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.

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 f_{d}. 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].

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 THAT CALCULATES THE BER OF M-QAM IN RAYLEIGH FADING
% n_bits: Input, number of bits
% M: Input, constellation size
% EbNodB: Input, energy per bit to noise power spectral density
% ber: Output, bit error rate
% Copyright RAYmaps (www.raymaps.com)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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 four modulation schemes 4-QAM, 16-QAM, 64-QAM and 256-QAM are shown in the figure above. 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.

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).

Lastly we explain some of the terms used above.

Rayleigh Fading

Rayleigh Fading is a commonly used term in simulation of Digital Communication Systems but it tends to differ in meaning in different contexts. The term Rayleigh Fading as used above means a single tap channel that varies from one symbol to the next. It has an amplitude which is Rayleigh distributed and a phase which is Uniformly distributed. A single tap channel means that it does not introduce any Inter Symbol Interference (ISI). Such a channel is also referred to as a Flat Fading Channel. The channel can also be referred to as a Fast Fading Channel since each symbol experiences a new channel state which is independent of its previous state (also termed as uncorrelated).

Gray Coding

When using QAM modulation, each QAM symbol represents 2,3,4 or higher number of bits. That means that when a symbol error occurs a number of bits are reversed. Now a good way to do the bit-to-symbol assignment is to do it in a way such that no neighboring symbols differ by more than one bit e.g. in 16-QAM, a symbol that represents a binary word 1101 is surrounded by four symbols representing 0101, 1100, 1001 and 1111. So if a symbol error is made, only one bit would be in error. However, one must note that this is true only in good signal conditions. When the SNR is low (noise has a higher magnitude) the symbol might be displaced to a location that is not adjacent and we might get higher number of bits in error.

Hard Decision

The concept of hard decision decoding is important when talking about channel coding, which we have not used in the above simulation. However, we will briefly explain it here. Hard decision is based on what is called “Hamming Distance” whereas soft decision is based on what it called “Euclidean Distance”. Hamming Distance is the distance of a code word in binary form, such as 011 differs from 010 and 001 by 1. Whereas the Euclidean distance is the distance before a decision is made that a bit is zero or one. So if the received sequence is 0.1 0.6 0.7 we get a Euclidean distance of 0.8124 from 010 and 0.6782 from 001. So we cannot make a hard decision about which sequence was transmitted based on the received sequence of 011. But based on the soft metrics we can make a decision that 001 was the most likely sequence that was transmitted (assuming that 010 and 001 were the only possible transmitted sequences).

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 THAT CALCULATES THE BER OF M-QAM IN AWGN
% n_bits: Input, number of bits
% M: Input, constellation size
% EbNodB: Input, energy per bit to noise power spectral density
% ber: Output, bit error rate
% Copyright RAYmaps (www.raymaps.com)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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.

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.

Lastly we explain some of the terms used above.

AWGN

All wireless receivers suffer from thermal noise which is a function of absolute temperature and bandwidth of the receiver. This noise is added to the received signal and makes detection of weak signals a major challenge. Just to given you an idea typical GSM receivers have a noise floor of -113 dBm. Therefore, if the received signal has a power of -100 dBm we get a signal to noise ratio (SNR) of 13 dB. In simulation this noise is usually modeled as a Gaussian Random Process. It is additive, as opposed to channel impairments which are multiplicative and has a flat spectrum (thus called White Noise).

Gray Coding

When using QAM modulation, each QAM symbol represents 2,3,4 or higher number of bits. That means that when a symbol error occurs a number of bits are reversed. Now a good way to do the bit-to-symbol assignment is to do it in a way such that no neighboring symbols differ by more than one bit e.g. in 16-QAM, a symbol that represents a binary word 1101 is surrounded by four symbols representing 0101, 1100, 1001 and 1111. So if a symbol error is made, only one bit would be in error. However, one must note that this is true only in good signal conditions. When the SNR is low (noise has a higher magnitude) the symbol might be displaced to a location that is not adjacent and we might get higher number of bits in error.

Hard Decision

The concept of hard decision decoding is important when talking about channel coding, which we have not used in the above simulation. However, we will briefly explain it here. Hard decision is based on what is called “Hamming Distance” whereas soft decision is based on what it called “Euclidean Distance”. Hamming Distance is the distance of a code word in binary form, such as 011 differs from 010 and 001 by 1. Whereas the Euclidean distance is the distance before a decision is made that a bit is zero or one. So if the received sequence is 0.1 0.6 0.7 we get a Euclidean distance of 0.8124 from 010 and 0.6782 from 001. So we cannot make a hard decision about which sequence was transmitted based on the received sequence of 011. But based on the soft metrics we can make a decision that 001 was the most likely sequence that was transmitted (assuming that 010 and 001 were the only possible transmitted sequences).

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.

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.

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.

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).

In the previous post we had introduced the Discrete Fourier Transform (DFT) as a method to perform the spectral analysis of a time domain signal. We now discuss an important property of the DFT, its spectral resolution i.e. its ability to resolve two signals with similar spectral content.

Initially one might think that increasing the sampling frequency would increase the spectral resolution but this totally incorrect. In fact if the sampling frequency is increased, keeping the number of time domain samples to be the same, the resolution actually decreases. So how do we calculate the spectral resolution. One simple way is to calculate the difference between two frequency bins as f_{s}/(N-1) or 1/(N-1)T_{s}. Simply put the resolution in the frequency domain is the inverse of the sample length in the time domain.

So let us now calculate the DFT of two closely spaced sine waves keeping the sampling frequency to be the same and changing the number of time domain samples (only the result for N=64 shown here). We again list down the code used to calculate the DFT.

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

We first do some quick math to find the spectral resolution.

f_{s}/(N-1)=1000/(64-1)=15.87 Hz

Now the two tones are space 20 Hz apart (100 Hz and 120 Hz), so we can predict that the two tones would be detected successfully. The result of the DFT operation on the composite signal is shown below.

It is observed that although the two tones are detected, they are not exactly at the desired frequencies (0.10 and 0.12). Secondly the amplitude of the two tones is different although the time domain signals had equal amplitude. Both these phenomenon are due to the fact that we only have a limited number of frequency bins (N=64) due to which the resulting spectrum is only an estimate of the true spectrum.

There are better techniques than DFT to separate two closely spaced sinusoids and these are known as super resolution spectral techniques and would be discussed some other time.