From the given bode frequency response, we can extract the gains and phases at the frequencies of the input (5 rad /sec and 10 rad/sec).
It is observed that, at 5 rad/sec, the magnitude (or) the gain is 5 dB (approx.)
Therefore, M1 = 5 dB = 1.778.
The phase at 5 rad/sec is -157.5degrees approximately.
Therefore Phi1 = -157.5 degrees.
Likewise,
At 10 rad/sec, the magnitude (or) the gain is -3 dB (approx.)
Therefore, M2 = -3 dB = 0.707.
The phase at 5 rad/sec is - 165 degrees approximately.
Therefore Phi2 = -165 degrees.
At zero frequency, i.e. DC gain is 20 dB
Therefore, Mdc = 20 dB = 10.
Now the output is given by,
V0(t) = Mdc*5 + M1*12 sin(5t+phi1) + M2*10 sin(10t+phi2)
= 10*5 + 1.778*12 sin(5t-157.5deg) + 0.707*10 sin(10t-165deg)
= 50 + 21.3 sin(5t-157.5deg) + 7.07 sin(10t-165deg).
From the frequency response, it is observed that the maximum slope during roll off is -40 dB /decade which indicates that there are two poles. Also that the time response has an overshoot which indicates that the system is underdamped.
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