How to find the value of gain K at which the root locus intersects with the imaginary axis

Question

Take a look at this closed loop system $$ T(s) = \frac{-K(s+1)^2}{(1-K)s^2 + 2(1-K)s + (2-K)} $$ The Routh Table is $$ \begin{bmatrix} s^2 & (1-K) & (2-K) \\ s^1 & 2(1-K) & 0 \\ s^0 &(2-K) & 0 \end{bmatrix} $$ Now we search for a complete row of zeros that yields the possibility for imaginary axis roots. From the table, there are two values of $K$ namely 1,2 and both are positive. The only thing I can conclude from this table is the system is unstable $1<K<2$ but I can't determine the value of K at which the root locus intersects with the imaginary axis. Any suggestion how one can use Routh table in this case? Is this a case where Rough table fails?

Answer 1

No need for Routh table because from

$$ (1-K)s^2+2(1-K)s+K-2=0 $$

we have

$$ s = \frac{-(K-1)\pm\sqrt{K-1}}{K-1} $$

or calling $\zeta = K-1$ and assuming that $K > 0$

$$ s = \frac{-\zeta\pm\sqrt{\zeta}}{\zeta} $$

and for $-1 < \zeta < 0$ both complex roots are inside the negative complex semiplane being stable. For $0\lt \zeta \lt 1$ the denominator has one root is inside the positive complex semiplane and the closed loop is unstable. Finally, for $1 < \zeta < \infty$ the system is stable.