= Vin 12 sin wt 2 C1 HH .002μF + R1 Vo 7.96kQ f) Draw the Frequency Response" in format. Cutoff of fc = 9997.17 the /Vo "log=log" = 0.707 Vin dB = 2010910 (0.707) = -3dB dB = 20/09 1060.0995) → -20dBINE dB = 2010910 (0.995) → OdB log₂x=y= by = x dB 1 Vo Vie Gain 0.707- -3 -20- Oel .01 100° cut off ۱۷ 0.1fC fc lofe
= Vin 12 sin wt 2 C1 HH .002μF + R1 Vo 7.96kQ f) Draw the Frequency Response" in format. Cutoff of fc = 9997.17 the /Vo "log=log" = 0.707 Vin dB = 2010910 (0.707) = -3dB dB = 20/09 1060.0995) → -20dBINE dB = 2010910 (0.995) → OdB log₂x=y= by = x dB 1 Vo Vie Gain 0.707- -3 -20- Oel .01 100° cut off ۱۷ 0.1fC fc lofe
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
Determine type of filter and the cut-off frequency (Fc).
High Pass Filter & Fc = 9997.17Hz
Determine the peak voltage at the cut-off frequency (Fc)
GIven --> Gain = 0.5, --> Vp = 0.5 x 12 = 6V
Find GAIN (Vo/Vi) at f = .1*Fc
gain = 0.0995
Find GAIN (Vo/Vi) at f = 10*Fc
gain = 0.99995
Determine the frequency at which GAIN = .4 (i.e., Vo = .4*Vi)
f = 4363.12Hz
Draw the “Frequency Response Plot” in “Log-Log” format
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