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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=y0log8log16
y = 6.02 dB
Now for section bc, slope is -20 dB/dec

20=166.02log ω1log 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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