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## Q. 19.6

What are the pole frequency and Q of the filter shown in Fig. 19-32? What is the cutoff frequency? Show the frequency response using a Multisim Bode plotter. ## Verified Solution

The $A_V, Q,$ and $f_p$ are:

$A_V=\frac{30 \ k\Omega }{51 \ k\Omega }+1=1.59$
$Q=\frac{1 }{3-A_V }=\frac{1}{3-1.59} =0.709$
$f_p=\frac{1}{2\pi RC}=\frac{1}{2\pi(47 \ kΩ)(330 \ pF)}=10.3 \ kHz$

It takes a Q of 0.77 to produce a ripple of 0.1 dB. Therefore, a Q of 0.709 produces a ripple of less than 0.003 dB. For all practical purposes, the calculated Q of 0.709 means that we have a Butterworth response to a very close approximation. The cutoff frequency of a Butterworth filter is equal to the pole frequency of 10.3 kHz. Notice in Fig. 19-32b that the pole frequency is at approximately 1 dB. This value is 3 dB down from the passband gain of 4 dB.