Al Khwarizmi was a Muslim scholar who worked under the patronage of emperor Al Mamun in the 9th century in Baghdad. Al Mamun like his father was interested in supporting learning and formed the House of Wisdom of which Al Khwarizmi was a part. It was here that Greek philosophical and scientific works were translated into Arabic. In recognition of his support Al Khwarizmi dedicated two of his works on algebra and astronomy to the great emperor. His treatise known as Hisab al-jabr w’al-muqabala was the most famous and important one and was translated into several other languages. Over the years al-jabr came to be known as algebra and Al Khwarizmi (Algoritmi in Latin) became algorithm.
According to Al Khwarizmi equations are linear or quadratic and are composed of units, roots and squares. To Al Khwarizmi a unit was a number, a root was x and a square was x2. He defined two basic operations to solve equations namely al-jabr (the process of removing negative terms from an equation) and al-muqabala (the process of reducing positive terms of the same power when they occur on both sides of an equation). All Al Khwarizmi mathematics was done entirely in words and no symbol was used e.g. he solves an equation x2+10x=39 as follows:
“… a square and 10 roots are equal to 39 units. The question therefore in this type of equation is about as follows: what is the square which combined with ten of its roots will give a sum total of 39? The manner of solving this type of equation is to take one-half of the roots just mentioned. Now the roots in the problem before us are 10. Therefore take 5, which multiplied by itself gives 25, an amount which you add to 39 giving 64. Having taken then the square root of this which is 8, subtract from it half the roots, 5 leaving 3. The number three therefore represents one root of this square, which itself, of course is 9. Nine therefore gives the square”
One of his masterpieces is his method of solving equations using simple geometry, as shown above. For example to solve the equation x2+10x = 39 he first creates a square with length of each side equal to x and area x2. He then adds 10x to it by creating four rectangles of area 10x/4 = 5x/2 each. Thus the lightly shaded area in figure (c) above represents x2+10x or 39. Now we add the areas of the four corners (darkly shaded) to get x2+10x+25 or 64 (39+25). So we find out that the side of the larger square is of length 8 (square root of 64). But we already know that the length of a side is 5/2+x+5/2 = x+5. So we have the final equation x+5 = 8 or x = 3.
Al Beruni was a great Muslim scientist of the eleventh century who had knowledge of many diverse fields such as astronomy, astrology, mineralogy etc. He was member of Mahmud of Ghazni’s court from 1017 to 1030. It is here that he had the opportunity to travel to India (as well as present day Pakistan) and write about it in his books. It is claimed that he even learned Sanskrit during his stay in India. However, Al Beruni is most well known for his experiment to calculate the radius of the earth at a location close to Katas Raj in Kallar Kahar region of Pakistan. Some claim that the location of his measurements was in fact Nandana fort which is about 40 km east of Katas Raj temples.
His method of calculation involved two steps.
Step 1: Calculate the inclination angle of a mountain at two locations with known separation between them. We can then use simple trigonometry to find the height of the mountain. The trick here is to realize that there are two right angled triangles with the same height, which is an unknown, and base lengths which are also unknown but can be factored out (the reader is encouraged to do the math himself).
Figure 1: Calculating the Height of a Mountain
Step 2: The second step involves calculating the angle of depression that the horizon makes as viewed from top of the mountain. Then with the height of the mountain already known the radius of the earth can be easily calculated as shown below.
Figure 2: Calculating the Radius of the Earth
As the reader might have noticed there are three angular calculations and one distance calculation involved. Distance is easy to measure but angle is not and an accurate Astrolabe is required for this purpose. The Astrolabe that Al Beruni used was accurate to two decimal places of a degree. The radius that he calculated was within 1% of the accepted radius of the earth today.
PS: The writer has been to Katas Raj temples several times and it his desire to go there again to find more information about the location of the experiment and to possibly re-enact the experiment.
A polymath is a person whose expertise spans a significant number of different subject areas; such a person is known to draw on complex bodies of knowledge to solve specific problems.
An astrolabe is an elaborate inclinometer, historically used by astronomers and navigators to measure the inclined position in the sky of a celestial body, day or night.
Nandana was a fort built at strategic location on a hilly range on the eastern flanks of the Salt Range in Punjab Pakistan. Its ruins, including those of a town and a temple, are present. It was ruled by the Hindu Shahi kings until, in the early 11th century, Mahmud of Ghazni expelled them from Nandana.
I have been asked many times that what is the area required for a solar array with a particular rating (such as 20 KW) and I have given a simple calculation method in one of my earlier post. But real life is never so simple. There are a bunch of inputs that we need to get an exact answer. In particular we want to know what is the required spacing between solar panels so that they are not in shade for a particular part of the day (typically 5 or 6 hours) and for this we need a number of inputs such as latitude, longitude, solar noon on the worst day of the year etc. Here we show the sequence of calculations for a typical city in Northern India. Similar calculations would hold true for many cities in Pakistan as well.
Let us first list down the data we have for New Dehli a typical city in Northern India (this data is available on ESRL website).
Time Zone: +5.5
Worst Solar Day: Dec 21
Start of Solar Charging: 09:19 am
Solar Noon: 12:19 pm
End of Solar Charging: 3:19 pm
Solar Azimuth: 135.5 degrees (Azimuth at 9:19 am, the time solar panel just comes out of the shade)
Solar Inclination: 22.33 degrees (Inclination at 9:19 am, the time solar panel just comes out of the shade)
Now lets assume that the length of the solar panel is 1.0 m and it is fixed at an angle of 30 degrees from the horizontal. Then the height can be calculated as:
h = 1.0 * sin(30) = 0.50
Now we are ready to calculate the distance between the panels.
D = h / tan(Inclination) = 0.50 / tan(22.33) = 1.22 m (this is the distance without Solar Azimuth Correction)
d = D * cos(180-Azimuth) = 1.22 * cos(44.5) = 0.87 (this is the distance with Solar Azimuth Correction)
So for a 1.0 m length panel we need to have a separation of about 0.87 m between the panels along the north-south line (panels facing south in the Northern hemisphere). To be on the safe side we can keep the distance a bit higher than this. We can estimate that the total area would be increased by a factor 2.0 (0.87 m is the separation of the panels and 0.87 m is the base of the triangle formed by the inclined solar panel) or we can simply multiply the area calculated by the simple calculations (for a panel lying flat on the ground) given in the earlier post by a factor of 1.74.
Note: These calculations are valid for solar panels placed on the ground, in straight lines, along the east-west line (facing south). For panels placed on inclined rooftops the calculations would be different (area required would be generally lower). Also note that solar panels are highly sensitive to shading and even if only 10% of the panel is in shade its performance without deteriorate appreciably. Lastly some installations have different panel tilts for winter and summer (latitude +/- 15 degrees) and this needs to be catered in the calculations.
Open Signal is a mobile application that collects the data about your wireless network (2G/3G/4G) and generates coverage maps and host of other reports. The data is collected in the background while the user is busy in his daily routines. But data can also be collected on the request of the user. This is much better than drive testing since the data is collected in real life scenarios and on thousands of different devices that are in use.
The app works while the user is indoor or outdoor, at rest or in motion, on land or on water, at sea level or on a mountain, in dry weather or in rain. Basically anywhere and anytime there are wireless signals available. There are currently 20 million users of the app (both Android and iOS combined) and this number is increasing. In Pakistan all major networks are supported including Jazz, Telenor, Zong and Ufone (both 2G/3G and 4G networks are supported).
As the data rates supported by wireless networks continue to rise the bandwidth requirements also continue to increase (although spectral efficiency has also improved). Remember GSM technology which supported 125 channels of 200KHz each, which was further divided among eight users using TDMA. Move on to LTE where the channel bandwidth could be as high as 20MHz (1.4MHz, 3MHz, 5MHz, 10MHz, 15MHz and 20MHz are standardized).
This advancement poses a unique challenge referred to as frequency selective fading. This means that different parts of the signal spectrum would see a different channel (different amplitude and different phase offset). Look at this in the time domain where the larger bandwidth means shorter symbol period causing intersymbol interference (as time delayed copies of the signal overlap on arrival at the receiver).
The solution to this problem is OFDM that divides the wideband signal into smaller components each having a bandwidth of a few KHz. Each of these components experiences a flat channel. To make the task of equalization simple a cyclic prefix (CP) is added in the time domain to make the effect of fading channel appear as circular convolution. Thus simplifying the frequency domain equalization to a simple division operation.
Shown below is the Python code that calculates the bit error rate (BER) of BPSK-OFDM which is the same as simple BPSK in a Rayleigh flat fading channel. However there is a caveat. We have inserted a CP which means we are transmitting more energy than simple BPSK. To be exact we are transmitting 1.25 (160/128) times more energy. This means that if this excess energy is accounted for the performance of BPSK-OFDM would be 1dB (10*log10(1.25)) worse than simple BPSK in Rayleigh flat fading channel.
Although we have shown the channel as a multiplicative effect in the figure above, this is only true for a single tap channel. For a multi-tap channel (such as the one used in the code above) the effect of the channel is that of a filter which performs convolution operation on the transmitted signal.
We have used a baseband model in our simulation and the accompanying figure. In reality the transmitted signal is upconverted before transmission by the antennas.
The above model can be easily modified for any modulation scheme such as QPSK or 16-QAM. The main difference would be that the signal would have a both a real part and an imaginary part, much of the simulation would remain the same. This would be the subject of a future post. For a MATLAB implementation of 64-QAM OFDM see the following post (64-QAM OFDM).
Serial to parallel and parallel to serial conversion shown in the above figure was not required as the simulation was done symbol by symbol (one OFDM symbol in the time domain represented 128 BPSK symbols in the frequency domain).
The channel model in the above simulation is quasi-static i.e. it remains constant for one OFDM symbol but then rapidly changes for the next, without any memory.
When wireless signals travel from a transmitter to a receiver they do so after reflection, refraction, diffraction and scattering from the environment. Very rarely is there a direct line of sight (LOS) between the transmitter and receiver. Thus multiple time delayed copies of the signal reach the receiver that combine constructively and destructively. In a sense the channel acts as an FIR (finite impulse response) filter. Furthermore since the transmitter or receiver may be in motion the amplitude and phase of these replicas varies with time.
There are several methods to model the amplitude and phase of each of these components. We look at one method called the “Smiths Fading Simulator” which is based on Clark and Gans model. The simulator can be constructed using the following steps.
1. Define N the number of Gaussian RVs to be generated, fm the Doppler frequency in Hz, fs the sampling frequency in Hz, df the frequency spacing which is calculated as df=(2*fm)/(N-1) and M total number of samples in frequency domain which is calculated as M=(fs/df).
2. Generate two sequences of N/2 complex Gaussian random variables. These correspond to the frequency bins up to fm. Take the complex conjugate of these sequences to generate the N/2 complex Gaussian random variables for the negative frequency bins up to -fm.
3. Multiply the above complex Gaussian sequences g1 and g2 with square root of the Doppler Spectrum S generated from -fm to fm. Calculate the spectrum at -fm and +fm by using linear extrapolation.
4. Extend the above generated spectra from -fs/2 to +fs/2 by stuffing zeros from -fs/2 to -fm and fm to fs/2. Take the IFFT of the resulting spectra X and Y resulting in time domain signals x and y.
5. Add the absolute values of the resulting signals x and y in quadrature. Take the absolute value of this complex signal. This is the desired Rayleigh distributed envelope with the required temporal correlation.
The Matlab code for generating Rayleigh random sequence with a Doppler frequency of fm Hz is given below.