Let us use the equations in Example 12.27 to design a band-pass filter of the form shown in Fig. 12.57 with a BW =2000 rad/s, (Vo/VS)(ω0) = -5, and Q = 3. Use C = 0.1 μF, and determine the center frequency of the filter.

Question

Let us use the equations in Example 12.27 to design a band-pass filter of the form shown in Fig. 12.57 with a BW =2000 rad/s, ([latex]V_o/V_S)(ω_0)[/latex] = -5, and Q = 3. Use C = 0.1 μF, and determine the center frequency of the filter.

Accepted Answer

Using the filter equations, we find that

[latex]\begin{aligned}\mathrm{BW} &=\frac{2}{R_{2} C} \\2000 &=\frac{2}{R_{2}(10)^{-7}} \\R_{2} &=10 \mathrm{k} \Omega \\\frac{\mathbf{V}_{o}}{\mathbf{V}_{S}}\left(\omega_{0} ight) &=-\frac{R_{2}}{2 R_{1}} \\-5 &=-\frac{10,000}{2 R_{1}} \\R_{1} &=1 \mathrm{k} \Omega\end{aligned}[/latex]

and

[latex]\begin{aligned}Q &=\frac{1}{2} \sqrt{\frac{R_{2}}{R_{1}}} \sqrt{1+\frac{R_{1}}{R_{3}}} \\3 &=\frac{1}{2} \sqrt{\frac{10,000}{1000}} \sqrt{1+\frac{1000}{R_{3}}}\end{aligned}[/latex]

or

[latex]R_{3}=385 \Omega[/latex]

Therefore, [latex]R=1 \mathrm{k} \Omega, R_{2}=10 \mathrm{k} \Omega, R_{3}=385 \Omega[/latex], and [latex]C=0.1 \mu \mathrm{F}[/latex] completely define the band-pass filter shown in Fig. 12.57. The center frequency of the filter is

[latex]\begin{aligned}\omega_{0} &=\frac{1}{C} \sqrt{\frac{1+R_{1} / R_{3}}{R_{1} R_{2}}} \\&=\frac{1}{10^{-7}} \sqrt{\frac{1+(1000 / 385)}{(1000)(10,000)}} \\&=6000 \mathrm{rad} / \mathrm{s}\end{aligned}[/latex]

Question 12.28: Let us use the equations in Example 12.27 to design a band-p...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions