BME 639 Lecture Notes - Lecture 11: Step Response

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Note that in this case, since there is no disturbance present, we do not have to use the superposition theorem, and can calculate the gain using the steady state error formula with the position constant: 0. 1 0. 1 1 which is the same result, of course. G (s) g(s) s3 56s2 305s 250 cl 1 g(s) 1 750k s3 56s2 305s 250. Apply the routh hurwitz criterion to the closed loop characteristic equation: K > (cid:882). (cid:885)(cid:885) s3 56s2 305s (750k 250) 0 s3 s2 s1 s0. The sufficient condition is computed from the routh array as: 0. The total mathematical condition for stability is: the total practical stability range is: Next, let"s consider the steady state error in the step response of the closed loop system. The system type is zero (no integrators in the open loop transfer function): Apply the definition of the position constant and calculate the steady state error: Ele/bme639 course notes winter 2012 lim lim 750.

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