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Question

An input of 1 sin 2t+0.3 sin 20t is applied to a first order insturment having a time constant of 0.2 sec. The output is

A
0.93 sin (2t0.23o)+0.073 sin (20t76o)
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B
1 sin (2t21.8o)+0.3 sin (20t2.3o)
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C
0.93 sin (2t21.8o)+0.073 sin (20t76o)
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D
1 sin (2t23o)+0.3 sin (20t2.3o)
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Solution

The correct option is C 0.93 sin (2t21.8o)+0.073 sin (20t76o)
output to input ratio for first order insturment
M=11+(ωτ2)

x(t)=1 sin 2t+0.3 sin 20t


Y(t)=11+(ω1τ)2sin(2ttan1(2×0.2)+0.31+(ω2τ)2sin(20ttan1(20×0.2)

=11+(2×0.2)2sin(2t21.8o)+0.31+(20×0.2)2sin(20t76o)

=0.93 sin (2t21.8o)+0.073 sin (20t76o)

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