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= 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
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
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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

= Vin 12 sin wt 2 C1 HH .002μF + R1 Vo 7.96kQ
Transcribed Image Text:= 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
Transcribed Image Text: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
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