Question

Sketch the root locus for the unity feedback system shown in Figure P8.3 for the following transfer functions: (Section: 8.4] K(s + 2)(8 + 6) a. G(s) = 52 + 8 + 25 K( +4) b. G(S) = FIGURE PR3 152 +1) C G(s) - K(s+1) K (n1)(x + 4) For each system record all steps to sketching the root locus: 1) Identify the # of branches of the system 2) Make sure your sketch is symmetric about the real-axis 3) Identify the real axis segments of the root locus. a. The Root locus can only exist to the LEFT of an odd number of poles & zeros b. This is because the angular contribution of the root locus is determined by KG(s)H(s) and the angular portion of this value must equal 180 deg so KG(s)H(s) = -1 4) Identify the starting & end points of the root locus a. The root locus begins at the finite and infinite poles of G(s)H(s) and ends at the finite and infinite zeros of G(s)H(s). b. Calculate the roots at infinity. # of poles - # of zeros = p -2. c. (Normally you'll have more finite poles than zeros.) When the # of poles is greater than the # of zeros, the difference of p-z equals the number of "zeros at infinity" and will also equal the number of asymptotes you'll have d. Ifz is greater than p, then you'll have "poles at infinity" 5) Find the real axis intercepto, and ea 6) Determine the Break-away and Break-in points of the root locus and show how you got these values. a. Use either differentiation or the transition method 7) Use Matlab to check your answers!

Answer 1

Answer 9 G150 = R15+2) (5+6) s2+85+25 Poles s285+25 50 Pales sa-4+3) Zeroef y s = -2,-6 S=-87 V8 ²-4x25 2 = -8£ 6J 2 ds 269.650 pjw = -4 +35 +j3 ext-3 - 269.69 - dd Hence, this is the root locus diagram of above transfer function, ar k15 +4) GIS) = S2+1 22+4=0 s=+2jt Zeroeg. 2+1 -0 =11€ Polea -

pjw 8+)2 *t51 Hence, this is the root focus diagram of above trangfer function, c) GIS) = k1S²+) Poleg ,s=oo, ²+1=0 iri setit Zeroes. JI the double Pole

GIS) = 1S+13 (5+4) Polea , S = -1,-1,-1,- 4 no. of root & terminate SP-22- 04 at enfinity Seps of pall - um of zeres centrolat = Polymboletic Point O s - - - - A poz 4 - 6 S À =-1.75 Asympotetic angle ♡ (29+0 1800 - 1294))1800 P-2 4-0 q=0, P= 45° q=1 or 45×3= 1350 9- 2 - 45x 5 = 2250 q=3 t = 45 x 7 = 315 to find breakawa point- to k (+1) 3 (Sta) 1 R = (Stig?1349) differentiate with reabect to kn work - (S+193 +3 (S+1)? (5+4) = 154122 ( 15 +1+345 +9))

mobile (S++32 (5+1+35+12) so (5+172 (45+13) so S = - 3 = -3.25 point there is two break away s = -3.25, d s =+1

od Rood locul Hence, this is the transfer function, root locus diagram of above