Determine the center frequency and bandwidth for the band-pass filter in Figure 18-26 if the inductor has a winding resistance of 15 Ω.

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Accepted Answer

Recall from Chapter 17 (Eq. 17-13 : [latex]Z= \sqrt{R^{2}+(X_{L}- X_{C})^{2}} \angle \pm an ^{-1} \left(\frac{ X_{tot}}{R} ight)[/latex]) that the resonant (center) frequency of a nonideal tank circuit is

[latex]f_{0}= \frac{\sqrt{1-(R^{2}_{W}C/L)} }{2\pi \sqrt{LC} }= \frac{\sqrt{1-(15 \ \Omega )^{2}(0.01 \ \mu F)/50 \ mH } }{2\pi \sqrt{(50 \ mH)(0.01 \ \mu F)} }= 7.12 \ kHz[/latex]

The Q of the coil at resonance is

[latex]Q=\frac{X_{L}}{R_{tot}}= \frac{2\pi f_{0}L}{R_{W}} = \frac{2\pi (7.12 \ kHz)(50 \ mH)}{15 \ \Omega } = 149[/latex]

The bandwidth of the filter is

[latex]BW=\frac{f_{0}}{Q}= \frac{7.12 \ kHz}{149} = 47.8 \ Hz[/latex]

Note that since Q > 10, the simpler formula,[latex]f_{0}= 1/(2\pi \sqrt{LC} )[/latex], could have been used to calculate [latex]f_{0}[/latex].

Question 18.10: Determine the center frequency and bandwidth for the band-pa...

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