The closed-loop system has one finite zero(s one"zero at infinity", one pole which not affected by K(s=-2), and one pole which is affected by K(s=-(K+4)) Note that the zeros in the feedback system can obscure certain complex frequencies in the output; these frequencies will never show up in the error signal. This is another reason to avoid pole-zero cancellation as a design technique, since it can cause unstable complex frequencies to become unobservable in the error signal Problem 1 Let the output of G(s)be y(t). Then, we can obtain the transfer function from a(t)to y(t)using Black,'s formula Y(s KG(s (s)1+KG(s) Thus, the closed loop poles are s satisfying 1+KG(s=0. To plot a root locus, you are sketching the location of s as a function of K such that 1+KG(s=0 Having said that, all the points on the locus satisfy the following criteria usually referred to as the angle criteria KG(sI) 1 G(1)= K There st is on the locus. If K>0, then ZG(s be an odd integer multiple of while for K <0, then zG(su be an even integer multiple of T Also, from Eqn(1), it is easy to see that st approaches to the poles of G(s)if K-0 and that sL approaches to the zeros of G(s)if K -o0. One thing to note here, however, is that by closing the loop in the feedback configuration shown in the problem, the number of the closed loop poles is either the same as that of G(s)or less if pole-zero cancellation takes place. Thus, to make the argument above applicable to any rational transfer functions, we need to have the same number of poles and zeros for G(s). This leads us to define poles or zeros at"infinity"as mentioned in the Home study exercise above. With all these concepts in your mind, let's go through the questions (a)Given G(s) s+1 G(s has a pole at-1 and no finite zero K>0: In this case, with the angle criteria, all the points, l on the locus should G(sn= odd integer multiple of 0, if you take any point on the real line segment(oo,1) ∠G(st) Also, this is the only segment that belongs to the locusThe closed-loop system has one finite zero (s = −1), one “zero at infinity”, one pole which is not affected by K (s = −2), and one pole which is affected by K (s = −(K + 4)). Note that the zeros in the feedback system can obscure certain complex frequencies in the output; these frequencies will never show up in the error signal. This is another reason to avoid pole-zero cancellation as a design technique, since it can cause unstable complex frequencies to become unobservable in the error signal. Problem 1 Let the output of G(s) be y(t). Then, we can obtain the transfer function from x(t) to y(t) using Black’s formula: Y (s) KG(s) = . X(s) 1 + KG(s) Thus, the closed loop poles are s satisfying 1 + KG(s) = 0. To plot a root locus, you are sketching the location of s as a function of K such that 1 + KG(s) = 0. Having said that, all the points on the locus satisfy the following criteria usually referred to as the angle criteria: 1 + KG(sl) = 0 1 G(sl) = − , (1) K where sl is on the locus. If K > 0, then G(sl) be an odd integer multiple of �, while for K < 0, then G(sl) be an even integer multiple of �. Also, from Eqn(1), it is easy to see that sl approaches to the poles of G(s) if | | K ∩ 0 and that sl approaches to the zeros of G(s) if | | K ∩ ∗. One thing to note here, however, is that by closing the loop in the feedback configuration shown in the problem, the number of the closed loop poles is either the same as that of G(s) or less if pole-zero cancellation takes place. Thus, to make the argument above applicable to any rational transfer functions, we need to have the same number of poles and zeros for G(s). This leads us to define poles or zeros at ”infinity” as mentioned in the Home study exercise above. With all these concepts in your mind, let’s go through the questions. (a) Given: 1 G(s) = . s + 1 G(s) has a pole at −1 and no finite zero. • For K > 0: In this case, with the angle criteria, all the points, sl on the locus should satisfy G(sl) = odd integer multiple of �. So, if you take any point on the real line segment (−∗, −1), G(sl) = 0 ±� = ±�. ���� − ���� phase due to zeros phase due to the pole Also, this is the only segment that belongs to the locus. 2