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For the Bode plat (magnitude, min-phase) shown below, - obtain I (JW)" 40 dB 200rad/sec 40 dB= 20 2= 20 log k Z dB = log k K=100 T(JW)=- slope = -60% -decade 2000 (1+− ) 3- -20 dB 100 (1 + 5000) " (1 + 200 per power ads -20 dB frakts out so must have numerator of porter of 2 -@ this frequency

For the Bode plat (magnitude, min-phase) shown below, - obtain I (JW)" 40 dB 200rad/sec 40 dB= 20 2= 20 log k Z dB = log k K=100 T(JW)=- slope = -60% -decade 2000 (1+− ) 3- -20 dB 100 (1 + 5000) " (1 + 200 per power ads -20 dB frakts out so must have numerator of porter of 2 -@ this frequency

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 Bode plat (magnitude, min phase) shown below,- obtain I (JW). 40 dB 200rad/sec 40 dB = 20 log k 2 dB = log k K=100 T(JW)== slope = -60t decade 2000 ·(1+-) ³ +00 (1 + $5000) " * (1 + 200 S rad/ -20 dB -@ this frequency levels out so must have numerator of porter of 2 -20 dB per power
Transcribed Image Text:For the Bode plat (magnitude, min phase) shown below,- obtain I (JW). 40 dB 200rad/sec 40 dB = 20 log k 2 dB = log k K=100 T(JW)== slope = -60t decade 2000 ·(1+-) ³ +00 (1 + $5000) " * (1 + 200 S rad/ -20 dB -@ this frequency levels out so must have numerator of porter of 2 -20 dB per power
Expert Solution
Step 1: Summarize the details given:

Given Data:

The Bode Plot of a system is given by,

Electrical Engineering homework question answer, step 1, image 1

To Find:

The transfer function.

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