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The algebraic equation F(s) = s 5 – 3s 4 + 5s 3 – 7s 2 + 4s + 20 is given. F(s) = 0 has
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- a single complex root with the remaining roots being real
- one positive real root and four complex roots, all with positive real parts
- one negative real root, two imaginary roots, and two roots with positive real parts
- one positive real root, two imaginary roots, and two roots with negative real parts
Correct Option: C
F(s) = s 5 – 3s4 + 5s 32 – 7s 2 + 4s + 20
We can solve it by making Routh Hurwitz array.
| s 5 | 1 | 5 | 4 |
| s 4 | – 3 | – 7 | 20 |
| s 3 | 8/3 | 20/3 | 0 |
| s 2 | 5 | 20 | 0 |
| s 1 | 0 | 0 | 0 |
| s 0 | 20 | 0 | 0 |
We can replace 1st element of s 1 by 10.
If we observe 1st column, sign is changing two times, so we have two poles on right half side of imaginary axis and 5s2 + 20 = 0.
So, s = ± 2 j and 1 pole on left side of imaginary axis.
