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Answers to quiz, test, and exam questions 699 Test: Part One 1. A 5. A 9. E 13. B 17. A 21. D 25. A 29. C 33. E 37. D 41. A 45. C 49. E 2. A 6. B 10. C 14. C 18. A 22. D 26. C 30. C 34. D 38. D 42. D 46. C 50. C 3. C 7. A 11. C 15. D 19. E 23. D 27. D 31. C 35. E 39. C 43. C 47. D 4. B 8. E 12. C 16. B 20. E 24. C 28. A 32. D 36. B 40. A 44. E 48. A

NOTE: The overall bit rate of the STS-N system is N STS-1. However, the maximum bandwidth that can be transported is STS-1, but N of them can be transported. This is analogous to a channelized T-1.

Exam-5 Illustration for Final Exam Question 78. 78. Figure Exam-5 shows a generic circuit that can be used for broadband RF power amplification. What is wrong with the circuit as shown (a) The resistor, R, should be replaced with a capacitor. (b) The capacitor, C, should be replaced with a resistor. (c) The +12-V power source should be replaced with a 12-V power source.

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Let s go back to our Fast Ethernet example mentioned earlier. In this case, 51.84 Mbps is inadequate for our purposes, because we have to transport the 100-Mbps Ethernet signal. For this, we need what is known as a concatenated signal. One thing you can say about SONET is that it doesn t hurt your polysyllabic vocabulary.

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Considering = 3.14 to three significant figures, this formula can be simplified to: fo = 0.159/(LC )1/2 The 1 2 power of a quantity represents the positive square root of that quantity. The preceding formulas are valid for series-resonant and parallel-resonant RLC circuits. The formula will also work if you want to find fo in megahertz (MHz) when L is given in microhenrys ( H) and C is in microfarads ( F). These values are far more common than hertz, henrys, and farads in electronic circuits. Just remember that millions of hertz go with millionths of henrys, and with millionths of farads.

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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

The Effects of R and G Interestingly, the value of R or G does not affect the resonant frequency in either type of circuit But these quantities are significant, nevertheless! The presence of nonzero resistance in a series-resonant circuit, or nonzero conductance in a parallel-resonant circuit, makes the resonant frequency less welldefined Engineers say that the resonant frequency response becomes more broad or less sharp In a series circuit, the resonant frequency response becomes more broad as the resistance increases In a parallel circuit, the resonant frequency response becomes more broad as the conductance increases The sharpest possible responses occur when R = 0 in a series circuit, and when G = 0 (that is, R = ) in a parallel circuit Problem 17-10 Find the resonant frequency of a series circuit with an inductance of 100 H and a capacitance of 100 pF.

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First, convert the capacitance to microfarads: 100 pF = 0000100 F Then find the product LC = 100 0000100 = 00100 Take the square root of this, getting 0100 Finally, divide 0159 by 0100, getting fo = 159 MHz Problem 17-11 Find the resonant frequency of a parallel circuit consisting of a 33- H coil and a 47-pF capacitor Again, convert the capacitance to microfarads: 47 pF = 0000047 F Then find the product LC = 33 0000047 = 000155 Take the square root of this, getting 00394 Finally, divide 0159 by 00394, getting fo = 404 MHz Problem 17-12 Suppose you want to design a circuit so that it has fo = 900 MHz You have a 33-pF fixed capacitor available.

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What size coil will be needed to get the desired resonant frequency Use the formula for the resonant frequency, and plug in the values This will allow you to use simple arithmetic to solve for L Convert the capacitance to microfarads: 33 pF = 0000033 F Then calculate as follows:.

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