The number of roots of s 3+ 5s 2+ 7s + 3 = 0 in the left half of the s-plane is
three
To determine the number of roots of a polynomial in the left half of the s-plane, we typically use the Routh-Hurwitz stability criterion. This method helps us understand the stability of a system represented by its characteristic equation.
The given polynomial equation is:
\( s^3 + 5s^2 + 7s + 3 = 0 \)
This is a third-order polynomial, which means it has a total of three roots. Our goal is to find how many of these roots lie in the left half of the s-plane.
The Routh-Hurwitz criterion involves constructing a Routh array from the coefficients of the polynomial. The first column of this array tells us about the location of the roots.
The general form of a third-order polynomial is \( a_3s^3 + a_2s^2 + a_1s + a_0 = 0 \).
Comparing this with our given equation \( s^3 + 5s^2 + 7s + 3 = 0 \), we have the coefficients:
Now, let's construct the Routh array:
| Row | Coefficient 1 | Coefficient 2 |
|---|---|---|
| \( s^3 \) | \( a_3 = 1 \) | \( a_1 = 7 \) |
| \( s^2 \) | \( a_2 = 5 \) | \( a_0 = 3 \) |
| \( s^1 \) | \( b_1 \) | \( 0 \) |
| \( s^0 \) | \( c_1 \) | \( 0 \) |
Calculate the elements for the \( s^1 \) row:
\( b_1 = \frac{(a_2 \times a_1) - (a_3 \times a_0)}{a_2} = \frac{(5 \times 7) - (1 \times 3)}{5} = \frac{35 - 3}{5} = \frac{32}{5} \)
Calculate the elements for the \( s^0 \) row:
\( c_1 = \frac{(b_1 \times a_0) - (a_2 \times 0)}{b_1} = \frac{(\frac{32}{5} \times 3) - (5 \times 0)}{\frac{32}{5}} = \frac{\frac{96}{5}}{\frac{32}{5}} = 3 \)
So, the complete Routh array is:
| Row | Coefficient 1 | Coefficient 2 |
|---|---|---|
| \( s^3 \) | \( 1 \) | \( 7 \) |
| \( s^2 \) | \( 5 \) | \( 3 \) |
| \( s^1 \) | \( \frac{32}{5} \) | \( 0 \) |
| \( s^0 \) | \( 3 \) | \( 0 \) |
Now, we examine the signs of the elements in the first column of the Routh array. The first column elements are:
All the elements in the first column are positive. There are no sign changes in the first column.
According to the Routh-Hurwitz criterion:
Furthermore, because there are no zero rows or zero elements in the first column that would indicate roots on the imaginary axis (jω-axis), we can conclude there are no roots on the jω-axis.
The total number of roots for a third-order polynomial is 3. Since there are no roots in the right half of the s-plane and no roots on the imaginary axis, all three roots must lie in the left half of the s-plane.
Based on the Routh-Hurwitz analysis, the polynomial \( s^3 + 5s^2 + 7s + 3 = 0 \) has all its roots in the left half of the s-plane. This implies that a system described by this characteristic equation would be stable.
Therefore, the number of roots in the left half of the s-plane is three.
Match List I with List II:
List I (Coefficients of s 2+ a 1s + a 2= 0) | List II (Nature of Roots) | ||
| (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) | a 1= 0 a 2≠ 0 | (IV) | Conjugate Complex (Real part negative) |
Choose the correct answer from the options given below:
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.