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The cut-off frequency of the constant k-low pass filter is?(a) 1/√LC(b) 1/(π√LC)(c) √LC(d) π√LCThe question was asked in an interview for job.My question is based upon Constant-K Low Pass Filter in portion Filters and Attenuators of Network Theory

Answer»

Right choice is (b) 1/(π√LC)

To explain I WOULD say: Z1/4Z2 = 0. Z1 = jωL and Z2 = 1/jωC. On solving the cut-off FREQUENCY of the constant k-low pass FILTER is fc = 1/(π√LC).



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