60 40 20 KG a)) -20 40 0 ZKG(Ga) -90 -180 -270 0.1 100 10 (rad/sec)...
20 100 10 Frequency (dis) -200 300 0.1 100 10 Frequency (radis) 20 tog M 40 Usng ufon Bde Plt f KG( Pictuve Closed oup systeu's chatctenic equati s Dis)e +kG (a): usng Magnitude Plot, Detaine the form of Thnfer fenctin from Stope of HL freueney, stofe t fow freguancy, Cut oft fetuancy an Sretch the Voot -locus C6) from the Bale Aot Pictuve, perevaine Gain mavgn (m) / and Phase margin lPM) C Prom the Bode Plht Pictun, Spetch...
40 20 20 -60 -80 -100 0.1 100 10 Frequency (radis) -50 -100 -150 200 250 -300 0.1 10 100 Frequency (rad/s) Phase (degrees) 20 log M O Pictuve UShg upon Bode Plot of KGIS) Closed lop syStem's chaactentic eguation is Des)= 1+KG(S) (4) uhg Magnitude Plot, Detamine the torm of Transfer fevction from slope of HigL fretueney, Slope t low freguency, Cut-aft fefuency an Sretch the Voot -locuS t(wro Miw) 20d8 Vampng atto =1) on a seand -order System)...
edit) cut of frequency -> cut-off frequency damping ratio =1 , w=0.3 -> M(w)=20db 40 20 60 -80 -100 0.1 10 100 Frequency (radis) 50 -100 -150 -200 -250 -300 0.1 10 100 Frequency (radis) 20 log M (saaap) ase USMg upon Bade Plt of KGIS) Pictuve Closed toop syste's chactec eguati is DCs) tkG) (a): ung Mapnitule Plot, Detane the torm of Transfer fctin tom slope f HL freuecy, Stofe t fow fresuancy, Cut of feuancy and Sketch +he...
Consider the system given below where K is a constant gain, Gp is the plant, and Ge is a compensator. The Bode Plots of a Gp is given below. Problem 1: Bode Diagram 20 2 40 -60 80 -100 90 135 180 a 225 270 101 10 Frequency (rad/s) 102 a. Looking at the low frequency behavior, determine its number of poles at origin. Explain. b. Looking at the high frequency behavior, determine the number of excess poles. Explain. C....