Tag Archives: Shannon Capacity

Shannon Capacity CDMA vs OFDMA

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')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Shannon Capacity of CDMA and OFDMA
Shannon Capacity of CDMA and 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.

Does Shannon Capacity Increase by Dividing a Frequency Band into Narrow Bins

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.

Soft Frequency Reuse in LTE

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.

7-Cell Frequency Reuse in GSM
7-Cell Frequency Reuse in GSM

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.

Frequency Planning and Power Allocation for SFR Scheme
Frequency Planning and Power Allocation for SFR Scheme

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.

MIMO Capacity in a Fading Environment

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
Shannon Capacity of a MIMO Channel

where NT is the number of transmit antennas, NR is the number of receive antennas, γ is the signal to interference plus noise ratio (SINR), INR is the NRxNR identity matrix and H is the NRxNT channel matrix. Furthermore, hij, 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 (NT=4, NR=4) where the channel is a simple AWGN channel and there is no fading. For this case hij=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 hij 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.

Complimentary Cumulative Distribution Function of Capacity
Complimentary Cumulative Distribution Function of Capacity

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 hij 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.

Shannon Capacity of LTE (Effective)

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)

LTE Capacity

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

Shannon Capacity of LTE (Ideal)

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.

LTE_capacity5

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).

 

Shannon Capacity of a GSM Channel in Fading Environment

In the previous post we calculated the Shannon Capacity of a 200kHz GSM channel in AWGN (Additive White Gaussian Noise). However, in a practical scenario the capacity is limited by time varying fading and interference. Let us consider a fading channel with four possible states corresponding to SNRs of 15dB, 10dB, 5dB and 0dB. The probability of these states is 0.50, 0.25, 0.15 and 0.10 respectively. The Shannon Capacity of such a channel is given as (assuming that the channel state information is known at the receiver):

C=Σ B*log2(1+SNRi)* p(SNRi)

C=B*(Σ log2(1+SNRi)* p(SNRi))

C=(200e3)*(log2(1+31.62)*0.50+log2(1+10.00)*0.25+log2(1+3.16)*0.15+log2(1+1)*0.10)

C=757.43kbps

Assuming that only one out of eight time slots is allocated to any user the Shannon Capacity of a GSM channel is reduced to 94.68kbps.

Note: The contribution of the high SNR states dominates the capacity of the channel. The contribution of the four states in terms of percentage capacity is given as 66.38%, 22.84%, 8.14% and 2.64%.

Shannon Capacity of a GSM Channel

We know that GSM bit rates can vary from a few kbps to a theoretical maximum of 171.2kbps (GPRS). But what is the actual capacity of a 200kHz GSM channel. We can use the Shannon Capacity Theorem to find this capacity.

C=B*log2(1+SNR)

or

C=B*log2(1+P/N)

The noise power can be found by using the following formula:

N=B*No=k*T*B=(1.38e-23)*(293)*(200e3)=8.08e-16W=-121dBm

Let us now assume a signal power 0f -90dBm. This gives us an SNR of 31dB or 1258.9 on linear scale. The capacity can thus be calculated as:

C=200e3*log2(1+1258.9)=2.06Mbps

This is the capacity if all time slots are allocated to a single user. If only one time slot is allocated to a user the capacity would be reduced to 257.48kbps.

CDMA vs FDMA Capacity (Mbps)

We saw previously that the channel capacity of a WCDMA system is severely limited by the Multiple Access Interference (MAI). Now let us consider the case where the 5MHz channel is divided equally among 20 users such that each user has a bandwidth of 250kHz. Keeping the transmit power for the narrow band signal the same as the wideband signal the signal to noise ratio would improve tremendously (since there is no MAI and the noise power is reduced by a factor of 20). Thus each narrowband channel would have a capacity of 2.99Mbps giving a combined capacity of 59.83Mbps for the 5MHz channel. The main drawback of an FDMA system is that the capacity of each narrowband channel is fixed at 2.99Mbps. However the capacity of a WCDMA user is dependent upon the number of users varying from 38Mbps for a single user to 5Mbps for two users (10Mbps combined) to 370kbps for 20 users (7.4Mbps combined). The higher capacity of FDMA when combined with dynamic allocation of channels builds the case for OFDMA based techniques.

Note: For the narrowband case the signal power is -80dBm and noise power is -116dBm resulting in an SNR of 36dB.