Direction:
| where T1 = | , T2 = | and T3 = | |||
| ω1 | ω2 | ω3 |
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The open-loop transfer function of the given Bode plot is—
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K(sT2 + 1) s(sT1 + 1)(sT3 + 1)
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K(sT2 + 1) (sT1 + 1)(sT3 + 1)
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K(sT1 + 1) (sT2 + 1)(sT3 + 1) -
K(sT1 + 1) s(sT2 + 1)(sT3 + 1)
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Correct Option: A
● The initial slope is – 20dB/d it means system is type 1.
● At ω1 there is – 20dB/d more means there is a pole i.e., 1/(1 + sT1)
where T1 = 1 ω1
● At ω2 there is increase in 20dB/d means a zero i.e., (1 + sT2),
where T2 = 1/ω2
● At ω3 there is – 20dB/d more means there is a pole i.e., 1/(1 + sT3)
where, T3 = 1/ω3
Now, the general form of the open-loop transfer function is
| G(s)H(s) = | |
| s(sT1 + 1)(sT3 + 1) |
