MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems--Fall 2003 PROBLEM SET 10 SOLUTIONS Home study exercise(E1)O&W 11.32 (a) H(s XO 1+KG(SH(s) (via Black's equation NISD D1 sD2(s)+K (multiply through by D(s)D, s)) The zeros of this closed-loop system are the roots of Ni(the zeros of H) and the roots of D2 (the poles of G) (b) If K=0, then the system is operating without feedback of any sort. Naturally the poles and zeros of the system without feedback are the poles and zeros of H(s) More formally, we can take limits of the above equation N1(s)D2(s)N1(s) 小(5)D2(8)+kM1()M2()=D2(s)D2()=D1()=H() (e)Check by simple substitution Q(s) (8)1+kG(s)f(s 1+K H(s 1+KG(SH(s (d)We are given the +2 (s+4)(s+2) In this simple example, we can find p and q by inspection p(s)=8+1,q(s)=s+2,B()-s+4,()=1 Root locus equation: H(sG(s) 8+4 K →s=-(K+4)
� MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 10 Solutions Home study exercise (E1) O&W 11.32 (a) Y (s) H(s) = (via Black’s equation) X(s) 1 + KG(s)H(s) N1(s)D2(s) = (multiply through by D1(s)D2(s)) D1(s)D2(s) + KN1(s)N2(s) The zeros of this closed-loop system are the roots of N1 (the zeros of H) and the roots of D2 (the poles of G). (b) If K = 0, then the system is operating without feedback of any sort. Naturally the poles and zeros of the system without feedback are the poles and zeros of H(s). More formally, we can take limits of the above equation: N1(s)D2(s) N1(s)D2(s) N1(s) lim = = = H(s) K�0 D1(s)D2(s) + KN1(s)N2(s) D1(s)D2(s) D1(s) (c) Check by simple substitution: ⎣ ⎤ ˆ p(s) H(s) Q(s) = q(s) · 1 + KGˆ(s)H(s) ˆ � N1(s)/p(s) p(s) D1(s)/q(s) = q(s) · � 1 + K N2(s)/q(s) N1(s)/p(s) ⎡ D2(s)/p(s) D1(s)/q(s) · · H(s) = 1 + KG(s)H(s) (d) We are given the following: s + 1 s + 2 H(s) = , G(s) = (s + 4)(s + 2) s + 1 In this simple example, we can find p and q by inspection: 1 ˆ ˆ p(s) = s + 1, q(s) = s + 2, H(s) = , G(s) = 1 s + 4 Root locus equation: Hˆ (s)Gˆ(s) = 1 = 1 =� s = −(K + 4) s + 4 −K 1
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 locus
The 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
For K<O: The angle criteria states that ZG(sn= even integer multiple of T So, if you take any point on the real line segment(-1, oo) (s) 0. phase due to zeros phase due to the pole Combining the two cases, the root loci for both cases are plotted as shown below. The solid line corresponds to the locus for K>0 case and the dashed line for K <0 case m (b)Given: G(s G(s has two poles at 5 and -3, and no finite zero For K>0: Evoking the angle criteria, we can first see that the real line segment(-3, 5) belongs to the locus. Now we would like to know at which point the locus branches out or bifurcates from the real line. At the branching point, the closed loop poles are of double pole. To find such a point, we can first find a value of K corresponding to the double pole 1+KG(s)=0 (s-5)(s+3 (s-5)(s+3)+K=0 s2-2s-15+K=0 (s-1)-1-15+K =0 completing the square So, when -1-15+K=0 or K= 16, the closed loop system has a double pole at 1. Because of the symmetry of the poles about s= 1, the locus branches out from s= 1 ong Res= l parallel to the imaginary axis. This
× • For K < 0: The angle criteria states that: G(sl) = even integer multiple of �. So, if you take any point on the real line segment (−1,∗), G(sl) = 0 0 = 0. ���� − ���� phase due to zeros phase due to the pole Combining the two cases, the root loci for both cases are plotted as shown below. The solid line corresponds to the locus for K > 0 case and the dashed line for K < 0 case. ≤m −1 ↓e (b) Given: 1 G(s) = . (s − 5)(s + 3) G(s) has two poles at 5 and −3, and no finite zero. • For K > 0 : Evoking the angle criteria, we can first see that the real line segment (−3, 5) belongs to the locus. Now we would like to know at which point the locus branches out or bifurcates from the real line. At the branching point, the closed loop poles are of double pole. To find such a point, we can first find a value of K corresponding to the double pole. 1 + KG(s) = 0 K 1 + = 0 (s − 5)(s + 3) (s − 5)(s + 3) + K s2 − 2s − 15 + K (s − 1)2 − 1 − 15 + K = = = 0 0 0 completing the square. So, when −1 − 15 + K = 0 or K = 16, the closed loop system has a double pole at 1. Because of the symmetry of the poles about s = 1, the locus branches out from s = 1 along ↓e{s} = 1 parallel to the imaginary axis. This can be seen in the figure below: 3
m For K <0: In this case, using the angle criteria, we can see that the real line segment (5, oo)belongs to the locus since both poles contributes 0 or even integer multiple of T Also, the real line segment(oo, -3)belongs to the locus since both poles contribute T or odd integer multiple of T; the net phase of G(sn) is even integer multiple of T Thus, the root loci containing both K>0(solid line) and K <0(dashed line)cases are shown below (c)Given G(s) has a double-pole at 0 and one finite zero at-1 0: With the angle criteria, we can see that the real line segment (oo,-1 belongs to the locus. Since both poles of G(s) do not belong to the segment, there will be a double-pole point in the segment. To find where it is, use the same technique as in(b)
1 × × 1 ≤m ↓e × −3 × 5 • For K < 0 : In this case, using the angle criteria, we can see that the real line segment (5,∗) belongs to the locus since both poles contributes 0 or even integer multiple of �. Also, the real line segment (−∗, −3) belongs to the locus since both poles contribute � or odd integer multiple of �; the net phase of G(sl) is even integer multiple of �. Thus, the root loci containing both K > 0 (solid line) and K < 0 (dashed line) cases are shown below: ≤m −3 5 ↓e (c) Given: s + 1 G(s) = 2 . s G(s) has a double-pole at 0 and one finite zero at −1. • For K > 0 : With the angle criteria, we can see that the real line segment (−∗, −1) belongs to the locus. Since both poles of G(s) do not belong to the segment, there will be a double-pole point in the segment. To find where it is, use the same technique as in (b), 4
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