Find the center frequency of the filter in Figure 18-33. Draw the output response curve showing the minimum and maximum voltages.

Question

Accepted Answer

The center frequency is

[latex]f_{0}= \frac{\sqrt{1-R^{2}_{W}C/L} }{2\pi \sqrt{LC} }= \frac{\sqrt{1-(8 \ \Omega )^{2}(150 \ pF)/5 \ \mu F } }{2\pi \sqrt{(5 \ \mu H)(150 \ pF)} }= 5.79 \ MHz[/latex]

At the center (resonant) frequency,

[latex]X_{L}=2\pi f_{0}L=2\pi (5.79 \ MHz)(5 \ \mu H)=182 \ \Omega[/latex]

[latex]Q=\frac{X_{L}}{R_{W}}= \frac{182 \ \Omega }{8 \ \Omega } = 22.8[/latex]

[latex]Z_{r}= R_{W}(Q^{2}+1)= 8 \ \Omega (22.8^{2}+ 1)= 4.17 \ k\Omega [/latex] ( purely resistive)

Next, use the voltage-divider formula to find the minimum output voltage magnitude.

[latex]V_{out(min)}= \left(\frac{R_{L}}{R_{L}+ Z_{r}} \right) V_{in}= \left(\frac{560 \ \Omega }{4.73 \ k\Omega } \right)10 \ V =1.18 \ V[/latex]

At zero frequency, the impedance of the tank circuit is [latex]R_{W}[/latex] because [latex]X_{C}= \infty[/latex] and [latex]X_{L}[/latex]= 0 Ω. Therefore, the maximum output voltage below resonance is

[latex]V_{out(max)}= \left(\frac{R_{L}}{R_{L}+ R_{W}} \right) V_{in}= \left(\frac{560 \ \Omega }{568 \ \Omega } \right)10 \ V =9.86 \ V[/latex]

As the frequency increases much higher than [latex]f_{0}[/latex],[latex]X_{C}[/latex] approaches 0 Ω , and [latex]V_{out}[/latex] approaches [latex]V_{in}[/latex](10 V). Figure 18-34 shows the response curve.

Question 18.12: Find the center frequency of the filter in Figure 18-33. Dra...

Related Answered Questions