## Chapter 19

## Q. 19.9

## Step-by-Step

## 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.