History of Western Civilization Essay Example for Free

History of Western Civilization Essay Civilizations of societies started long before the times Jesus in the Middle East especially in the ancient Greek and Roman Empires. In the 18th century most European countries followed the suit and underwent changes from pre- industrialized to industrialized nations characterized by social and economic changes. There were a lot of technological innovations which led to increased energy production and thus large scale production of most products with ease. Industrialization has over the years been linked to some factors which enhance fast changes for instance cheap skilled labor, political stability, presence of raw materials and available markets of the products produced. Industrial revolution was experienced in England to the northwest and the midlands. People used to make their own furniture, clothes and equipments until industrialization took over whereby they could buy goods already made from the industries. History of Western Civilization England underwent several civilizations in which there was increased agricultural output (Agrarian revolution) which led to increased population density and eventually led to the industrial revolution. New techniques were initiated and allowed farmers to produce more yield than there before. As time passed machines and other equipments were produced by skilled personnel who sold them to those who were specialized in agriculture. The business to make machines and other related equipments boomed and led formation of a class of people who worked in the industries as they expanded. These machines were very advantageous because they substituted the human job since it did not involve a lot of work and therefore people did not get tired fast. Those who worked in the industries earned more than those who were specialized on agriculture. Most of these industries were located in the urban centers and they would obtain their raw materials from the farmers in the country side. Industrialization in England led to numerous increases in population leading to less arable land and therefore some of the people migrated to other countries where they introduced the new techniques for agriculture and industry (Landes, 1969). However, most of the skilled personnel were not allowed to immigrate to the England’s new colonies so that England would remain superior in technological advances. Industrialization in England was characterized by many factors which made this country to be first to be industrialized. Agrarian, commercial and cottage revolutions greatly enhanced the fast social and economic changes in this country leading to a lot of extra money which facilitated the improvement of infrastructure. The industries spread all over England and Europe and since the number of workers were reducing as a result of immigration to England colonies, they resulted in taking people from the countries that were still not in the process of social and economic change especially in Africa and made them to work in their farms and industries as slaves. As more and more industries were built the private sector was also changing as money for expansion and setting up of more industries was needed. This led to the emergence of banks and other financial institutions which financed all these industrialization activities. These institutions gave loans to people to expand and introduce more industries and then repay later with the profits obtained. At around this time of industrialization most people used wood as a source of energy for the industries. The number of industries increased enormously and therefore the number of trees reduced drastically and made the people to think of alternative source of energy. They came up with coal which was used in the industries as an alternative and could be even used in homes. The coal mines were usually flooded with water at times and a method of pumping water out of the mines was innovated but the coal would not generate power. This method was not embraced and therefore coal was not a lasting solution and more innovations needed to be initiated for instance in the textile industry. Textile industry in England was the first industry to be mechanized and utilized wool which obtained from large scale sheep farms in the country. Weaving was usually done in cottages by skilled people after which the yarn was taken to the industries where it would be made in to clothes. England by then conquered many colonies where they would grow cotton in addition to importing cotton and used to substitute the wool. At around 1773 John Kay introduced a machine he named the flying shuttle which made it possible for a person to easily weave. In the textile industry machines that utilized water for energy were introduced. This machinery could reduce the cost of production and at the same time increase the rate of production. These measures were taken when the cost of production started to sky rocket reducing the profits of the merchants and the merchants did not want to raise the goods prices in a quest to maintain their customers. Transportation in England was favored by the fact that it had many rivers and natural harbors which greatly reduced transportation costs because rivers covered most areas and therefore goods could be transported to most parts of the country. Canals were also built on rivers which were not naturally navigable and also facilitated easy transportation of raw materials from farms in England and from distant colonies and transportation of finished products from the industries to the consumers. Tram ways which were pulled by horses were also used especially when transporting goods inland until when trains were made and used instead because they relatively faster. Due to the limitation of the road transport some people decided to borrow loans from the financial institutions ton repair them and then use toll fees to repay the loans. This would also ensure easy transportation of goods in areas where the rivers and coastline was far away. Improvement of roads led to introduction of stage coaches which would transport even people from one area to another. Transport was boosted with the introduction and improvement of railway. The first type of railway was wooden and did not last long until when iron plates were put on top of the wood and the railway became more durable. Another factor that facilitated faster industrialization of England is its isolation from the European mainland. This helped this country to evade the wars that were occurring at this time for instance the Napoleon war. They concentrated more on development and industrialization of their country till the wars were over. After the war the British took advantage of selling their products to the other European countries and the Americas at relatively cheap prices thereby making a lot of profits. The private sector with the government support also helped a lot by offering loans to people for expansion and initiation of new industries and for the repair of roads which enhanced faster movement of goods between places. Conclusion Civilization of England in many ways has influenced the modern world with only improvement of the machineries that were made during the period of industrialization. The support from the government and relative stability of the country made it to develop even faster without any worries of attacks during the war of Napoleon in the mainland. Advancements in technology led to improvement and construction of new transport systems and infrastructure which also facilitated movement of raw materials and finished products to the several destinations within England and other countries. Civilization in England led to migration of people to the urban centers where they would be able to obtain jobs in the urban centers. This situation is the same even in the modern days where especially young people move to the towns and cities in search of better jobs in the industries. References Jackson, S. Western Civilization: A Brief History. 4th Edition. Cengage Learning, 2007. Jacob, F. The Development of Western Civilization : A study in Ethical, Economic and Political Kenneth, P. Steven, T. The world that trade created: society, culture, and the world economy, 1400- the present. 2nd Edition. M. E Sharpe, 1999. Richard, B. Society and Economy in Modern Britain 1700-1850. Routledge, 1991. Evolution. The University of Chicago Press, 1906. Roy, P. David, L. The Cambridge History of Science: Eighteenth-century science. 2nd Edition. Cambridge University Press, 2003.

Capital Punishment Essay -- essays research papers

Susan Smith purposely drove her car off into a lake with her 2 children strapped to the back seats. Think of how they must have felt as the cold water started to fill the car, and then ultimately drowned them. Barbaric is the word I would use to describe her actions. But yet, the jury rejected the death penalty and chose a 30-year sentence instead because capital punishment was not enforced in the state. Broken up from the death of his 2 children, Mr. Smith said, â€Å"Me and my family are disappointed that the death penalty was not the verdict.† I am going to convince you that capital punishment has to be enforced in all states. I will tell you about the problem, the solution to the problem, and what it would be like if my solution is implemented. The problem is that the current criminals that commit these brutal crimes are not fearful enough of the consequences and punishment they have to pay. Life in prison is too easy on the convicts. We have to be harder on them. Another problem is capital punishment is not imposed in enough states. All of the states need to adopt it in the United States for it to be effective. Another dilemma is that it is not fair that an individual who took the life of another receives heating, shelter, clothing, 3 meals a day, and indoor plumbing while a homeless person who does not cause any harm to anybody else receives nothing. An additional problem is the current prisons are very low on cell space. The criminals that have life in prison are takin... Capital Punishment Essay -- essays research papers Susan Smith purposely drove her car off into a lake with her 2 children strapped to the back seats. Think of how they must have felt as the cold water started to fill the car, and then ultimately drowned them. Barbaric is the word I would use to describe her actions. But yet, the jury rejected the death penalty and chose a 30-year sentence instead because capital punishment was not enforced in the state. Broken up from the death of his 2 children, Mr. Smith said, â€Å"Me and my family are disappointed that the death penalty was not the verdict.† I am going to convince you that capital punishment has to be enforced in all states. I will tell you about the problem, the solution to the problem, and what it would be like if my solution is implemented. The problem is that the current criminals that commit these brutal crimes are not fearful enough of the consequences and punishment they have to pay. Life in prison is too easy on the convicts. We have to be harder on them. Another problem is capital punishment is not imposed in enough states. All of the states need to adopt it in the United States for it to be effective. Another dilemma is that it is not fair that an individual who took the life of another receives heating, shelter, clothing, 3 meals a day, and indoor plumbing while a homeless person who does not cause any harm to anybody else receives nothing. An additional problem is the current prisons are very low on cell space. The criminals that have life in prison are takin...

Dc Power Supply Design

Abstract: The main aim of this assignment is to design a pre amplifier circuit with an NPN transistor to be used in a simple public address (PA) system. The pre amplifier is fed from a microphone that produces an average output voltage of 10 mV rms. The amplifier is to operate over a frequency range of 300 Hz to 5 kHz and should have an adjustable volume control. The expected gain of the amplifier is 100.First we are going to design an amplifier for given specifications, model the operation of the circuit using h-parameter and r-parameter model, use computer aided design software to analyze the circuit performance and demonstrate the working of the circuit by hardware implementation. Then, we will plot the frequency response of the circuit and analyze the effect of the emitter bypass capacitor. Finally we will compare the mid-band gain, bandwidth and lower cutoff frequency obtained from the simulation result and the hardware implementation with the designed values. Chapter 1Introduct ion: Bipolar Junction Transistor (BJT) is a three terminal device with three regions (Emitter, Base and Collector) and two PN junctions (Emitter-Base junction and Base-Collector junction). Since there are two junctions that means there are four possible ways of biasing a transistor. If both junctions are forward biased then the transistor will operate in the saturation region. If both junctions are reverse biased then the transistor will operate in the cut of region. These two conditions of operation are used when the transistor is needed to work as a switch.To use a transistor as an amplifier, the emitter base junction should be forward biased and the collector base junction should be reverse biased. Amplifier is an electronic circuit that can amplify signals applied to its input terminal. If an AC signal is given to a transistor amplifier it will produce an AC base current. This AC base current will produce a much larger AC collector current since IC=? IB. The AC collector current produces an AC voltage across the load resistor RL, thus producing an amplified, but inverted, reproduction of the AC input voltage in the active region of operations.DC load line is a sloping straight line connecting all the operating points of a transistor biasing drawn on the output characteristics of the transistor and the intersection point gives the Quiescent point (Q-point). A prober Q-point should be in the middle of the DC load line. Selecting a good Q-point prevents the transistor from going into the cutoff or the saturation region and gives more stability. A fixed bias (i. e. base bias) circuit or a voltage divider bias circuit can be used for this assignment but a voltage divider circuit is more efficient.The main disadvantage in a fixed bias circuit is that ? ac depends on temperature, which means ? ac is not stable. And when ? ac changes, IC will change(IC=? IB) and VCE will change. The changes in IC and VCE make the Q-point unstable. Whereas in voltage divider bias c ircuit, IC is independent of ? ac and hence the Q-point is more stable. Voltage divider bias is widely used because reasonably good stability reached with a single power supply. Chapter 2 Problem Description: The problem is to design and fabricate a pre amplifier circuit with an NPN transistor to be used in a simple public address (PA) system.The input of the pre amplifier circuit is taken from a microphone that produces an average output voltage of 10 mV rms. The amplifier is to operate over a frequency range of 300 Hz to 5 kHz. Also, it should have an adjustable volume control. The expected voltage gain of the amplifier is 100. Design Specifications: Voltage gain = 100 Lower cut off frequency = 300Hz Vin = 10mV (rms) RL = 10k? DC power supply = 10V to 15V Type of transistor – NPN We will begin our assignment by selecting a suitable transistor. Then we will decide on a DC voltage supply and assume a prober Q-point (IC, VCE) to carry out the design.We will start the design by calculating the values of Resistors RC and RE and the voltage divider resistors R1 and R2. After that we will calculate the values of the two coupling capacitors (C1 and C2) and the emitter bypass capacitor (CE) for the required cut off frequency. After finishing the mathematical model we will simulate the circuit using OrCAD to analyze the circuit performance. Then, after finishing the simulation, we will assemble the circuit using approximate values of the calculated ones. Finally, we will compare the simulation results with the hardware results.The results we will be focusing on are the voltage gain, the cutoff frequency and the Bandwidth. Chapter 3 Circuit Diagram and Design: Av = 100 FL = 300Hz Av = 100 FL = 300Hz Figure 1 – Circuit Diagram Step1 – Selection of Transistor, Supply Voltage (VCC) and Collector Current (IC): The selected transistor should have a minimum current gain (? ) that is equal to or greater than the desired voltage gain. Therefore, we will us e Q2N2222 in this assignment. Since the output voltage swing is not specified in this assignment, we will choose 12V as our voltage supply. We will choose IC as 4 mA. Transistor: Q2N2222Supply Voltage: VCC = 12 V Collector Current: IC = 4 mA * To carry out the design we need to draw the dc equivalent circuit. Figure 2 – DC Equivalent Circuit Step2 – Design of Collector Resistor (RC) and Emitter Resistor (RE): VCE = 50% VCC = 50% ? 12 = 6 V VE = 10% VCC = 10% ? 12 = 1. 2 V VRC = VCC – VE – VCE = 12 – 6 – 1. 2 = 4. 8 V RC = VRCIC = 4. 8 V4 mA = 1. 2 k? RE = VEIE = VEIC = 1. 2 V4 mA = 300 ? , since IC ? IE Step3 – Design of Voltage Divider R1 and R2: ? = 100 (data sheet) R2 = ? RE10= 100? 30010 = 3 k? VB = VBE + VE = 0. 7 + 1. 2 = 1. 9 V VB = VCCR2R1+R2 R1 = VCCR2VB+R2 = 100? 3k1. +3k = 16 k? * Now we need to draw the ac equivalent circuit. Figure 3 – AC Equivalent Circuit Step4 – Design of RE1 and RE2: RE = RE1 + RE2 Rout = Rc || RL = 1. 2? 101. 2+10= 1 k? r'e = 26mIE = 26mIC = 6. 5 ? AV = Routr'e+RE1 r'e+RE1= RoutAv = 1k100 = 10 ? RE1 =10 – r'e = 10 – 6. 5 = 3. 5 ? RE2 = RE – RE1 = 300 – 3. 5 = 296. 5 ? Step5 – Design of Coupling Capacitors C1 and C2: hie = Rin (base) = ? (r'e+RE1) = 100 ? (3. 5 + 6. 5) = 1 k? Rin (tot) = R1 || R2 || Rin (base) = 1116+13+11 = 716. 4 ? XC1 = Rin(tot)10 = 716. 410 = 71. 64 ? C1 = 12? fLXC1 = 12 300? 71. 64 = 7. 4  µF XC2 = RC + RL = 1. 2 + 10 = 11. k? C2 = 12? fLXC2 = 12 300? 11200 = 47. 4 nF Step6 – Design of Bypass Capacitor CE: R’S = R1 || R2 = 16. 09? 316. 09+3 = 2. 5 k? Re = RE2 ||{ R’S ? + (r'e+RE1)} = 296. 5 ||{ 2500 100+ (6. 5+3. 5)} = 296. 5? 35296. 5+35 = 31. 3 ? XCE = Re10 = 31. 310 = 3. 13 ? CE = 12? fLXCE = 12 300? 3. 13 = 169. 5  µF Av = 100 FL = 300Hz Av = 100 FL = 300Hz Figure 4 – Circuit Diagram with values Simulation Results: With CE: Mid-band gain, AV = 99. 8 Lower Cutoff Frequency, FL = 334 Hz Higher Cutoff Frequency, FH = 20. 6 MHz Bandwidth, BW = FH – FL = 20. 6 M – 334 = 20. 6 MHz Without CE: Mid-band gain, AV = 3. 5Lower Cutoff Frequency, FL = 305 Hz Higher Cutoff Frequency, FH = 46 MHz Bandwidth, BW = FH – FL = 46 M – 305 = 46 MHz (Circuit Diagram and Frequency Response are enclosed along with this report) Chapter 4 Hardware Fabrication and Testing Details: During circuit assembling process we tried to find the nearest values to the calculated ones. These are the values we used: RC = 1. 2 k? we selected1. 2 k? RE1 = 3. 5 ? we selected4. 5 ? RE2 = 296. 5 ? we selected270 ? R1 = 16 k? we selected15 k? R2 = 3 k? we selected2. 2 k? C1 = 7. 4  µF we selected10  µF C2 = 47. 4 nF we selected47 nF CE = 169. 5  µF we selected147  µF Procedure: . Assembled the circuit on a breadboard and connected a DC power supply of 12V. 2. Applied a sine wave of 10 mV amplitude and 100 Hz frequency to the input. 3. Observed the output waveform in the CRO and noted down the amplitude. 4. Increased the input signal frequency in steps, without changing its amplitude, and noted down the output amplitude at each step. 5. Calculated the voltage gain of the amplifier by the equation, AV = Vout/Vin found the voltage gain in dB by the equation, AV (dB) = 10 log (AV). 6. Plotted the frequency response curve and found the frequencies (fL and fH) for which the gain reaches 0. 07 of mid band gain. 7. Found the frequency range between fL and fH which gives the bandwidth of the amplifier. Hardware Results: With CE: Frequency (Hz)| Vout (mV)| AV| AV (dB)| log f| 100| 182| 18. 2| 25. 20| 2. 0| 500| 662| 66. 2| 36. 42| 2. 7| 1 k| 750| 75. 0| 37. 50| 3. 0| 5 k| 784| 78. 4| 37. 89| 3. 7| 10 k| 786| 78. 6| 37. 91| 4. 0| 50 k| 786| 78. 6| 37. 91| 4. 7| 100 k| 786| 78. 6| 37. 91| 5. 0| 500 k| 786| 78. 6| 37. 91| 5. 7| 1 M| 786| 78. 6| 37. 91| 6. 0| 2 M| 784| 78. 4| 37. 89| 6. 3| 5 M| 770| 77. 0| 37. 73| 6. 7| 10 M| 728| 72. 8| 37. 24| 7. 0| 50 M| 344| 34. 4| 30. 73| 7. 7| 100 M| 182| 18. 2| 25. 0| 8. 0| Mid-band gain, AV = 78. 6 Lower Cutoff Frequency, FL = 2. 6 B = 398 Hz Higher Cutoff Frequency, FH = 7. 35 B = 17. 78 MHz Bandwidth, BW = FH – FL = 17. 78 M – 398 = 17. 78 MHz Without CE: Frequency (Hz)| Vout (mV)| AV| AV (dB)| log f| 100| 12| 1. 2| 1. 58| 2. 0| 500| 32| 3. 2| 10. 10| 2. 7| 1 k| 36| 3. 6| 11. 13| 3. 0| 5 k| 38| 3. 8| 11. 60| 3. 7| 10 k| 38| 3. 8| 11. 60| 4. 0| 50 k| 38| 3. 8| 11. 60| 4. 7| 100 k| 38| 3. 8| 11. 60| 5. 0| 500 k| 38| 3. 8| 11. 60| 5. 7| 1 M| 38| 3. 8| 11. 60| 6. 0| 2 M| 38| 3. 8| 11. 60| 6. 3| 5 M| 38| 3. 8| 11. 60| 6. 7| 10 M| 36| 3. 6| 11. 13| 7. 0| 50 M| 26| 2. 6| 8. 0| 7. 7| 100 M| 18| 1. 8| 5. 10| 8. 0| Mid-band gain, AV = 78. 6 Lower Cutoff Frequency, FL = 2. 55 B = 356 Hz Higher Cutoff Frequency, FH = 7. 6 B = 39. 81 MHz Bandwidth, BW = FH – FL = 39. 81 M – 356 = 39. 81 MHz (Frequency responses of the circuit with and without CE are enclosed along with this report) (Frequency responses of the circuit with and without CE are enclosed along with this report) Chapter 5 Discussion and Conclusion: * First of all, there are several ways and various methods to design a common emitter amplifier or so-called RC coupled amplifier that are completely different than the one we used.We did not choose this method because it is the best method, actually, there is no such a thing called the best method. There are simple ways and there are more accurate ways. It depends on the primary assumptions, the design specifications and the thumb rules used. Simply, the method we used achieved the design requirements and accomplished desired results. * An Amplifier is a circuit that is capable of amplifying signals applied to its input terminal. The main component in any amplifier circuit is usually a transistor.Since the transistor configuration we used is a common emitter configuration, the circuit is called a Common Emitter Amplifier. Unlike other configurat ions, CE amplifier exhibit high voltage gain and high current gain. Generally, the process of a common emitter amplifier can be explained in three steps. First, the AC input signal produces an AC base current. Then, This AC base current will produce a much larger AC collector current since IC=? IB. After that, The AC collector current produces an AC voltage across the load resistor RL, thus producing an amplified, but inverted, reproduction of the AC input voltage. To use a transistor as an amplifier it should be operated in the active region (linear region). To set a transistor in the active region both junctions, Emitter-Base junction and Base-Collector junction, should be forward biased. Since changes in in temperature and other factors during the amplification process may drive the transistor into the cutoff or the saturation region, the Q-point should be in the middle of the active region to enhance the stability of the amplifier. * We preferred using a voltage divider bias cir cuit over other biasing circuits because in this kind of biasing circuits, IC is independent of ? nd therefore the Q-point is more stable. Voltage divider bias circuit is widely used because of the good stability reached with a single power supply. * C1 and C2 are called coupling capacitors. They pass ac from one side to another and block dc from appearing at the output side. In addition to that, C1 act as a high pass filter on the input signal and its value must be chosen so that it does not attenuate the frequencies which are to be amplified. Similarly, C2 also must be prevented from attenuating the output signal. * The bypass capacitor CE provides an effective short to the ac signal round the emitter resistor RE2, thus keeping only RE1 seen by the ac signal between the emitter and ground. Therefore, with the bypass capacitor, the gain of the amplifier is maximum and equal to AV=Routr'e+RE1 . Without the bypass capacitor, both RE1 and RE2 are seen by the ac signal between the emit ter and ground and effectively add to r'e in the voltage gain formula. Hence, AV=Routr'e+RE1+RE2 . * r'e is a dynamic resistor that depends on temperature. If AV was dependent only on r'e, and RE1 was not there (i. e. AV=Routr'e ), AV will be unstable over changes in temperature because when r'e increases, the gain decreases and vice versa.In order to minimize the effect of r'e without reducing the voltage gain to its minimum value we partially bypassed the total emitter resistance RE. This is known as swamping which is a compromise between having a bypass capacitor across RE and not having a bypass capacitor at all. RE1 should be at least ten times greater than r'e to minimize the effect of it. In our design RE1 is less than r'e and hence it will not do anything other than slightly reducing the gain to be about 100. In other words, in our design RE1 is somehow useless. * At lower frequencies, a capacitor will act as an open circuit.At higher frequencies, a capacitor will act as a s hort circuit. That is because the capacitive reactance is inversely proportional to the frequency (XC=1/2? fC). In an RC coupled amplifier circuits, at lower frequencies, more voltage drops across C1 and C2 because their reactance is very high. This higher signal voltage drop reduces the voltage gain of the amplifier. Similarly, at lower frequencies, the reactance of the bypass capacitor (CE) increases and this reactance in parallel with RE1 create an impedance that reduces the voltage gain.This is why RC coupled amplifier circuits have less voltage gain at lower frequencies than they have at higher frequencies. However, at higher frequencies, the reactance of the internal transistor junction capacitance goes down and when it becomes small enough, a portion of the output signal voltage is fed back out of phase with the input, thus effectively reducing the voltage gain. * Our hardware implementation results and simulation results were different. Obviously, that is because we did not find the exact values for our design. There was a notable difference between the design values and the values we have selected, especially for R2.The cutoff frequency (fL=398 Hz) is somehow acceptable but the mid band gain (AV=78. 6) is a little bit less than the desired one. Increasing the value of R2 could have solved the problem. It could have increased the voltage gain and reduced the cutoff frequency. * One of the aims of the design is to have an adjustable volume control. There are several ways to do this. One of them, and I think it’s the best, is by using a variable resistor in place of RE1 (i. e. a 100 ? variable resistor). Basically, this resistor is inversely proportional to the voltage gain (AV=Routr'e+RE1 ).Reducing the value of RE1 will increase the voltage gain, thereby increasing the volume and vice versa. References: 1. Theodore F. Bogart, Jefferey S. Beasley and Guilermo Rico (2004). Electronic Devices and Circuits. India: Pearson Education, Inc. 2. Thomas L . Floyd (2005). Electronic Devices. 7th ed. India: Pearson Education, Inc. 3. HyperPhysics  (2004)  Common Emitter Amplifier,[online] Available at: http://hyperphysics. phy-astr. gsu. edu/hbase/electronic/npnce. html [Accessed: 20th Nov 2011]. 4. Scribd  (2006)  Common Emitter Amplifier,  [online] Available at: http://www. cribd. com/doc/27767944/Common-Emitter-Amplifier [Accessed: 25th Nov 2011]. 5. Visionics  (2005)  RC Coupled Amplifier,  [online] Available at: http://www. visionics. ee/curriculum/Experiments/RC%20Ampr/RC%20Coupled%20Amplifier1. html [Accessed: 1st Nov 2011]. 6. SSIT  (2006)  Analog Electronic Circuits,  [online] Available at: http://www. ssit. edu. in/dept/assignment/aeclabmanual. pdf [Accessed: 5th Nov 2011]. 7. Edutalks  (2007)  RC Coupled Amplifier,  [online] Available at: http://www. edutalks. org/electronics%20lab%20manual%201. pdf [Accessed: 7th Nov 2011].

Ternary Operator - Java Definition

The ternary operator ?: earns its name because its the only operator to take three operands. It is a conditional operator that provides a shorter syntax for the if..then..else statement. The first operand is a boolean expression; if the expression is true then the value of the second operand is returned otherwise the value of the third operand is returned: boolean expression ? value1 : value2 Examples: The following if..then..else statement: boolean isHappy true; String mood ; if (isHappy true) { mood Im Happy!; } else { mood Im Sad!; } can be reduced to one line using the ternary operator: boolean isHappy true; String mood (isHappy true)?Im Happy!:Im Sad!; Generally the code is easier to read when the if..then..else statement is written in full but sometimes the ternary operator can be a handy syntax shortcut.