Spring 2017 MIMO Communication Systems Solution of Homework Assignment #5

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1 Spring 217 MIMO Communication Systems Solution of Homework Assignment #5 Problem 1 (2 points Consider a channel with impulse response h(t α δ(t + α 1 δ(t T 1 + α 3 δ(t T 2. Assume that T 1 1 µsecs and T 2 2 µsecs. You want to design a multicarrier system for the channel, with subchannel bandwidth B N B c /2. If raised cosine pulses with β 1 are used, and the subcarriers are separated by the minimum bandwidth necessary to remain orthogonal, then what is the total bandwidth occupied by a multicarrier system with 8 subcarriers? Assuming a constant SNR on each subchannel of 2 db, what is the maximum constellation size for MQAM modulation that can be sent over each subchannel with a target BER of 1 3, assuming M is restricted to be a power of 2. Also find the corresponding total data rate of the system. Sol: Since we know T c 2µsec, B c 1/T c 5 KHz and thus B N B c /2 25 KHz. So B NB N KHz. For SNR2 db and BER1 3, we have P e.2e 1.5γ/(M 1 < 1 3, which gives M < and we should choose M 16. Thus, Problem 2 (25 points R NR N N log 2 M/T N N log 2 M B N 1 + β 4 bps. Consider a high-speed data signal with bandwidth.5 MHz and a data rate of.5 Mbps. The signal is transmitted over a wireless channel with a delay spread of 1 µsec. (a If multicarrier modulation with nonoverlapping subchannels is used to mitigate the effects of ISI, approximately how many subcarriers are needed? What is the data rate and symbol time on each subcarrier? (We do not need to eliminate the ISI completely. So T s T m is enough Sol: Since delay spread T m T s is good enough and T m 1 5, we need every subcarrier channel has T s T m, i.e. B N 1/T m. we know B N B and thus 1/T N b N/T m. In other words, we need N Tm T b Thus, the minimum number of subcarriers that does not induce ISI is 5. If N 5, the data rate on each subcarrier is.5/5.1 Mbps and symbol time on each subacrrier is 1 µsec assuming BPSK is used. Assume for the remainder of the problem that the average received SNR (γ s on the nth subcarrier is 1/n (linear units and that each subcarrier experiences flat Rayleigh fading (so ISI is completely eliminated. Assume for the remainder of the problem that the average received SNR (γ s on the nth subcarrier is 1/n (linear units and that each subcarrier experiences flat Rayleigh fading (so ISI is completely eliminated. (b Suppose BPSK modulation is used for each subcarrier. If a repetition code is used across all subcarriers (i.e. a copy of each bit is sent over each subcarrier then what is the BER 1

2 after majority decoding? What is the data rate of the system? Sol: The average BER of the nth subcarrier for flat Rayleigh fading is given by P b (γ s 1 ( γs 1 1 (for large γ s, 2 γ s + 1 4γ s where γ s 1. If there are N subcarriers and a repetition code is used across all n subcarriers, the BER of this multicarrier channel is P all b N n1 P b (γ s 1 2 N N n1 ( γs 1 γ s + 1 N! 4 N if N is not very large. For example, if N 5, then we have P all b 5! Note that this repetition coding over multicarrier can achieve diversity order of N. The data rate in the system does not change and it is still.5 Mbps. (c Suppose you use adaptive loading (i.e. use different constellations on each subcarrier such that the average BER on each subcarrier does not exceed 1 3 (this is averaged over the fading distribution, do not assume that the TX and RX adapt power or rate to the instantaneous fade values. Find the MQAM constellation that can be transmitted over each subcarrier while meeting this average BER target. What is the total data rate of the system with adaptive loading? Sol: We know the BER of BPSK is P b 1 ( γs γ s Hence, the minimum required averaged SNR on each subcarrier is γ s For QAM, the average BER can be found in Table 6.1 and it is given by ( P b α M.5β M γ s 1. 2 log 2 M 1 +.5β M γ s For γ s and M 4, we have α M β M 1 and thus P b < 1 3. Thus, M 4 is good. For γ s and M 8, we have α M 4 and β M 3 and 7 thus P b > 1 3. So M 8 is not appropriate and it should not be used. Therefore, we can only use BPSK and 4-QAM for adaptive loading on the subcarriers. For the subcarriers with a high SNR, they can carry higher constellation size so that we can assign 4-QAM starting with the first carrier. Since γ s , n 4.2 and n thus we can use just 4 subcarriers. So for the first carrier, we can use M 4 since we have γ s 1 and P b < 1 3. For second carrier, we still can use M 4 since γ s 5 and P b < 1 3. For the third carrier, we should use BPSK because γ s 1/3 and P b < 1 3. For the forth carrier, we should use BPSK because γ s 25 and P b < 1 3. The total data rate is Problem 3 (3 points R.1 (2 log log Mbps. 2

3 Consider an one-dimensional cellular system deployed along a highway. The system has square cells of length 2R 2 Km, as shown in the figure below. This problem focuses on the downlink transmission from the base station to the mobiles. Assume that each cell has two mobiles located as is indicated in the figure, so the mobiles in each cell have the exact same location relative to their respective base stations. Assume that the total transmit power at each base station is P t 1 W, which is evenly divided between the two users in its cell. The total system bandwidth is 1 KHz, and the noise density at each receiver is 1 16 W/Hz. The propagation follows the model P r P t K(d /d 3, where d 1 m and K 12. All interference should be treated as AWN, and interference outside the first ring of interfering cells can be neglected. The system uses a frequency division strategy, with the bandwidth allocated to each base station evenly divided between the two users in its cell. Figure 1: One-Dimensional Cellular System with Square Cells (a For a reuse distance D 2 (frequencies reused every other cell, what bandwidth is allocated to each user in the system. Sol: Since each cell use the total bandwidth 1 KHz, two user share 1 KHz and thus each user can have 5 KHz. (b Compute the minimum reuse distance D required to achieve a 1 3 BER for BPSK modulation in fast Rayleigh fading. Sol: The average BER of BPSK is P b 1 and thus for P 4γ b 1 3 we have γ 25. Let the number of intermediate cells within the reuse distance D be n. So D 2R(n+1 and thus the bandwidth is shared by 2(n + 1 users. Thus, the SIR of a user can be written as SINR P t K(d /R 3 σ d B d /(2n P t K(d / D 2 + R σ d R 3 B d / ((2n + 2P t Kd 3 + 2(1/ 4(n γ 25, 3 where B d is the bandwidth and σ d is the power spectrum density of noise. For R 1m, σ d 1 16 W/Hz, B d 1 KHz, P t 5W, d 1 and K 1, we have n 3 and SINR Thus, D 8R 16km. (c Neglecting any fading or shadowing, use the Shannon capacity formula for user rates to compute the area efficiency of each cell under frequency division, where frequencies are reused every other cell (D 2. Sol:Since D 2 and each user has bandwidth of 5 KHz, the rate of a user is C u B u log 2 (1 + SINR 5k log 2 ( kbps. The area spectrum efficiency is C u /(2R 2 /B u.68 bps/hz/km 2. 3

4 Problem 4 (3 points Consider a CDMA system with the following SINR SIR Nc 1 χ i + N, where is the processing gain of the system, χ i s represent intracell interference and follow a Bernoulli distribution with probability α p(χ i 1 equal to the voice activity factor, and N characterizes intercell interference, and is assumed to be aussian with mean.247 N c and variance.78n c. The probability of outage is defined as the probability that the SIR is below some target : P out p(sir <. (a Show that Sol: P out p ( Nc 1 χ i + N > ] Nc 1 P out P Nc 1 χ i + N < SIR P χ i + N >. ] (b Find an analytical expression for P out. Sol: Since X N c 1 χ i, X Bi(α, N c 1. P X + N > P X + N > ] ] N c 1 n P n + N > ] X n PX n] ( Nc 1 PX n] α n (1 α Nc n 1. n N c 1 n N c 1 n N c 1 n P N > ] n X n PX n] N.247 P > / n.247n c.78nc.78nc ( /SIR n.247n c Q PX n]..78nc ] X n (c Using the analytical expression from part (b, compute the outage probability for N c 3 users, 12, α.5 and a target SIR of 5 (7 db. Sol: Using Matlab, we can find the outage probability is.258. (d Assume now that Nc is sufficiently large so that the random variable N c 1 χ i can be approximated as a aussian random variable. Under this approximation, find the distribution of N c 1 χ i + N as a function of N c and an analytical expression for outage 4

5 probability based on this approximation. Sol: If x can be approximated as aussian, then Problem 5 (2 points x N ((N c 1α, (N c 1α(1 α x + N N (.247N c + (N c 1α,.78N c + (N c 1α(1 α P x + N > ] ( / (.247N c + (N c 1α Q.78Nc + (N c 1α(1 α Consider a cellular system with hexagonal cells of radius R 2. Km. Suppose the minimum distance between cell centers using the same frequency must be D 5 Km to maintain the required SINR. (a Find the required reuse factor N and the number of cells per cluster. Sol: N 1 ( 2 D 1 ( R , take N 3 (b If the total number of channels for the system is 12, find the number of channels that can be assigned to each cell. Sol: The number of channels assigned in each cell is N c 12 N (c Sketch two adjacent cell clusters and show a channel assignment for the two clusters with the required reuse distance. Sol: Choose i 1, j 2. The red areas use the same channel. Figure 2: Cell clusters for Problem 5 5

6 Problem 6 (Coverage Probability in Small Cell Networks, 45 points An interference-limited small cell network consists of many small-sized base stations that are randomly distributed in a large circular region. In this simulation problem, we want to study how the interference of the base stations impacts on the coverage performance of a base station in a small cell network. Consider the base station X is serving one user located at the origin, called it reference user. If all base stations use the same transmit power, the coverage probability of the reference user for its base station is defined as H d α ] p cov P θ] P θ E ] e θi d α, (1 I where d is the constant distance from base station X to the origin, I M H i X i α is the interference power generated by M interfering base stations that are uniformly distributed in the circular region, X i denotes the distance from base station X i to the origin and all H i s are i.i.d. exponential random variables with unit mean and variance. Assume the small cell network is a circular area of radius R and its center is at the origin. (a The density of the base stations in the circular area is given by λ M πr 2 (base stations/m 2, which is the number of base stations per unit square meter. For the network parameters α 3, d 2 m and R 1 km, plot the diagram of coverage probability p cov versus base stations density λ for three different SIR thresholds θ.1,.5 and.1 provided that the typical density range is 1 5 λ 1 3. The theoretical approximated formula of the coverage probability can be shown as p cov ( 1 d2 θ 2 α R 2 R 2 d 2 θ 2 α 1 dx 1 + x α 2 πr 2 λ exp ( 2π2 d 2 θ 2 α λ α sin(2π/α as R 1. (2 Draw the simulation results and their corresponding theoretical results given by the formula in above in the same plot and check if they are very much the same or not. Sol: The simulation result is given in Figure 3: The simulation results perfectly match with the theoretical results given in (2. (b Now consider the base station has the channel state information (CSI of the user and thus it can perform channel-aware scheduling in the downlink transmission. Namely, the base station will transmit to its user whenever H δ, and this scheduling scheme can avoid transmitting whenever the user s channel is in a deep fade. Under this scheduling algorithm, the coverage probability of the user becomes H d α ] p cov P θ I H δ. (3 Using the same network parameters given in part (a and δ.5, plot the diagram of coverage probability p cov versus base station density λ for three different SIR thresholds θ.1,.5 and.1. Do you see p cov with scheduling is better than that simulated in part (a? Sol: The simulation results are given in Fig. 4 (c (Optional Derive the coverage probability in (2. Hint: (i First, condition on all node distances X i and then use the i.i.d. property 6

7 1.95 Coverage Probability θ.1 (Simulated Result θ.5 (Simulated Result θ.1 (Simulated Result.75 θ.1 (Theoretical Result θ.5 (Theoretical Result θ.1 (Theoretical Result Base Station Density (BSs/m 2 x 1 4 Figure 3: Coverage probability without channel-aware scheduling 1.95 Coverage Probability θ.1 (simulated result θ.5 (simulated result θ.1 (simulated result.75 θ.1 (theoretical result, no scheduling θ.1 (theoretical result, no scheduling θ.1 (theoretical result, no scheduling Base Station Density (BSs/m 2 x 1 4 Figure 4: Coverage probability with channel-aware scheduling between all H i s to find Ee θdα X i α H i X i ]. (ii Since all X i are also i.i.d and their pdf is f X (r 2r (why?, Ee θdα R 2 X i α H i ] R Ee θdα r α H i ]f X (rdr. (iii Use the following approximation identities to simplify your results and you should have the coverage probability shown in (2. R a u α 2 + a du a u α 2 + a du 2πa, for a > and R 1. α sin(2π/α ( 1 + x k mx e km, for x 1. 2 α 7

8 Ans: According to the result in (6, the outage probability can be rewritten as M p cov E e θdα H i X i α] M E e θdα H i X i α] since all H i s and X i s are independent. Also we know E e θdα H i X i α] ] ] X i α θd α E 1 E θd α + X i α θd α + X i α 1 π R 2 θd α dx R 2 x α/2 + θd α Since all {H i X j α } are i.i.d., the coverage probability is further given by p cov (1 πr2 R 2 dx x α/2 + θd α πr 2 λ ( exp 2π2 d 2 θ 2 α λ, as R 1. α sin(2π/α The Matlab codes for this problem are given in the following. % This code is used to simulate the coverage probability of % a small cell network. clear all ; d 2; % transmission distance alpha 3; % path loss exponent theta.1.5.1]; lambda.1:.2:.5; R 8; % The radius of a cell hsamp 2; % Fading channel gain samples M round (( pi*r.^2 * lambda ; % Number of BSs delta.5; % Threshold of channel - aware scheduling DropNum 5; 1./(1+ x.^( alpha /2; PcovThe zeros ( length ( theta, length ( lambda ; for l 1: length ( theta a( d ^2* theta (l ^(2/ alpha /R ^2; b integral (fun,,1/ a; PcovThe (l,:(1 - b*a.^(( pi*r.^2 * lambda ; end Pcov zeros ( length ( theta, length ( lambda ; for k 1: length ( lambda PcovSum zeros (3,1; Io ; Spow ; H ; for m 1: DropNum Hs exprnd (1,1, hsamp ; HHs( find (Hs > delta ; Spow H.* d ^(- alpha ; rr* sqrt ( rand (M(k,1; 8

9 %r 1+(R -1* sqrt ( rand (M(k,1; ang -pi+pi* rand (M(k,1; Xr.* cos ( ang +j*r.* sin ( ang ; hdelta length (H; th exprnd (1,M(k, hdelta ; Iosum (th.*(( abs (X.^( - alpha * ones (1, hdelta,1; SIR Spow./ Io; for l 1: length ( theta % Coverage probability PcovSum (l PcovSum (l+ length ( find (SIR > theta (l / hdelta ; end end Pcov (:,k PcovSum / DropNum ; CovProbPPP figure (1; plot ( lambda (1: k, Pcov (:,1:k, lambda (1: k, PcovThe (:,1:k; xlabel ( Base Station Density ( BSs /m ^2 ; ylabel ( Coverage Probability ; grid ; saveas ( CovProbPPP, CovProbPPP2, jpg ; end 9