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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 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 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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(438)

where h represents the impulse response matrix of the linear system Because w(t) is a random variable, so is y(t) The Fourier transform of the impulse response yields the system transfer function

fo = 0159/(LC )1/2 900 = 0159/(L 0000033)1/2 9002 = 01592/(0000033 L) 810 = 00253/(0000033 L)

H(j ) =

Problem 17-13 Suppose a circuit must be designed to have fo = 455 kHz A coil of 100 H is available What size capacitor is needed Convert the frequency to megahertz: 455 kHz = 0455 MHz Then the calculation proceeds in the same way as with the preceding problem:

(439)

The Energy Spectrum of a time-invariant linear system is |H(j )| = H(j )H(j )

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fo = 0159/(LC )1/2 0455 = 0159/(100 C )1/2 04552 = 01592/(100 C ) 0207 = 00253/(100 C ) 0207 100 C = 00253 207 C = 00253 C = 00253/207 = 000122 F = 1220 pF In practical circuits, variable inductors and/or variable capacitors are often placed in tuned circuits, so that small errors in the frequency can be compensated for The most common approach is to design the circuit for a frequency slightly higher than fo, and to use a padder capacitor in parallel with the main capacitor (Fig 17-13)

(440)

(Cp ) allow limited adjustment of the resonant frequency in a series LC circuit (as shown at A), or in a parallel LC circuit (as shown at B)

where H(j ) represents the complex conjugate transpose of H(j ) If w (t) denotes the mean of the system input, then the mean of the system output is y (t) = E y(t)

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89 Figure Exam-6 shows a radar display and a single blip, indicating the position of a target What is the approximate azimuth of this target (a) 122 (b) 212 (c) 238 (d) 148 (e) More information is needed to answer this 90 Which of the following is an e-mail address in the proper format (a) wwwsciencewriternet (b) membersauthorsguildnet/stangib (c) sciencewriter@tahoecom (d) http://googlecom (e) All of the above 91 A robot can be guided along a specific route by means of a current-carrying wire beneath the surface The robot detects, and follows, the magnetic field produced by the current, thus following the wire This is called (a) an epipolar system (b) a fluxgate magnetometer (c) a servo system (d) an electromagnetic router (e) an embedded path

h(t )E w( ) d h(t ) w ( )d (441)

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