Question

5. Consider the feedback system in Figure 4 where! G(s) = 26+10% Figure 4 The Bode plot of G is shown in Figure 5. Boda Diagram Magnitude (dB) -100- -156 -135 -root -225 10 Frequency radici Figure 5: Bode plot of G (a) [2 marks] Find the phase margin, gain margin and gain crossover frequency (approximate as needed) for the case when C(s) = 1. PM = GM = wc =

Answer 1

© 0 [cs] FGg) Given Ges) = 20 s(Stic)2 8 The overall closed loop transfer function is c(s)G(5) ST+G(9) C(5) Bode diagram Given Bode plot for GCS) 15 magnitude (06)-> -150 F . - -'. l&D + . phase cu -225+ - 270 t frequency (raa sec) a). Given CCs) = Openloop transfer function (s) GC) CH(s) = 1 unity flb system) Gos) = R0 s(sto) Gain crossover frequency (w): The frequency out which gaen of GCs). 00 magnitude of G[5) is odB from the graph, The magnitude is OdB at We = Qul rad/sec ( We oorrad see Since it's a log scale phase Margin.cpm): = (180 + ) - It's allowabie phase log that can To LG(jw) be added to bring the system lo: LaCiul lowe into verge of instability. Gciw jw cien LGciw) = 0-90-etan lehenen

Ø = -90-Tan ( ) 2.90-1-145 D = -91.145 PM = 180 + Tom = 88.897

Gain Margin (GM) = ) : x = ( Giulu- X = Gew : The amount of gain ne System reaches Verge of instability that can be added before the System reaches verge of Wpe = phase cross over frequency The frequenty at which LGW) = -180° Ris WPL From the bode diagram, wpe = lorad /sec Ano Xş laciwilow-wpc iho teste loco(w. Plot lewens x = lower times herco le ore * Too Gain margina = 100 = zolog (1x) de ė 2010g Clou) do Gmis 40dB ] . PM = 88. BID! GM = 40dB We = obrad /sec = 0.2 rad/sec

Given cc5) = k, k>0 than or equal to 6 for unit Yamp (1) Steady State error is less input Gi ) phase margin is greater than or equal to 45 (s) G(s): - Gr's OK 5 (5+10)2 C) to be satisfied: Velocity error constant Ckyy = It S. (s) G(s) 50 it 8 20k so' s(s+10) 2 kr = the Steady state error + 6+* Kw = 60 >| > 500 JT lea> 83-33] Cü) to be satisfied : euge - Gain cross over frequency PM = 180 +0= 45 3-135 LCCW)G(jw) .. *-135 "wewge L o 20KV 2-135 (ju +107) w-wage 0-90- 2 tang -135 Tante woone) a 22:59 CNOL 4•14 Lrad /sec At this auger lGcjww-wge! 20K jWgc (jwget 10) July, toz

2012 Wgc (wige 7100) 201 4.142 (CU-142+100) 20K 485.26 K* 24.26 To achieve both the specifications, there is a possibility of Me K>, 8333]