Question

In the circuit given below, R1 = 8 and R, = 8 . Determine the center frequency and bandwidth of the bandpass filter. 1F :) =17 R, {v, The value of center frequency of the bandpass filter is rad/s. The value of bandwidth of the bandpass filter is rad/s.

Answer 1

Ans.

The below image shows the low pass and high pass parts of the filter :

Source Low-pass filter section Ri High-pass filter section c+

To find the cut-off frequency we can find the transfer function and find its poles is magnitude gives the cut-off frequencies :

In laplace domain the circuit is:

Source Low-pass filter section Ri High-pass filter section 1/sC2 1/sC,

We get :

$V_{0}=\frac{V_{i}}{8+\frac{1/s*(8+1/s)}{1/s+(8+1/s)})}*(\frac{\frac{1}{s}}{\frac{1}{s}+8+\frac{1}{s}})*8$

simplifying the equation we get :

VO 8s * 6432 +24s +1 8s 6482 +245 +1 V 7 V

The denominator is :

$64s^{2}+24s+1$

the magnitude of roots poles of s gives the cut-off frequencies:

The roots are   -0.3273 and   -0.0477

Thus cut-off frequencies are :

$\\\omega_{H}=0.3273rad/s\\ \omega_{L}=0.0477rad/s$

Thus Center frequency is :

$\\\omega_{C}=\sqrt{\omega_{H}*\omega_{L}} = 0.1249rad/s$

Bandwidth :

$\\BW = \omega_{H}-\omega_{L}= 0.2796rad/s$