Question
Download Solution PDFMatch List I with List II:
List I (Coefficients of s2 + a1s + a2 = 0) |
List II (Nature of Roots) |
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(A) | a\(_1^2\) > 4a2 | (I) | Negative real and equal |
(B) | a\(_1^2\) = 4a2 | (II) | Conjugate Imaginary |
(C) | a\(_1^2\) < 4a2 | (III) | Negative Real and Unequal |
(D) |
a1 = 0 a2 ≠ 0 |
(IV) | Conjugate Complex (Real part negative) |
Choose the correct answer from the options given below:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation
The locus of the open loop poles of the root locus is represented as:
\(s=-\zeta\omega_n \pm j\omega _n\sqrt {1-\zeta^2} \)
The solution of s2+a1s+a2 = 0 is given by:
\(s={-a_1± j√{a_1^2-4(1)(a_2)}}\)
Nature of roots
Case 1: When roots are negative, real, and equal
The imaginary part must be zero
a\(_1^2\) - 4a2 = 0
a\(_1^2\) = 4a2
Case 2: When roots are conjugate imaginary
The imaginary part must exist and the real part should be zero.
a1 = 0 and a2 ≠ 0
Case 3: Negative Real and Unequal
a\(_1^2\) > 4a2
Case 4: Conjugate Complex
Both real and imaginary parts must exist
a\(_1^2\) < 4a2
Hence, the correct answer is option 3.
Last updated on Jun 6, 2025
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