0 votes 0 votes 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 $f_{3\:dB}=\frac{1}{(R_1+R_2)C}\: rad/s$ Continuous-time Signals gate2012-ec continuous-time-signals digital-filter-design-techniques + – Milicevic3306 asked Mar 25, 2018 • edited Nov 18, 2020 by soujanyareddy13 Milicevic3306 16.0k points 127 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.