Wireless channels: path loss models

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1 Wireless channels: path loss models
Αν. Καθηγητής Γεώργιος Ευθύμογλου Module Title

2 Εισαγωγή Path-loss models for various propagation environments
Link Budget Analysis Calculation of received SNR Statistical description of wireless channels Frequency Response of fading channels Module Title

3 Modulation and coding schemes (MCSs)
Q: where can I use 64-QAM? Adaptive physical layer: use different modulation and coding scheme depending on the BS-user distance!!!

4 Available MCSs The MCS used depends on the received SNR at the receiver side. Example for 3.5MHz bandwidth: ID MCS Received power (dBm) SNR (dB) Throughput Mb/s 1 BPSK 1/2 -91 6.4 1.33 2 QPSK 1/2 -88 9.4 2.67 3 QPSK 3/4 -86 11.2 4.00 4 16-QAM ½ -81 16.4 5.33 5 16-QAM 3/4 -79 18.2 8.00 6 64-QAM 2/3 -74 22.7 10.07 7 64-QAM 3/4 -73 24.4 12.00

5 Average received power at distance d
Received power of desired signal at distance d based on free space path-loss only, ), is expressed by the well-known Friis equation, given as (L is for hardware losses, Gt is transmit antenna gain, Gr is receive antenna gain) Free space path loss: In dB, where fMHZ is carrier frequency in MHz and dKm is distance in Km.

6 Average received power at distance d
Simplified path-loss model with depending on propagation environment (Average) Received SNR at distance d Modified Friis equation (modified free-space path loss) with the path loss exponent n that varies with the environments. In dB, we have

7 Path loss model: Modified Friis equation
Path loss exponent varies between 2 – 6 depending on the propagation environment.

8 Path loss (PL) models and link budget (LB)
LB is used to determine the distance from the BS, d, that a specific received power and SNR at the receiver side is achieved: Many statistical and empirical propagation models of path-loss (PL) exist. (Average) Received SNR at distance d

9 Example 1 of link budget Consider a base station transmitting to a mobile station in free space. Distance between base station and mobile station: 8000 m Transmitter frequency: 1.5 GHz ( 0.2 m) Base station transmitting power, Pt =10 W (10 dBW) Total system losses: 8 dB Mobile receiver noise figure F = 5 dB Mobile receiver antenna temperature 290 K Mobile receiver bandwidth BW = 1.25 MHz Antenna gains are 8 dB and 0 dB for the base station and mobile station, respectively. Calculate the received signal power at the mobile receiver antenna and signal-to-noise ratio (SNR) of the received signal.

10 Example 1 of link budget Free space path loss at distance 8000 m Received power Noise power: Received SNR

11 Example 1 of link budget The noise power is related to a noise temperature Te by the equation where k is Boltzmann’s constant k=1.38*10-23 and BW is for bandwidth. The Noise Figure of the receiver is related to the noise temperature: where T0 is room temperature 2900 Kelvin. Assuming ANTENNA temperature TA= T0, we have where N0 is the power spectral density of thermal noise at room temperature.

12 Example 2 of link budget Calculate the received power at a distance of 3 km from the transmitter if the path-loss exponent is 4. Assume the transmitting power of 4 W at 1800 MHz, a shadow effect of 10.5 dB, and the power at reference distance (d0=100 m) of -32 dBm. What is the allowable path loss?

13 Empirical propagation path loss models
Various path loss models exist. These models depend on the different terrain types of network operation: urban, suburban, rural where PLurban > PLsuburban > PLrural Empirical propagation models of path-loss (PL) include: Erceg’s model – SUI models

14 Empirical propagation path loss models
Empirical propagation models of path-loss (PL) include: 2. Okumura-Hata model

15 Empirical propagation path loss models
Comparison between Cost 231 and Hata-Okumura model

16 Erceg’s PL model According to the SUI model, the path loss at distance d is given by where is the fixed part of path loss attenuation at distance d and s is a random variable with lognormal distribution that accounts for shadowing and other phenomena. Its value ranges between 8.2 dB and 10.6 dB. Parameter γ is given by where is the height of the base station antenna in meters. The values of α, b, c for each terrain category are provided in the link budget computation below.

17 Erceg’s PL model Two correction terms are needed in this model to account for different frequencies and CPE antenna heights. where f is the frequency of operation and is the height (in m) of the antenna CPE. The value of path loss exponent for outdoor environments is a function of terrain (urban, suburban, rural, and foliage type). The link budget will be used in the capacity dimensioning to estimate the distances from the BS where a particular data rate can be provided based on the assigned MCS. The link budget equations are implemented in Excel software, as shown below:

18 Frequency correction factor, dB 1,458 Wavelength, m 0,086
Center frequency, GHz 3,50E+00 Frequency correction factor, dB 1,458 Wavelength, m 0,086 Reference distance, do, m 100 Base station antenna height, m 25 User station antenna height, m 3 NLOS OUTDOOR PATH LOSS MODEL (Erceg et al, 2001) NLOS Terrain Category A B C a 4,6 4,0 3,6 b 0,0075 0,0065 0,0050 c 12,6 17,1 20,0 Path Loss Exponent, g, (dB/decade) 4,92 4,52 4,28 Standard deviation of shadow fading (dB), s 10,6 9,6 8,2 Target Cell area Coverage (%) 90 K = Shadow_Margin_dB/Shadow_Std_devn_dB 0,67 0,63 Shadow fade allowance S (dB) 7,1 6,4 5,1 Path loss at do (dB) 83,3 User antenna height correction (dB) -1,90 -3,52 Total modifications (dB) -0,44 -2,06 Shadowing + Total modifications (dB) 6,66 5,94 3,06 NLOS INDOOR ADJUSTMENTS NLOS INDOOR ADJUSTMENTS Outdoor-to-Indoor additional median loss (dB) 10,00 Composite std. dev. for lognormal fading 12,00 11,00 Indoor SS antenna height m 3,00 Indoor SS antenna height adjustment (dB) 0,00 Additional Shadow fade allowance (dB) 0,94 0,93 1,13 Total additional indoor loss (dB) 10,94 10,93 11,13 EXAMPLE PATH LOSS FOR GIVEN TARGET DISTANCE Target distance, d, m 2500 Median Path Loss, dB 151,6 146,1 141,0 NLOS Outdoor Total Path Loss Budget (dB) 158,7 152,5 NLOS Indoor Total Path Loss Budget (dB) 169,6 163,4 157,3

19 Link Budget Analysis Example: at d=2500 m, the path loss in the excel sheet πεδίο Β39 is determined as follows: 158,7=$B$23+10*B13*LOG10($B$37/$B$4)+$B$2+B24+B17 that is, according to the path loss equation: (in the excel sheet: Xf is at B2, Xh is at B24, shadowing is at B17)

20 SNR determination from Link Budget Analysis
Example: the receiver noise power N in the excel sheet πεδίο Β15 is determined as follows: (N0 = kT0 = -174 dBm/Hz) = *LOG($B$4* )+$B$13+$B$14 that is, according to the equation (in dB): (in the excel sheet: noise figure NF is at B13 and losses is at B14) The SNR (in dB) at distance d is then determined as follows:

21 SNR determination from Link Budget Analysis
Example: The SNR (in dB) at distance d=2500 is then determined as follows: The received power is

22 Link budget analysis Note that sensitivity received power Pr(d), that is minimum required power to obtain a required SNR value, can be obtained as For BPSK ½ MCS For 64-QAM ¾ MCS

23 Link budget analysis Example: Outdoor CPE in Type A terrain Assuming that we use BPSK ½ we need to have minimum received power Pr(d) = -97dBm at the receiver. How much path loss is allowed? is found to be 153dB, obtained in cell B30, using the equation B30=$B$9-B19+$B$20-$B$21+B25+B26+B27-B28-B29

24 Link budget analysis At which distance d are we going to have the above path loss ? Assuming that we use the Erceg’s path loss model, the distance where we have the maximum allowed path loss is obtained in cell B36 using:

25 Link budget analysis BS Tx Parameters SS Rx Parameters DOWNLINK
Modulation BPSK QPSK 16-QAM 64-QAM Code Rate 1/2 3/4 2/3 Channel bandwidth (MHz) 3,5 BS Tx Parameters Average Tx power (dBm) 30 Tx antenna gain (dBi) 18 Other Tx power gain dB Transmit Power EIRP (dBm) 48,0 SS Rx Parameters Required SNR (dB) 6,4 9,4 11,2 16,4 18,2 22,7 24,4 Rx Noise Figure (dB) 4,0 Implementation Loss (dB) 1,0 Rx. noise power (dBm) -103,6 Required received power (dBm) -97,2 -94,2 -92,4 -87,2 -85,4 -80,9 -79,2 Rx sensitivity spec (dBm) -91,0 -88,0 -86,0 -81,0 -79,0 -74,0 -73,0 Rx. perf. improvement over std (dB) 6,0 Rx sensitivity (dBm) -97,0 -94,0 -92,0 -87,0 -85,0 -80,0 Outdoor SS Rx antenna gain (dBi) 18,0 Outdoor SS Rx antenna gain reduction (dB) 0,0 Indoor SS Rx antenna gain (dBi) Indoor SS antenna gain reduction (dB)

26 Link budget analysis Overall Link Budget Range analysis LOS (km)
Diversity gain (MRC, STC, MIMO etc) (dB) 0,0 Beamforming gain (AAS, STC, MIMO etc) (dB) Other gains (dB) Wideband/Temporal Fade Margin (dB) 10,0 Other adjustments (dB) Total Allowed Outdoor SS Loss+Shadow Fade (dB) 153,0 150,0 148,0 143,0 141,0 136,0 135,0 Total Allowed Indoor SS Loss+ShadowFade (dB) 138,0 131,0 129,0 124,0 123,0 Range analysis LOS (km) 305 216 171 96 77 43 38 Near LOS (km) Category A NLOS Outdoor (km) 1,9 1,7 1,5 1,2 1,1 0,9 0,8 NLOS Indoor (km) 0,7 0,6 0,5 0,4 0,3 Category B 2,6 2,2 2,0 1,4 1,0 Category C 3,6 3,1 2,8 2,1

27 Example design of WLAN In a WLAN the minimum SNR required is 12 dB for an office environment. The background noise at the operational frequency is –115 dBm. If the mobile terminal transmit power is 100 mW, what is the coverage radius of an access point if there are three floors between the mobile transmitter and the access point? Solution Transmit power of mobile terminal 10 log10 (100) = 20 dBm Receiver sensitivity=background noise + minimum SNR = =-103 dBm Maximum allowable path loss = transmit power – receiver sensitivity = 20 – (- 103 ) =123 dB

28 Example design of WLAN Indoor path loss model (d0 = 1 m)

29 Statistical description of wireless channels

30 Wireless channel models
There are two distinct fading models for the wireless channel

31 Received signal Frequency Flat fading channel (narrowband systems) Frequency selective channel (wideband systems) Each multipath component is typically associated with different time delay and attenuation, the shortest of which is the LOS path.

32 Narrowband wireless multipath channel
Έστω ότι το σήμα εκπομπής είναι ένα απλό ημιτονοειδές σήμα στη συχνότητα του φέροντος (μη διαμορφωμένο σήμα στην fc): Το λαμβανόμενο σήμα δίνεται ως (αγνοώντας για λίγο τον θόρυβο)

33 Narrowband wireless multipath channel
όπου ai είναι η εξασθένιση (attenuation) της i-th multipath συνιστώσας θi είναι η αλλαγή φάσης (phase-shift) της i-th multipath συνιστώσας Θεωρούμε ότι οι Ν1 συνιστώσες φτάνουν σχεδόν ταυτόχρονα, οπότε αυτό που παρατηρείται στο δέκτη είναι μόνο το συνισταμένο σήμα. Οι όροι ai και θi είναι τυχαίες μεταβλητές (random variables). Η παραπάνω έκφραση μπορεί να αναλυθεί στις δύο ορθογώνιες συνιστώσες {cos, -sin} του φέροντος: Επίσης αν εισάγουμε δύο τυχαίες διαδικασίες (random processes) X1(t) και X2(t) μπορούμε να γράψουμε:

34 Narrowband wireless multipath channel
Αν το Ν1 είναι αρκετά μεγάλο (μεγάλος αριθμός από scattered waves είναι παρόντα), και χρησιμοποιώντας το Central Limit Theorem, μπορούμε να προσεγγίσουμε τα X1(t) και X2(t) με Gaussian random variables με zero mean και variance σ2. Η προηγούμενη σχέση γίνεται: όπου το πλάτος της διάλειψης του καναλιού (channel fading) R(t) δίνεται ως και η ισχύς της διάλειψης του καναλιού

35 PDF of sum of Gaussian random variables
If X1, ..., Xk are k independent Normal random variables ~G(0, 1), then the sum of their squares, is distributed according to the chi-squared distribution with k degrees of freedom. This is usually denoted as The chi-squared distribution has one parameter: k — a positive integer that specifies the number of degrees of freedom (i.e. the number of Xi’s). PDF is given by

36 PDF of sum of 2 Gaussian random variables
For k=2, is the exponential distribution For the Rayleigh distribution follows chi-squared with 2 degrees of freedom that is, exponential distribution: Z = R2 (dZ = 2R dR)

37 Statistical description of fading
Όταν οι X1(t) και X2(t) είναι Gaussian random variables with equal variances σ2, η R(t) είναι Rayleigh distributed random variable με μέση ισχύ Ω = Ε[R2] = 2 σ2 : Η φάση του σήματος λήψης θ(t) δίνεται ως όπου αφού οι X1(t) και X2(t) είναι Gaussian random variables μπορεί να δειχτεί ότι η θ(t) είναι uniform distributed random variable με PDF:

38 Received signal in a flat fading channel
In a frequency flat fading channel (or narrowband system), the CIR (channel impulse response) reduces to a single impulse (just one path from tx to receiver) scaled by a time-varying complex coefficient: h(t)~Gauss(0, 1/2)+ j Gauss(0, 1/2) The received (equivalent lowpass) signal is of the form α(t) is the magnitude of h(t) (α(t) has Rayleigh pdf) h(t) is constant for many symbol intervals Phase θ(t) varies slowly and can be tracked

39 BER vs. Average SNR (cont.)
Fading h varies with time  SNR γ varies with time Let us define instantaneous SNR and average SNR:

40 BER vs. Average SNR (cont.)
Since using we get Rayleigh distribution Exponential distribution

41 Fading Models Rayleigh: Η κατανομή Rayleigh χρησιμοποιείται συχνά σε multipath fading μοντέλα χωρίς LOS, δηλαδή χωρίς απευθείας διαδρομή. Σε αυτή την περίπτωση το πλάτος του καναλιού διάλειψης R είναι random variable με PDF (Σχήμα 1a): όπου δηλώνει τη μέση ισχύ της διαδρομής λήψης, που είναι το άθροισμα όλων των ανακλωμένων διαδρομών που φτάνουν με την ίδια καθυστέρηση. Επομένως, το στιγμιαίο SNR ανά σύμβολο, είναι τώρα μία random variable με PDF chi-squared που δίνεται από όπου

42 PDF of SNR with Rayleigh fading
Proof : finding the pdf of PDF of R

43 Rayleigh amplitude pdf and Exponential power
Pdf of random variable R that follows Rayleigh distribution Pdf of γb =R2Eb/N0 Average received SNR Es/N0 = Pr(d) / N depends on distance d!!! according to link budget.

44 FADING in wireless channels
Whenever relative motion exists between transmitter and receiver, there is a Doppler shift in the received signal. The maximum Doppler shift fd is given by Fading power για fd = 50 Hz.

45 FADING in wireless channels
Fading power για fd = 160 Hz.

46 FADING in wireless channels
Fading power για fd = 160 Hz.

47 Coherence time of channel
The coherence time, Tc, describes the expected time duration over which the impulse response of the channel stays relatively constant. The coherence time is approximately inversely proportional to Doppler spread If the transmitted symbol interval, Ts, exceeds Tc, then the channel will change during the symbol interval and symbol distortion will occur. If signal symbol time Ts <<Tc the channel does not change during the symbol interval. This case is called slow fading.

48 Frequency Response of fading channels

49 Discrete time signal modeling
In all communication systems, we assume a sampling frequency Fs, which is related to the transmit bandwidth In the time domain, the signal samples are separated in time by the sampling time Usually we assume that the multipath components appear in time instants that are multiples of Ts. The frequency response of a multipath channel is given by with , where τmax is the maximum delay spread of the channel, and ω=2πf/Fs.

50 Ψηφιακή γωνιακή συχνότητα
NOTE: για διακριτού χρόνου σήματα (με συχνότητα δειγματοληψίας Fs), οι αναλογικές συχνότητες στο διάστημα [0-Fs) εμφανίζονται ως αντίστοιχες ψηφιακές γωνιακές στο [0-2π). Π.χ. η συχνότητα 20Hz με δειγματοληψία 200Hz εμφανίζεται με ψηφιακή γωνιακή συχνότητα 0.2π Προκύπτει ότι η συχνότητα Fs ισοδυναμεί με τη ψηφιακή γωνιακή συχνότητα 2π: Fs ↔ 2π

51 Frequency response of channel
The frequency response of the channel is usually sampled in N frequencies in the frequency interval [0 – Fs). The frequency spacing is then given by Fs/N. Using digital frequencies, which map the interval [0 – Fs)[0, 2π), the frequency spacing in digital radian frequency is 2π/Ν. The sampled frequency response at ω=2πk/N, k=0,1,…, N-1, is given by the Discrete Fourier Transform (DFT)

52 Frequency response of 1 path channel
Assuming L=1 path channel, h(n) = h(0), it is easy to see that it results in a frequency flat channel As we will show later, each path h(n) is the summation of signal replicas with random amplitude and phase that arrive almost at the same delay. The resultant signal is statistically modeled as a complex Gaussian signal and its amplitude follows the Rayleigh probability density function (pdf), as we saw earlier.

53 Frequency flat channel
h_flat = (randn(1,1)+i*randn(1,1))/sqrt(2); % complex Gaussian N=64; H = fft(h_flat, N); % frequency response of channel response b = abs(H); % magnitude of frequency response plot(b, '-o') plot([1:N], b, '-o') xlabel('subcarrier index') ylabel('frequency response‘)

54 Frequency selective channel: time domain
p=[0.5, 0.3, 0.2, 0.1]; % multipath power profile h(1) = sqrt(p(1))*(randn(1,1)+i*randn(1,1))/sqrt(2); h(2) = sqrt(p(2))*(randn(1,1)+i*randn(1,1))/sqrt(2); h(3) = sqrt(p(3))*(randn(1,1)+i*randn(1,1))/sqrt(2); h(4) = sqrt(p(4))*(randn(1,1)+i*randn(1,1))/sqrt(2); h = [h(1), 0, h(2), h(3), 0, 0, h(4)]; %multipath delay profile stem(abs(h)); xlabel('path number, n') ylabel('abs(h)')

55 Frequency selective channel: frequency response
N=64; H = fft(h, N); % frequency response of last channel response b = abs(H); % magnitude of frequency response plot(b, ‘-o’) xlabel('subcarrier index') ylabel('frequency response')

56 PDF of H(k), k=0,1,…,N-1 Παρατηρήστε ότι η απόκριση συχνότητας της κρουστικής απόκρισης ενός πολυδιαδρομικού καναλιού είναι όπου L είναι το μήκος της κρουστικής απόκρισης τmax είναι η μέγιστη καθυστέρηση του καναλιού (maximum delay spread) H δειγματοληψία της Η(ω) για ω=2πk/N, , είναι ο DFT{h(n)} δηλαδή

57 Multipath channel in the frequency domain
Show that the multipath channel results in selective fading clear; Fs = 10*10^6; % sampling frequency AND signal bandwidth L=3; % multipath fading N=64; % total number of carriers – FFT length p = [0.5, 0.3, 0.2]; % declare power delay profile h(1) = sqrt(p(1))*(randn(1,1)+i*randn(1,1))/sqrt(2); h(2) = sqrt(p(2))*(randn(1,1)+i*randn(1,1))/sqrt(2); h(3) = sqrt(p(3))*(randn(1,1)+i*randn(1,1))/sqrt(2);

58 Multipath channel in the frequency domain
h = [h(1), 0, h(2), 0, h(3)]; % consider the multipath delay profile H = fft(h, N); % frequency response of last channel response b = abs(H); % magnitude of frequency response plot(b, ‘-o’) xlabel('subcarrier index') ylabel('frequency response')

59 Coherence bandwidth of channel
We use the coherence bandwidth, which is a range of frequencies over which two frequency components have a strong potential for amplitude correlation, to define whether the channel fading is flat or frequency selective. The coherence bandwidth (Bc) between two frequency envelopes is given as Frequency components of a signal separated by more than Bc will fade independently. A channel is a frequency-selective channel if Bc < Bw, where Bw is the signal bandwidth.

60 PDF of H(k), k=0,1,…,N-1 Παρατηρήστε ότι η απόκριση συχνότητας της κρουστικής απόκρισης ενός πολυδιαδρομικού καναλιού είναι όπου L είναι το μήκος της κρουστικής απόκρισης τmax είναι η μέγιστη καθυστέρηση του καναλιού (maximum delay spread) H δειγματοληψία της Η(ω) για ω=2πk/N, , είναι ο DFTΝ{h(n)} δηλαδή

61 PDF of H(k), k=0,1,…,N-1 clear; Fs = 10*10^6; % sampling frequency AND signal bandwidth L=3; % multipath fading N=64; % total number of carriers – FFT length bins=50; carrier_no = 22; % select any sub-carrier form 1-N (=64) p = [0.5, 0.3, 0.2]; % declare power delay profile NN=10000; for m=1:NN % find the amplitude of the frequency response h(1) = sqrt(p(1))*(randn(1,1)+i*randn(1,1))/sqrt(2); h(2) = sqrt(p(2))*(randn(1,1)+i*randn(1,1))/sqrt(2); h(3) = sqrt(p(3))*(randn(1,1)+i*randn(1,1))/sqrt(2);

62 PDF of H(k), k=0,1,…,N-1 h = [h(1), 0, h(2), 0, h(3)]; % consider a multipath delay profile % τ = [ ] nsec, since Ts=1/Fs = 100 nsec H = fft(h, N); % channel frequency response H0 = abs(H); % magnitude of frequency response H0_no(m)=H0(carrier_no); % pick amplitude of carrier_no m end figure(2) % Make histogram of H0_no [nfreq xoutfreq]=hist (H0_no, bins); bar(xoutfreq, nfreq / (NN* (max(xoutfreq) /bins) ) ) axis ([ ] )

63 PDF of H(k), k=0,1,…,N-1 g = findobj (gca, 'Type' , 'patch' ) ; set (g, 'FaceColor', 'r' , 'LineStyle', ':' , 'EdgeColor', 'w' ) hold on y = raylpdf ( [0 : 0.1 : 3] , 1/sqrt(2)) ; x=0 : 0.1 : 3 ; plot (x, y, '* ' )

64 Multipath channel in the frequency domain
Show that the multipath channel results in selective fading clear; Fs = 10*10^6; % sampling frequency AND signal bandwidth L=3; % multipath fading N=64; % total number of carriers – FFT length p = [0.5, 0.3, 0.2]; % declare power delay profile h(1) = sqrt(p(1))*(randn(1,1)+i*randn(1,1))/sqrt(2); h(2) = sqrt(p(2))*(randn(1,1)+i*randn(1,1))/sqrt(2); h(3) = sqrt(p(3))*(randn(1,1)+i*randn(1,1))/sqrt(2);

65 Multipath channel in the frequency domain
h = [h(1), 0, h(2), 0, h(3)]; % consider the multipath delay profile H = fft(h, N); % frequency response of last channel response b = abs(H); % magnitude of frequency response plot(b, ‘-o’) xlabel('subcarrier index') ylabel('frequency response')

66 Multipath channel in the frequency domain
Παρατηρήστε ότι η απόκριση συχνότητας της κρουστικής απόκρισης του καναλιού με δειγματοληψία στις ψηφιακές γωνιακές συχνότητες ω = 2πk/N, , (ισοδύναμα αναλογικές συχνότητες f= kFs/N, ) είναι ο DFTΝ{h(n)}, έχει διαφορετικά πλάτη στα διάφορα k (υποφέροντα)  frequency selective fading !!! Αυτό χρησιμοποιείται για την ανάθεση υποφερόντων σε διαφορετικούς χρήστες στο σύστημα πολλαπλής πρόσβασης χρηστών OFDMA.

67 Multipath channel in the frequency domain
Κάθε χρήστης θα έχει μία απόκριση καναλιού, όπως φαίνεται στο παρακάτω σχήμα: subchannel frequency magnitude carrier channel

68 Resource (frequency bins) allocation
Ο Σταθμός Βάσης γνωρίζει την απόκριση συχνότητας του καναλιού κάθε χρήστη και προσπαθεί να δώσει υποφέροντα k {0,1,…, N-1} σε κάθε χρήστη στα οποία το πλάτος έχει υψηλές τιμές. User 1 User K frequency magnitude Base Station - has knowledge of each user’s channel state information thru ideal feedback from the users User 2 . . .