The University of Texas at Arlington
Department of Electrical Engineering
Project Report
EE 5364 – Information theory and coding
Spring 2011
Submitted by:
CHIRAG PANDYA:
1000722427
VAIBHAV PATANKAR: 1000734804
CONTENTS
1) Introduction
2) Problem Statement
3) Channel Capacity
4) Channel Side Information
5) Antenna Diversity
6) Power Control Schemes
7) MATLAB Simulation Results
8) Conclusion
9) MATLAB Code
10) References
1. INTRODUCTION:
Mobile phone technology has become an integral part of our lives today. We use it for trivial
tasks such as making calls, listening to music and web browsing as well as for more significant
purposes like video conferencing etc. Mobile phone technology is growing at incredibly faster
rate. In 2008 there were 4,100 million mobile cellular subscriptions in the world. IS-95 and
CDMA 2000 are well known mobile phone technologies – both use CDMA (Code Division
Multiple Access) as an underlying channel access method.
Comparison of IS-95 and CDMA 2000
2. PROBLEM STATEMENT:
Rate versus Power Control in IS-95 and CDMA2000
CDMA2000 1×EV-DO (Enhanced Version Data Optimized) is a cellular data standard operating
on the 1.25MHz bandwidth. In this project, you (a team up to 3 graduate students) are requested
to evaluate and compare rate versus power control in IS-95 (a CDMA standard for 2G wireless
communications) and CDMA2000 1×EV-DO.
Rayleigh fading channel is assumed with Doppler shift 100Hz (the channel gain will be
provided). Please evaluate rate versus power control for the following different scheme and write
a technical report to compare these approaches.
Using channel inversion power control for IS-95
Using equal power allocation for CDMA 2000
Using waterfilling power control for CDMA2000
For each scheme, please use SISO, SIMO, MISO, and MIMO for IS-95 and CDMA2000. Please
choose M=2, and M=4.
Consider two cases:
Full Channel Side Information (CSI)
Channel Side Information at the Receiver (CSIR)
3. CHANNEL CAPACITY:
Let X represent the space of signals that can be transmitted, and Y the space of signals received,
during a block of time over the channel. Let
be the conditional distribution function of Y given X. Treating the channel as a known statistic
system, pY | X(y | x) is an inherent fixed property of the communications channel (representing the
nature of the noise in it). Then the joint distribution
of X and Y is completely determined by the channel and by the choice of
the marginal distribution of signals we choose to send over the channel. The joint distribution
can be recovered by using the identity
Under these constraints, next maximize the amount of information, or the message, that one can
communicate over the channel. The appropriate measure for this is the mutual
information I(X;Y), and this maximum mutual information is called the channel capacity and is
given by
AWGN channel:
If the average received power is
[W] and the noise power spectral density is N0 [W/Hz], the
AWGN channel capacity is
[bits/Hz],
4. CHANNEL SIDE INFORMATION:
CSIR is the scheme in which receivers can track the channel but the transmitter does not have
access to the channel realizations. All the above cases are with respect to CSIR.
Full CSI is the scheme with tracking of channels of all the users at the receiver and all the
transmitters, here we can allocate power dynamically to the users as a function of the channel
states, the sum of capacities for such scheme is as follows:
Subject to powers being non-negative and the power constraint on each user:
The optimal Power allocation policy is (where λ is chosen to meet the sum power constraint
equation as above):
The resulting sum capacity is:
where k*(h) is the index of the user with the strongest channel at the joint channel state h
5. ANTENNA DIVERSITY:
Antenna diversity, also known as space diversity, is any one of several wireless diversity
schemes that use two or more antennas to improve the quality and reliability of a wireless link.
Often, especially in urban and indoor environments, there is no clear line-of-sight (LOS)
between transmitter and receiver. Instead the signal is reflected along multiple paths before
finally being received. Each of these bounces can introduce phase shifts, time delays,
attenuations, and distortions that can destructively interfere with one another at the aperture of
the receiving antenna. Antenna diversity is especially effective at mitigating
these multipath situations. This is because multiple antennas offer a receiver several observations
of the same signal. Each antenna will experience a different interference environment. Thus, if
one antenna is experiencing a deep fade, it is likely that another has a sufficient signal.
Collectively such a system can provide a robust link. While this is primarily seen in receiving
systems (diversity reception), the analog has also proven valuable for transmitting systems
(transmit diversity) as well.
The antenna system characterization with respect to the number of antennas used at transmitter
and receiver can be described as follows:
1.
2.
3.
4.
SISO : Single-Input Single-Output
SIMO : Single-Input Multiple-Output
MISO : Multiple-Input Single-Output
MIMO: Multiple-Input Multiple-Output
5.1 SISO: Single-Input Single-Output
In this case transmitter and receiver both have single antenna. The signal is transmitted using
single channel. The basic SISO communication system can be described as follows
SISO Communication System
Here are as shown in figure the channel used has the channel gain h1. This model is the very
basic simulation model of communication system. Here we used channel with the gain h1 as
follows:
𝒉𝟏 = |𝒈|𝟐 Where, g is the channel coefficient.
5.2 SIMO: Single-Input Multiple-Output
In this case transmitter have single antenna and receiver have multiple antennas. In here we
consider two and four antennas at receiver side. The signal is transmitted using single channel.
Below figure shows the basic SIMO communication system.
SIMO Communication Systems
Here we used channel with the gain as h11, h12, h13, h14.
5.3 MISO: Multiple-Input Single-Output
In this case transmitter has multiple antennas (2 antennas and 4 antennas in our case) and
receiver has one antenna. In here we consider two and four antennas at receiver side. The signal
is transmitted using single channel. Below figure shows the basic MISO communication system.
MISO Communication Systems
Here are we used channel with the gain as h11, h21, h31, h41.
5.4 MIMO: Multiple-Input Multiple-Output
In this case transmitter and receiver both have multiple antennas (2 antennas and 4 antennas in
our case). In here we consider two and four antennas at receiver side. The signal is transmitted
using single channel. Below figure shows the basic MIMO communication system.
MIMO Communication Systems
6. POWER CONTROL SCHEMS:
6.1 Equal Power Allocation:
Usually, when transmitter doesn’t know the channel, Equal Power Allocation is used. This was
given by Dr. Alamouti. In this scheme if we have ‘n’ number of channels available for
communication, the total power will be given by “P/n” where P shows the total power. In this
case, transmitter doesn’t know the channel and hence the power allocation is independent from
each other irrespective of the individual channel performance.
Let us think of a MISO (as shown in fig) system having 2 transmitter antennas and 1 receiver
antenna. We are using the equal power allocation system for assignment of the power to each
channel.
For this system,
Total Power = P
Gain of the 1st channel = h1
Gain of the 2nd channel = h2
Power assigned to each of the channel = P/2
Power spectral density for noise for each channel = N0
SNR = P/N0 and
||h||2 = ||h1||2 + ||h2||2
The channel capacity for this system would be,
C = log(1 + ||h||2 P/2N0 ) bits/s/Hz
Therefore,
C = log(1 + ||h||2 SNR/2 ) bits/s/Hz
Which shows when the transmitter doesn’t have the channel information; it transmits the power
in all directions rather than concentrating on a specific direction.
Now instead of two if we have L number of transmitting antenna, the transmitted power would
be P/L. And hence, the above equation will become generalized,
C = log(1 + ||h||2 SNR/L ) bits/s/Hz
So, outage probability would be, Pout(R) = Pr {log(1 + ||h||2 SNR/L ) < R}
The disadvantage of this scheme is that we don’t worry about the behaviour of the channel and
just allocate the power equally to all the channels. But for Rayleigh distributed random variables
which are i.i.d., h1, h2,......,hL, this scheme is optimum and gives best outage probability. Now,
we will discuss another power allocation scheme which is Channel Inversion scheme in the next
section.
6.2 Channel Inversion Scheme:
Now, when the transmitter knows the channel, we can control the power which is going to be
supplied as per the channel gain. In this case, we keep received SNR as a constant entity and
hence vary power to do so. This technique is called channel inversion scheme. Here the
transmitted power is an inverse of the channel gain. So, if the channel gain is high, we need to
transmit small amount of power and if the channel gain is low, we need to transmit high amount
of power making the system moderate.
Let us take the same example as above where,
Total Power = P = sum(Pi)
Gain of the 1st channel = h1
Gain of the 2nd channel = h2
Power assigned to channel 1 = 1/||h1||2
Power assigned to channel 1 = 1/||h2||2
Power spectral density for noise for each channel = N0
SNR for each channel= constant and
||h||2 = ||h1||2 + ||h2||2
The channel capacity for this system would be,
C(i) = log(1 + ||h||2 / ||hi||2 N0 ) bits/s/Hz
Hence, in this case our capacity is constant as the SNR is a constant entity. So, this is the sub
optimal scheme by which we can get moderate performance depending upon the channel
performance. Now in the very next section, we will focus on optimum power allocation scheme
which is called “Waterfiling”.
6.3 Waterfilling Scheme:
We can find the optimum power allocation for the channel using Waterfilling scheme. When the
channel is known, we take N0/|hn2| values plotted as the subcarrier indices i.e. n=0,1,.....,N-1.
Then we find the graph shown in fig. 2. So, as shown in fig.2 we are supposed to fill the “water”
into the empty vessels up to some defined level which is the height of the vessel. Some bars in
the graph are higher than the height of the vessel. In this case we pour nothing. Actually
speaking, the water poured in the vessel is the assigned power to the channel and the vessel is
defined as Noise to Carrier ratio. So, when the noise is maximum, i.e. the channel performance is
poor, we transmit minimum amount of the power and when the noise is very less, i.e. the channel
performance is very high, we transmit maximum power. And hence, transmitter takes advantage
of the better channel conditions.
Waterfilling Schemes
Now if we take the same example we used earlier then,
Total Power = P = sum(Pi)
Gain of the 1st channel = h1
Gain of the 2nd channel = h2
Power assigned to individual channel Pi = ((1/ lambda) – (N0/|hi|2))+
Where, 1/ lambda = height of the vessel
Power spectral density for noise for each channel = N0
The channel capacity for this system would be,
C(i) = log(1 + ||h||2 Pi / N0 ) bits/s/Hz
So, this scheme is precisely opposite to the channel inversion scheme. By doing dynamic power
allocation this scheme ensures spending more power when channel is good and hence to boost
the received power.
7. MATLAB SIMULATION RESULTS:
7.1 Channel Inversion Power Control in IS-95:
SISO [1x1]:
Channel Inversion : SISO
10
9
8
Capacity [C] --->
7
6
5
4
3
2
1
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
SIMO [1x2 and 1x4]:
Channel Inversion for SIMO : M=2 & M=4)
6
SIMO:1x2
SIMO:1x4
5
Capacity [C] --->
4
3
2
1
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
MISO [2x1 and 4x1]:
Channel Inversion for MISO : M=2 & M=4)
5
MISO:2x1
MISO:4x1
4.5
4
Capacity [C] --->
3.5
3
2.5
2
1.5
1
0.5
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
MIMO [2x2and4x4]:
Channel Inversion for MIMO : M=2 & M=4
15
MIMO: 2x2
MIMO: 4x4
Capacity [C] --->
10
5
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
Comparison between All communication Systems for Channel Inversion:
Channel Inversion for SISO, SIMO, MISO and MIMO for M=2)
6
SISO
SIMO:1x2
MISO:2x1
MIMO:2x2
5
Capacity [C] --->
4
3
2
1
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
7.2 Equal Power Allocation in CDMA:
MISO [2x1 and 4x1]:.
Equal Power Allocation for MIMO: M=2 & M=4
14
MIMO: 2x2
MIMO: 4x4
12
Capacity [C] --->
10
8
6
4
2
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
MIMO [2x2 and 4x4]:
Equal Power Allocation for MIMO: M=2 & M=4
14
MIMO: 2x2
MIMO: 4x4
12
Capacity [C] --->
10
8
6
4
2
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
Comparison of MISO and MIMO Systems for Equal Power Allocation System:
Equal Power Allocation for MISO and MIMO : M=2 & M=4)
14
MISO:2x1
MISO:4x1
MIMO:4x4
MIMO:2x2
12
Capacity [C] --->
10
8
6
4
2
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
7.3 Water Filling Algorithm for CDMA-2000:
Water Filling Algorithm for MIMO: M=2 & M=4
20
MIMO: 2x2
MIMO: 4x4
18
16
Capacity [C] --->
14
12
10
8
6
4
2
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
Comparison among Channel Inversion, Equal Power Allocation and Water
Filling Algorithm:
Comparision of all techniques for MIMO: M=2
8
Equal Power Allocation
Channel Inversion
Water Filling Algorithm
7
Capacity [C] --->
6
5
4
3
2
1
0
-15
-10
-5
0
SNR(in dB) --->
5
10
15
8. CONCLUSION:
We have studied Channel Inversion Power Control method used in IS-95, Equal Power
Allocation and Water-Filling methods both used for power control in CDMA-2000
considering Full Channel Side Information (CSI) and Channel Side Information at Receiver
(CSIR) cases using M=2 and M=4. We conclude from all the MATLAB simulation results
that the Water-Filling algorithm gives the maximum channel capacity hence the
transmission rate among all power control methods. The Equal Power Allocation method
gives the lowest performance and the Channel Inversion gives the moderate channel
capacity at given transmission rate.
9. MATLAB CODE:
% Channel Inversion:
clc;
clear all;
close all;
load g11.mat; load g12.mat;
load g21.mat; load g22.mat;
%$$$$$$$$$$$$$$$$$$$$$
% SISO System : 1X1
%$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
h1 = g22(1,1)'*g22(1,1);
p1 = 1/(h1);
N1 = p1/SNR;
C_SISO(index) = log(1+(1/N1));
% Convert in to Watt
% h1 = |g|^2
% By Channel Inversion
% SNR = P/N => N = P/SNR
% C = log(1 + h*p/N)
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
% SISO System : for 5000 Iterations
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:1:15;
C_SISO = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10); % Convert in to Watt
C_SISO = 0;
C_TEMP = 0;
for i = 1:5000
% Iteration for 5000 coefficient
h(i) = g11(1,i)'*g11(1,i);
% h = |g|^2
p(i) = 1/h(i);
% Channel Inversion
N(i) = p(i)/SNR;
% SNR = P/N => N = P/SNR
C_TEMP(i) = log(1+(1/N(i))); % C = log(1 + h*p/N)
C_SISO = C_TEMP(i) + C_SISO;
end
C_FINAL_SISO(index)= C_SISO/5000; % Average of all 5000 coefficients of channel
end
%$$$$$$$$$$$$$$$$$$$$$$
% SIMO System : M = 2
%$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
C_SIMO_1x2 = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% h = |g|^2
h11 = g11(1,2)'*g11(1,2);
h12 = g12(1,1)'*g12(1,1);
% Channel Inversion
p11 = 1/(h11);
p12 = 1/(h12);
% SNR = P/N => N = P/SNR
N11 = p11/SNR;
N12 = p12/SNR;
% C = log(1 + h*p/N) => C = log(1 + 1/N)
C1 = log(1+(1/N11));
C2 = log(1+(1/N12));
% Take the max value as the system capacity
C_SIMO_1x2(index) = max(C1,C2);
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
% SIMO System : for 5000 Iterations
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
index=0;
C_FINAL_SIMO = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10); % Convert in to Watt
C1_SIMO = 0; C2_SIMO = 0; C_TEMP = 0;
% Iteration for 5000 coefficient
for i = 1:5000
h11(i) = g22(1,i)'*g22(1,i); % Channel gain : h = |g|^2
h12(i) = g22(1,i)'*g22(1,i);
p11(i) = 1/(h11(i)); % Channel Inversion
p12(i) = 1/(h12(i));
N1(i) = p11(i)/SNR; % SNR = P/N => N = P/SNR
N2(i) = p12(i)/SNR;
C1_TEMP(i) = log(1+(1/N1(i))); % C = log(1 + h*p/N) => C = log(1 + 1/N)
C2_TEMP(i) = log(1+(1/N2(i)));
C1_SIMO = C1_TEMP(i) + C1_SIMO; % for 5000 values
C2_SIMO = C2_TEMP(i) + C2_SIMO;
end
C1_FINAL = C1_SIMO/5000;
C2_FINAL = C2_SIMO/5000;
% Taking the final channel capacity as the max of both values
C_FINAL_SIMO(index) = max(C1_FINAL,C2_FINAL);
end
%$$$$$$$$$$$$$$$$$$$$$$$
% SIMO System : M = 4
%$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% h = |g|^2
h11 = g11(1,2)'*g11(1,2);
h12 = g12(1,2)'*g12(1,2);
h13 = g21(1,2)'*g21(1,2);
h14 = g21(1,2)'*g22(1,2);
% Channel Inversion
p11 = 1/(h11);
p12 = 1/(h12);
p13 = 1/(h13);
p14 = 1/(h14);
% SNR = P/N => N = P/SNR
N11 = p11/SNR;
N12 = p12/SNR;
N13 = p13/SNR;
N14 = p14/SNR;
% C = log(1 + h*p/N) => C = log(1 + 1/N)
C1 = log(1+(1/N11));
C2 = log(1+(1/N12));
C3 = log(1+(1/N13));
C4 = log(1+(1/N14));
C = [C1 C2 C3 C4];
% Take the max value as the system capacity
C_SIMO_1x4(index) = max(C);
end
%$$$$$$$$$$$$$$$$$$$$$$$$
% MISO System : M = 2
%$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
C_MISO_2x1 = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% h = |g|^2
h_MISO = [g11(1,1); g21(1,1)]' * [g11(1,1); g21(1,1)];
p_MISO = 1/(h_MISO);
% Channel Inversion
N_MISO = p_MISO/SNR;
% SNR = P/N => N = P/SNR
% C = log(1 + h*p/N) => C = log(1 + 1/N)
C_MISO_2x1(index) = log(1+(1/N_MISO));
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
% MISO System : for 5000 Iterations
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
C_FINAL_MISO = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10); % Convert in to Watt
C_MISO = 0;
% Iteration for 5000 coefficient
for i = 1:5000
% Common coefficient for both channel g = [x y]
g_MISO = [g11(1,i) g12(1,i)];
h_MISO = g_MISO*g_MISO'; % h = |g|^2
p_MISO = 1/(h_MISO);
% Channel Inversion
N_MISO = p_MISO/SNR;
% SNR = P/N => N = P/SNR
C_TEMP = log(1+(1/N_MISO)); % C = log(1 + h*p/N)
C_MISO = C_TEMP + C_MISO; % Adding all 5000 values
end
% Taking average of 5000 values
C_FINAL_MISO(index) = C_MISO/5000;
end
%$$$$$$$$$$$$$$$$$$$$$$$
% MISO System : M = 4
%$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
C_MISO_4x1 = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% h = |g|^2
h_MISO = [g11(1,1); g12(1,1); g21(1,1); g22(1,1)]' * [g11(1,1); g12(1,1); g21(1,1); g22(1,1)];
p_MISO = 1/(h_MISO);
% Channel Inversion
N_MISO = p_MISO/SNR;
% SNR = P/N => N = P/SNR
% C = log(1 + h*p/N) => C = log(1 + 1/N)
C_MISO_4x1(index) = log(1+(1/N_MISO));
end
%$$$$$$$$$$$$$$$$$$$$$$$
% MIMO System : 2X2
%$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
C_MIMO_2x2 = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% Using SVD to convert 2X2 matrix into parallel channels
h_MIMO = [g11(1,1) g12(1,1); g21(1,1) g22(1,1)];
[m n] = eig (h_MIMO);
% h = |g|^2
h11 = n(1,1)'*n(1,1);
h22 = n(2,2)'*n(2,2);
% Channel Inversion
p11 = 1/(h11);
p22 = 1/(h22);
% SNR = P/N => N = P/SNR
N1 = p11/SNR;
N2 = p22/SNR;
% C = log(1 + h*p/N) => C = log(1 + 1/N)
C1 = log(1+(1/N1));
C2 = log(1+(1/N2));
% Adding all two parallel channels capacities
C_MIMO_2x2(index) = C1 + C2;
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
% MIMO System : for 5000 Iterations
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
C_FINAL_MIMO = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10); % Convert in to Watt
C_MIMO=0;
% Iteration for 5000 coefficient
for i = 1:5000;
% Doing SVD to convert MIMO to Parallel
h_MIMO = [g11(1,i) g12(1,i); g21(1,i) g22(1,i)];
[m b] = eig (h_MIMO);
h11 = b(1,1)'*b(1,1);
% Parallel Channel gain
h22 = b(2,2)'*b(2,2);
% Channel Inversion
p11 = 1/(h11);
p22 = 1/(h22);
% SNR = P/N => N = P/SNR
N1 = p11/SNR;
N2 = p22/SNR;
% C = log(1 + h*p/N)
C1_TEMP = log(1+(1/N1));
C2_TEMP = log(1+(1/N2));
% Adding all capacities for 5000 iterations
C_MIMO = C1_TEMP + C2_TEMP + C_MIMO;
end
C_FINAL_MIMO(index) = C_MIMO/5000; % Taking average over 5000 values
end
plot([-20:20], C_FINAL_MIMO,'g');
XLABEL('SNR');
YLABEL('Channel Capacity');
LEGEND('Equal Power Allocation','AWGN Capacity','Channel Inversion without
constraint','Channel Conversion with contraint','Water Filling');
%$$$$$$$$$$$$$$$$$$$$$$
% MIMO System : 4x4
%$$$$$$$$$$$$$$$$$$$$$$
SNR_db=-15:1:15;
C_MIMO_4x4 = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% Using SVD to convert 4X4 into parallel channel
h_MIMO_4X4 = [g11(1,1) g11(1,2) g11(1,3) g11(1,4); g12(1,1) g12(1,2) g12(1,3) g12(1,4);
g21(1,1) g21(1,2) g21(1,3) g21(1,4); g22(1,1) g22(1,2) g22(1,3) g22(1,4)];
[m b] = eig(h_MIMO_4X4);
% h = |g|^2
h11 = b(1,1)'*b(1,1);
h22 = b(2,2)'*b(2,2);
h33 = b(3,3)'*b(3,3);
h44 = b(4,4)'*b(4,4);
%Channel Inversion
p11= 1/h11;
p22= 1/h22;
p33= 1/h33;
p44= 1/h44;
% SNR = P/N => N = p/SNR
N1 = p11/SNR;
N2 = p22/SNR;
N3 = p33/SNR;
N4 = p44/SNR;
% C = log(1 + h*p/N) => C = log(1 + 1/N)
C1_TEMP = log(1+(1/N1));
C2_TEMP = log(1+(1/N2));
C3_TEMP = log(1+(1/N3));
C4_TEMP = log(1+(1/N4));
% Adding all four parallel channels capacities
C_MIMO_4x4(index)= C1_TEMP + C2_TEMP + C3_TEMP + C4_TEMP;
end
%
Equal Power Allocation:
clc;
clear all;
close all;
load g11.mat; load g12.mat;
load g21.mat; load g22.mat;
%$$$$$$$$$$$$$$$$$$$$$$$$$
% MISO System : M = 2
%$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
No_Antenna = 2;
C_MISO_2x1 = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% Channel coefficient for 2X1 MISO and h = |g|^2
h1 = ([g11(1,1); g21(1,1)]'*[g11(1,1); g21(1,1)]);
% SNR = P/N => P = N/SNR
pwr = SNR*N1;
% C = log(1 + h*p/N)
C_MISO_2x1(index) = log(1+(h1 * pwr/No_Antenna));
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
% MISO System : for 5000 Iterations
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
No_Antenna = 2;
C_MISO = 0; C_MISO_FINAL = 0; C_MISO_TOTAL = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10); % Convert in to Watt
for i = 1:5000 % Iteration for 5000 values
% Channel coefficient for 2X1 MISO and h = |g|^2
h = ([g11(1,i); g21(1,i)]'*[g11(1,i); g21(1,i)]);
% SNR = P/N => P = N/SNR
pwr = SNR*N1;
% C = log(1 + h*p/N)
C_MISO = log(1+(h * pwr/No_Antenna));
% Adding 5000 values
C_MISO_TOTAL = C_MISO_TOTAL + C_MISO;
end
% Average over 5000 values
C_MISO_FINAL(index) = C_MISO_TOTAL/5000;
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$
% MISO System : M = 4
%$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
N1 = 1;
No_Antenna = 2;
C_MISO_4x1 = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% Channel coefficient for 4X1 MISO and h = |g|^2
h1 = ([g11(1,1); g12(1,1); g21(1,1); g22(1,1)]' * [g11(1,1); g12(1,1); g21(1,1); g22(1,1)]);
% SNR = P/N => P = N/SNR
pwr = SNR*N1;
% C = log(1 + h*p/N)
C_MISO_4x1(index) = log(1+(h1 * pwr/No_Antenna));
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$
% MIMO System : M = 2
%$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
C_MIMO_2x2 = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% Using SVD to convert 2X2 matrix into parallel channels
h_MIMO = [g11(1,1) g12(1,1); g21(1,1) g22(1,1)];
[m g] = eig (h_MIMO);
% Parallel Channel gain
h11 = g(1,1)'*g(1,1);
h22 = g(2,2)'*g(2,2);
% SNR = P/N => P = N/SNR
pwr11 = SNR / N1;
pwr22 = SNR / N1;
% C = log(1 + h*p/N)
C1 = log(1+(h11 * pwr11/No_Antenna));
C2 = log(1+(h22 * pwr22/No_Antenna));
% Adding both channel capacities
C_MIMO_2x2(index) = C1 + C2;
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
% MIMO System : for 1000 Iterations
%$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
C_MIMO_TOTAL = 0; C_MIMO_AVG = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10); % Convert in to Watt
% Iteration for 5000 coefficient
for i = 1:5000
% Channel coefficient for 2X2 MIMO
h_MIMO = [g11(1,i) g12(1,i); g21(1,i) g22(1,i)];
% Doing SVD to convert MIMO to Parallel
[m b] = eig (h_MIMO);
h11 = b(1,1)'*b(1,1);
% Parallel Channel gain
h22 = b(2,2)'*b(2,2);
% SNR = P/N => P = N/SNR
pwr11 = SNR * N1;
pwr22 = SNR * N1;
% C = log(1 + h*p/N)
C1 = log(1+(h11 * pwr11/No_Antenna));
C2 = log(1+(h22 * pwr22/No_Antenna));
% Adding 5000 values
C_MIMO_TOTAL = C1 + C2 + C_MIMO_TOTAL;
end
% Average over 5000 values
C_MIMO_AVG(index) = C_MIMO_TOTAL/1000;
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$
% MIMO System : M = 4
%$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:15;
C_MIMO_4x4 = 0;
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% Using SVD to convert 4X4 matrix into parallel channels
h_MIMO_4X4 = [g11(1,1) g11(1,2) g11(1,3) g11(1,4); g12(1,1) g12(1,2) g12(1,3) g12(1,4);
g21(1,1) g21(1,2) g21(1,3) g21(1,4); g22(1,1) g22(1,2) g22(1,3) g22(1,4)];
[m b] = eig (h_MIMO_4X4);
% h = |g|^2
h11 = b(1,1)'*b(1,1);
h12 = b(2,2)'*b(2,2);
h21 = b(3,3)'*b(3,3);
h22 = b(4,4)'*b(4,4);
% SNR = P/N => P = N/SNR
pwr11 = SNR * N1;
pwr12 = SNR * N1;
pwr21 = SNR * N1;
pwr22 = SNR * N1;
% C = log(1 + h*p/N)
C11 = log(1+(h11 * pwr11/No_Antenna));
C12 = log(1+(h12 * pwr12/No_Antenna));
C21 = log(1+(h21 * pwr21/No_Antenna));
C22 = log(1+(h22 * pwr22/No_Antenna));
% Adding all four channel capacties
C_MIMO_4x4(index) = C11 + C12 + C21 + C22;
end
%
Water Filling Algorithm:
%$$$$$$$$$$$$$$$$$$$$$$$$$$
% MIMO System : 2X2
%$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db = -15:1:15;
Sigma_n = 1; % assume
B = 1;
% assume
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% Using SVD to convert 2X2 matrix into parallel channels
h_MIMO = [g11(1,5) g12(1,2); g21(1,5) g22(1,1)];
[m b] = eig (h_MIMO);
% Parallel Channel gain
h11 = b(1,1)'*b(1,1);
h22 = b(2,2)'*b(2,2);
% Sigma^2 = h => Channel coefficient
Sigma1_Square = h11;
Sigma2_Square = h22;
% SNR = P / Sigma_n^2 => P = SNR * Sigma_n^2
pwr = SNR * Sigma_n * Sigma_n;
% Lamda_i = (Sigma_i^2 * P) / Sigma_n^2
Lamda_1 = (Sigma1_Square * pwr) / (Sigma_n*Sigma_n);
Lamda_2 = (Sigma2_Square * pwr) / (Sigma_n*Sigma_n);
Lamda_0 = (2*Lamda_1*Lamda_2)/(Lamda_1+Lamda_2+(Lamda_1*Lamda_2));
% Capacity for Water-filling case:
C = [B * log(Lamda_i/Lamda_0)]
% Here : n = 2 and B = 1
C_MIMO_2X2_WF(index) = log(Lamda_1/Lamda_0) + log(Lamda_2/Lamda_0);
end
%$$$$$$$$$$$$$$$$$$$$$$$$$$
% MIMO System : 4X4
%$$$$$$$$$$$$$$$$$$$$$$$$$$
SNR_db= -15:1:15;
Sigma_n = 1; % assume
B = 1;
% assume
for index = 1:length(SNR_db)
SNR = 10^(SNR_db(index)/10);
% Convert in to Watt
% Using SVD to convert 2X2 matrix into parallel channels
h_MIMO = [g11(1,1) g11(1,3) g11(1,4) g11(1,5); g11(1,7) g11(1,8) g11(1,9) g11(1,10);
g11(1,11) g11(1,12) g11(1,13) g11(1,14); g11(1,6) g11(1,15) g11(1,16) g11(1,17)];
[m b] = eig(h_MIMO);
% Parallel Channel gain
h11 = b(1,1)'*b(1,1);
h22 = b(2,2)'*b(2,2);
h33 = b(3,3)'*b(3,3);
h44 = b(4,4)'*b(4,4);
% Sigma^2 = h => Channel coefficient
Sigma1_Square = h11;
Sigma2_Square = h22;
Sigma3_Square = h33;
Sigma4_Square = h44;
% SNR = P / Sigma_n^2 => P = SNR * Sigma_n^2
pwr = SNR*Sigma_n*Sigma_n;
% Lamda_i = (Sigma_i^2 * P) / Sigma_n^2
Lamda_1 = (Sigma1_Square * pwr)/(Sigma_n*Sigma_n);
Lamda_2 = (Sigma2_Square * pwr)/(Sigma_n*Sigma_n);
Lamda_3 = (Sigma3_Square * pwr)/(Sigma_n*Sigma_n);
Lamda_4 = (Sigma4_Square * pwr)/(Sigma_n*Sigma_n);
% Here n = 4
Lamda_0= 4/[1+(1/Lamda_1)+(1/Lamda_2)+(1/Lamda_3)+(1/Lamda_4)];
% Capacity for Water-filling case:
C = [B * log(Lamda_i/Lamda_0)]
C_MIMO_4X4_WF(index) = log(Lamda_1/Lamda_0) + log(Lamda_2/Lamda_0) +
log(Lamda_3/Lamda_0) + log(Lamda_4/Lamda_0);
end
10. REFERENCES:
[1] Class Notes by Lie Xu
[2] Thomas Cover and Joy Thomas, Elements of Information Theory, 2nd Edition, John Wiley &
Sons, Inc, 2006
[3] David Tse and Pramod Viswanath, Fundamentals of Wireless Communication, Cambridge
University Press, 2005.
[4] CDMA Development Group: http://www.cdg.org/technology/3g.asp
[5] http://www.articlesbase.com/communication-articles/modern-mobile-phone-technology213572.html
[6] http://www.buzzle.com/articles/todays-mobile-phones-modern-technology-changedlives.html
[7] http://users.ece.utexas.edu/~mkountouris/publications/conferences/VTC_Spring_2006.pdf
[8] http://paros.stanford.edu/Publications/Chris/NgSubmitted.pdf
[9] http://www.cl.cam.ac.uk/research/dtg/publications/public/ja316/PowerAlloc_ver8.pdf
[10] http://persons.unik.no/arehj/publications/vtc08_Behrouz.pdf
[11] http://www.stanford.edu/class/ee360/Ahmad/Lin_f.pdf
[12] http://ns01.net.i.kyoto-u.ac.jp/~kazunori/paper/wpmc2006_2.pdf