Hw3 Mimo Cap

8/6/2019 Hw3 Mimo Cap

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ELECTRICAL ENGINEERING DEPARTMENT, KFUPM

MIMO Capacity and gain of

Optimal power allocation

using Water-Filling

algorithmEE 575 Information Theory

Assignment # 3

Submitted by:

Raza Umar

Student ID: g200905090

5/23/2010

In this assignment, capacity of parallel Gaussian channels has been compared for equal power allocation

and optimal power allocation based on water-filling algorithm. Mean capacity comparisons,

Complementary CDF comparisons and Outage probability comparisons have been analyzed for SISO,

SIMO, MISO, MIMO and MIMO using Water-Filling power allocation.


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Table of ContentsProblem 1: Parallel Gaussian Channels .................................................................................................... 4

Part (1a) Total capacity of parallel channels for equal power distribution ........................ ................... 4

Part (1b) Water-filling algorithm ..................... ............................ ................... ............................ ........ 5

Part (1c) Total capacity of parallel channels for optimal power distribution ......................... .............. 5

Problem 2 + Problem 3: MIMO Capacity + Water filling for 4x4 MIMO Channel ............................ ........... 6

Part (2a+3a) Mean capacity comparisons for MIMO channels ........................ ...................... .............. 6

Part (2b+3b) Complementary CDF comparison for flat fading channels ........................... ................... 8

Part (2c+3c) Outage probability vs. SNR for flat adding channels .......................... ................... ........... 9

Appendix A ............................................................................................................................................ 10

Appendix B ............................................................................................................................................ 13


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Problem 1: Parallel Gaussian Channels

Given received signal:

Where,

~ 0, 1 0 0 00 7 0 00 0 5 00 0 0 3

Total Power transmission=Pt=20

Part (1a) Total capacity of parallel channels for equal power distribution

As we have 4 parallel channels

M=4=Nt=Nr

N1=1, N2=7, N3=5, N4=3

For equal power distribution:

204 5 , 1 , 2 , , 4

max: ; 12 log 1

12 log 1 2.8888 /


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Part (1b) Water-filling algorithm

Matlab implementation of Water filling algorithm is available in Appendix A.

Part (1c) Total capacity of parallel channels for optimal power distribution

For, N1=1, N2=7, N3=5, N4=3

Water level obtained from Matlab code = v=9 &

P1=8, P2=2, P3=4, P4=6 using,

With optimal power allocation;

12

log 1

2.9827 /

Hence, 0.0939 /


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Problem 2 + Problem 3: MIMO Capacity + Water filling for 4x4 MIMO

Channel

Matlab Implementation for MIMO Capacity and Water filling for 4x4

MIMO Channel is available in Appendix B.

Part (2a+3a) Mean capacity comparisons for MIMO channels


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Water filing Capacity Gain

-10 -5 0 5 10 15 20 25 300

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9WF gain in capacity

SNR in dB --->

WFgaininbps/Hz

--->


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Part (2b+3b) Complementary CDF comparison for flat fading channels

0 5 10 150.9

0.91

0.92

0.93

0.94

0.95

0.96

0.97

0.98

0.99

1

Mean Capacity bps/Hz --->

1-OutageProbability

--->

Complementary CDF comparisons (vs capacity) at SNR=10dB

4x4 MIMO WF

4x4 MIMO

4x1 MISO

1x4 SIMO

1X1 SISO


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Part (2c+3c) Outage probability vs. SNR for flat adding channels

2 4 6 8 10 12 14 16 18 20

10-5

10-4

10

-3

10-2

10-1

100

SNR in dB --->

Outa

geProbability

Outage probability vs SNR for 4 bps/Hz

4x4 MIMO

4x1 MISO

1x4 SIMO

1X1 SISO

4x4 MIMO WF


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Appendix A

%% Function: wfill% This routine optimally allocates the power among "m" channel using water% filling algorithm% Input:% 1. Pt: Total Power budget% 2. m: Total available parallel channels% 3. N: Un-correlated Noise variances ( a row vector of lenght m)% Output:% 1. v: water level% 2. P: power levels corresponding to "m" (arranged in the same order as% N) according to water filling algo.

function [v P] = wfill(Pt,m,N) % Optimum power allocation function

res=1; % resolution of step sizeP=zeros(1,m); % initialize transmitted power over each parallel channel to be

0[N_sorted,index]=sort(N); % Noise power sorted in ascending orderstep=(N_sorted(2)-N_sorted(1))/res;N_sorted_temp=N_sorted;for p=1:length(N_sorted_temp)-1if ((max(N_sorted_temp)-min(N_sorted_temp))>Pt)

m=m-1;N_sorted_temp=N_sorted_temp(1:end-1);

end

end

if step>Pt/2step=Pt/2;

endif step


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N_sorted(i)=N_sorted(i)+step;P(i)=P(i)+step;

if(sum(abs((N_sorted(1).*ones(1,q))-N_sorted(1:q))step)

step =step_old;j=0;

endif (sum(P)>Pt)

k=k+1;N_sorted(1:i)=N_sorted(1:i)-step;P(1:i)=P(1:i)-step;% find how many channels(out of m) are at same levelcheck = abs(N_sorted-N_sorted(1).*ones(1,length(N)))2)display(['Warning: Power alltoment exceeding budget ',num2str(k),'rd

time']);sum(P)

end

if iepsilon)

%if water has gone above the next level, set the step such that when water%is added to the next level, its level becomes the same as previous level

if ((z~=1)&&(step~=N_sorted(i)-N_sorted(i+1)))step_old=step;step=N_sorted(i)-N_sorted(i+1);

z=0;endif (step< 0.001)

j=1;endi=i+1;elseif i>1 % if water is below next entry of N,

%re-initiazlize it to point at 1st entry and start re-filling wateri=1;

endend

else % if i is pointing at the last entry of N, re intialize it to

point at 1st entry


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i=1;endif (abs(sum(P)-Pt) < epsilon2)

if i>1&&(N_sorted(i) < N_sorted(i-1))N_sorted(1:i-1)=N_sorted(1:i-1)-step;P(1:i-1)=P(1:i-1)-step;step=(step*(i-1))/i;

N_sorted(1:i)=N_sorted(1:i)+step;P(1:i)=P(1:i)+step;

j=0;i=1;z=1;

endend

if (abs(sum(P)-Pt) < epsilon2)flag_vector2=abs(N_sorted(1).*ones(1,length(find(P)))-

N_sorted(1:length(find(P))))< epsilon2.*ones(1,length(find(P)));flag_vector=find(flag_vector2-ones(1,length(find(P))));if isempty(flag_vector)

break;end

endif iter>1000

display('Warning: wf took more than 1000 iterations');end

endv= N_sorted(1); % water levelN_sorted;

N;P=P(index);if(abs(sum(P)-Pt) > epsilon2)

display('Warning: Power above budget');end


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Appendix B

%% Main function to calculate MIMO channel capacity and associated figures%% of merit like complementary CDF & Outage probability% Objective: (1)This function compares Mean channel capacity for different% MIMO realizations (SISO, SIMO,MISO, MIMO)as a function of SNR% (2) compares complementary CDF for different MIMO realizations% (SISO, SIMO,MISO, MIMO)as a function of capacity in bps/Hz% (3) compares Outage probability for different MIMO realizations% (SISO, SIMO,MISO, MIMO)as a function of SNR% Author: Raza Umar as part of EE 575 Information Theory Assignment% Date: May 08, 2010

%% >>>>>>>>>>>>>>>>>>>>>>>> ... CLEANING ...


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count11_norm=cumsum(count11)/max(cumsum(count11));if(isempty(find(abs(range11-c_outage)>>>>>>>>>>>>>>>>>>> ... MIMO (4x4) ...


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%%

%%complementary CDF at SNR=10dB

c44_all=abs(c44_i(21,:)); % channel realizations at 10dB SNRrange_44=0:0.1:max(c44_all);count44=histc(c44_all,range_44);count44_norm=cumsum(count44)/max(cumsum(count44));comp_CDF_44=1-count44_norm;%%MIMO WFc44_all_wf=abs(c44_i_wf(21,:)); % channel realizations at 10dB SNRrange_44_wf=0:0.1:max(c44_all_wf);count44_wf=histc(c44_all_wf,range_44_wf);count44_norm_wf=cumsum(count44_wf)/max(cumsum(count44_wf));comp_CDF_44_wf=1-count44_norm_wf;

%%%outage prob

j1= 1;for i1=find(SNR_dB==2):find(SNR_dB==20)

c44_all=abs(c44_i(i1,:));range44=0:0.1:max(c44_all);count44=histc(c44_all,range44);count44_norm=cumsum(count44)/max(cumsum(count44));if(isempty(find(abs(range44-c_outage)>>>>>>>>>>>>>>>>> ... SIMO (1x4) ...


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SNR_i= SNR(i1); % for one specific value of SNR

h14=1/sqrt(2).*complex(randn(Nr*Nt,ch_realizations),randn(Nr*Nt,ch_realizatio

ns));% complex normal r.v. having var=1/2 per dim.h14_mag_sq=abs(h14).^2;h14_mag_sq_sum=sum(h14_mag_sq);c14_i(i1,1:ch_realizations)=log2(1+SNR_i.*h14_mag_sq_sum); %

instantaneous cap.c14(i1)=mean(c14_i(i1,:)); % mean capacity

end

%complementary CDF at SNR=10dBc14_all=c14_i(21,:); % channel realizations at 10dB SNRrange_14=0:0.1:max(c14_all);count14=histc(c14_all,range_14);count14_norm=cumsum(count14)/max(cumsum(count14));comp_CDF_14=1-count14_norm;

%outage probabilityj1= 1;for i1=find(SNR_dB==2):find(SNR_dB==20)

c14_all=abs(c14_i(i1,:));range14=0:0.1:max(c14_all);count14=histc(c14_all,range14);count14_norm=cumsum(count14)/max(cumsum(count14));if(isempty(find(abs(range14-c_outage)>>>>>>>>>>>>>>>>>>> ... MISO (4x1) ...


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%complementary CDF at SNR=10dBc41_all=c41_i(21,:); % channel realizations at 10dB SNRrange_41=0:0.1:max(c41_all);

count41=histc(c41_all,range_41);count41_norm=cumsum(count41)/max(cumsum(count41));comp_CDF_41=1-count41_norm;

%outage probabilityj1= 1;for i1=find(SNR_dB==2):find(SNR_dB==20)

c41_all=abs(c41_i(i1,:));range41=0:0.1:max(c41_all);count41=histc(c41_all,range41);count41_norm=cumsum(count41)/max(cumsum(count41));if(isempty(find(abs(range41-c_outage)>>>>>>>>>>>>>>>>>> ... Plotting ...


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',SNR_dB(find(SNR_dB==2):find(SNR_dB==20)),outage14, 'gs-

',SNR_dB(find(SNR_dB==2):find(SNR_dB==20)),outage11, 'b:');legend('4x4 MIMO WF','4x4 MIMO','4x1 MISO','1x4 SIMO','1X1

SISO','Location','SouthEast');title('Outage probability vs SNR for 4 bps/Hz');xlabel('SNR in dB --->');ylabel('Outage Probability --->');axis([2 20 1e-6 1]);

% MIMO water-filling capacity gainfigure('Name','capacity gain of water-filling');plot(SNR_dB,abs(c44_wf)'-abs(c44));grid on;title('WF gain in capacity');xlabel('SNR in dB --->');ylabel('WF gain in bps/Hz --->');

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