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Homework_12_solutions

# Homework_12_solutions - EE 2130 Homework 12 Name{9 1 In the...

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Unformatted text preview: EE 2130 Homework 12 Name: {9% 1. In the series resonant circuit shown, determine the resonant frequency fo, the bandwidth BW, and the halfvpower frequencies fH and fL when the output is taken across the resistor. Also determine the phasor voltages V; and VG at resonance. L = 159.2mH I _ w ”A V“; R Vow Ma Wmdwwgw) ,Eni-___5’_____4‘:5_).. Mc- HM; . . ijC+(JUJf[—Cﬁ£/ /%J§W7§€¢f'égiﬂf) 3. For a parallel resonant RLC circuit, we ﬁnd that f0 = 1/2m/LC just as in the series case. We ﬁnd that the quality factor is the reciprocal of the series case, i.e., Q3, 2 l/QS = R/ZrthL = erﬁC R. For the parallel resonant circuit below, determine the L and C values if R = 10 kﬂ, f0 = 1 MHZ, and BW = 100 kHz. Also determine I L and I C at resonance and compare these values to the source current. 4. Consider the second order ﬁlter circuit shown below. Real inductors have a smali amount of series resistance associated with them and this has been modeled in the schematic. R1=10§l I .. z/A/% £7”ng 0 J -M: f g (:2- RE giggly. Q92”? 73W: IE3 :: £05.. ..__— . Q5 (262-3 /éO//? A bit of work (PH spare you the details!) 3110 3 flat the voltfe gain transfer function is given by H(f)=V°ut=£W ﬂak \$56026)" :5 7&4} Wéd “1'2””3W‘1‘f/fez Peak/H%;VOIQ°3(JW1€R3¢)5 3562615 where R3 = R1 + R2. We’ve seen these terms before. The numerator is a quadratic zero and the denominator is a quadratic pole of the transfer function. Determine the resonant frequency (f0) and the quality factor (Q5). Then draw the straight line approximations (in dB) for the magnitudes of the two transfer functions (one for the zero and one for the pole) on the ﬁrst of the provided semi-log graphs. Compute the band width of the peak for each quadratic term and the height of each peak, then sketch them in as weil. Finally, on the second provided graph, sketch in the composite transfer function magnitude in decibels. Remember that [H [d3 = ZOIogIHﬂ + 20£ogEH2 ]. Name this filter type. 4o — .. e1 1 W Ill iii. 20 ‘iJ i . ! g 0 . . _ "Ti l . 8 —20 ii i I _i i “‘“E (D . . a» H . l l = -40 m4 =— -. . s t .. _. A 78% - t l E l , Ila i.E+co 1.5+01 1.E+02 1.5+03 1.E+04 1.E+05 1.E+06 I Frequency (Hz) 10W: Ma 6914 : 351‘? » 7g; : 3/0, 1% 22mm: I T1 ka OIII .i I1 :2 11 7M 11!! ,JL 1.E+00 1.E+01 1.E+02 1.E+03 1.E+O4 1.E+05 1.E+06 Frequency (Hz) rmmm: Voltage Gain (dB) lb 0 Filter Typ_e: ...
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