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The Bode plot of a transfer function G (s) is shown in the figure below.
The gain (20 log| G(s)|) is 32 dB and – 8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all ω. Then G(s) is
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39.8 s -
39.8 s2 -
32 s -
32 s2
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Correct Option: B
⇒ 20 log k = 32
k = 101.6 = 39.8
⇒ As slope is – 40db/decade so two pols at ons’m
| so , T(s) = | ||
| 52 |
