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MCQ Practice Question

For a polynomial to be Hurwitz, all of its roots must lie in which region, allowing simple j-omega-axis roots only under the standard special conditions?

AClosed left-half plane
BOpen right-half plane
COnly positive real axis
DOnly at the origin

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Correct Answer: A. Closed left-half plane

Explanation: A Hurwitz polynomial has no roots in the open right-half plane; permitted imaginary-axis roots are simple under the standard network-synthesis definition.

Reference: Standard network theory: Hurwitz polynomial

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