In a previous post we have seen that MIMO fading capacity is much higher than AWGN capacity with multiple antennas. How is this possible? How can randomness added by a fading channel help us? In this post we try to find the reason for this. Let’s assume the following signal model for a Multi Input Multi Output antenna system.

x=Hs+w

Here s is the N_{T} by 1 signal vector, w is the N_{R} by 1 noise vector and H is the N_{R} by N_{T} channel matrix. The received signal vector is represented by x which has dimensions of N_{R} by 1. In expanded form this can be written as (assuming N_{T} =4 and N_{R} =4):

Recently I came across a post from T-Mobile in which they claim to have achieved a download speed of 5.6 Gbps over a 100 MHz channel resulting in a Spectral Efficiency of more than 50 bps/Hz. This was achieved in an MU-MIMO configuration with eight connected devices having an aggregate of 16 parallel streams i.e. two parallel streams per device. The channel used for this experiment was the mid-band frequency of 2.5 GHz.

We have previously discussed Shannon Capacity of CDMA and OFMDA, here we will discuss it again in a bit more detail. Let us assume that we have 20 MHz bandwidth for both the systems which is divided amongst 20 users. For OFDMA we assume that each user gets 1 MHz bandwidth and there are no guard bands or pilot carriers. For CDMA we assume that each user utilizes full 20 MHz bandwidth. We can say that for OFDMA each user has a dedicated channel whereas for CDMA the channel is shared between 20 simultaneous users.

We know that Shannon Capacity is given as

C=B*log2(1+SNR)

or in the case of CDMA

C=B*log2(1+SINR)

where ‘B’ is the bandwidth and SINR is the signal to noise plus interference ratio. For OFDMA the SNR is given as

SNR=Pu/(B*No)

where ‘Pu’ is the signal power of a single user and ‘No’ is the Noise Power Spectral Density. For CDMA the calculation of SINR is a bit more complicated as we have to take into account the Multiple Access Interference. If the total number of users is ‘u’ the SINR is calculated as

SINR=Pu/(B*No+(u-1)*Pu)

The code given below plots the capacity of CDMA and OFDMA as a function of Noise Power Spectral Density ‘No’.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% CAPACITY OF CDMA and OFDMA
% u - Number of users
% Pu - Power of a single user
% No - Noise Power Spectral Density
%
% Copyright RAYmaps (www.raymaps.com)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear all
close all
u=20;
Pu=1;
No=1e-8:1e-8:1e-6;
B=20e6;
C_CDMA=u*B*log2(1+Pu./(B*No+(u-1)*Pu));
B=1e6;
C_OFDMA=u*B*log2(1+Pu./(B*No));
plot(No,C_CDMA/1e6);hold on
plot(No,C_OFDMA/1e6,'r');hold off
xlabel('Noise Power Spectral Density (No)')
ylabel('Capacity (Mbps)')
legend('CDMA','OFDMA')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

We see that the capacity of OFDMA is much more sensitive to noise than CDMA. Within the low noise region the capacity of OFDMA is much better than CDMA but as the noise increases the capacity of the two schemes converges. In fact it was seen that as the noise PSD is further increased the two curves completely overlap each other. Therefore it can be concluded that OFDMA is the preferred technique when we are operating in the high SNR regime.

Somebody recently asked me this question “Does Shannon Capacity Increase by Dividing a Frequency Band into Narrow Bins”. To be honest I was momentarily confused and thought that this may be the case since many of the modern Digital Communication Systems do use narrow frequency bins e.g. LTE. But on closer inspection I found that the Shannon Capacity does not change, in fact it remains exactly the same. Following is the reasoning for that.

Shannon Capacity is calculated as:

C=B*log2(1+SNR)

or

C=B*log2(1+P/(B*No))

Now if the bandwidth ‘B’ is divided into 10 equal blocks then the transmit power ‘P’ for each block would also be divided by 10 to keep the total transmit power for the entire band to be constant. This means that the factor P/(B*No) remains constant. So the total capacity for the 10 blocks would be calculated as:

C=10*(B/10)*log2(1+P/(B*No))

So the Shannon Capacity for the entire band remains the same.

PS: The reason for the narrower channels is that for a narrow channel the channel appears relatively flat in the frequency domain and the process of equilization is thus simplified (a simple multiplication/division would do).

Note: ‘No’ is the Noise Power Spectral Density and ‘B*No’ is the Noise Power.

Frequency Reuse is a well known concept that has been applied to wireless systems over the past two decades e.g. in GSM systems. As the name suggests Frequency Reuse implies using the same frequencies over different geographical areas. If we have a 25MHz band then we can have 125 GSM channels and 125*8=1000 time multiplexed users in a given geographical area. Now if we want to increase the number of users we would have to reuse the same frequency band in a geographically separated area. The technique usually adopted is to use a fraction of the total frequency band in each cell such that no two neighbor cells use the same frequency. Typically the frequency band is divided into 3 or 7 cells.

The division of the frequency band in to smaller chunks reduces the system capacity e.g. one cell with 25 MHz bandwidth would have much higher capacity then 7 cells having 3.5 MHz each. To overcome this problem a frequency reuse of 1 has been proposed i.e. each cell has the full system bandwidth (nearly). The problem of co-channel interference at the cell boundaries is resolved by dedicating a small chunk of the available spectrum for the cell edges.

In Soft Frequency Reuse (SFR) the cell area is divided into two regions; a central region where all of the frequency band is available and a cell edge area where only a small fraction of the spectrum is available. The spectrum dedicated for the cell edge may also be used in the central region if it is not being used at the cell edge. The lack of spectrum at the cell edge may result in much reduced Shannon Capacity for that region. This is overcome by allocating high power carriers to the users in this region thus improving the SINR and the Shannon Capacity.

Note:
1. The Signal to Interference and Noise Ratio is given as:
SINR=Signal Power/(Intercell Interference+Intracell Interference+AWGN Noise)
2. Typically the term capacity was used to describe the number of voice channels (or users) that a system can support. But with modern digital communication systems it usually refers to the Shannon Capacity that can be achieved (in bits/sec/Hz).

[1] Yiwei Yu, Eryk Dutkiewicz, Xiaojing Huang, Markus Mueck and Gengfa Fang, “Performance Analysis of Soft Frequency Reuse for Inter-cell Interference Coordination in LTE Networks”, ISCIT 2010.

The Shannon Capacity of a channel is the data rate that can be achieved over a given bandwidth (BW) and at a particular signal to noise ratio (SNR) with diminishing bit error rate (BER). This has been discussed in an earlier post for the case of SISO channel and additive white Gaussian noise (AWGN). For a MIMO fading channel the capacity with channel not known to the transmitter is given as (both sides have been normalized by the bandwidth [1]):

Shannon Capacity of a MIMO Channel

where N_{T} is the number of transmit antennas, N_{R} is the number of receive antennas, γ is the signal to interference plus noise ratio (SINR), I_{NR} is the N_{R}xN_{R} identity matrix and H is the N_{R}xN_{T} channel matrix. Furthermore, h_{ij}, an element of the matrix H defines the complex channel coefficient between the ith receive antenna and jth transmit antenna. It is quite obvious that the channel capacity (in bits/sec/Hz) is highly dependent on the structure of matrix H. Let us explore the effect of H on the channel capacity.

Let us first consider a 4×4 case (N_{T}=4, N_{R}=4) where the channel is a simple AWGN channel and there is no fading. For this case h_{ij}=1 for all values of i and j. It is found that channel capacity of this simple channel for an SINR of 10 dB is 5.36bits/sec/Hz. It is further observed that the channel capacity does not change with number of transmit antennas and increases logarithmically with increase in number of receive antennas. Thus it can be concluded that in an AWGN channel no multiplexing gain is obtained by increasing the number of transmit antennas.

We next consider a more realistic scenario where the channel coefficients h_{ij} are complex with real and imaginary parts having a Gaussian distribution with zero mean and variance 0.5. Since the channel H is random the capacity is also a random variable with a certain distribution. An important metric to quantify the capacity of such a channel is the Complimentary Cumulative Distribution Function (CCDF). This curve basically gives the probability that the MIMO capacity is above a certain threshold.

It is obvious (see figure above) that there is a very high probability that the capacity obtained for the MIMO channel is significantly higher than that obtained for an AWGN channel e.g. for an SINR of 9 dB there is 90% probability that the capacity is greater than 8 bps/Hz. Similarly for an SINR of 12 dB there is a 90% probability that the capacity is greater than 11 bps/Hz. For a stricter threshold of 99% the above capacities are reduced to 7.2 bps/Hz and 9.6 bps/Hz.

In a practical system the channel coefficients h_{ij} would have some correlation which would depend upon the antenna spacing. Lower the antenna spacing higher would be the antenna correlation and lower would be the MIMO system capacity. This would be discussed in a future post.

The MATLAB code for calculating the CCDF of channel capacity of a MIMO channel is given below.

clear all
close all
Nr=4;
Nt=4;
I=eye(Nr);
g=15.8489;
for n=1:10000
H=sqrt(1/2)*randn(Nr,Nt)+j*sqrt(1/2)*randn(Nr,Nt);
C(n)=log2(det(I+(g/Nt)*(H*H')));
end
[a,b]=hist(real(C),100);
a=a/sum(a);
plot(b,1-cumsum(a));
xlabel('Capacity (bps/Hz)')
ylabel('Probability (Capacity > Abcissa)')
grid on

[1] G. J. Foschini and M. J. Gans,”On limits of Wireless Communications in a Fading Environment when Using Multiple Antennas”, Wireless Personal Communications 6, pp 311-335, 1998.