s2+2s+4+K(s+2)=s2+(K+2)s+2K+4=0. (a) For K>0, for all st on the locus ZH(st) is an odd integer multiple of T. Thus, we can first see that the real line segment(oo, 2) belongs to the locus. Then, we would like to know at which point on that segment there is a double-pole. On that point, Eqn(5)has a double root I.e. s-+(K+2)s+2K+4=0 completing the square +2K+4 +2K+4 0 This tells us that there are two points where double pole occurs; one is when K=6>0 and the other is when K=-2<0. Here, we need to look at only the positive case. The negative one will be used in part(b). For K=6, the double-root is at -#2=-4 (b)For K<0, using the angle criteria, we can see that the real line segment (2, oo) belongs to the locus and the double-pole point is at -k*2=0 where K=-2 as computed in(a) (c)Since the closed-loop sy tem is st ill just a second order system, the condition that the closed loop impulse response does not exhibit any oscillatory behavior simply means that the closed- loop system is critically damped or overdamped. Since H(s) is underdamped i. e, its poles are not on the real axis. at the smallest k to be found the denominator takes the following form s-+ 2Swn8+w n = 0, $ =1 critically damped where wn is the undamped natural frequency of the closed loop system. Thus, the required condition is nothing but the double-pole condition we found in(a). Since K is constrained to be positive, the smallest K we are looking for is K= 6or 2 s2 + 2s + 4 + K(s + 2) = s + (K + 2)s + 2K + 4 = 0. (5) (a) For K > 0, for all sl on the locus H(sl) is an odd integer multiple of �. Thus, we can first see that the real line segment (−∗, −2) belongs to the locus. Then, we would like to know at which point on that segment there is a double-pole. On that point, Eqn (5) has a double root, i.e., s2 + (K + 2)s + 2K + 4 = 0 completing the square � K + 2 �2 �K + 2 �2 s + 2 − 2 + 2K + 4 = 0 � �2 K + 2 ∩ − 2 + 2K + 4 = 0 ∩ K = 6 −2, . This tells us that there are two points where double pole occurs; one is when K = 6 > 0 and the other is when K = −2 < 0. Here, we need to look at only the positive case. The negative one will be used in part (b). For K = 6, the double-root is at − K+2 = −4. 2 (b) For K < 0, using the angle criteria, we can see that the real line segment (−2,∗) belongs to the locus and the double-pole point is at − K+2 = 0 where K = −2 as computed in (a). 2 (c) Since the closed-loop system is still just a second order system, the condition that the closed￾loop impulse response does not exhibit any oscillatory behavior simply means that the closed￾loop system is critically damped or overdamped. Since H(s) is underdamped i.e., its poles are not on the real axis, at the smallest K to be found, the denominator takes the following form: 2 s2 + 2��ns + �n = 0 2 s2 + 2�ns + �n = 0, � = 1 critically damped (s + �n) 2 = 0, where �n is the undamped natural frequency of the closed loop system. Thus, the required condition is nothing but the double-pole condition we found in (a). Since K is constrained to be positive, the smallest K we are looking for is K = 6. 10