Question

One system has been designed to control the power distribution in a resident area in Bestari Jaya. Given an open loop transfer function of the system as follows: 250 K G(s) = (5 + 5)(s + 50) As an engineer that incharge of that area, you were asked to analyse the system. In order to do so, the use of bode plot can be adopted. a) By using the straight line approximation method, draw the bode plot for magnitude and phase for the open loop system when K = 10. b) From your drawing in a), determine the i. value of static error constant, Ky. ii. gain margin iii. phase margin iv. gain crossover frequency V. phase crossover frequency c) If an extra pole is added to the system, determine the phase value where the phase plot ends. Explain your answer

Answer 1

6(S) = 250k - S (5+5) (S+50) for Kelo G) 200x10 - S(5+5) (S+60) Converting into standard form GO = 2500 S5(5+1) 50 (5+) S(5 +0 (5 +1) for Bode magnitude plot Corner frequency and In Giren G s the Wzo with s Cu 5 with s tl a pole Contributes Odß below -20 dB / dee alcove corner frequency corner frequencies are these there are the poles We 50 with S +1 the initial gain given by 20 log 10 = 20 dB

at start as for s= 0 corner frequency is o dB initially has -20 dB at we 5 rad -20 dB adeled and it becomes - Ho dB dec dec at w 250 rad -20 dB adeled it becomes - 60 dB dec dec the value at we 5 calculated by - 20 lag lot 20 lag A =- 20 dB - log (51 dec. 20 log A 2 6.02 dB using this slope method the values at we go rad canke found also the wat which Gol= odB is found

Bodle Magnitude plot is Zodb 6 dB ;-20 dB/dec -40dB /der OdB Yazol W-5 (W=70g = 50 w -34 dB GodB/do

the phase of Gol is 168 z Llo - LS +45 +1) + LS + LAS = 210- (Ls +LS +1 + 2 + 1 ] LGS. 0-19.+ tent.com + that com ] LGS 2-[got tant wt tan to ] 50 at w = 5 LOOI (90 + 46° + tan! 5 ] = -140.7° at W250 LGUIz got fent 50 + 45° = -219.28

for C6 (8) 2-180° - Coot tant w t tant to ]=- =-180° I goo tent cottant w Go fent w two =9. for this 1- W2 = 0 5 w=5x60 = 15.8 1 rad.

phase plat 5 دقی وینی) 8-16 د له - 140.70 - 180° 219.28 - - 1270

(bo cis Static error constant Ku when the magnitude plot proceeds with the initial slope of Lower doequency portion at point where it cut the drequency axis and become dB that frequences has a relation with K as I won Ku where n= no. of pales at zero here na - Wzku to find - Ku 0 dB - 2o log 10 =-20dB log co dec log 10 a log to We lorad Ku zlo

gain margin is equal to gain lielaw Odß line when phase becomes - 180 that is phase crossover requency which 1508 I rad calculated Cain at W2 15 8 1 dal at this portion slope is - 40dB = - 12 dB Bolo dec o ct at we 7.06 magnitude is o dB than gains at W2 16.81 by 20 log X- dB 10. 81 7.06 - 40 dB dec

20 lag x = gain - 14 dB Gain Margin 0 - (-14dB) = 14dB lain phase Margin we add the is the phase at phase gain adoled to we get when cross over frequency to 1800 o dB gain Cross over frequency at which gain becomes which is wz7006 rad phase at 706 rad X GU1 = -(got tant 3.06 + fant 7-06) go 2 -162.73 phose Margin = 180+/assluge 180- 182.73 2 276 26 degree lu2 ( gain phase cross over at w = 7.0 Grad Cross over at Cz 15-81 rad

(c) when a estro GO pole = is adoled to the Go 10 at waa SCS +) ($_+) (5+) than the phase 200) = out tentot tant wt tento] at the end was 164 = -[go + tentaf + tautas ttant ]

L = -(got got 90° +90°) LGS) - - 360° pole is added at the end the phase as cehen becomes one more - 360°
in finding kv in section b the relations w^n=k gives static error constant Kv when n =1 , n=2 it gives ka