The characteristic equation of a system is s +5s³ + 10s² + 10s + 80 = 0. Complete the R-H table of thesystem. Put a zero on blank space. Do not leave any space blank.Routh-Hurwitz Table:83,s²S7Which of the following statements is TRUE?O The system is stable since there are no sign changes along the first column.O The system is marginally stable since one element along first column is zero.O The system is unstable because there is a change in sign along the first column, from row 2 (from the top) to row 3.O The system is unstable because there is a change in sign along the first column, from row 3 (from the top) to row 4.O The system is unstable because there are changes in sign along the first column, from row 2 (from the top) to row 3, and from row 3 to row 4O The system is unstable because there are changes in sign along the first column, from row 3 (from the top) to row 4, and from row 4 to row 5.

Characteristic Equation :-s4 + 10s2 + 10s + 80=0

The characteristic equation of a system is sª + 5s³ + 10s² + 10s +80 = 0. Complete the R-H table of the system. Put a zero on blank space. Do not leave any space blank. Routh-Hurwitz Table: 84/ حدى 82/ S7 so Which of the following statements is TRUE?

The characteristic equation of a system is sª + 5s³ + 10s² + 10s +80 = 0. Complete the R-H table of the system. Put a zero on blank space. Do not leave any space blank. Routh-Hurwitz Table: 84/ حدى 82/ S7 so Which of the following statements is TRUE?

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.39P

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Question

The characteristic equation of a system is s +5s³ + 10s² + 10s + 80 = 0. Complete the R-H table of the
system. Put a zero on blank space. Do not leave any space blank.
Routh-Hurwitz Table:
83,
s²
S7
Which of the following statements is TRUE?
O The system is stable since there are no sign changes along the first column.
O The system is marginally stable since one element along first column is zero.
O The system is unstable because there is a change in sign along the first column, from row 2 (from the top) to row 3.
O The system is unstable because there is a change in sign along the first column, from row 3 (from the top) to row 4.
O The system is unstable because there are changes in sign along the first column, from row 2 (from the top) to row 3, and from row 3 to row 4
O The system is unstable because there are changes in sign along the first column, from row 3 (from the top) to row 4, and from row 4 to row 5.

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