Question

The root locus of a plant is given in the following figure. The root locus crosses imaginary axis at $\omega =4\sqrt{2}$ rad/s with gain k =384. It is observed that the point s = -1.5 +j1.5 lies in the root locus. The gain K at s = -1.5 + j1.5 is computed as

• A. 11.3
• B. 21.2
• C. 41.25
• D. 61.2

Answer 1

The correct option is C 41.25

Open-loop gain in pole-zero form
$k=k^{\prime }\left[\frac{{\pi }_{j=1}^n(s+p_j)}{{\pi }_{i=1}^n(s+Z_i)}\right]$
$=K^{\prime }s(s+4)(s+8)$
At $\omega =4\sqrt{2}$ rad/sec and k = 384
$384=k^{\prime }\left(4\sqrt{2}\right)(4+4\sqrt{2}j)(8+4\sqrt{2}j)$
$\Rightarrow k^{\prime }=1$
At, s = -1.5 + j1.5
$k=|(-1.5+j1.5)(2.5+j1.5)(6.5+j1.5)|$
$=41.25$