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The characteristic polynomial of a linear system is given as s4 + 3s3 + 5s2 + 6s + K + 10 = 0. What should be the condition on K so that the system is stable?
- a)-10 < K < -4
- b)K > 10
- c)K > -4
- d)K > -10
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The characteristic polynomial of a linearsystem is given as s4+ 3s3+ 5...
Concept:
The characteristic equation for a given open-loop transfer function G(s) is
1 + G(s) H(s) = 0
To find the closed system stability by using RH criteria we require a characteristic equation. Whereas in remaining all stability techniques we require open-loop transfer function.
The nth order general form of CE is
RH table shown below:
Necessary condition: All the coefficients of the characteristic equation should be positive and real.
Sufficient Conditions for stability:
1. All the coefficients in the first column should have the same sign and no coefficient should be zero.
2. If any sign changes in the first column, the system is unstable.
And the number of sign changes = Number of poles in right of s-plane.
Calculation:
Characteristic equation: s4 + 3s3 + 5s2 + 6s + K + 10 = 0
By applying Routh tabulation method,
The system to become stable, the sign changes in the first column of the Routh table must be zero.
- 4 - K > 0 and K + 10 > 0
4 + K < 0 and K + 10 > 0
K < - 4 and K > - 10
⇒ - 10 < K < - 4
Most Upvoted Answer
The characteristic polynomial of a linearsystem is given as s4+ 3s3+ 5...
To determine the stability condition for the linear system, we need to look at the coefficients of the characteristic polynomial. The general form of the characteristic polynomial for an nth-order linear system is:
P(s) = s^n + a_(n-1)s^(n-1) + a_(n-2)s^(n-2) + ... + a_1s + a_0
In this case, the characteristic polynomial is given as:
P(s) = s^4 + 3s^3 + 5s^2 + 6s + K + 10 = 0
For the system to be stable, all the roots of the characteristic polynomial must have negative real parts (or be on the left-half of the complex plane). One way to determine the stability condition is by using the Routh-Hurwitz criterion.
The Routh-Hurwitz criterion states that for a polynomial of the form:
P(s) = s^n + a_(n-1)s^(n-1) + a_(n-2)s^(n-2) + ... + a_1s + a_0
The necessary and sufficient condition for stability is that all the coefficients in the first column of the Routh array have the same sign.
In this case, the Routh array would be:
| 1 5 K + 10 |
| 3 6 0 |
| 12 K + 10 |
| K + 10 |
| 0 |
For stability, all the elements in the first column of the Routh array should have the same sign. Therefore, the condition for stability is:
1 > 0
5 > 0
K + 10 > 0
From the third row, we have K + 10 > 0. Solving this inequality, we get:
K > -10
Therefore, the condition for stability is K > -10.
So, the correct option is:
a) K > -10
P(s) = s^n + a_(n-1)s^(n-1) + a_(n-2)s^(n-2) + ... + a_1s + a_0
In this case, the characteristic polynomial is given as:
P(s) = s^4 + 3s^3 + 5s^2 + 6s + K + 10 = 0
For the system to be stable, all the roots of the characteristic polynomial must have negative real parts (or be on the left-half of the complex plane). One way to determine the stability condition is by using the Routh-Hurwitz criterion.
The Routh-Hurwitz criterion states that for a polynomial of the form:
P(s) = s^n + a_(n-1)s^(n-1) + a_(n-2)s^(n-2) + ... + a_1s + a_0
The necessary and sufficient condition for stability is that all the coefficients in the first column of the Routh array have the same sign.
In this case, the Routh array would be:
| 1 5 K + 10 |
| 3 6 0 |
| 12 K + 10 |
| K + 10 |
| 0 |
For stability, all the elements in the first column of the Routh array should have the same sign. Therefore, the condition for stability is:
1 > 0
5 > 0
K + 10 > 0
From the third row, we have K + 10 > 0. Solving this inequality, we get:
K > -10
Therefore, the condition for stability is K > -10.
So, the correct option is:
a) K > -10
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The characteristic polynomial of a linearsystem is given as s4+ 3s3+ 5s2+ 6s + K +10 = 0. What should be the condition on Kso that the system is stable?a)-10 < K < -4b)K > 10c)K > -4d)K > -10Correct answer is option 'A'. 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 The characteristic polynomial of a linearsystem is given as s4+ 3s3+ 5s2+ 6s + K +10 = 0. What should be the condition on Kso that the system is stable?a)-10 < K < -4b)K > 10c)K > -4d)K > -10Correct answer is option 'A'. 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 The characteristic polynomial of a linearsystem is given as s4+ 3s3+ 5s2+ 6s + K +10 = 0. What should be the condition on Kso that the system is stable?a)-10 < K < -4b)K > 10c)K > -4d)K > -10Correct answer is option 'A'. Can you explain this answer?.
The characteristic polynomial of a linearsystem is given as s4+ 3s3+ 5s2+ 6s + K +10 = 0. What should be the condition on Kso that the system is stable?a)-10 < K < -4b)K > 10c)K > -4d)K > -10Correct answer is option 'A'. 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 The characteristic polynomial of a linearsystem is given as s4+ 3s3+ 5s2+ 6s + K +10 = 0. What should be the condition on Kso that the system is stable?a)-10 < K < -4b)K > 10c)K > -4d)K > -10Correct answer is option 'A'. 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 The characteristic polynomial of a linearsystem is given as s4+ 3s3+ 5s2+ 6s + K +10 = 0. What should be the condition on Kso that the system is stable?a)-10 < K < -4b)K > 10c)K > -4d)K > -10Correct answer is option 'A'. Can you explain this answer?.
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