Question 9.6: Determine the center frequency, maximum gain, and bandwidth ...

f_0=\frac{1}{2 \pi C} \sqrt{\frac{R_1+R_3}{R_1 R_2 R_3}}=\frac{1}{2 \pi(0.01 ~\mu \mathrm{F})} \sqrt{\frac{68 ~\mathrm{k} \Omega+2.7~ \mathrm{k} \Omega}{(68 ~\mathrm{k} \Omega)(180 ~\mathrm{k} \Omega)(2.7 ~\mathrm{k} \Omega)}}=736~ \mathrm{~Hz}

A_0=\frac{R_2}{2 R_1}=\frac{180~ \mathrm{k} \Omega}{2(68 ~\mathrm{k} \Omega)}=1.32

Q=\pi f_0 C R_2=\pi(736 \mathrm{~Hz})(0.01~ \mu \mathrm{F})(180~ \mathrm{k} \Omega)=4.16

B W=\frac{f_0}{Q}=\frac{736 \mathrm{~Hz}}{4.16}=177 \mathrm{~Hz}

PRACTICE EXERCISE

If R_2 in Figure 9–19 is increased to 330 kΩ, how does this affect the gain, center frequency, and bandwidth of the filter?