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6. An electrical signal v,(t)-5+12sin(5t) +10sin 10t) is plugged into a black box with an unknown electric circuit. The frequ

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Answer #1

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).

Bode Disgro 40 20 -20 40 45 ·135 10 10 10 10 10 Frequency rdisec

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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