# 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
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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.

# 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.

# Inside Qualcomm Snapdragon S4

We have previously looked at the antennas inside a cell phone. Now we look at another important component of a cell phone; the mobile station modem (MSM). One of the most popular MSM in cell phones today is the Qualcomm Snapdragon S4. The details of this MSM are given in the table below.

As can be seen from the above table this small chipset (can easily fit on a fingertip) packs a punch as far as processing power is concerned. It supports a number of wireless standards from GSM/GPRS to LTE and from CDMA 2000 to TD-SCDMA. One of its close competitors is the NVIDIA Tegra 3 which has four ARM Cortex A9 cores (compared to Snapdragon’s two).

Qualcomm Snapdragon – S4

# Implementing a Non-Uniformly Spaced Tapped Delay Line Channel Model

Question:
Since you are good on fundamentals I would like to ask you a question that puzzles me. LTE channels models are defined at irregular time intervals as shown in [1].

The EPA, EVA and ETU channel taps can best be described as being sampled at multiples of 10 nsec. However, LTE signal is sampled at multiples of 3.84 MHz (Ts=260.416667 nsec). So how does one perform convolution operation.

Empirical multipath channel is usually characterized as a τ-spaced tapped delay line (TDL), whose power delay profile (PDP) is either uniformly spaced, or more frequently, spaced with arbitrary time delay(s). Converting a τ-spaced empirical model to a more tractable T-spaced (sampling time-spaced) statistical model will greatly facilitate the link-level simulation and performance evaluation of the broadband digital wireless system. The practical approaches to accomplish this task could be either approximation or interpolation,

1. Approximation
a. Ceil or Floor a τ-spaced tap to the neighbouring sampling time.
b. Split the energy of a τ-spaced tap to the adjacent sampling times.
2. Interpolation
An Ideal Bandlimited (sinc) Interpolator can fulfil the goal.

It should be noted that although T-spaced statistical model is simpler for analysis, τ-spaced empirical model represents the channel more accurately.

[1] http://www.steepestascent.com/content/mediaassets/html/LTE/Help/PropagationConditions.html

# WINNER-II Path Loss Model

In simple terms the path loss is the difference between the transmitted power and the received power of a wireless communication system. This may range from tens of dB to more than a 100 dB e.g. if the transmitted power of a wireless communication system is 30 dBm and the received power is -90 dBm then the path loss is calculated as 30-(-90)=120 dB. Path loss is sometimes categorized as a large scale effect (in contrast to fading which is a small scale effect).

According to the WINNER-II model the path loss can be calculated  as:

Here d is the separation between the transmitter and receiver in meters, fc is the frequency in GHz, A is the path loss exponent, B is the intercept and C is the frequency dependent parameter. X is the environment specific parameter such as path loss due to a wall. PLfree is the path loss in a free space line of sight environment (here A=20, B=46.4 and C=20).

The table below describes the different environments defined in the WINNER-II model. Once an environment is selected the path loss parameters A, B and C can be selected from the table further down e.g. A1 is the in-building scenario with A=18.7, B=46.8 and C=20 for the LOS case. The PL for a T-R separation of 100 m and frequency of 2 GHz is calculated as:

PL=18.7*log10(100)+46.8+20*log10(2/5)=76.42 dB

A separate equation for the path loss is given where the parameters A, B and C are not sufficient to describe the scenario.

Note:

1. Here CG is the concept group that developed the particular scenario. This is either Local Area (LA), Metropolitan Area (MA) or Wide Area (WA).

2. For more details visit:

L. Hentilä, P. Kyösti, M. Käske, M. Narandzic , and M. Alatossava. (2007, December.) MATLAB implementation of the WINNER Phase II Channel Model ver1.1 [Online]. Available: https://www.ist-winner.org/phase_2_model.html

# 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.

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.

# QAM Theoretical BER in AWGN

Quadrature Amplitude Modulation (QAM) is an important modulation scheme as it allows for higher data rates and spectral efficiencies. The bit error rate (BER) of QAM can be calculated through Monte Carlo simulations. However this becomes quite complex as the constellation size of the modulation schemes increases. Therefore a theoretical approach is sometimes preferred. The BER for Gray coded QAM, for even number of bits per symbol, is shown below.

Gray coding ensures that a symbol error results in a single bit error. The code for calculating the theoretical QAM BER for k even (even number of bits per symbol) is given below. The formula for calculating the BER for k odd is different, however, the formula given below can be used a first estimate.

```EbNodB=-6:2:24
EbNo=10.^(EbNodB/10);
k=8;
M=2^k;
x=sqrt(3*k*EbNo/(M-1));
Pb=(4/k)*(1-1/sqrt(M))*(1/2)*erfc(x/sqrt(2));
semilogy(EbNodB,Pb)```

Note:
1. Each additional bit/symbol requires about 2dB extra in SNR to achieve the same BER.
2. 4-QAM is essentially QPSK modulation.

# 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

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 &gt; 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.

# CDMA vs OFDMA

 Property CDMA OFDMA 1. Channel bandwidth Full system bandwidth Variable system bandwidth to accommodate users with different data rates, 1.25, 2.50, 5.00, 10.00, 15.00 and 20.00 MHz, actual transmission bandwidth is a bit lower than this 2. Frequency-selective scheduling Not possible A key advantage of OFDMA, although it requires accurate real-time feedback of channel conditions from receiver to transmitter 3. Symbol period Very short—inverse of the system bandwidth Very long—defined by subcarrier spacing and independent of system bandwidth 4. Equalization Complicated time domain equalization Simple frequency domain equalization 5. Resistance to mulitpath Rake receiver can combine various multipath components Highly resistant to multipath due to insertion of cyclic prefix (CP) 6. Suitability for MIMO MIMO is not suited to a wideband frequency selective channel MIMO is suited to the independent narrowband flat fading channels that the subcarriers provide 7. Resistance to narrowband interference Resistant to narrow band interference Some subcarriers to be affected by narrowband interference 8. Separation of users Scrambling and orthogonal spreading codes Frequency and time although scrambling and spreading can be added as well

Reference: Agilent 3GPP Long Term Evolution System Overview, Product Development and Test Challenges Application Note.

# Peak to Average Power Ratio (PAPR)

Peak to Average Power Ratio (PAPR) as the name suggests is the ratio of peak signal power to the average signal power and has received considerable attention in the context of multicarrier signals like OFDM which exhibit a high PAPR. The down side of this high PAPR is that the power amplifier in the transmitter is operated at a relatively lower power level so that the peaks in the signal are not distorted by the saturating amplifier. This is called the amplifier backoff and it plays an important part in wireless system design.

Power Amplifier Input Output Behavior

The reason for this high PAPR is that when multiple sinusoids are added together in a multicarrier transmission the resulting signal exhibits constructive and destructive behavior. The higher the number of these sinusoids higher is the PAPR. The following example illustrates this behavior.

Constructive and Destructive Behavior of Multiple Carriers

It is observed that the PAPR of a signal composed of two sinusoids is greater than that of a single sinusoid. Similarly the PAPR of a signal composed of three carriers is even higher. The PAPR for the case of a single tone, two tones and three tones is 2.00, 3.10 and 4.15 respectively. Or on a logarithmic scale 3.01 dB, 4.91 dB and 6.18 dB respectively. So if the power amplifier in a wireless system starts saturating at 24 dBm then the average signal power of three tone system must not exceed 24-6.18=17.82 dBm. In practical systems techniques are adopted that decrease the PAPR so that the power amplifier can operate close to its maximum limit. One simple technique clips the peaks in the signal, while another adds tones at the unused frequencies such that the total PAPR is reduced.

Note:

1. The three tones in the above example are at ‘f’, ‘2f’ and ‘3f’ Hz.

2. Another metric that we will discuss in a future post is the cubic metric (CM).