Solve for the voltage output using the transfer function below. please label your step-by-step on how your solve for the circuit and please dont reupload or copy from other bartleby post because I couldn't understand them. The answer from the picture is from my other bartleby post. I dont understand how the circled equation arrived. please explain that equation or make your simple computation for the voltage out where I could understand it. voltageOutR1v 330.46mV14voltageln80280Ω1VC11kHz|0 2µFO 2µF0 0°Vout (s)Vin(s)39,062,500G(s) =s2+18,750s+39,062,500 (2) The transfer function of the above circuit is written as:H (s) =v,(s)V,()PRR, CC,+s(RC+RC,+R}Cj)+lH (jw)VA(ja)%3DVII-(m* RR, CC;]²+/lRC+RC; +R;C}fwhere2nf = 2n × 1 × 103= 2000n rad/ sec|V: (jo)| = 1 V (given)Putting the values:IV, ja)[I-(2000z)²x80x80x2x10*x2x101²+/[2000= (80×2x10“+80x2x10+80x2x10)]?|V, (ja)| =|V. (ja)| = 0. 3305 V = 330. 5 mV0.0106+3.015So, the final output voltage obtained from the calculation is 330.5 mV.

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Answered: Solve for the voltage output using the…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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Solve for the voltage output using the transfer function below.

please label your step-by-step on how your solve for the circuit and please dont reupload or copy from other bartleby post because I couldn't understand them.

The answer from the picture is from my other bartleby post. I dont understand how the circled equation arrived. please explain that equation or make your simple computation for the voltage out where I could understand it.

voltageOut
R1
v 330.46mV
1
4
voltageln
802
80Ω
1V
C1
1kHz
|0 2µF
O 2µF
0 0°
Vout (s)
Vin(s)
39,062,500
G(s) =
s2+18,750s+39,062,500

(2) The transfer function of the above circuit is written as:
H (s) =
v,(s)
V,()
PRR, CC,+s(RC+RC,+R}Cj)+l
H (jw)
VA(ja)
%3D
VII-(m* RR, CC;]²+/lRC+RC; +R;C}f
where
2nf = 2n × 1 × 103
= 2000n rad/ sec
|V: (jo)| = 1 V (given)
Putting the values:
IV, ja)
[I-(2000z)²x80x80x2x10*x2x101²+/[2000= (80×2x10“+80x2x10+80x2x10)]?
|V, (ja)| =
|V. (ja)| = 0. 3305 V = 330. 5 mV
0.0106+3.015
So, the final output voltage obtained from the calculation is 330.5 mV.

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