CHAPTER 16 APPLICATIONS OF LOGARITHMS TO TRANSMISSION PROBLEMS 205. The Transmission Unit When signal power is transmitted along a transmission line, there is a power loss or attenuation; if an amplifier is used in the circuit, there may be a power gain. This loss or gain of power, resulting in a decrease or increase in the intensity of the signal, is measured in terms of the decibel (db). The decibel is a measure of power ratio and is probably the most widely used unit in communications. The formula for measuring transmission loss or gain is: P where is the ratio of the two powers being compared (par. 206). 206. Converting Power Ratio to Decibels When converting a power ratio into its decibel expression, represent the larger power as P1 and the smaller power as P2, regardless of whether the larger power is the input or output. Thus, the power ratio will always be greater than 1, and its logarithm will be a positive number. Prefix a plus sign to the answer if the power change is a gain (the power output greater than the power input); prefix a minus sign if the power change is a loss. Example 1: The input power to a transmission line is 10 milliwatts, and the output power is 2.46 milliwatts. Express the power change in db. 207. Converting Decibels to Power Ratio To find the power ratio when the gain or loss is expressed in decibels, reverse the procedure given in paragraph 206. If the number of decibels is positive, the circuit has a power gain and the output power is greater than the input power. If the number of decibels is negative, the circuit has a power loss and the output power is less than the input power. Insert the power change in decibels in the formula given in paragraph 200 and divide by 10; then find the antilog of both sides of the equation (par. 118) to obtain the power ratio. Example 1: A circuit is known to have a power change of +12 db. Since the number of decibels has a minus sign, the circuit attenuates power. The output power is less than the input power by a ratio of 1 to 316.2. 208. Review Problems-Transmission Problems a. A network has a loss of 16 decibels. What power ratio correspond to this loss? b. The input to a powerline 50 miles long is 210 milliwatts. The power delivered at the end of the line is 40 microwatts. What is the attenuation in decibels per mile? c. A power of 10 milliwatts is required to drive an audiofrequency (af) amplifier. The output of the amplifier is 120 milliwatts. What is the gain in decibels? d. What is the ratio of the output power to the input power if there is a power gain of 14 decibels? CHAPTER 17 MISCELLANEOUS ELECTRICAL PROBLEMS 209. Efficiency Efficiency is the ratio of output to input and usually is expressed in percent (ch. 2). Generators, motors, and other electrical devices often are rated according to their efficiency. To express efficiency in percent, write the ratio of output to input as a fraction, convert to a decimal, and then convert the decimal to a percent (par. 4). Another application of percent is the overload rating of motors, generators, etc. In this application, the amount of power, that can be applied to or taken from an electrical device, above the rated output, is expressed as a percent of the rated output. Example 1: What is the percent of overload capacity of a generator that Overload = maximum power-rated power Example 2: Find the maximum output of a generator that is rated at 1,500 watts, and has a 10 percent overload capacity. 0.10 1,500 150 watts 1,650 watts maximum output. 211. Tolerances A tolerance is an allowance for variations from the standard or specified value. In the manufacture of resistors, for example, the resistance is permitted to be within a specified percentage of the standard value. This percentage is indicated in the color code of the resistors. 212. Transformer Relationships a. General. In a transformer, relationships exist between the currents, voltages, impedances, and number of turns of wire in the windings. These relationships are expressed by equations containing ratios involving these quantities. b. Relationship Between Voltage and Number of Turns. This relationship is expressed by the following equation: E. = N where E, is the voltage across the primary winding, N, is the number of turns on the primary winding, E, is the voltage across the secondary winding, and N, is the number of turns on the secondary winding (fig. 87). The equation may also be written: The ratios N,/N, and N./N, are called the turns ratios and may be expressed as a single factor. Find the voltage across the secondary winding of a trans- ratio from primary to secondary is 1 to 4. c. Relationship Between Current and Number of Turns. This relationship is expressed by the following equation: = N. where I, is the primary current, I, is the secondary current, and N, and N, the number of turns on the primary and secondary as before. The equation may be written: Example: Find the primary current in a transformer if the secondary d. Relationship Between Current and Voltage. By combining the relationships given in b and c above, a relationship can be derived between primary and secondary currents as follows: Since voltage multiplied by current equals power, the last form of the equation states that the power absorbed by the primary winding is equal to the power delivered to the secondary winding. This is true in an ideal transformer which has no loss, and is essentially true in an actual transformer which has very little loss; efficiencies of 98 percent are common in actual transformers. ( |