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. Letâ€™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):

In a previous post we had discussed MIMO capacity in a fading environment and compared it to AWGN capacity. It sometimes feels unintuitive that fading capacity can be higher than AWGN capacity. If a signal is continuously fluctuating how is it possible that we are able to have reliable communication. But this is the remarkable feature of MIMO systems that they are able to achieve blazing speeds over an unreliable channel, at least theoretically. It has been shown mathematically that an NxN MIMO channel is equivalent to N SISO channels in parallel.

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.

There is a 3 way race for 5G leadership in the US between T-Mobile(+Sprint), Verizon and AT&T. There are competing claims for the number of 5G subscribers, coverage area and download speeds. But let us look where the 5G industry stands today compared to the expectations a few years back. More than 80% of US population lives in urban areas which comprise of 2% of the total land area of about 10 million squared kilometers. That is 80% of the population lives in an area of about 200,000 squared kilometers.

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.