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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 0.000047 F. Then find the product LC 33 0.000047 0.00155. Take the square root of this, getting 0.0394. Finally, divide 0.159 by 0.0394, getting fo 4.04 MHz. There are times when you might know the resonant frequency fo that you want, and you need to find a particular inductance or capacitance instead. The next two problems illustrate this type of situation.

including residential. I am writing this in my home office on a PC that is connected to a 100-Mbps Ethernet LAN that ties together three PCs and a couple of printers, and the total cost of the network, including the router and firewall that protects the machines from intrusion due to the always-on connection through the cable modem, is less than $200. Most businesses use Ethernet today because most businesses have normal traffic flows office automation traffic and the like. For businesses that experience constant, bandwidth-intensive traffic, such as that found in engineering firms, architectural enterprises, or businesses with other graphics-heavy traffic, Token Ring may be a better choice, although some will argue. Businesses that already have a large installed base of IBM hardware may also be good candidates for Token Ring, since it integrates well (for obvious reasons) into IBM environments. Even still, Ethernet is ruling the roost.

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Problem 17-12

In a circuit containing resistance and capacitive reactance, the characteristics are two-dimensional in a way that is analogous to the situation with the RL plane from the previous chapter The resistance ray and the capacitive-reactance ray can be placed end to end at right angles to make a quarter plane called the RC plane (Fig 14-5) Resistance is plotted horizontally, with increasing values toward the right Capacitive reactance is plotted downward, with increasingly negative values as you go down The combinations of R and X C in this RC plane form impedances You ll learn about impedance in greater detail in the next chapter Each point on the RC plane corresponds to one and only one impedance Conversely, each specific impedance coincides with one and only one point on the plane.

Another variation on the LAN theme that is experiencing a great deal of attention today is 802.11, the set of standards that address wireless LAN considerations. IEEE 802.11 is a wireless LAN standard developed by the IEEE s 802 committee to specify an air interface between a wireless client and a base station, as well as among a variety of wireless clients. First discussed in 1990, the standard has evolved through six draft versions and won final approval on June 26, 1997.

Problem 17-13

Any impedance that consists of a resistance R and a capacitive reactance X C can be written in the form R + jX C Remember that X C is always negative or zero Because of this, engineers will often write R jX C instead If an impedance is a pure resistance R with no reactance, then the complex impedance is R j 0 (or R + j 0; it doesn t matter if j is multiplied by 0!) If R = 3 with no reactance, you get an impedance of 3 j 0, which corresponds to the point (3,j 0) on the RC plane If you have a pure capacitive reactance, say X C = 4 , then the complex impedance is 0 j 4, and this is at the point (0, j 4) on the RC plane.

All 802 standards address themselves to both the Physical (PHY) and MAC layers. At the PHY layer, IEEE 802.11 identifies three options for wireless LANs: diffused infrared, direct sequence spread spectrum (DSSS), and frequency hopping spread spectrum (FHSS). Although the infrared PHY operates at a baseband level, the other two radios operate at 2.4 GHz, part of the Industrial, Scientific, and Medical (ISM) band. The ISM band can be used for operating wireless LAN devices and does not require an end-user license. All three PHYs specify support for 1-Mbps and 2-Mbps data rates.

A circuit must be designed to have fo 455 kHz. A coil of 100 H is available. What size capacitor is needed Convert kHz to MHz: 455 kHz 0.455 MHz. Then the calculation proceeds in the same way as with the preceding problem: fo 0.159/(LC)1/2 0.455 0.159/(100 C)1/2 0.4552 0.1592/(100 C) 0.207 0.0253/(100 C) 0.207 100 C 0.0253 20.7 C 0.0253 C 0.0253/20.7 0.00122 F 1220 pF

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