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Problem 16-5 Suppose a resistor, a coil, and a capacitor are connected in series with R = 50 , XL = 22 , and XC = 33 . What is the net impedance, Z Consider the resistor to be part of the coil, obtaining two complex vectors, 50 + j 22 and 0 j33. Adding these gives the resistance component of 50 + 0 = 50, and the reactive component of j22 j33 = j11. Therefore, Z = 50 j11. Problem 16-6 Consider a resistor, a coil, and a capacitor that are connected in series with R = 600 , XL = 444 , and XC = 444 . What is the net impedance, Z Again, imagine the resistor to be part of the inductor. Then the complex impedance vectors are 600 + j444 and 0 j444. Adding these, the resistance component is 600 + 0 = 600, and the reactive component is j444 j444 = j0. Thus, Z = 600 + j0. This is a purely resistive impedance, and you can rightly call it 600 . Problem 16-7 Suppose a resistor, a coil, and a capacitor are connected in series. The resistor has a value of 330 , the capacitance is 220 pF, and the inductance is 100 H. The frequency is 7.15 MHz. What is the complex impedance of this series RLC circuit at this frequency First, calculate the inductive reactance. Remember that XL = 6.28f L and that megahertz and microhenrys go together in the formula. Multiply to obtain the following:

A classic business quote, imminently applicable to the optical networking world, observes in its success lie the seeds of its own destruction. As the marketplace clamors for longer transmission distances with minimal amplification, more wavelengths per fiber, higher bit rates, and increased signal power, a rather ugly collection of transmission impairments, known as fiber nonlinearities, rises to challenge attempts to make them happen. These impairments go far beyond the simple concerns brought about by loss and dispersion; they represent a significant performance barrier. The special relationship that exists between transmission power and the refractive index of the medium gives rise to four service-affecting optical nonlinearities: self-phase modulation (SPM), cross-phase modulation (XPM), FWM, and intermodulation.

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Test: Part two 349 13. A complex impedance is represented by 34 j23. The absolute-value impedance is: A. 34 . B. 11 . C. 23 . D. 41 . E. 57 . 14. A coil has an inductance of 750 H. The inductive reactance at 100 kHz is: A. 75.0 . B. 75.0 k . C. 471 . D. 47.1 k . E. 212 . 15. Two waves are 180 degrees out of phase. This is a difference of: A. 1/8 cycle. B. 1/4 cycle. C. 1/2 cycle. D. A full cycle. E. Two full cycles. 16. If R denotes resistance and Z denotes absolute-value impedance, then R/Z is the: A. True power. B. Imaginary power. C. Apparent power. D. Absolute-value power. E. Power factor. 17. Two complex impedances are in series. One is 30 + j50 and the other is 50 j30. The net impedance is: A. 80 + j80. B. 20 + j20. C. 20 j20. D. 20 + j20. E. 80 + j20. 18. Two inductors, having values of 140 H and 1.50 mH, are connected in series. The net inductance is: A. 141.5 H. B. 1.64 H.

jXL = j(6.28 7.15 100) = j4490 Next, calculate the capacitive reactance using the formula XC = 1/(6.28f C ). Convert 220 pF to microfarads to obtain C = 0.000220 F. Then calculate: jX C = j[1/(6.28 7.15 0.000220)] = j101 Now, lump the resistance and the inductive reactance together, so one of the impedance vectors is 330 + j4490. The other is 0 j101. Adding these gives Z = 330 + j4389; this rounds off to Z = 330 + j4390.

350 Test: Part two C. 0.1415 mH. D. 1.64 mH. E. 0.164 mH. 19. Which of the following types of capacitor is polarized A. Mica. B. Paper. C. Electrolytic. D. Air variable. E. Ceramic. 20. A toroidal-core coil: A. Has lower inductance than an air-core coil with the same number of turns. B. Is essentially self-shielding. C. Works well as a loopstick antenna. D. Is ideal as a transmission-line transformer. E. Cannot be used at frequencies below about 10 MHz.

Self-Phase Modulation (SPM)

The ability of diodes to conduct with forward bias, and to insulate with reverse bias, makes them useful for switching in some electronic applications. Diodes can perform switching operations much faster than any mechanical device. One type of diode, made for use as an RF switch, has a special semiconductor layer sandwiched in between the P-type and N-type material. The material in this layer is called an intrinsic (or I-type) semiconductor. The intrinsic layer (or I layer) reduces the capacitance of the diode, so that it can work at higher frequencies than an ordinary diode. A diode with an I-type semiconductor layer sandwiched in between the P- and N-type layers is called a PIN diode (Fig. 20-5). Direct-current bias, applied to one or more PIN diodes, allows RF currents to be effectively channeled without using relays and cables. A PIN diode also makes a good RF detector, especially at very high frequencies.

21. The efficiency of a generator: A. Depends on the driving power source. B. Is equal to output power divided by driving power. C. Depends on the nature of the load. D. Is equal to driving voltage divided by output voltage. E. Is equal to driving current divided by output current.

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