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Question

# The asymptotic log-magnitude curve for open loop transfer function is sketched below, Open loop transfer function is

A
T(s)=100(s+8)(s+4)s2(s+1.268)
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B
T(s)=16(s+1.268)(s+4)s2(s+8)
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C
T(s)=10(s+1.268)(s+8)s2(s+4)
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D
T(s)=8(s+1.268)(s+8)s2(s+4)
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Solution

## The correct option is B T(s)=16(s+1.268)(s+4)s2(s+8)From the above Bode plot, For section de, slope is -20 dB/dec ∴ −20=y−0log8−log16 y = 6.02 dB Now for section bc, slope is -20 dB/dec ∴ −20=16−6.02log ω1−log 4 ω1=1.268 rad/sec To find value of gain K y = mx + c 16 = -40 log 1.268 + 20 log K K = 10.14 From all the result, transfer function is, T(s)=10.14(s1.268+1)(s4+1)s2(s8+1) T(s)=16(s+1.268)(s+4)s2(s+8)

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