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1. In the circuit below find the expression of the frequency response H and plot the corresponding Bode plot for R1=12, R;=2N and C=1µF. Is it a high pass or a low pass filter? What is the gain in the pass-band? Find the -3dB cut-off frequency using the graph and by computing it. R1 + + Vout V;

1. In the circuit below find the expression of the frequency response H and plot the corresponding Bode plot for R1=12, R;=2N and C=1µF. Is it a high pass or a low pass filter? What is the gain in the pass-band? Find the -3dB cut-off frequency using the graph and by computing it. R1 + + Vout V;

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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For the solution provided, i do not understand, when finding w = infinity,

why do we take into account the imaginary numbers only (R2 /(R1+R2))? 

Pls provide the derivation of the formula as shown in the solution

T8.1 : cutoff frequency 1. In the circuit below find the expression of the frequency response H and plot the corresponding Bode plot for R1=1N, R2=2N and C=lµF. Is it a high pass or a low pass filter? What is the gain in the pass-band? Find the -3dB cut-off frequency using the graph and by computing it. + R1 + Vout R2 Vị 53kHz gain 2/3 (or -3.5dB) high-pass Answer:
Transcribed Image Text:T8.1 : cutoff frequency 1. In the circuit below find the expression of the frequency response H and plot the corresponding Bode plot for R1=1N, R2=2N and C=lµF. Is it a high pass or a low pass filter? What is the gain in the pass-band? Find the -3dB cut-off frequency using the graph and by computing it. + R1 + Vout R2 Vị 53kHz gain 2/3 (or -3.5dB) high-pass Answer:
Voltage divider: R2 V. = V;- R1 + R2 + 1 juC V.(jw) V:(jw) jwR2C H(jw) := 1+ jw(R1 + R2)C Hlow jwR2C jwR2C H(jw) = R2 2 1+ jw(R1 + R2)C Hhigh R1+ R2 2 → 20 log = -3.5dB 3 Cutoff frequency 1 1 fo = 2т (Ri + R2)С wo = = 53kHz (R1 + R2)C 27 321µF 20dB/decade -3.5dB 90° -6.5dB 102 103 104 105 106 102 103 104 105 106 fo = 53kHz fo: = 53kHz
Transcribed Image Text:Voltage divider: R2 V. = V;- R1 + R2 + 1 juC V.(jw) V:(jw) jwR2C H(jw) := 1+ jw(R1 + R2)C Hlow jwR2C jwR2C H(jw) = R2 2 1+ jw(R1 + R2)C Hhigh R1+ R2 2 → 20 log = -3.5dB 3 Cutoff frequency 1 1 fo = 2т (Ri + R2)С wo = = 53kHz (R1 + R2)C 27 321µF 20dB/decade -3.5dB 90° -6.5dB 102 103 104 105 106 102 103 104 105 106 fo = 53kHz fo: = 53kHz
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