Poles Location Analysis using Routh-Hurwitz Criterion

The question requires us to find the number of poles located in the Right-Half Plane (RHP) and the Left-Half Plane (LHP) for a given system's closed-loop transfer function. Determining the location of these poles is crucial for assessing the system's stability. We will use the Routh-Hurwitz stability criterion, a standard method in control systems engineering.

Identifying the Characteristic Equation

The provided closed-loop transfer function is:

The poles of the system are the roots of the denominator polynomial, known as the characteristic equation. Setting the denominator to zero gives us:

The highest power of 's' is 5, indicating that the system has a total of 5 poles.

Constructing the Routh Array

The Routh-Hurwitz criterion involves constructing an array (the Routh array) from the coefficients of the characteristic equation. The general form is \( a_n s^n + a_{n-1} s^{n-1} + \dots + a_1 s + a_0 = 0 \). For our equation \( s^5 - s^4 + 4s^3 - 4s^2 + 3s - 2 = 0 \), the coefficients are: \( a_5 = 1, a_4 = -1, a_3 = 4, a_2 = -4, a_1 = 3, a_0 = -2 \).

Step 1: Form the first two rows of the Routh Array

The first row consists of the coefficients with odd indices, and the second row consists of the coefficients with even indices.

Step 2: Calculate the elements of the \(s^3\) row

The first element (\(b_1\)) is calculated using the determinant of the top-left 2x2 matrix formed by the \(s^5\) and \(s^4\) rows, divided by the element above it (\(a_4\)).

\( b_1 = - \frac{\det \begin{pmatrix} 1 & 4 \\ -1 & -4 \end{pmatrix}}{-1} = - \frac{(1)(-4) - (4)(-1)}{-1} = - \frac{-4 + 4}{-1} = - \frac{0}{-1} = 0 \)

The second element (\(b_2\)) is calculated similarly:

\( b_2 = - \frac{\det \begin{pmatrix} 1 & 3 \\ -1 & -2 \end{pmatrix}}{-1} = - \frac{(1)(-2) - (3)(-1)}{-1} = - \frac{-2 + 3}{-1} = - \frac{1}{-1} = 1 \)

Since \(b_1\) is zero, we have a special case. For the Routh criterion, when an element in the first column is zero, we replace it with a small positive number \(\epsilon\). This allows us to continue the array calculation and analyze the behavior as \(\epsilon\) approaches zero.

Step 3: Calculate the elements of the \(s^2\) row

Using the \(s^4\) and \(s^3\) rows:

\( c_1 = - \frac{\det \begin{pmatrix} -1 & -4 \\ \epsilon & 1 \end{pmatrix}}{\epsilon} = - \frac{(-1)(1) - (-4)(\epsilon)}{\epsilon} = - \frac{-1 + 4\epsilon}{\epsilon} = \frac{1 - 4\epsilon}{\epsilon} \)

\( c_2 = - \frac{\det \begin{pmatrix} -1 & -2 \\ \epsilon & 0 \end{pmatrix}}{\epsilon} = - \frac{(-1)(0) - (-2)(\epsilon)}{\epsilon} = - \frac{2\epsilon}{\epsilon} = -2 \)

The \(s^2\) row is:

Step 4: Calculate the elements of the \(s^1\) row

Using the \(s^3\) and \(s^2\) rows:

\( d_1 = - \frac{\det \begin{pmatrix} \epsilon & 1 \\ \frac{1 - 4\epsilon}{\epsilon} & -2 \end{pmatrix}}{\frac{1 - 4\epsilon}{\epsilon}} = - \frac{\epsilon(-2) - 1(\frac{1 - 4\epsilon}{\epsilon})}{\frac{1 - 4\epsilon}{\epsilon}} = - \frac{-2\epsilon - \frac{1}{\epsilon} + 4}{\frac{1 - 4\epsilon}{\epsilon}} \)

As \(\epsilon \to 0^+\), the term \(\frac{1}{\epsilon}\) dominates. \( d_1 \approx - \frac{-1/\epsilon}{1/\epsilon} = -(-1) = 1 \).

The \(s^1\) row is:

Step 5: Calculate the elements of the \(s^0\) row

Using the \(s^2\) and \(s^1\) rows:

\( e_1 = - \frac{\det \begin{pmatrix} \frac{1 - 4\epsilon}{\epsilon} & -2 \\ 1 & 0 \end{pmatrix}}{1} = - \frac{(\frac{1 - 4\epsilon}{\epsilon})(0) - (-2)(1)}{1} = - \frac{2}{1} = -2 \)

The \(s^0\) row is:

Analyzing the Routh Array for Stability

The completed Routh array is shown below:

Step 6: Count sign changes in the first column

We examine the signs of the elements in the first column: \(1, -1, \epsilon, \frac{1 - 4\epsilon}{\epsilon}, 1, -2\).

Assuming \(\epsilon\) is a small positive number (\(\epsilon \to 0^+\)):

• The first element is \(1\) (Positive).
• The second element is \(-1\) (Negative). The sign changes from \(+\) to \(-\). (1st sign change)
• The third element is \(\epsilon\) (Positive). The sign changes from \(-\) to \(+\). (2nd sign change)
• The fourth element \(\frac{1 - 4\epsilon}{\epsilon}\) is approximately \(\frac{1}{\epsilon}\), which is a large positive number. The sign remains \(+\). No sign change.
• The fifth element is \(1\) (Positive). No sign change.
• The sixth element is \(-2\) (Negative). The sign changes from \(+\) to \(-\). (3rd sign change)

There are exactly 3 sign changes in the first column of the Routh array.

Determining Poles in RHP and LHP

The Routh-Hurwitz stability criterion states that:

• The number of poles in the Right-Half Plane (RHP) is equal to the number of sign changes in the first column of the Routh array.
• The total number of poles is equal to the degree of the characteristic polynomial (which is 5 in this case).
• The number of poles in the Left-Half Plane (LHP) is the total number of poles minus the number of poles in the RHP.

Based on our analysis:

• Number of poles in RHP = 3 (due to 3 sign changes).
• Total number of poles = 5.
• Number of poles in LHP = Total poles - RHP poles = \( 5 - 3 = 2 \).

Therefore, the system has 3 poles in the RHP and 2 poles in the LHP.

Conclusion

The analysis shows that there are 3 poles in the RHP and 2 poles in the LHP. This corresponds to the third option provided.