Example 4.8 Comparing Responses of Three-Pole Systems PROBLEM: Find the step response of each of the transfer functions shown in Eqs. (4.62) through (4.64) and compare them. 24.542 T₁(s): = (4.62) s² + 4s + 24.542 245.42 T₂(s) = = (4.63) (s+10) (s²+4s + 24.542) 73.626 T3(s) = = (4.64) (s+3)(s²+4s+ 24.542) SOLUTION: The step response, C;(s), for the transfer function, Ti(s), can be found by multiplying the transfer function by 1/s, a step input, and using partial-fraction expansion followed by the inverse Laplace transform to find the response, ci(t). With the details left as an exercise for the student, the results are c₁(t) 11.09e2 cos(4.532t - 23.8°) (4.65) C2(t) = 1 -0.29e-10r - 1.189e-2 cos(4.532t — 53.34°) (4.66) €3(t) = 11.14e3 +0.707e-2 cos(4.532t+78.63°) (4.67)

Example 4.8 Comparing Responses of Three-Pole Systems PROBLEM: Find the step response of each of the transfer functions shown in Eqs. (4.62) through (4.64) and compare them. 24.542 T₁(s): = (4.62) s² + 4s + 24.542 245.42 T₂(s) = = (4.63) (s+10) (s²+4s + 24.542) 73.626 T3(s) = = (4.64) (s+3)(s²+4s+ 24.542) SOLUTION: The step response, C;(s), for the transfer function, Ti(s), can be found by multiplying the transfer function by 1/s, a step input, and using partial-fraction expansion followed by the inverse Laplace transform to find the response, ci(t). With the details left as an exercise for the student, the results are c₁(t) 11.09e2 cos(4.532t - 23.8°) (4.65) C2(t) = 1 -0.29e-10r - 1.189e-2 cos(4.532t — 53.34°) (4.66) €3(t) = 11.14e3 +0.707e-2 cos(4.532t+78.63°) (4.67)

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.15P

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