=>|VO/Vin|=2/4.123= =0.4851 = 20log(0.4851) = -6.28dB. 9. How is the higher order filters formed?
A. Using first order filter
B. Using second order filter
C. Connecting first and second order filter in series
D. Connecting first and second order filter in parallel
Answer:A
=>|VO/Vin|=2/4.123= =0.4851 = 20log(0.4851) = -6.28dB. 9. How is the higher order filters formed?
A. Using first order filter B. Using second order filter C. Connecting first and second order filter in series D. Connecting first and second order filter in parallel
Explanation: Higher filters are formed by using the first and second order filters. For example, a third order low pass filter is formed by cascading first and second order low pass filter. 10. State the disadvantage of using higher order filters?