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Automatic Control Systems

Q1:
For the system of the given figure the closed loop poles are located at

A s = 0 and s = -2

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; }
s = 0, s = -l ± j3

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; }
s = -1 ± j3

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; }
s = -2 and s = -1 ± j3

ANS: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; }

s = -1 ± j3

Transfer function = . .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; } The roots of equation s2 + 2s + 4 = 0 are at -1 ± 3.



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