e, complete the square and find the corresponding gain K and the bifurcation point 1+KG(s) 1+S+1 Ks+K 0 2 +K 2 K K(1 4 K 0.4. This result tells us that at K=0 and K= 4 we have double-pole and the corresponding locations are at s=0 and s= How can we know the shape of the path that locus takes after branching out of s=0 and merging at s=-2? In the case in our hand, it is not hard to determine. Since the closed loop poles for the K between 0 and 4 are a complex conjugate pair, we can assume that s=0+ jw for a given K. Then, using one of the equations above (o+ju)+k(o+ju)+K (o+2jow-w)+Ko+jKw+K=0 For real part +Ko+K For imaginary part: 2ow+Kw =0-K=-20 o(combining the two equations above) As you can see, the last equation is nothing but an equation of a circle whose radius is and whose center is located at(o, w)=(1, 0). Thus, the locus is of the ciro 0<K<4. Then at s=-2 the locus bifurcates to the zero at s=-l and s=-oo on the real axis One thing to note here is that, in general, it is extremely difficult to obtain expressions of the locus for higher order systems For K<0: In this case, we can see that the real line segments (-1, 0) and(0, oo) belong to the locus. On the locus, one of the poles will move to the zero at s=-1 as K decreases or-K increases, while the other pole moves out to oo along the positive real axis Thus, the loci containing both K>0(solid) and K <0(dashed) cases are shown below:i.e., complete the square and find the corresponding gain K and the bifurcation point. 1 + KG(s) = 0 s + 1 1 + K s2 = 0 s2 + Ks + K = 0 � K �2 �K �2 s + 2 + K − 2 = 0 � K � ∩ K 1 − 4 = 0 � K = 0, 4. This result tells us that at K = 0 and K = 4 we have double-pole and the corresponding locations are at s = 0 and s = 4 = −2. −2 How can we know the shape of the path that locus takes after branching out of s = 0 and merging at s = −2 ? In the case in our hand, it is not hard to determine. Since the closed loop poles for the K between 0 and 4 are a complex conjugate pair, we can assume that s = ξ + j� for a given K. Then, using one of the equations above; (ξ + j�) 2 + K(ξ + j�) + K = 0 (ξ2 + 2jξ� − �2) + Kξ + jK� + K = 0 For real part: ξ2 − �2 + Kξ + K = 0 For imaginary part: 2ξ� + K� = 0 ∩ K = −2ξ ξ2 − �2 − 2ξ2 − 2ξ = 0 (combining the two equations above) ξ2 + �2 + 2ξ = 0 (ξ + 1)2 + �2 = 1, As you can see, the last equation is nothing but an equation of a circle whose radius is 1 and whose center is located at (ξ, �) = (−1, 0). Thus, the locus is of the circle for 0 < K < 4. Then, at s = −2 the locus bifurcates to the zero at s = −1 and s = −∗ on the real axis. One thing to note here is that, in general, it is extremely difficult to obtain expressions of the locus for higher order systems. • For K < 0 : In this case, we can see that the real line segments (−1, 0) and (0,∗) belong to the locus. On the locus, one of the poles will move to the zero at s = −1 as K decreases or −K increases, while the other pole moves out to ∗ along the positive real axis. Thus, the loci containing both K > 0 (solid) and K < 0 (dashed) cases are shown below: 5