Question
An LTI system is shown in the figure where G(s) = 100/(s2+0.1s+100) The steady state output of the system, to the input r(t) , is given as y(t)= a+bsin(10t+ θ). The values of ‘ a ’ and ‘b ’ will be
Options :
a = 100, b = 1
a = 10, b = 1
a = 1, b = 10
a = 1, b = 100
Answer :
a = 1, b = 10
Solution :
Given : G(s) = 100/(s2+0.1s+100)
y(t)= a+bsin(10t+ θ)
Using super position theorem,
Case 1 : r(t) =1, ω=0 rad/sec
The steady state response,
y1(t) = 100/(s2+0.1s+100) × 1
y1(t) = 100/((jω)2+0.1jω+100) × 1
y1(t) = 100/100 =1
y1(t)I ω=0 = 1
Case 2 : r(t) =0.1sin10t, ω=10 rad/sec
The steady state response,
y2(t) = 100/((jω)2+0.1jω+100) × 0.1sin10t
y2(t) = 10sin10t/(-100+j+100) = 10sin(10t- 90°)
Total steady state response,
y(t)= y1(t) + y2(t) = 10sin(10t- 90°) .......(i)
Given response, y(t)= a+bsin(10t+ θ) ......(ii)
Compare equation (i) and (ii),
a =1 and b = 10
Hence, the correct option is (C).
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