Stability and Routh Hurwitz Criterion Questions and Answers

Answer: B

Apply the Routh array to \(s^5 + 6s^4 + 3s^3 + 2s^2 + s + 1\).

\[ \begin{array}{c|ccc} s^5 & 1 & 3 & 1 \\ s^4 & 6 & 2 & 1 \\ s^3 & 2.667 & 0.833 & \\ s^2 & 0.125 & 1 & \\ s^1 & -20.5 & & \\ s^0 & 1 & & \end{array} \]

The first column is \(1, 6, 2.667, 0.125, -20.5, 1\). The sign changes twice: from \(0.125\) to \(-20.5\) and from \(-20.5\) to \(1\). The number of sign changes equals the number of roots in the right half plane, so there are 2 roots with a positive real part, and the system is unstable. The answer is option B.

Answer: B

Apply the Routh array to \(s^4 + 2s^3 + 3s^2 + 4s + 5 = 0\).

\[ \begin{array}{c|ccc} s^4 & 1 & 3 & 5 \\ s^3 & 2 & 4 & \\ s^2 & 1 & 5 & \\ s^1 & -6 & & \\ s^0 & 5 & & \end{array} \]

The \(s^2\) row is \((2 \times 3 - 4)/2 = 1\) and \((2 \times 5)/2 = 5\), and the \(s^1\) entry is \((1 \times 4 - 2 \times 5)/1 = -6\). The first column is \(1, 2, 1, -6, 5\), which has two sign changes, so two roots are in the right half plane and the system is unstable. The options stable, partially stable and stable at origin all need no sign change in the first column. The answer is Unstable, option B.

Answer: A

The system is stable for \(0 \lt k \lt 390\).

Build the Routh array for \(s^{3} + 13s^{2} + 30s + k\):

\[ \begin{aligned} s^{3}&: \quad 1 \quad 30 \\ s^{2}&: \quad 13 \quad k \\ s^{1}&: \quad \frac{13 \times 30 - k}{13} \\ s^{0}&: \quad k \end{aligned} \]

Every entry of the first column must be positive. The \(s^{0}\) row needs \(k \gt 0\), and the \(s^{1}\) row needs \(390 - k \gt 0\), so \(0 \lt k \lt 390\). At \(k = 390\) the \(s^{1}\) row vanishes and the system oscillates.

The other ranges are built from the coefficients themselves (13 and 30) or drop a factor of ten (39); none comes from the stability condition.

Answer: D

Both (A) and (R) are true and (R) is the correct explanation of (A).

The Assertion is true: by the Hurwitz condition, a necessary condition for stability is that every coefficient of the characteristic equation is present and of the same sign, taken as positive.

The Reason is true and explains it: if a coefficient is zero or negative while the others are positive, the equation must have at least one root on the imaginary axis or in the right half of the s-plane, so the system is not stable. That is exactly why all coefficients must be positive with none missing. The condition is necessary but not sufficient; for third and higher order a Routh test is still needed.

Answer: A

The correct matching is a-ii, b-i, c-iii, d-iv.

• (a) In the Routh array of s⁵ + s⁴ + 2s³ + 2s² + 3s + 15, the s³ row starts with zero (2 - 2 = 0); replacing it by a small ε gives two sign changes in the first column, so two roots lie in the right half plane: unstable (ii).
• (b) The Routh array of the sixth-order polynomial produces a row of zeros whose auxiliary polynomial has roots on the imaginary axis, with no sign change in the first column: marginally stable (i).
• (c) With GH = 10/s(s + 2)(s + 5), the characteristic equation is s³ + 7s² + 10s + 10 = 0; since \(7 imes 10 = 70 \gt 10\), all roots are in the left half plane: stable (iii).
• (d) G(s) = (s + 5)/(s² + 4s + 9) has phase 0° at zero frequency and approaches -90° at high frequency, since it has one more pole than zero (iv).

The other codes call the polynomial with a right-half-plane pair marginally stable or the third-order loop unstable, which the Routh test rules out.

Answer: A

The Routh-Hurwitz criterion is used to determine the stability of a control system.

From the coefficients of the characteristic equation of a linear time-invariant system, the Routh array is built. The number of sign changes in its first column equals the number of roots in the right half of the s-plane, so the system is stable only if there are no sign changes. This tells whether the system is stable without solving for the roots.

It is not a tool for analysing power distribution in grids, optimising communication bandwidth or designing filter responses, although it can check that a designed filter or controller is stable.