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GATE ECE 2012  Question: 41
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The circuit shown is a
low pass filter with $f_{3\:dB}=\frac{1}{(R_1+R_2)C}\: rad/s$
high pass filter with $f_{3\:dB}=\frac{1}{R_1C}\: rad/s$
low pass filter with $f_{3\:dB}=\frac{1}{R_1C}\: rad/s$
high pass filter with $f_{3\:dB}=\frac{1}{(R_1+R_2)C}\: rad/s$
gate2012ec
continuoustimesignals
digitalfilterdesigntechniques
asked
Mar 25, 2018
in
Continuoustime Signals
by
Milicevic3306
(
15.8k
points)
edited
Nov 18, 2020
by
soujanyareddy13

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