Vin = 6 sin wt 2 L1 mm .3183H R1 Vo 200ΚΩ f) = Draw the "Frequency Response Plot" in Log -Log Format". cut off of fc = 100,003- ¿B = 2010910 (0,707) = -3dB dB = 20log10 (0.995) = 0 dB Vo = 0.707 Vin dB = 20log10 (0.0995) = -20 dB = Vo Vin dB Gain 1 0-0.995 -3-0.707- -20 0.1 0.0995 -40-0.01 Low pass 1 -Filter Oolfc fc 10fc Freq (Hz)

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Determine type of filter and the cut-off frequency (Fc). = It is a low pass filter and fc = 100.003kHz
Determine the peak voltage at the cut-off frequency (Fc)  = Vp = 4.242V
Find GAIN (Vo/Vi) at  f = .1*Fc --> = 0.995 gain
Find GAIN (Vo/Vi) at  f = 10*Fc  --> = 0.0995 gain
Determine the frequency at which GAIN = .4  (i.e., Vo = .4*Vi)  --> f = 229.136kHz
Draw the  “Frequency Response Plot” in “Log-Log” format

Vin = 6 sin wt
2
L1
mm
.3183H
R1
Vo
200ΚΩ
Transcribed Image Text:Vin = 6 sin wt 2 L1 mm .3183H R1 Vo 200ΚΩ
f) = Draw the "Frequency Response Plot" in
Log -Log Format".
cut off of fc = 100,003-
¿B = 2010910 (0,707) = -3dB
dB = 20log10 (0.995) = 0 dB
Vo
= 0.707
Vin
dB = 20log10 (0.0995) = -20 dB =
Vo
Vin
dB Gain
1
0-0.995
-3-0.707-
-20 0.1
0.0995
-40-0.01
Low pass
1
-Filter
Oolfc
fc
10fc
Freq (Hz)
Transcribed Image Text:f) = Draw the "Frequency Response Plot" in Log -Log Format". cut off of fc = 100,003- ¿B = 2010910 (0,707) = -3dB dB = 20log10 (0.995) = 0 dB Vo = 0.707 Vin dB = 20log10 (0.0995) = -20 dB = Vo Vin dB Gain 1 0-0.995 -3-0.707- -20 0.1 0.0995 -40-0.01 Low pass 1 -Filter Oolfc fc 10fc Freq (Hz)
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