The magnitude plot in the figure represents the transfer function of a preamplifier. Identify the correct H(s). Take C= 2282. H (dB) 40 20 5 Multiple Choice 50 OH(s)- 20 dB/decade OH(s)- 2282(+500) (8+50)(8+22820) 2282(a+50) (8+500)(8+22820) OH(s)- 22820(+50) (+500)(8+2282) OH(s)- 22820(+500) (8+50)(+2282) 500 20 dB/decade
The magnitude plot in the figure represents the transfer function of a preamplifier. Identify the correct H(s). Take C= 2282. H (dB) 40 20 5 Multiple Choice 50 OH(s)- 20 dB/decade OH(s)- 2282(+500) (8+50)(8+22820) 2282(a+50) (8+500)(8+22820) OH(s)- 22820(+50) (+500)(8+2282) OH(s)- 22820(+500) (8+50)(+2282) 500 20 dB/decade
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Question
![The magnitude plot in the figure represents the transfer function of a preamplifier. Identify the correct H(s). Take C= 2282.
H (dB)
40
20
5
Multiple Choice
о
H(s) =
H (s)-
H(s) =
H(s) =
50
20 dB/decade
2282(8+500)
(8+50)(+22820)
2282(8+50)
(8+500)(a +22820)
22820(8+50)
(8+500)(8+2282)
22820(8+500)
(8+50)(8+2282)
500
20 dB/decade](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7aead205-e276-4ec2-92c3-64b463e0a132%2F4122bb23-1dca-4eed-b12d-0527d36a1e25%2F38wv9k_processed.png&w=3840&q=75)
Transcribed Image Text:The magnitude plot in the figure represents the transfer function of a preamplifier. Identify the correct H(s). Take C= 2282.
H (dB)
40
20
5
Multiple Choice
о
H(s) =
H (s)-
H(s) =
H(s) =
50
20 dB/decade
2282(8+500)
(8+50)(+22820)
2282(8+50)
(8+500)(a +22820)
22820(8+50)
(8+500)(8+2282)
22820(8+500)
(8+50)(8+2282)
500
20 dB/decade
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Follow-up Question
![A parallel resonant circuit with quality factor 120 has a resonant frequency of 5.0 x 106 rad/s. Calculate the bandwidth and half-power
frequencies.
The bandwidth is
krad/s.
The lower half-power frequency is
The higher half-power frequency is
x
106 rad/s.
x
106 rad/s.](https://content.bartleby.com/qna-images/question/7aead205-e276-4ec2-92c3-64b463e0a132/f53cda4f-28a8-4a38-87d1-09d609a81f2a/llzqlql_thumbnail.png)
Transcribed Image Text:A parallel resonant circuit with quality factor 120 has a resonant frequency of 5.0 x 106 rad/s. Calculate the bandwidth and half-power
frequencies.
The bandwidth is
krad/s.
The lower half-power frequency is
The higher half-power frequency is
x
106 rad/s.
x
106 rad/s.
Solution
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