Sizing Up a Solar System for a Cellular Base Station

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.

 

Posted in Fundamentals, LTE, Solar Energy, WiMAX | Tagged , , , , | Leave a comment

Calculate Solar Panel Tilt

1. Find the direction of magnetic North and consequently magnetic South.

2. Adjust for magnetic declination to find exact true South.

3. Point solar panels towards true South.

4. Find optimum tilt angle based on the latitude and the season.

Enter the value of latitude below to find the panel tilt in degrees.



Winter

Latitude

*+ = Degrees

Spring and Fall

Latitude

*- = Degrees

Summer

Latitude

*- = Degrees

 


Note:
1. The result above is the angle in degrees from the horizontal.
2. If you do not know the latitude of your city you can look it up here.

Posted in Fundamentals, Solar Energy | Tagged , , , , , , | 4 Comments

How to Calculate the Surface Area Required by Solar Panels

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/m2 and the Conversion Efficiency is 18%.  Plugging these number in the above equation we get:

1000 Watts = Total Area x 1000 Watts/m2 x 0.18

or

Total Area =   5.56 m2

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 m2 solar panel to produce 1000 Watts of electrical energy.

Posted in Fundamentals, Solar Energy | Tagged , , , , , | Leave a comment

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.

Posted in Capacity, Fundamentals, LTE | Tagged , , , , | 3 Comments

Uniform, Gaussian and Rayleigh Distribution

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.

Posted in Channel Modeling, Fundamentals | Tagged , , , | 1 Comment

Fading Model – From Simple to Complex

1. The simplest channel model just scales the input signal by a real number between 0 and 1 e.g. if the signal at the transmitter is s(t) then at the receiver it becomes a*s(t). The effect of channel is multiplicative (the receiver noise on the other hand is additive).

2.   The above channel model ignores the phase shift introduced by the channel. A more realistic channel model is one that scales the input signal as well rotates it by a certain angle e.g. if s(t) is the transmitted signal then the received signal becomes a*exp(jθ)*s(t).

3. In a realistic channel the transmitter, receiver and/or the environment is in motion therefore the scaling factor and phase shift are a function of time e.g. if s(t) is the transmitted signal then the received signal is a(t)*exp(jθ(t))*s(t). Typically in simulation of wireless communication systems a(t) has a Rayleigh distribution and θ(t) has a uniform distribution.

4. Although the above model is quite popular, it can be further improved by introducing temporal correlation in the fading envelope. This can be achieved by the Smith’s simulator which uses a frequency domain approach to characterize the channel. The behavior of the channel is controlled by the Doppler frequency fd. Higher the Doppler frequency greater is the variation in the channel and vice versa [1].

5. Finally the most advanced wireless channel model is one that considers the channel to be an FIR filter where each tap is defined by the process outlined in (4). The channel thus performs convolution on the signal that passes through it. In the context of LTE there are three channel models that are defined namely Extended Pedestrian A (EPA), Extended Vehicular A (EVA) and Extended Typical Urban (UTU) [2].

Note: As an after thought I have realized that this channel model becomes even more complicated with the introduction of spatial correlation between the antennas of a MIMO system [3].

Posted in Fundamentals | Tagged , , , | 2 Comments

M-QAM Bit Error Rate in Rayleigh Fading

We have previously discussed the bit error rate (BER) performance of M-QAM in AWGN. We now discuss the BER performance of M-QAM in Rayleigh fading. The one-tap Rayleigh fading channel is generated from two orthogonal Gaussian random variables with variance of 0.5 each. The complex random channel coefficient so generated has an amplitude which is Rayleigh distributed and a phase which is uniformly distributed. As usual the fading channel introduces a multiplicative effect whereas the AWGN is additive.

The function “QAM_fading” has three inputs, ‘n_bits’, ‘M’, ‘EbNodB’ and one output ‘ber’. The inputs are the number of bits to be passed through the channel, the alphabet size and the Energy per Bit to Noise Power Spectral Density in dB respectively whereas the output is the bit error rate (BER).

function[ber]= QAM_fading(n_bits, M, EbNodB)
 
% Transmitter
k=log2(M);
EbNo=10^(EbNodB/10);
x=transpose(round(rand(1,n_bits)));
h1=modem.qammod(M);
h1.inputtype='bit';
h1.symbolorder='gray';
y=modulate(h1,x);
 
% Channel
Eb=mean((abs(y)).^2)/k;
sigma=sqrt(Eb/(2*EbNo));
w=sigma*(randn(n_bits/k,1)+1i*randn(n_bits/k,1));
h=(1/sqrt(2))*(randn(n_bits/k,1)+1i*randn(n_bits/k,1));
r=h.*y+w;
 
% Receiver
r=r./h;
h2=modem.qamdemod(M);
h2.outputtype='bit';
h2.symbolorder='gray';
h2.decisiontype='hard decision';
z=demodulate(h2,r);
ber=(n_bits-sum(x==z))/n_bits
return

The bit error rates of the four modulation schemes 4-QAM, 16-QAM, 64-QAM and 256-QAM are shown in the figure below. All modulation schemes use Gray coding which gives a few dB of margin in the BER performance. As with the AWGN case each additional bit per symbol requires about 1.5-2 dB in signal to ratio to achieve the same BER.

M-QAM Bit Error Rate in Rayleigh Fading

M-QAM Bit Error Rate in Rayleigh Fading

Although not shown here similar behavior is observed for higher order modulation schemes such as 1024-QAM and 4096-QAM (the gap in the signal to noise ratio for the same BER is increased to about 5dB).

Posted in BER Performance, Fundamentals, LTE, WiMAX | Tagged , , , | 24 Comments

M-QAM Bit Error Rate in AWGN

Quadrature Amplitude Modulation has been adopted by most wireless communication standards such as WiMAX and LTE. It provides higher bit rates and consequently higher spectral efficiencies. It is usually used in conjunction with Orthogonal Frequency Division Multiplexing (OFDM) which provides a simple technique to overcome the time varying frequency selective channel.

We have previously discussed the formula for calculating the bit error rate (BER) of QAM in AWGN. We now calculate the same using a simple Monte Carlo Simulation.

function[ber]= QAM_AWGN(n_bits, M, EbNodB)
 
% Transmitter
k=log2(M);
EbNo=10^(EbNodB/10);
x=transpose(round(rand(1,n_bits)));
h1=modem.qammod(M);
h1.inputtype='bit';
h1.symbolorder='gray';
y=modulate(h1,x);
 
% Channel
Eb=mean((abs(y)).^2)/k;
sigma=sqrt(Eb/(2*EbNo));
w=sigma*(randn(1,n_bits/k)+1i*randn(1,n_bits/k));
r=y+w';
 
% Receiver
h2=modem.qamdemod(M);
h2.outputtype='bit';
h2.symbolorder='gray';
h2.decisiontype='hard decision';
z=demodulate(h2,r);
ber=(n_bits-sum(x==z))/n_bits
return

The above function basically has three inputs and one output. The inputs are the number of bits to be passed through the channel, the size of the constellation and the signal to noise ratio in dB. The output is the bit error rate (BER). The simulation can be divided into three section namely the transmitter, the channel and the receiver. In this simulation we have used Gray coding which gives us about a dB of improvement at low to medium signal to noise ratio.

M-QAM Bit Error Rate in AWGN
M-QAM Bit Error Rate in AWGN

As seen above the BER obtained through our simulation matches quite well with the BER obtained through the theoretical formula. Each additional bit per symbol required about 2dB extra in signal to noise ratio to achieve the same bit error rate.

Posted in BER Performance, Fundamentals, LTE, WiMAX | Tagged , , | 6 Comments

Fast Fourier Transform – Code

Most of us have used the FFT routine in MATLAB. This routine has become increasingly important in simulation of communication systems as it is being used in Orthogonal Frequency Division Multiplexing (OFDM) which is employed in 4G technologies like LTE and WiMAX. We would not go into the theoretical details of the FFT, rather, we would produce the MATLAB code for it and leave the theoretical discussion for a later time.

The underlying technique of the FFT algorithm is to divide a big problem into several smaller problems which are much easier to solve and then combine the results in the end.

%%%%% INITIALIZATION %%%%%
clear all
close all

fm=100;
fs=1000;
Ts=1/fs;
nn=1024;
isign=1;

t=0:Ts:(nn-1)*Ts;

x_c=exp(1i*2*pi*fm*t);
x(1:2:2*nn-1)=imag(x_c);
x(2:2:2*nn)=real(x_c);

%%%%% BIT REVERSAL %%%%%
n=2*nn;
j=1;
for i=1:2:n-1
    if(j>i)
        temp=x(j);
        x(j)=x(i);
        x(i)=temp;

        temp=x(j+1);
        x(j+1)=x(i+1);
        x(i+1)=temp;
    end
    m=nn;
    while (m >= 2 && j > m)
        j=j-m;
        m=m/2;
    end
    j=j+m;
end

%%%%% DANIELSON-LANCZOS ALGO %%%%%
mmax=2;
while (n > mmax)
    istep=2*mmax;
    theta=isign*(2*pi/mmax);
    wtemp=sin(0.5*theta);
    wpr=-2*wtemp*wtemp;
    wpi=sin(theta);
    wr=1.0;
    wi=0.0;
    for m=1:2:mmax-1
        for i=m:istep:n
            j=i+mmax;
            tempr=wr*x(j)-wi*x(j+1);
            tempi=wr*x(j+1)+wi*x(j);
            x(j)=x(i)-tempr;
            x(j+1)=x(i+1)-tempi;
            x(i)=x(i)+tempr;
            x(i+1)=x(i+1)+tempi;
        end
        wtemp=wr;
        wr=wr*wpr-wi*wpi+wr;
        wi=wi*wpr+wtemp*wpi+wi;
    end
mmax=istep;
end
F=x(2:2:n)+1i*x(1:2:n-1);
plot((0:fs/(nn-1):fs),abs(F))

In the above example we have calculated an ‘nn’ point complex FFT of an ‘nn’ point complex time domain signal. The algorithm takes the input in a special arrangement where the ‘nn’ point complex input signal is converted into ’2*nn’ real sequence where the imaginary components are placed in odd elements and real components are placed in even elements of the input sequence. A similar arrangement works for the output sequence.

Shown below is the FFT of a complex exponential with a frequency of 100 Hz. The plot is shown from 0Hz to 1000 Hz which is the sampling frequency. A signal with multiple frequencies would have to be passed through a Low Pass Filter (LPF) so that the signal components above 500 Hz (fs/2) are filtered out. When the FFT of a real signal is performed an image frequency is produced between 500 Hz to 1000 Hz.

Fast Fourier Transform of a Complex Exponential

Fast Fourier Transform of a Complex Exponential

Here we have discussed the case of complex input sequence. Simplifications can be made for a real sequence or for special signals such as pure sine and cosine waves. We will discuss these in later posts.

 [1] Numerical Recipes in C, The Art of Scientific Computing, Cambridge University Press.

Posted in Fundamentals | Tagged , , , , | Leave a comment

Detecting Sinusoids in Noise

We have previously discussed the problem of detecting two closely spaced sinusoids using the Discrete Fourier Transform (DFT). We assumed that the data set we got was pure i.e. there was no noise. However, in reality this is seldom the case. There is always some noise, corrupting the signal. Let us now see how it effects the detection problem.

We consider Additive White Gaussian Noise (AWGN) as the corrupting source. The noise power is set equal to the power of the two sinusoids i.e. we have an SNR of 0 dB. This is quite a severe case, the noise power is usually a few dB below the signal power. We are also bounded by the number of samples, N=64, giving us a resolution of 15.87 Hz.

clear all
close all
 
fm1=100;
fm2=120;
fs=1000;
Ts=1/fs;
N=64;
t=0:Ts:(N-1)*Ts;
 
x1=sqrt(2)*cos(2*pi*fm1*t);
x2=sqrt(2)*cos(2*pi*fm2*t);
wn=randn(1,N);
x=x1+x2+wn;
W=exp(-j*2*pi/N);
 
n=0:N-1;
k=0:N-1;
 
X=x*(W.^(n'*k));
 
plot(k/N,20*log10(abs(X)))
xlabel ('Normalized Frequency')
ylabel ('X(dB)')

The data is plotted on a logarithmic scale so that we can compare the signal and noise  power levels.

DFT of Two Tones in Noise

DFT of Two Tones in Noise

It is observed that we still have two peaks around the required frequency bins but there are also a number of false peaks. These peaks are around 10 dB lower than the signal peaks and should not cause a false detection. So the signal to noise ratio of 0 dB in the time domain is translated to a signal to noise ratio of about 10 dB in the frequency domain (this can be realized using an appropriate filter).

Posted in Fundamentals | Tagged , , , , | 1 Comment