Question

The first tow rows of Routh's table of a third-order characteristic eqation are
$\begin{array}{ccc}{s}^3& 3& 3\\ s^2& 4& 4\end{array}$
It can be inferred that the system has

• A. one real pole in the right-half of s-plane
• B. a pair of complex conjugate poles in the right-half of s-plane
• C. a pair of real poles symmetrically placed around s = 0
• D. a pair of complex conjugate poles on the imaginary axis of the s-plane

Answer 1

The correct option is D a pair of complex conjugate poles on the imaginary axis of the s-plane

Give first two rows of Routh's table of a third order characteristic equation are

This is the special case of Routh's array because first two row's having same elements so it will produce a row of zeros permaturely.

Since a row of zeros appear prematurely, we since a row equation using the form the auxiliary equation using the coefficient $s^2$ row.
$A\left(s\right)=4s^2+4=0$ ...(i)
$\Rightarrow \frac{dA\left(s\right)}{ds}=8s=0$
From which the coefficient 8 and 0 replace the zeros in the s'row of the original table.
Since these are no sign changes in the firest column of the entire Routh's tabulation, the equation (i) does not have any root in the right-half s-plane. Solving the auxiliary equation, we get the two roots at s = j and s = -j which are also the roots of original characteristic equation. Thus, the equation has two roots onthe $j\omega$-axis, and the system is marginally stable.