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Consider a first-order time-delay system given by ke H(s)= 1+TS where kis the static gain, t, is the time delay and T is the time constant. Q1) Show that by applying the Laplace inverse, the output y(f) can be expressed as y(t) =0 VOt,

Consider a first-order time-delay system given by ke H(s)= 1+TS where kis the static gain, t, is the time delay and T is the time constant. Q1) Show that by applying the Laplace inverse, the output y(f) can be expressed as y(t) =0 VOt,

Power System Analysis and Design (MindTap Course List)
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 6.16P
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Consider a first-order time-delay system given by ke H(s)= 1+TS where kis the static gain, t, is the time delay and T is the time constant. Q1) Show that by applying the Laplace inverse, the output y can be expressed as y(1) = 0 Vo<t<t, ry(0 +y{d)= ku(t=t,) VI>t, Q2) show by integration that Q3) Let e = [T,k,kt,] and w = y)dn, write the Equation in 02) in the following form w) = ¢ (t)e.where the expression for o (t) is to be specified. 04) Identlfy the three parameters 1, k and t, using the optimization technique in the sense of the Least square method. giving that the step response y(t) measured at each sampling time t=D0.1.2.3,4.5,6.7.8,9 is given as y(t)=[000.26 0.46 0,560.65 0.67 0.720.730.72].
Transcribed Image Text:Consider a first-order time-delay system given by ke H(s)= 1+TS where kis the static gain, t, is the time delay and T is the time constant. Q1) Show that by applying the Laplace inverse, the output y can be expressed as y(1) = 0 Vo<t<t, ry(0 +y{d)= ku(t=t,) VI>t, Q2) show by integration that Q3) Let e = [T,k,kt,] and w = y)dn, write the Equation in 02) in the following form w) = ¢ (t)e.where the expression for o (t) is to be specified. 04) Identlfy the three parameters 1, k and t, using the optimization technique in the sense of the Least square method. giving that the step response y(t) measured at each sampling time t=D0.1.2.3,4.5,6.7.8,9 is given as y(t)=[000.26 0.46 0,560.65 0.67 0.720.730.72].
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