Answer:
The answer is "Option B".
Explanation:
Given equation:
[tex]G(s) =\frac{K(s + 1)}{2s^2 + (K-1)s + (K-1)}\\\\[/tex]
if
[tex]\to 2s^2 + (K-1)s + (K-1)=0[/tex]
Calculating by the Routh's Hurwitz table:
[tex]\to s^2 \ \ \ \ \ 2 \ \ \ \ \ \ K-1 \\\\\to s^2 \ \ \ \ \ K-1 \ \ \ \ \ \ \\\\\to s^0 \ \ ( \frac{(K-1)(K-1)(-2) (0)}{K-1} \\\\ \ \ \ \ = (K-1) )[/tex]
Form the above table:
[tex]\to K-1 > 0 \\\\ \to K > 1[/tex]
In the above, the value of k is greater than 1.
