam ); q(m)= temp / NumDrops ; %% Plot Section plot ( lambda,q, bo, lambda,qth, b - ); leg ( Simulation Result, Theortical Result ); xlabel ( Intensity ( Nodes /m ^2) ); ylabel ( Outage Probability ); grid ; (b) Now we assume every RX can feedback its channel gain information to its TX. If every TX use the same power and only transmits when its channel gain is greater than some threshold β (i.e., their transmitting probability is q = P[H o β]), then run a computer simulation to draw a figure of p out (λ) vs. λ by using distance d = 2, threshold θ = 1, β = 1, and α = 4. Compare the result here with the result obtained in (a). Can you see which one is better? Explain why. Sol: In (a), we calculate the outage probability no matter what the channel gain(h ) is. In (b), we only calculate when the channel fading H is good enough. Since the impact of the thinning and scheduling to the interference are the same, scenario (b) further assures the quality of received signal compared to (a). See the Simulation results in Fig. 3 and Matlab code in below. %% Parameter Setting lambda = 1e -4:1e -4:1e -3; SimRad =5; SimArea = SimRad ^2* pi; NumSam =1 e4; NumDrops =1 e4; 5

6 .6.5 Simulation Result without Scheduling Theortical Result without Scheduling Simulation Result with Scheduling Outage Probability Intensity (Nodes/m 2 ) Figure 3: Simulation result of outage probability with/without scheduling. D =2; alpha =4; theta =1; thin =.3679; % Thinning Parameter q lambda_thin = lambda * thin ; q= zeros (1, length ( lambda )); qth =q; %% Theoretical Result num =2* pi ^2* lambda * thin *D ^2* theta ^(2/ alpha ); den = alpha * sin (2* pi/ alpha ); qth =1- exp (- num./ den ); %% Simulation code for m =1: length ( lambda_thin ) MeanTX = round ( lambda_thin (m)* SimArea ); NumTX = poissrnd ( MeanTX,1, NumDrops ); temp =; temp_sch =; for k =1: NumDrops Rad = sqrt ( rand (1, NumTX (k )))* SimRad ; Ang = rand (1, NumTX (k ))*2* pi; Xt=Rad.* cos ( Ang )+j* Rad.* sin ( Ang ); H= exprnd (1,1, NumSam ); S=H*D^(- alpha ); 6

7 Ht= exprnd (1, NumTX (k), NumSam ); I=abs (Xt ).^( - alpha )* Ht; SIR =S./ I; temp = temp +( length ( find (SIR < theta ))/ NumSam ); % Simulation with scheduling. Ss=S( find (H >1)); SIRs =Ss./ I (1: length (Ss )); temp_sch = temp_sch +... ( length ( find (SIRs < theta ))/ length (Ss )); q(m)= temp / NumDrops ; q_sch (m)= temp_sch / NumDrops ; %% Plot Section plot ( lambda,q, bo, lambda,qth, b - ); hold on plot ( lambda, q_sch, r-o ); leg ( Simulation Result without Scheduling,... Theortical Result without Scheduling,... Simulation Result with Scheduling ); xlabel ( Intensity ( Nodes /m ^2) ); ylabel ( Success Probability ); grid ; (c) Consider the same situation in part (a) again, but now every TX use the channel inversion power control scheme, i.e., each TX use the reciprocal of its channel gain as its transmit power. For example, transmitter X o uses transmit power P o = 1 H o to transmit to its RX Y o. All TXs use the same transmit power control scheme. Thus, the outage probability in above becomes [ ] 1 p out (λ) P < θ, d α I o where I o is given by I o = X i Φ (H 1 i ) H i X i α and note that H i and H i are i.i.d. Run a computer simulation to draw a figure of p out (λ) vs. λ for this channel-inversion power control scheme by using the same parameters in (a), i.e., distance d = 2, threshold θ = 1, β = 1, and α = 4. Compare the simulation result here with the result in part (a). Which one is better? Can you explain the phenomenon you observe? Sol: By using channel inversion power control scheme, the power eliminates the effect of channel gain and leads to a constant received signal power. In the view of the interfering part, the power of the interference is indepent of the channel gain to the reference point and can be possibly larger than 1 due to the property of exponential distribution. Hence, the SIR after inversion power control scheme is less than the one without power control, yet we can observe that the channel gain benefits the SIR. See the Simulation results in Fig. 4 and Matlab code in below. 7

8 %% Parameter Setting lambda = 1e -4:1e -4:1e -3; SimRad =5; SimArea = SimRad ^2* pi; NumSam =1 e1; NumDrops =1 e3; D =2; alpha =4; theta =1; thin =.3679; % Thinning Parameter q lambda_thin = lambda * thin ; q= zeros (1, length ( lambda )); qth =q; %% Theoretical Result num =2* pi ^2* lambda * thin *D ^2* theta ^(2/ alpha ); den = alpha * sin (2* pi/ alpha ); qth =1- exp (- num./ den ); %% Simulation code for m =1: length ( lambda_thin ) MeanTX = round ( lambda_thin (m)* SimArea ); NumTX = poissrnd ( MeanTX,1, NumDrops ); temp =; temp_pc =; for k =1: NumDrops Rad = sqrt ( rand (1, NumTX (k )))* SimRad ; Ang = rand (1, NumTX (k ))*2* pi; Xt=Rad.* cos ( Ang )+j* Rad.* sin ( Ang ); H= exprnd (1,1, NumSam ); S=H*D^(- alpha ); % Channel fading from nodes to reference RX Ht= exprnd (1, NumTX (k), NumSam ); % Channel fading between the reference pair Hi= exprnd (1, NumTX (k), NumSam ); I=abs (Xt ).^( - alpha )* Ht; SIR =S./ I; temp = temp +( length ( find (SIR < theta ))/ NumSam ); % inversion channel power control. P =1./ H; Pi =1./ Hi; Spc =P.* H*D^(- alpha ); Ipc = abs (Xt ).^( - alpha )*( Ht.* Pi ); SIRpc = Spc./ Ipc ; temp_pc = temp_pc +... ( length ( find (SIRpc < theta ))/ NumSam ); q(m)= temp / NumDrops ; q_pc (m)= temp_pc / NumDrops ; %% Plot Section plot ( lambda,q, b-o, lambda,qth, b - ); 8

9 hold on plot ( lambda,q_pc, r-o ); leg ( Simulation Result in (a),... Theortical Result in (a),... Simulation Result in (c) ); xlabel ( Intensity ( Nodes /m ^2) ); ylabel ( Outage Probability ); grid ;.7.6 Simulation Result in (a) Theortical Result in (a) Simulation Result in (c).5 Outage Probability Intensity (Nodes/m 2 ) Figure 4: Simulation result of outage probability with/without channel inversion power control. 9