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Automatic Control Systems - Engineering

Q1:

For the system of the given figure, the closed loop poles are at

A 0 and -4

B .root { background-image: url(/_files/images/chemical-engineering/common/root.gif); background-position: top left; background-repeat: no-repeat; padding-left: 10px; padding-top: 1px; }
-2 ± j6

C .root { background-image: url(/_files/images/chemical-engineering/common/root.gif); background-position: top left; background-repeat: no-repeat; padding-left: 10px; padding-top: 1px; }
2 ± j6

D .root { background-image: url(/_files/images/chemical-engineering/common/root.gif); background-position: top left; background-repeat: no-repeat; padding-left: 10px; padding-top: 1px; }
-3 ± j8

ANS:D - .root { background-image: url(/_files/images/chemical-engineering/common/root.gif); background-position: top left; background-repeat: no-repeat; padding-left: 10px; padding-top: 1px; }

-3 ± j8

.root { background-image: url(/_files/images/chemical-engineering/common/root.gif); background-position: top left; background-repeat: no-repeat; padding-left: 10px; padding-top: 1px; } For the characteristic equation s2 + 4s + 10 = 0, the roots are -2 ± j 6.