HW7-S10Soln

HW7-S10Soln - 20 . 1 53 . 1 1 ln 1 e) The natural frequency...

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Solution HW 7 Problem 1 a) For the 1 st order system, the static sensitivity and time constant will fully characterize the system. b) Static sensitivity: 4 0.001 2 10 / 5 o o V K V C C From the step response plot for the 1 st order system, the response is 10.00063 V when t=0.1 s . 10.00063 10 100% 63% 10.001 10 So the time constant is about 0.1s. Problem 2. a) The system is 2 nd order system, since it exhibits some oscillation around the steady state. b) Static sensitivity: 1 K c) The oscillation period d T is about 0.065sec. When we determine d T , we estimate the time instant at which the step response is equal to the steady state output. d) The %OS method will be used to estimate the damping ratio, since the maximum overshoot is marked in the step response.
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Unformatted text preview: 20 . 1 53 . 1 1 ln 1 e) The natural frequency is s rad T d d n / 7 . 98 20 . 1 065 . / 2 1 / 2 1 2 2 2 a) The frequency response function of this system is 2 2 2 2 7 . 98 1 7 . 98 20 . 2 1 1 2 ) ( j j K j G n n The differential equation is ) ( ) ( ) ( 2 ) ( 2 2 t x K t y t y t y n n n Substituting the numerical value into the above equation, we get ) ( 9742 ) ( 9742 ) ( 39 ) ( t x t y t y t y Problem 3 Problem is solved by reading values off the graphs: a) gain = 40 dB so K=100. b) n is 50 rad/sec from the phase plots (-90 deg) c) Amplification method: 100 0.1 500 2 Half power method: 54 44 0.1 50 2 Slope of phase angle method: 55 45 1 0.105 2.55 0.65 50...
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This note was uploaded on 10/24/2010 for the course MA 303 taught by Professor Staff during the Spring '08 term at Purdue.

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HW7-S10Soln - 20 . 1 53 . 1 1 ln 1 e) The natural frequency...

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