Calculate the gain, center frequency, Q, and bandwidth of the bandpass filter in Fig. 36.35.
Question 36.12: Calculate the gain, center frequency, Q, and bandwidth of th...
A=\frac{-R_2}{2R_1}=\frac{-36 k\Omega }{2(18 k \Omega )}=-1
f_0=\frac{1}{2\pi C\sqrt{R_1R_2} } =\frac{1}{2\pi(8.2 nF)\sqrt{(18 k\Omega)(36 k \Omega)}}
=762.85 Hz
Q=0.5 \sqrt{\frac{R_z}{R_1} }=0.5\sqrt{\frac{36 k\Omega}{18 k\Omega} } =0.707BW=\frac{f_0}{Q}=\frac{762.85}{0.707}=1079 Hz
Related Answered Questions
This is a noninverting first-order low-pass filter...
The feedback fraction is
B=\frac{18k\Omega ...
Recall that the duty cycle is defined as the pulse...
With Formula (36-4),B=\frac{R_1}{R_1+R_2 }[...
This is one way to create a 60-Hz clock, a square-...
With Formulas (36-32) f_0=\frac{1}{2\pi RC}...
This is an inverting first-order low-pass filter. ...
With the equations shown in Fig. 36-25, we can ana...