=> RF= (AF-1)×R1 =(1.586-1)×10kΩ =5860 =5.86kΩ. 6. Consider the following circuit and calculate the low cut-off frequency value?
A. 178.7Hz
B. 89.3Hz
C. 127.65Hz
D. 255.38Hz
Answer:A
=> RF= (AF-1)×R1 =(1.586-1)×10kΩ =5860 =5.86kΩ. 6. Consider the following circuit and calculate the low cut-off frequency value?
A. 178.7Hz B. 89.3Hz C. 127.65Hz D. 255.38Hz
Explanation: The low cut-off frequency for the given filter is fL =1/√[2π√(R2×R3×C2×C3)]=178.7Hz. 7. Determine voltage gain of second order high pass butterworth filter.