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## Q. 19.9

What are the pole frequency and Q of the filter shown in Fig. 19-36a? What is the cutoff frequency? Show the frequency response using a Multisim Bode plotter. ## Verified Solution

The Q and pole frequency are:

$Q=0.5\sqrt{\frac{R_1}{R_2} } =0.5\sqrt{\frac{24 \ kΩ}{12 \ kΩ } }=0.707$

$f_P=\frac{1}{2\pi R\sqrt{R_1R_2} } =\frac{1}{2\pi (4.7 \ nF )\sqrt{(24 \ kΩ))(12 \ kΩ) } }= 2 \ kHz$

Since Q = 0.707, the filter has a Butterworth second-order response and:
$f_c = f_p = 2$ kHz
The filter has a high-pass response with a break at 2 kHz, and it rolls off at 40 dB per decade below 2 kHz. Figure 19-36(b) shows the Multisim frequency response plot.