Question

G(s) Y(s) s+2 1. (25 points) A system has G(S) = 21ac11: (a) Find the two points that define each real-axis segment of the root locus. (b) Find the maximum value of the gain K for the closed-loop to be stable. If there are root loci that cross the imaginary axis, also find the corresponding frequency of the closed-loop roots that lie on the imaginary axis. (c) Find the angle of departure from the complex poles. (d) Find the location of the breakaway point and its corresponding gain (K) value.

Answer 1

Please go through the pictures sequentially.....

At the end I attached matlab simulation for verification (for the value of k=1)....

PAGE NO. 1 DATE: Roat locus + RTS) -ki -Gls) 7189 - sa Guls) St2. topen-loop-gain) s²265113. (since Hils) = 1) Zeroes=1 >> S= -2 -poles = 2 –6 $V 36-52 2 s -6 I VICE 2 S = -6 + uj =-3+27 -3+2; 3 کام 2 х + X -3-2; centroid a-Ereal part of poles - Ereal part of -P-2 = -3-3-62) - - 6+2 Zeroes. 2- 1 (29+1) 1802 PI 2-1-1=0 a = - 4 angle of asymptates o P-2 whineq = 0, So q=0 (0+1180° 2-1 Now for may value of k k=) for Routh's Array characteristic eq =) It Gy(s) = 0 o = 180°

PAGE NO. 2 pare! 1t K(5+2) s? +654.13 s? +65-+23+ K(542)=0 K_-52-65-13 (5+2) s² 1 1.3.2K. s! 6tk 0 13+2K. SO So, for stabilty 6+KXO__and_1372ko k> 6. -and k >-6.5 KEG-6-,00) Now, for breakaway points K = Es2-65=13)/ (+2) dk ds dk (5+2) (-25-6)-(3-65-13) (1) ds (5+2)2 (5+2)(– 25–6) +5+65413 = 0 -252-65us-12+5+85+13=0 -s? - Ust! Sztus-ICO 52-40/16+y 2 S =-ut 2v5 =0 2 s =-2 I/5 SF-21 2:23 S7-4.23 0.23 I -2 for S= 4:23 K-74-23) — 6X6-4.23) - 13 -4.23+2

PAGENO. 3 DATE: K--(4:23)2 +6x4.23 – 13 -2:23. -17.89.-13 +2538 -2.23 30-89_+25:38_-78.51-3181 -2.23 +2:23 K =(+ve) 5= -4:23 (breakaway point) k= sabunpuokvirno) Sog no.11- for S = 23. -66.2317 610.23)-13 2.33 (ks-ve) so S=0-23leave angle of departure - 3+2j So tomo = 2 2 6= tant (2) 0=63.43'appox) * Σ φz od ed = 1800 Q = EQp 1809 -( torent tant (2) 0d=1809 – 90° +63:43 2436.43

PAGE NO.4 DATE: rin Root locus jw 24307 colzero) Assumed kkor -3 The vroot locus doesn't imaginaryaxis cross the Please Upvate]

5:37 AM 3.4KB/s & Metall >> y tf([0,1,2], [1,7,15]). y = S + 2 s^2 + 7 s + 15 Continuous-time transfer function. >> rlocus (y)

5:37 AM 0.2KB/s & alla 15 Figures SHOW DATA CURSOR Figure 1: rlocus(y) Root Locus Root Locus 1.5 0.5 Imaginary Axis (seconds) Imaginary Axis (seconds) -0.5 -1 -1.5 1 -6 -9 -8 -7 -5 -4 -3 -2. - 1 0 Road Acicis (Swecorats45)