Match List I with List II: List I (Coefficients of s 2+ a 1s + a 2= 0) List II (Nature of Roots) a 1= 0 a 2≠ 0 Choose the correct answer from the options given below:(A) a \(_1^2\) > 4a 2 (I) Negative real and equal (B) a \(_1^2\) = 4a 2 (II) Conjugate Imaginary (C) a \(_1^2\) < 4a 2 (III) Negative Real and Unequal (D) (IV) Conjugate Complex (Real part negative)
(A) - (III), (B) - (I), (C) - (IV), (D) - (II)
The question asks us to match different conditions on the coefficients \(a_1\) and \(a_2\) of a quadratic equation \(s^2 + a_1s + a_2 = 0\) with the nature of its roots. This is a fundamental concept in algebra, especially relevant in areas like control systems where the roots of characteristic equations determine system stability.
A general quadratic equation is given by \(as^2 + bs + c = 0\). The roots of this equation are given by the quadratic formula:
\(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
The nature of the roots is determined by the value of the discriminant, \(\Delta = b^2 - 4ac\).
In the given equation \(s^2 + a_1s + a_2 = 0\), we have \(a=1\), \(b=a_1\), and \(c=a_2\). So, the discriminant is \(\Delta = a_1^2 - 4(1)(a_2) = a_1^2 - 4a_2\).
Let's analyze each condition provided in List I based on the discriminant and the quadratic formula:
This condition means \(a_1^2 - 4a_2 > 0\). The discriminant \(\Delta > 0\).
When the discriminant is positive, the quadratic equation has two distinct real roots.
The roots are \(s = \frac{-a_1 \pm \sqrt{a_1^2 - 4a_2}}{2}\).
For the roots to be "Negative Real and Unequal" as mentioned in List II option (III), they must be real and distinct (which \(\Delta > 0\) guarantees) and both must be negative.
Assuming \(a_1 > 0\) and \(a_2 > 0\), the condition \(a_1^2 > 4a_2\) leads to two distinct negative real roots. This matches List II option (III).
This condition means \(a_1^2 - 4a_2 = 0\). The discriminant \(\Delta = 0\).
When the discriminant is zero, the quadratic equation has exactly one real root (which is a repeated root).
The root is \(s = \frac{-a_1 \pm \sqrt{0}}{2} = \frac{-a_1}{2}\).
For the root to be "Negative real and equal" as mentioned in List II option (I), it must be real and negative. Since \(\Delta = 0\), the roots are equal and real. For the root \(\frac{-a_1}{2}\) to be negative, we need \(-a_1 < 0\), which means \(a_1 > 0\). If \(a_1 > 0\), then \(a_1^2 > 0\), and since \(a_1^2 = 4a_2\), it implies \(4a_2 > 0\), so \(a_2 > 0\).
Assuming \(a_1 > 0\), the condition \(a_1^2 = 4a_2\) leads to a negative real and equal root. This matches List II option (I).
This condition means \(a_1^2 - 4a_2 < 0\). The discriminant \(\Delta < 0\).
When the discriminant is negative, the quadratic equation has two distinct complex conjugate roots.
The roots are \(s = \frac{-a_1 \pm \sqrt{a_1^2 - 4a_2}}{2} = \frac{-a_1 \pm \sqrt{-(4a_2 - a_1^2)}}{2} = \frac{-a_1 \pm i\sqrt{4a_2 - a_1^2}}{2}\).
These roots are in the form \(\alpha \pm i\beta\), where the real part is \(\alpha = -\frac{a_1}{2}\) and the imaginary part is \(\beta = \frac{\sqrt{4a_2 - a_1^2}}{2}\).
List II option (IV) mentions "Conjugate Complex (Real part negative)". The roots are indeed complex conjugates. For the real part \(\left(-\frac{a_1}{2}\right)\) to be negative, we need \(-a_1 < 0\), which implies \(a_1 > 0\). Also, since \(a_1^2 < 4a_2\) and \(a_1^2 \ge 0\), it must be that \(4a_2 > a_1^2 \ge 0\), which implies \(a_2 > 0\).
Assuming \(a_1 > 0\), the condition \(a_1^2 < 4a_2\) leads to conjugate complex roots with a negative real part. This matches List II option (IV).
Substitute \(a_1 = 0\) into the equation \(s^2 + a_1s + a_2 = 0\). This gives \(s^2 + 0 \cdot s + a_2 = 0\), which simplifies to \(s^2 + a_2 = 0\).
So, \(s^2 = -a_2\).
Given the options in List II, it is implied that for condition (D), the case where \(a_2 > 0\) leading to conjugate imaginary roots is intended. This matches List II option (II).
Based on the analysis above, the matching is as follows:
| List I (Coefficients Condition) | List II (Nature of Roots) | Reason |
|---|---|---|
| (A) \(a_1^2 > 4a_2\) | (III) Negative Real and Unequal | \(\Delta > 0\). Roots are real and distinct. If \(a_1 > 0, a_2 > 0\), roots are negative. |
| (B) \(a_1^2 = 4a_2\) | (I) Negative real and equal | \(\Delta = 0\). Roots are real and equal. If \(a_1 > 0\), root is negative. |
| (C) \(a_1^2 < 4a_2\) | (IV) Conjugate Complex (Real part negative) | \(\Delta < 0\). Roots are complex conjugates. If \(a_1 > 0\), real part is negative. |
| (D) \(a_1 = 0, a_2 \ne 0\) | (II) Conjugate Imaginary | \(s^2 + a_2 = 0\). If \(a_2 > 0\), roots are \(\pm i\sqrt{a_2}\) (pure imaginary). |
This matching corresponds to: (A) - (III), (B) - (I), (C) - (IV), (D) - (II).
| Discriminant \(\Delta = b^2 - 4ac\) | Nature of Roots |
|---|---|
| \(\Delta > 0\) | Two distinct real roots |
| \(\Delta = 0\) | One real root (repeated) |
| \(\Delta < 0\) | Two distinct complex conjugate roots |
In many engineering and physics applications, particularly in the analysis of linear time-invariant (LTI) systems (like electrical circuits, mechanical systems, control systems), the behavior and stability of the system are determined by the roots of a characteristic equation, which is often a polynomial equation like the quadratic one discussed here. The roots are also called poles of the system.
The conditions on \(a_1\) and \(a_2\) in the question, especially the implied conditions \(a_1 > 0\) and \(a_2 > 0\) leading to negative real parts or pure imaginary roots, are relevant to stability analysis for second-order systems represented by the characteristic equation \(s^2 + a_1s + a_2 = 0\).
For a standard second-order system \(s^2 + 2\zeta\omega_n s + \omega_n^2 = 0\), where \(\zeta\) is the damping ratio and \(\omega_n\) is the natural frequency, we have \(a_1 = 2\zeta\omega_n\) and \(a_2 = \omega_n^2\).
A closed-loop control system has a characteristic equation given by s 3 + 2.4s 2 + 1.8s + 0.5 = 0. Find out the value of a, b, c, and d using the Routh Hurwitz criterion.
s 3 | 1 | 1.8 |
s 2 | 2.4 | 0.5 |
s 1 | a | c |
s 0 | b | d |
Determine the stability of system:
S 3+ S 2+ S + 4
Which of the following is NOT the advantage of Routh-Hurwitz criterion of control systems?
The characteristic equation of given system is 6s + K = 0. Determined the range of K for which the system to be stable.
The number of roots of s 3+ 5s 2+ 7s + 3 = 0 in the left half of the s-plane is