Question

find the bode plot of H(s) = 40(S^2/(s+100)^2)

This is the solution of the teacher: I dont get the phasor digrams to well to well, like how do we know when to start at 0,90,180 etc. I understsnd amplitude. for phasors I also understsnd if we have a pole of 10 for example: we start decresing by -45deg at 1 and take that -45deg away at 100. I just don't understand where to start on the y axis of the phasor diagram. Is it also true that a zero increases by 20db on the amplitude diagram and 45deg on the phasor diagram, and the pole the opposite. Ofc if there is s^2 it will become 40db and 90deg

sun these and was Awe gel kodb d) H(s) = 408² (5+700) Toore {1+sa) (145) +2blog -47d6 250 250 1440) 140 Babi + 104 10 07 702 103 -20 100 101 764

Answer 1

H(s). 405 (54100) HG) 40st His) - 0.0045 HGL) - 0.004 (3) [1+io.ow] factor corner fequency slope Net slope 0.004 40 40 (دو) 6 -40 [1tjo.ru] w. %. Too 20log (0.04) -47 dB How] 1 20 + Mode I 1 to o 20 -48 $ 180-2 ton [0.016] כי فی (คง 40+ 100 lo 90 168.57 To Lo Lot 103 Ib.. 11.42