Question

3. (28 pts.) The unity feedback system with K(5+3) G(s) = (s + 1)(s + 4)(s + 10) is operating with 12% overshoot ({=0.56). (a) the root locus plot is below, find the settling time (b) find ko (c) using frequency response techniques, design a lead compensator that will yield a twofold improvement in K, and a twofold reduction in settling time while keeping the overshoot at 12%; the Bode plot is below using the margin command and using the gain which provides the twofold improvement in Ky, that is uses K=197.4 Note: there is a blank page on the back. Value of settling time = Value of Kp- The lead compensator transfer function = Hels ) Figure 1 File Edit View Insert Tools Desktop Window Help Bode Diagram Gm - Inf dB (at Inf rad's). Pm = 48.4 deg (at 12.2 rad/s) Magnitude (dB) Phase (deg) System G Frequency rad's) 122 Phase (deg):-131 Frequency (rad's)

Answer 1

Sol: a) from the Root locus plot to 12% ove shoot in dominant poles are su= -6:08+8.995, there are in the from Sd= -swn I Waj The settling time ts= 4 = 4 = 0.6579 seconds. . swn 608 The gain for Se= -6.08 +8.995 is K(5+3) la 1 is +i) (s+u)(Sto) ! b) @ss - 6-08 +P.99 J K = 98.4 Kp = Glo) where Gis) = K (3+3) (5+1) (3+4) (S+10) = 7.38 98.4 * 3 s 14 4х10 - {= 0.56 implies phare malgin of 56 The phase mayin with K= 197.4 is 48.40, but Required margin is 56° ii fm = 56°48.4 +5° = 12.6° The attemation Required is S ka ir sin em = 0.6418 Itsinom The Attuation point in the mogintude Cenue is -20 1880 Ita] = -1.9257 dB

to . new settling time ts= 0.6579 = 0.329 seconds na 46 to s= 0.56 in Wea46 head is in the from Gels) = K (st llot) - [stilat * = We sa = 36.8517 as a way = 57.4193 Ka 197.4 = 307.5725 and DSG , is) = 307.5725 (5+ 36.8517) (5+574193)