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.

We have previously looked at frequency domain fading simulators i.e. simulators that define the Doppler components in the frequency domain and then perform an IDFT to get the time domain signal. These simulators include Smith’s Simulator, Young’s Simulator and our very own Computationally Efficient Rayleigh Fading Simulator. Another technique that has been widely reported in the literature is Sum of Sinusoids Method. As the name suggests this method generates the Doppler components in the time domain and then sums them up to generate the time domain fading envelope. There are three parameters that define the properties of the generated signal.

1) Number of sinusoids – Higher the number better the properties of the generated signal but greater the computational complexity
2) Angle of arrival – This can be generated statistically or deterministically, spread from –pi to pi.
3) Phase of the arriving wave – This is uniformly distributed between –pi and pi.

The MATLAB code below gives three similar sum of sinusoids techniques for generating a Rayleigh faded envelope [1].

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% SUM OF SINUSOIDS FADING SIMULATORS
% fd - Doppler frequency
% fs - Sampling frequency
% ts - Sampling period
% N - Number of sinusoids
%
% www.raymaps.com
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear all
close all
fd=70;
fs=1000000;
ts=1/fs;
t=0:ts:1;
N=100;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Method 1 - Clarke
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
x=zeros(1,length(t));
y=zeros(1,length(t));
for n=1:N;n
alpha=(rand-0.5)*2*pi;
phi=(rand-0.5)*2*pi;
x=x+randn*cos(2*pi*fd*t*cos(alpha)+phi);
y=y+randn*sin(2*pi*fd*t*cos(alpha)+phi);
end
z=(1/sqrt(N))*(x+1i*y);
r1=abs(z);
plot(t,10*log10(r1))
hold on
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Method 2 - Pop, Beaulieu
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
x=zeros(1,length(t));
y=zeros(1,length(t));
for n=1:N;n
alpha=2*pi*n/N;
phi=(rand-0.5)*2*pi;
x=x+randn*cos(2*pi*fd*t*cos(alpha)+phi);
y=y+randn*sin(2*pi*fd*t*cos(alpha)+phi);
end
z=(1/sqrt(N))*(x+1i*y);
r2=abs(z);
plot(t,10*log10(r2),'r')
hold on
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Method 3 - Chengshan Xiao
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
x=zeros(1,length(t));
y=zeros(1,length(t));
for n=1:N;n
phi=(rand-0.5)*2*pi;
theta=(rand-0.5)*2*pi;
alpha=(2*pi*n+theta)/N;
x=x+randn*cos(2*pi*fd*t*cos(alpha)+phi);
y=y+randn*sin(2*pi*fd*t*cos(alpha)+phi);
end
z=(1/sqrt(N))*(x+1i*y);
r3=abs(z);
plot(t,10*log10(r3),'g')
hold off
xlabel('Time(sec)')
ylabel('Envelope(dB)')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

All the three techniques given above are quite accurate in generating a Rayleigh faded envelope with the desired statistical properties. The accuracy of these techniques increases as the number of sinusoids goes to infinity (we have tested these techniques with up to 1000 sinusoids but realistically speaking even 100 sinusoids are enough). If we want to compare the three techniques in terms of the Level Crossing Rate (LCR) and Average Fade Duration (AFD) we can say that the first and third technique are a bit more accurate than the second technique. Therefore we can conclude that a statistically distributed angle of arrival is a better choice than a deterministically distributed angle of arrival. Also, if we look at the autocorrelation of the in-phase and quadrature components we see that for the first and third case we get a zero order Bessel function of the first kind whereas for the second case we get a somewhat different sequence which approximates the Bessel function with increasing accuracy as the number of sinusoids is increased.

The above figures show the theoretical Bessel function versus the autocorrelation of the real/imaginary part generated by method number two. The figure on the left considers 20 sinusoids whereas the figure on the right considers 40 sinusoids. As can be seen the accuracy of the autocorrelation sequence increases considerably by doubling the number of sinusoids. We can assume that for number of sinusoids exceeding 100 i.e. N=100 in the above code the generated autocorrelation sequence would be quite accurate.

[1] Chengshan Xiao, “Novel Sum-of-Sinusoids Simulation Models for Rayleigh and Rician Fading Channels,” IEEE TRANSACTIONS ON WIRELESS COMMUNICATIONS, VOL. 5, NO. 12, DECEMBER 2006.

We have been using a wireless signal model in our simulations without going into the details of noise calibration for simulation. In this article we discuss this. Lets assume the received signal is given as

r(t)=s(t)+n(t)

where r(t) is the received signal s(t) is the transmitted signal and n(t) is the Additive White Gaussian Noise (AWGN). Channel fading is ignored at the moment. Signal to noise ratio for simulation of digital communication systems is given as

ρ=Eb/No (1)

Where Eb is the energy per bit and No is the noise Power Spectral Density (PSD). We also know that for the case of Additive White Gaussian Noise the noise power is given as [Tranter]

σ^2=(No/2)*fs

No=2*(σ^2)/fs

Where σ is the standard deviation of noise and fs is the sampling frequency. Substituting in equation 1 we get

ρ=Eb/No=Eb/(2*σ^2/fs )

Eb/No=(Eb*fs)/(2*σ^2)

σ^2=(Eb*fs)/(2*Eb/No )

σ=√((Eb*fs)/(2*Eb/No ))

If the energy per bit and the sampling frequency is set to 1 the above equation reduces to

σ=√(1/(2*Eb/No ))

The simulation software can thus calculate the noise standard deviation (or variance) for each value of Eb/No in the simulation cycle. The following piece of MATLAB code generates AWGN with the required power and adds it to the transmitted signal.

s=sign(rand-0.5); % Generate a symbol
sigma=1/sqrt(2*EbNo); % Calculate noise standard deviation
n=sigma*randn; % Generate AWGN with the required std dev
r=s+n; % Add noise to the signal

How can we assume that energy per bit and sampling frequency is equal to one and are we breaking some discrete time signal processing rule here. This will be discussed in a later post.

In the previous posts we had discussed generation of a correlated Rayleigh fading sequence using Smith’s method [1] and Young’s modification of Smith’s method [2]. The main contribution of Young was that he proposed a mechanism where the number of IDFTs was reduced by half. This was achieved by first adding two length N IID zero mean Gaussian sequences filtered by the filter F[k] and then performing the IDFT on the resulting complex sequence.

This was different to Smith’s method where the IDFT was performed simultaneously on two branches and then the outputs of these branches were added in quadrature to achieve the desired sequence with Rayleigh distributed envelope and Uniformly distributed phase. Another problem with Smith’s method was that the outputs of the two arms after performing IDFTs was assumed to be real which is not always the case in implementation and depends upon the combination of Doppler frequency (fm) and length of Gaussian sequence (N).

Young’s technique is shown graphically in the above figure. Also shown is the definition of filter F[k] which depends upon N, fm and km (please note that the fm in the above equation is normalized by the sampling frequency fs). Here km = N*(fm/fs). We propose three modifications to Young’s technique which significantly reduces computation and at the same time maintains the statistical properties of the generated sequence. The modifications we propose are.

1. First modification has to do with the generated Gaussian sequence. It is observed that the filter F[k], at very high sampling rates, is mostly zero and there are very few points which have some non-zero value. So when we multiply the Gaussian sequence with the filter we mostly get zeros at the output. So we propose that the filter response in the frequency domain must be calculated first and the the Gaussian random sequence must be generated for only those points where the filter F[k] is non-zero e.g. for a sampling frequency of 7.68 MHz (standard sampling frequency for a BW of 5 MHz in LTE) and Doppler frequency of 70 Hz (corresponding to medium Doppler case in LTE) the filter F[k] has 99.9982% zeros in its frequency response and it would be a highly wasteful to calculate Gaussian RVs for all those values.

2. Secondly according to Clarke [3] the Doppler Spectrum measurements have “Marked disagreement at very low frequencies and at frequencies in the region of the sharp cut-off associated with the maximum Doppler frequency shift. At very low frequencies the spectral energy is always observed to be higher than that predicted by theory”. He goes on to add “The reason for this is that neither theoretical model takes into account the large scale variations in total energy which result from the changing topography between transmitter and mobile receiver”. This suggests that the Classical Doppler Spectrum might not be the best choice under all scenarios. This has also been noted in [4] where a flat fading model is evaluated in terms of its Level Crossing Rate and Average Fade Duration. Such a flat spectrum is especially suited to indoor scenarios as noted in [5] and [6].

We propose a filter that gives equal weight to all the frequencies up to the maximum Doppler frequency. So our filter is a box-type filter which applies a constant scaling factor to all the frequencies in the pass-band and zeros out all the frequencies in the stop-band. So in fact the Gaussian sequence that is generated in the in-phase arm may directly be added with the Gaussian sequence from the other arm without applying the frequency domain filter and then IDFT of the complex sequence is taken. We will look into the deviation from ideality that this causes later.

3. The third modification that we propose is in the implementation of the IDFT. Here again we take into consideration that the complex sequence being fed to the IDFT is filled with zeros (as we noted earlier 99.9982% zeros for 7.68 MHz and even more for higher frequencies) so we can avoid a lot multiplications and summations. The IDFT is defined below and also given is our modification to it.

Further improvement in computation time is achieved by implementing the above as a matrix multiplication. The matrix multiplication is implemented as H*X where H is the IDFT coefficients of size N x 2(km+1) and X is a vector of size 2(km+1) x 1 upon which the IDFT has to be performed.

Now let us look at the output sequence generated by using the above techniques. We consider the case of Medium Doppler Frequency of 70 Hz (EVA channel) as defined by LTE specifications. Sampling frequency is fixed at 10 kHz giving a normalized Doppler frequency of 0.007. This was done due to limitation of memory on the machine. The author also experimented with a sampling frequency of 7.68 MHz but this did not yield enough samples for statistically accurate results. We did use a sampling frequency of 7.68 MHz for our bit error rate simulation which is shown in the end.

It is observed for fm=70 Hz the envelope of the output sequence using the proposed technique matches quite well with the envelope of the output sequence generated by the ideal filter proposed by Young. Also the phase and envelope of the sequence generated using the proposed technique has the desired distribution. Some of the other metrics that we can look at are the level crossing rate (LCR) and average fade duration (AFD) as well as the Auto Correlation of the real and imaginary parts of the complex sequence generated.

Parameter

Young

Modified

LCR (ideal)

48.1086

48.1086

LCR (sim)

48.1506

39.4348

AFD (ideal)

0.0018

0.0018

AFD (sim)

0.0018

0.0022

If we look at the results for LCR and AFD we see that the simulated results match reasonably well with the results predicted by theory. These results correspond to 100 snapshots of the fading sequence. It was important to take the average of several snapshots as results varied with each simulation run. Sometimes Young’s technique produced more accurate results while at other times the proposed technique was better. Again the limitations of computer memory and processing power dictated the length of the sequence that could be generated.

In general Young’s technique produced better results than our proposed technique. It was found that product of LCR and AFD for both cases matched quite well with the theoretical value. So the total time spent in a fade state per second was equal in both the cases. In the proposed method the duration of a single fading event was higher, whereas the number of fading events per second was lower. This can be attributed to the fact that in our proposed technique higher weighting is given to lower frequency components and the fading sequence is smoothed out by these low frequency components. One technique to overcome this is spectral broadening as suggested by [4] but this is not the subject here and we postpone its discussion to another article.

Auto Correlation of Real Part fm=70Hz

The Auto Correlation of the real and imaginary parts of the generated sequences are also calculated for a Doppler frequency of 70 Hz. It is found that the Auto Correlation sequence for the two techniques matches quite well. However, the Auto Correlation sequence deviates from the theoretical value as calculated the by Bessel function of the first kind and zero order. Since we have used a flat spectral mask the Auto Correlation function resembles the sinc(x) function which is the same as zeroth order Spherical Bessel Function of the first kind (which is related to 1/2 order Bessel Function of the first kind).

It was found that when the Rayleigh fading sequence is generated by the program provided in Young’s thesis the shape of the Auto Correlation function depends upon the sampling frequency. At a normalized Doppler frequency of 0.05 and N=2^16 Young’s technique produces quite accurate results. We also measured the mean squared error (MSE) between the two Auto Correlations sequences and found it to be a function of the Normalized Doppler Frequency. It was found that as the Normalized Doppler Frequency was increased from 0.00007 to 0.0007 the MSE error dropped from 0.0277 to 0.0041. The relationship between the Normalized Doppler Frequency and MSE, for a fixed sequence length, seems to be resembling an exponential function. For more accurate results, at higher sampling frequencies, the number of samples would have to be increased considerably. In fact it was found that if the variable km (km=N*fm/fs) is maintained at around 20 the error between the two correlation sequences is less than 1% for all possible sampling rates.

We also compared the bit error rate (BER) performance of different QAM modulation schemes using both the techniques for fading envelope generation and found these to be matching quite well. A single tap was used which results in a flat fading channel. This is a simplistic channel model but it gives us some confidence that the proposed approach does have the desired statistical properties. A good test of a temporally correlated Rayleigh fading sequence is to test it on a system that implements interleavers and channel coders whose performance strongly depends upon factors such as the LCR and AFD e.g. a certain forward error correction (FEC) code might work well in high LCR and low AFD as this distributes out errors in different blocks and allows the code to correct them. In simulations done so far (not shown here) we have found that for a 1/2 rate convolutional encoder with Hard Viterbi Decoding the BER for the two schemes matches quite well. In general the results for correlated fading are much worse than uncorrelated fading.

In future we would also like to evaluate the bit error rate (BER) performance of an M-QAM OFDM system with Frequency Selective Rayleigh fading as described by the LTE fading channels EPA, EVA and ETU. This is probably a good scenario to compare the accuracy of the two techniques used to generate Rayleigh fading sequences above. One challenge in this regard is that the LTE channel taps are described in increments of 10 nsec whereas the LTE signal sampling rate is defined on a different scale (minimum Ts=32.5521 nsec corresponding to a sampling rate of 30.72 Msps). So we would have to do sample rate conversion to implement a time varying frequency selective Rayleigh fading channel.

[1] John I. Smith, “A Computer Generated Multipath Fading Simulation for Mobile Radio”, IEEE Transactions on Vehicular Technology, vol VT-24, No. 3, August 1975.

[2] David J. Young and Norman C. Beaulieu, “The Generation of Correlated Rayleigh Random Variates by Inverse Discrete Fourier Transform”, IEEE Transactions on Communications vol. 48 no. 7 July 2000.

[3] R. H. Clarke, A Statistical Theory of Mobile Radio Reception”, Bell Systems Technical Journal 47 (6), pp 957–1000, July 1968.

[4] Rosmansyah, Y.; Saunders, S.R.; Sweeney, P.; Tafazolli, R., “Equivalence of flat and classical Doppler sample generators,” Electronics Letters , vol.37, no.4, pp.243,244, 15 Feb 2001.

[5] JTC (Joint Technical Committee T1 RIP1.4 and TIA TR46.3.3/TR45.4.4 on Wireless Access): “Draft final report on RF channel characterization”. Paper no. JTC(AIR)/94.01.17-238R4, January 17, 1994.

[6] ETSI (European Telecommunications Standards Institute): “Universal mobile telecommunications system (UMTS); selection procedures for the choice of radio transmission technologies of the UMTS (UMTS 30.03 version 3.2.0)”. Technical Report, TR 101 112 V3.2.0 (1998-04), http://www.etsi.org, 1998.

We have previously discussed the Bit Error Rate of M-QAM in Rayleigh Fading using Monte Carlo Simulation. We now turn our attention to calculation of Bit Error Rate (BER) of M-QAM in Rayleigh fading using analytical techniques. In particular we look at the method used in MATLAB function berfading.m. In this function the BER of 4-QAM, 16-QAM and 64-QAM is calculated from series expressions having 1, 3 and 5 terms respectively. These are given below (M is the constellation size and must be a power of 2).

Although using these expressions we get very accurate BER but it is not that simple to calculate (the expressions become even more complicated for higher constellation sizes such as 256-QAM). Therefore we try to simplify these expressions by using only the first term in each expression. To our surprise the results match quite well with the results using the exact formulae. There is very minor difference at low signal to noise ratios but that can be easily bargained for the ease of calculation.

So here is our program for calculating the BER using the approximate method.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FUNCTION TO CALCULATE THE BER OF M-QAM IN RAYLEIGH FADING
% M: Input, Constellation Size
% EbNo: Input, Energy Per Bit to Noise Power Spectral Density
% ber: Output, Bit Error Rate
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [ber]= BER_QAM_fading (M, EbNo)
k=log2(M);
EbNoLin=10.^(EbNo/10);
gamma_c=EbNoLin*k;
if M==4
%4-QAM
ber = 1/2 * ( 1 - sqrt(gamma_c/k./(1+gamma_c/k)) );
elseif M==16
%16-QAM
ber = 3/8 * ( 1 - sqrt(2/5*gamma_c/k./(1+2/5*gamma_c/k)) );
elseif M==64
%64-QAM
ber = 7/24 * ( 1 - sqrt(1/7*gamma_c/k./(1+1/7*gamma_c/k)) );
else
%Warning
warning('M=4,16,64')
ber=zeros(1,length(EbNo));
end
semilogy(EbNo,ber,'o-')
xlabel('EbNo(dB)')
ylabel('BER')
axis([0 24 0.001 1])
grid on
return
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

So we see that the results match quite well with the results previously obtained through simulation. We will next tackle the problem of simplifying the expression for higher order modulations such as 256-QAM in both Rayleigh and Ricean channels.

Many operators are thinking of moving from the main grid to alternative energy sources such as wind and solar. This is especially true in third world countries where electricity is not available 24/7 and is also very expensive. This has forced operators to switch their base stations to diesel generators (which is also a costly option).

In this article we do a rough estimation of the size a solar system required to run a cellular base station. We start with the assumption that 20 Watts of power are transmitted from a single antenna of base station. For a 3 sector site there are 3 antennas giving us total transmitted power of 60 Watts. Now if 50% of the power is lost in cables and connections we would have to boost up the transmitted power to 120 Watts.

We know that power amplifiers are highly in-efficient (depending upon the load) and a large amount of power is lost in this stage. So we assume an efficiency of 12 % giving us a total input power of 1000 Watts. Another 500 Watts are given to Air Conditioning (200 W), Signal Processing (150 W) and Rectifier (150 W). So the combined AC input to the base station is 1500 Watts. Now we turn our attention to sizing up the solar system.

If we assume that the BS is continuously consuming 1500 Watts over a 24 hour period we have a total energy consumption of 36 kWh. If the solar panels receive peak sun hours of 5 hours/day we would require solar panels rated at 7200 Watts. This could mean 72 solar panels of 100 Watts each or 36 solar panels of 200 Watts each or any other combination. It must be noted that we have not considered any margins for cloudy days when peak sun hours would be reduced. Also, we have not considered any reduction in power consumption when there is no load (or very less load) on the BS.

Next we calculate the amount of batteries required. We assume that the batteries are rated at 200 AH and 12 V. This gives us a total energy storage capacity per battery of 2.4 kWh. So the number of batteries required is calculated as 36 kWh/2.4 kWh = 15. It must be noted that some of the energy would be consumed in real-time and the actual number of batteries required would be lesser. Furthermore we would need an inverter of at least 1500 Watts and charge controller of 125 Amps.

You have estimated the size of the solar system that you need and are ready to get the equipment from the market to install it. But wait, are you sure you have enough space in your garden or your backyard or your rooftop to install the solar panels? How can you do a rough estimate of the area required by the solar panels? Here is a quick and easy way to go about it.

Lets assume that you want to install 10 solar panels rated at 100 Watts each and having a conversion efficiency of 18%. The total power output of the solar system can be calculated as:

Total Power Output=Total Area x Solar Irradiance x Conversion Efficiency

We know the required Total Output Power is 1000 Watts (10 panels x 100 Watts), the Solar Irradiance for a surface perpendicular to the Sun’s rays at sea level on a clear day is about 1000 Watt/m^{2} and the Conversion Efficiency is 18%. Plugging these number in the above equation we get:

1000 Watts = Total Area x 1000 Watts/m^{2} x 0.18

or

Total Area = 5.56 m^{2}

I you are going to install all the panels in one line you would need a space of approximately 1 m x 5.56 m (each panel having a size of 1 m x 0.556 m) on your rooftop. There you go. You have a rough estimate of the space required by the solar panels of your system.

Note:

1. Do remember that solar panels are usually installed at an angle to the earth surface and this may change the results somewhat.

2. Imagine a solar panel has a conversion efficiency of 100% i.e. it converts all the solar energy into electrical energy then all you would need is a 1 m^{2} solar panel to produce 1000 Watts of electrical energy.

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.

It is sometimes important to know the relationship between various distributions. This can be useful if there is a function available for one distribution and it can be used to derive other distributions. In the context of Wireless Communications it is important to know the relationship between the Uniform, Gaussian and Rayleigh distribution.

According to Central Limit Theorem the sum of a large number of independent and identically distributed random variables has a Gaussian distribution. This is used to model the amplitude of the in-phase and quadrature components of a wireless signal. Shown below is the model for the received signal which has been modulated by the Gaussian channel coefficients g1 and g2.

r=g1*a1*cos(2*pi*fc*t)+g2*a2*sin(2*pi*fc*t)

The envelope of this signal (sqrt(g1^2+g2^2)) as a Rayleigh distribution. Now if you only had a function for Uniform Distribution you can generate Rayleigh Distribution using the following routine.

clear all
close all
M=10000;
N=100;
for n=1:M;
x1=rand(1,N)-0.5;
x2=rand(1,N)-0.5;
y1=mean(x1);
y2=mean(x2);
z(n)=sqrt(y1^2+y2^2);
end
hist(z,20)

Note: Here a1 and a2 can be considered constants (at least during the symbol duration) and its really g1 and g2 that are varying.