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.Continue reading 5G Data Rates and Shannon Capacity
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
or in the case of CDMA
where ‘B’ is the bandwidth and SINR is the signal to noise plus interference ratio. For OFDMA the SNR is given as
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
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:
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:
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.
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 ):
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.
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
 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 in AWGN can be calculated by using the Shannon Capacity formula:
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).
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))
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%.