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Phase margin in degree of G(s) = 
Using asymptotic Bode plot is _____
Using asymptotic Bode plot is _____Correct answer is between '42,48'. Can you explain this answer?
Verified Answer
Phase margin in degree of G(s) =Using asymptotic Bode plot is _____Cor...
Phase margin = 180° + ∠GH(jω) at ω = ωgc
ωgc = Gain cross over frequency
ωgc: Gain crossover frequency: It is defined as the frequency at which the gain of the system is unity.
Now, see the magnitude expression (1) of the system, it looks very complicated to calculate ωgc. Here comes, the role of the Bode plot.
Points to Remember:
Before drawing the Bode plot, always convert the transfer function into its time constant form.
Pole-zero form:
Time-constant form:
Bode plot:
Poles are at s = -0.1, -1 and -10
Initially, since it is a type zero system, the slope of the Bode plot is 0 dB/decade.
Magnitude, M = 20 log10 k dB
M = 20 log10 10 = 20 dB
From Bode plot magnitude is 0 dB at ω = 1
0 dB = 20 log10 M
M = 1 at ω = 1
Hence ωgc = 1
At ω = ωgc = 1, we get:
= - 45° - 84.289 – 5.710 = -135°
Phase margin, PM = 180° + ∠GH(jω) at ω = ωgc
PM = 180° - 135°
PM = 45°
ωgc = Gain cross over frequency
ωgc: Gain crossover frequency: It is defined as the frequency at which the gain of the system is unity.
Now, see the magnitude expression (1) of the system, it looks very complicated to calculate ωgc. Here comes, the role of the Bode plot.
Points to Remember:
Before drawing the Bode plot, always convert the transfer function into its time constant form.
Pole-zero form:
Time-constant form:
Bode plot:
Poles are at s = -0.1, -1 and -10
Initially, since it is a type zero system, the slope of the Bode plot is 0 dB/decade.
Magnitude, M = 20 log10 k dB
M = 20 log10 10 = 20 dB
From Bode plot magnitude is 0 dB at ω = 1
0 dB = 20 log10 M
M = 1 at ω = 1
Hence ωgc = 1
At ω = ωgc = 1, we get:
= - 45° - 84.289 – 5.710 = -135°
Phase margin, PM = 180° + ∠GH(jω) at ω = ωgc
PM = 180° - 135°
PM = 45°
Most Upvoted Answer
Phase margin in degree of G(s) =Using asymptotic Bode plot is _____Cor...
Phase margin = 180° + ∠GH(jω) at ω = ωgc
ωgc = Gain cross over frequency
ωgc: Gain crossover frequency: It is defined as the frequency at which the gain of the system is unity.
Now, see the magnitude expression (1) of the system, it looks very complicated to calculate ωgc. Here comes, the role of the Bode plot.
Points to Remember:
Before drawing the Bode plot, always convert the transfer function into its time constant form.
Pole-zero form:
Time-constant form:
Bode plot:
Poles are at s = -0.1, -1 and -10
Initially, since it is a type zero system, the slope of the Bode plot is 0 dB/decade.
Magnitude, M = 20 log10 k dB
M = 20 log10 10 = 20 dB
From Bode plot magnitude is 0 dB at ω = 1
0 dB = 20 log10 M
M = 1 at ω = 1
Hence ωgc = 1
At ω = ωgc = 1, we get:
= - 45° - 84.289 – 5.710 = -135°
Phase margin, PM = 180° + ∠GH(jω) at ω = ωgc
PM = 180° - 135°
PM = 45°
ωgc = Gain cross over frequency
ωgc: Gain crossover frequency: It is defined as the frequency at which the gain of the system is unity.
Now, see the magnitude expression (1) of the system, it looks very complicated to calculate ωgc. Here comes, the role of the Bode plot.
Points to Remember:
Before drawing the Bode plot, always convert the transfer function into its time constant form.
Pole-zero form:
Time-constant form:
Bode plot:
Poles are at s = -0.1, -1 and -10
Initially, since it is a type zero system, the slope of the Bode plot is 0 dB/decade.
Magnitude, M = 20 log10 k dB
M = 20 log10 10 = 20 dB
From Bode plot magnitude is 0 dB at ω = 1
0 dB = 20 log10 M
M = 1 at ω = 1
Hence ωgc = 1
At ω = ωgc = 1, we get:
= - 45° - 84.289 – 5.710 = -135°
Phase margin, PM = 180° + ∠GH(jω) at ω = ωgc
PM = 180° - 135°
PM = 45°
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Phase margin in degree of G(s) =Using asymptotic Bode plot is _____Correct answer is between '42,48'. Can you explain this answer?
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Phase margin in degree of G(s) =Using asymptotic Bode plot is _____Correct answer is between '42,48'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Phase margin in degree of G(s) =Using asymptotic Bode plot is _____Correct answer is between '42,48'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Phase margin in degree of G(s) =Using asymptotic Bode plot is _____Correct answer is between '42,48'. Can you explain this answer?.
Phase margin in degree of G(s) =Using asymptotic Bode plot is _____Correct answer is between '42,48'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Phase margin in degree of G(s) =Using asymptotic Bode plot is _____Correct answer is between '42,48'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Phase margin in degree of G(s) =Using asymptotic Bode plot is _____Correct answer is between '42,48'. Can you explain this answer?.
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