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Engineering Science - Research Paper Example

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The paper "Engineering Science" tells us about resonant frequency. Using the formula, the calculated value of the resonant frequency is 1/ 6.283 * 10-3 = 159.2 HZ. Due to limitations in the resolution of the measurement equipment, the table indicates a band of frequencies where the maximum current seems to indicate resonance…
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Engineering Science
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Engineering Science Assignment III Compare the actual resonant frequency obtained from the graph with thecalculated value. Using the formula, the calculated value of the resonant frequency is 1/ 6.283 * 10-3 = 159.2 HZ. Due to limitations in the resolution of the measurement equipment, the table indicates a band of frequencies where the maximum current seems to indicate resonance. However, we can approximate the frequency of resonance from the chart by finding the median frequency of the peak current values for each value of R. R=0 Median Frequency = 155 HZ R=500 Median Frequency = 155 HZ R=1000 Median Frequency = 160 HZ R=1500 Median Frequency = 160 HZ The measured values from the table were all within range and close to the calculated value. The discrepancies arise from the accuracy of the components, resolution of the testing equipment, and the limitations of the test setup. 2. What is the condition for resonance in an electrical circuit In a series LC circuit, resonance occurs when the Inductive Reactance (XL) is equal to the Capacitive Reactance (XC). Because XC and XL are 180 degrees out of phase the combined impedance is equal to zero. This is the point where the waveform encounters its least resistance and it will tend to oscillate at this frequency. We can calculate the values at resonance from the experiment to verify this theory. XC = Capacitive Reactance in Ohms f = Frequency at resonance (Calculated at 159.2 HZ) C= 1ufd XC = 1 / (6.283*159.2*1*10-6)) = 1/ (1000.25*10-6) = 1000.25 Ohms XL = 2pifL XL = Inductive Reactance in Ohms pi = 3.1415 f = Frequency at resonance (Calculated at 159.2 HZ) L = 1 H XL = 6.283*159.2*1 = 1000.25 Ohms As can be seen, the capacitive reactance equals the inductive reactance at 159.2 HZ and thus sets up the conditions for resonance at this frequency. 3. Discuss the effect of changing the value of the resistor. The most obvious effect of changing the resistor value is to reduce the current flow at resonance. The current flow when R=0 ohms was .92ma. However, when the resistance was increased to 1500 ohms, the resonance current was reduced to .57ma. The current at 16HZ and 1600HZ remained nearly constant at R=0 ohms and R=1500 ohms. This is due to the capacitive reactance rising at 16HZ and the inductive reactance rising at 1600 HZ to about 10000 ohms. The increased reactance of the capacitor and inductor at these frequencies diminished the effect of the resistor. Another effect was to alter the shape of the curve as the frequency varies from 16 to 1600 HZ. When R=0 ohms, the higher peak at resonance resulted in a sharper rate of increase and a more defined peak. When R=1500 ohms, the curve was more flattened and had a less well defined peak. 4. Discuss the term damping and its effect on the amplitude of the current and the resonant frequency. Damping is a term that describes the loss of energy in a circuit, thus causing the decay in amplitude of a series RLC resonant circuit. In an ideal series LC circuit at resonance, the resistance would theoretically be 0 ohms. Once excited, this ideal circuit would oscillate forever because there is no resistance to dissipate the energy. As resistance is added to the circuit, the resistor dissipates the energy in the form of heat and the energy in the circuit decays to 0. A circuit can be critically damped, under damped, or over damped. A critically damped circuit will decay the energy in one cycle of the resonant frequency. An under damped circuit will have less resistance and the oscillation will decay more slowly. An over damped circuit will have a higher resistance value and the oscillation will decay to 0 in less than one cycle. The damping resistor has no effect on the resonant frequency. Though the resonant frequency stays constant, the Q factor of the circuit is reduced at higher resistance values and results in a less well defined resonant peak. 5. Discuss the implication of resonance on the components of an electrical system such as the one in figure 1 at low and high frequencies. As seen in the formula for resonance, the frequency is inversely related to the value of the inductor and the capacitor. Inductive reactance increases at higher frequencies, while capacitive reactance decreases with increased frequency. fo = Resonant Frequency L = Inductance ( in henries ) C = Capacitance ( in farads ) wo = Radians per Second = 6.283 * f To raise the resonant frequency, it becomes necessary to lower the value of the capacitor, the inductor, or both. As noted in the formulas for reactance, capacitive reactance is inversely proportional to the frequency whilst inductive reactance is directly proportional. If we raise the resonant frequency by lowering the value of the inductor, the increase in frequency will result in a lower capacitive reactance. The lower value of the inductor will likewise result in a lower inductive reactance (since it is directly proportional to XL). At the point that these two values are equal, resonance will again occur, this time at a higher frequency. Raising the value of the capacitor or inductor will result in a lower resonant frequency. As we lower the frequency, the inductor's reactance becomes less. Increasing the value of the capacitor will likewise result in a lower capacitive reactance. This theory holds true whether we increase the capacitor, inductor, or both. In figure 1, at 1600 HZ, the XC has decreased and the XL has increased. 16 HZ has the opposite effect. The XC has increased and the XL has decreased 6. Explain the term Q Factor and its relationship with resistance. The Q Factor, often referred to as the quality factor, is a measurement of a circuit's ability to reach a well defined resonance and have a sharp resonant peak. As can be seen in the graphs from the experiment, the change from R=0 ohms to R=1500 ohms caused the peak to be lower with a more gradual rise toward the peak. The circuit with 1500 ohms is said to have a lower Q Factor. In an ideal and theoretical circuit with R=0 ohms, the Q would approach infinity. However, due to resistance in the wiring, inductor, and other stray resistance, this is not a real possibility. The Q factor is useful in electrical circuit design as it defines the selectivity of the circuit. A high Q circuit is very precise as to its resonance. This is used in filters where a specific or narrow band of frequencies must be eliminated or passed. Likewise, by lowering the Q, we can design a circuit or filter that has a wider bandpass for applications such as audio or video transmission. 7. Determine the Q Factor for each resistor value and discuss the effect it has on the shape of the curve. Theory describes Q as equal to XC or XL divided by the series resistance. In a theoretical circuit with no stray resistance the Q would be as follows: (Reactance=1000 ohms) R = 0 Q = 1000 / 0 = infinity (no curve, pure spike) R=500 Q = 1000 / 500 = 2 (sharp well defined curve) R=1000 Q = 1000 / 1000 = 1 (Curve is more rounded) R=1500 Q = 1000 / 1500 = .67 (Curve begins to flatten) However, in reality there is stray resistance whose value may not be readily apparent or available. An alternate method used to measure the Q of an active circuit may be used. The Q Factor is calculated as follows: fc = Center Frequency f2 = High 3db Cutoff (.707 * ma ) f1 = Low 3db Cutoff (.707 * ma ) As can be seen in the formula, as the resistance is lowered, the 3db rolloff points get closer to the resonant frequency. As f2 and f1 approcah fc, the change in frequncy is lowered and the Q rises proportionally. From the graph (chart), at R=0 ohms, the 3DB points are .707*.92 = .65 ma. Approximating gives us f2 = 450 HZ and f1 = 58 HZ. Solving for Q = 159.2 / 392 = .41. This indicates that there is approximately 2500 ohms of stray resitance in the circuit. Adding this value to R allows us to calculate the Q as follows R=0 Q= 1000 / 2500 = .40 R=500 Q = 1000 / 3000 = .33 R=1000 Q = 1000/ 3500 = .29 R=1500 Q = 1000 / 4000= .25 By looking at the chart, we see that when the resistance is increased to 1500 ohms, the 3db points are .57ma*.707=.40ma. f2=725 HZ and f1=37 HZ. Solving for Q=159.2 / 688 = .23. This is well within the theoretical value of .25. Using the 3db method for R=500 ohms and R=1000 ohms yeilds results of .35 and .29 respectively. Both are well within the accuracy and resolution limitations of the setup. The experiment has verified that the addition of series resitance lowers the Q factor and reduces the circuits selectivity. It has also verified that the circuit has approximatley 2500 stray ohms of resistance. 8. Resonance is an important phenomenon in both electrical and mechanical systems. Discuss both the advantages and disadvantages of such a phenomenon on such systems. Resonance, the frequency at which a systems tends to naturally oscillate, can be used to design systems and when harnessed correctly can serve useful functions. However, unwanted resonance can result in undesirable electrical signals or cause premature mechanical failure. Mechanical resonance is used in musical instruments to create sound such as a piano or guitar string. Vibrating separators use resonance to maximize the device efficiency and lower input power requirements. However, when unwanted vibration sets up at or near the resonant frequency, it can cause excessive movement, wear, and lead to mechanical failure. Resonance is used in electrical circuitry to create highly tuned oscillators and filters. By correct configuration, a filter can eliminate signals outside the bandpass of the resonant frequency. They can also be used to eliminate only one given frequency. However, when series capacitance and induction appear in a circuit, it introduces the possibility of introducing or passing unwanted signals. It can also create a situation where the resonance gets carried away and drowns out the useful signal. In audio systems it can result in "parasitic feedback", while at radio frequencies it can cause oscillators to become "microphonic" and overly sensitive. Adding the correct damping resistance and adjusting the Q can help eliminate the unwanted characteristics of resonance as well as maximize the desirable effects. Conclusion Michael Faraday and Joseph Henry could not have conceived the wide uses we put the capacitor and inductor to today. Their early 19th century experiments paved the way for a large portion of today's electrical and electronic systems. The theory of oscillation and having a frequency where resonance occurs has not changed since their time and the series RLC circuit is a simple and elegant design. Understanding the infinite peak, and therefore the infinite Q, of an ideal theoretical circuit is an aid to understanding the formulas. The inverse effect of the capacitor and inductor value on resonant frequency can be seen in the formula. The effect of series resistance on Q can be seen in the formula for Q factor, as well as measured in a test setup. With a minimum number of measurements and a few formulas, we can calculate and predict the effect of altering the values in a circuit. This allows the designer to very closely approach real world operation, or predict the outcome of a modification. By adding series resistance, we can minimize unwanted effects of resonance or decrease the resistance to allow for a more sharply tuned signal. By inserting various values of R, L, and C, the designer can enhance the desired effect while eliminating the harmful effects of resonance. Read More
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