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.

Peak LTE data rate can be calculated using the following parameters:

1 Time-slot=0.5 ms (i.e 1 Sub-frame = 1 ms)
1 Time-slot=7 Modulation Symbols (when normal CP length is used)
1 Modulation Symbol=6 bits; if 64 QAM is used as modulation scheme

Data rate for a single carrier=Number of symbols per time slot*Bits per symbol/Duration of a time slot=7*6/0.5e-3=84kbps

If 1200 carriers (100RBs) are used then the aggregated throughput would be=1200*84kbps=100.8Mbps

If 4×4 MIMO is used then the capacity would increase four fold to=403.2Mbps

With 3/4 channel coding the data rate would be reduced to=302.4Mbps

Note:

1. A Resource Block (RB)=12 Carriers

2. Actual data rate would depend upon the instantaneous channel conditions and number of users sharing the resources e.g. going down from 64QAM to QPSK in adverse channel conditions would reduce the data rate from 302.4Mbps to 100.8Mbps. Changing the code rate from 3/4 to 1/3 would further reduce the data rate to 44.8Mbps.

In the previous post we calculated the Shannon Capacity of LTE as a function of bandwidth. We now calculate the capacity as a function of SNR (bandwidth fixed at 20MHz and signal power varied). We also use the concept of effective bandwidth to get a more realistic estimate of the capacity. The modified Shannon Capacity formula is given as:

C=B_eff*log2(1+SNR)

where

B_eff=Bandwidth*eff1*eff2*eff3*eff4

eff1=0.9=due to adjacent channel leakage ratio and practical filter issues

eff2=0.93=due to cyclic prefix

eff3=0.94=due to pilot assisted channel estimation

eff4=0.715=due to signalling overhead

B_eff=0.57*B

Therefore

C=0.57*B*log2(1+SNR)

Note: This is the capacity in a SISO channel with no fading.

Shannon Capacity of LTE in AWGN can be calculated by using the Shannon Capacity formula:

C=B*log2(1+SNR)

or

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

The signal power P is set at -90dBm, the Noise Power Spectral Density No is set at 4.04e-21 W/Hz (-174dBm/Hz) and the bandwidth is varied from 1.25MHz to 20MHz.

It is seen that the capacity increases from about 10Mbps to above 70Mbps as the bandwidth is varied from 1.25MHz to 20MHz (keeping the signal power constant). It must be noted that this is the capacity with a single transmit and single receive antenna (MIMO capacity would obviously be higher).