Compute the transfer function H(s) for the asymptotic magnitude Bode di- gram sketched below. All the zeros and the poles should be considered with a negative real part. 130 120 €110 Magnitude (dB) 100 90 80 102 Page 3 101 10⁰ Bode Diagram 10¹ 10² 10³ 104 Angular Frequency (rad/s) 105

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Compute the transfer function H(s) for the asymptotic magnitude Bode di
gram sketched below. All the zeros and the poles should be considered with a negative
real part.
Magnitude (dB)
130
120
110
100
90
80
10¹2
10-1
Page:
Bode Diagram
10⁰ 10⁰¹ 10² 10³
Angular Frequency (rad/s)
104
105
Transcribed Image Text:Compute the transfer function H(s) for the asymptotic magnitude Bode di gram sketched below. All the zeros and the poles should be considered with a negative real part. Magnitude (dB) 130 120 110 100 90 80 10¹2 10-1 Page: Bode Diagram 10⁰ 10⁰¹ 10² 10³ Angular Frequency (rad/s) 104 105
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for the follow up answer i have a couple of questions, first what is the general formula to find k? i dont understand where the values came from to find k i also dont understand what a pole was added and what a zero was added mean and so how the wc1 wc2 and wc3 values were calculated can you please elaborate on those points? PLEASE DO NOT RE-SOLVE BUT CHECK THE SOLUTION AND EXPLAIN BASED ON THE GIVEN SOLUTION. thank you

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Follow-up Question

both this and the follow up are incorrect as the answer should be H(s)=10 ((s+10)(s+1000))/(s+1) = 10^5 ((s/10+1)(s/1000+1)) / (s+1) please resolve in GREAT detail and only put in your answer if it is IDENTICAL TO THE ONE I GAVE so please re-solve in detail and ofcourse i'll give thumbs up if answer is correct. thank you.

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Follow-up Question

i didnt understand how the slopes were calculated and how the transfer function was initially written/concluded can you please solve it in a more detailed manner and the final answer given to me is H(s)=10 ((s+10)(s+1000))/(s+1) = 10^5 ((s/10+1)(s/1000+1)) / (s+1)     are they identical? and please resolve in a more detailed manner as i didnt understand the stuff i stated above. Thank you!

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