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GATE201241
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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
signalsandsystems
asked
Mar 25, 2018
in
Others
by
Milicevic3306
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points)
retagged
Apr 13
by
soujanyareddy13
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