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Question

The inverse Laplace transform of the signal X(s)=1n[1+ω2s2] is

A
2(1cosωt)tu(t)
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B
None of the above
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C
2(1+cosωt)tu(t)
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D
2(1cosωt)u(t)
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Solution

The correct option is A 2(1cosωt)tu(t)
Given, X(s)=1n[1+ω2s2]

Let x(t)=L1[X(s)]=L1[ln(1+ω2s2)]

L[x(t)]=ln[1+ω2s2]=ln[s2+ω2s2]

=ln[s2+ω2]lns2

L[tx(t)]=dds[ln(s2+ω2)lns2]

=1s2+ω2.2s+1s22s=2s2ss2+ω2

tx(t)=L1[2s2ss2+ω2]

=(22cosωt)u(t)=2(1cosωt)u(t)

x(t)=2(1cosωt)t u(t)

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