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: 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: 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 K0 case and the dashed line 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 case and the dashed line 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(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 (solid line) and 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 00(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 0 (solid) and K < 0 (dashed) cases are shown below: 5
Sn Problem 2 (a) First let output of the plant G(s) s+10 be y(t. Then, by Black's formula, the clos transfer function from the input a(t)to y(t) can be found as Y G(sK(s 1+G(s)K(s) X(s)8(s+10)+K The error signal e(t)is the difference between r(t) and y(t). Thus from the linearity of Laplace rIms e(t)= (t-y(t) E(s (s)-Y(s) So, with Eqns(2) and 3), we can obtain the transfer function from the input (t) to the error
× ≤m 2 −2 −1 ↓e Problem 2 1 (a) First let output of the plant G(s) = s(s+10) be y(t). Then, by Black’s formula, the closed transfer function from the input x(t) to y(t) can be found as: Y (s) = G(s)K(s) X(s) 1 + G(s)K(s) K = s(s+10) 1 + K s(s+10) � Y (s) X(s) = K s(s + 10) + K . (2) The error signal e(t) is the difference between x(t) and y(t). Thus from the linearity of Laplace transforms: e(t) = x(t) − y(t) E(s) = X(s) − Y (s). (3) So, with Eqns (2) and (3), we can obtain the transfer function from the input x(t) to the error e(t): 6
e(s X(s E s(s+1)+K (s+10)+K s(s+10) X(s)82+10s+K Thus, using Eqn(4), E(s)for the unit step input s(t=u(t) is E s2+10s+K s(s+10)I +10s+K s2+10s+K so if K 0. Then, both poles of E(s)are in the left half plane and the order of the numerator is strictly less than that of the denominator (the latter condition is usually referred to as being strictly proper ). Thus we can apply the final value theorem to compute the steady state tracking error: lim e(t) s+10 +10s+K Thus, regardless of the value of K, as far as it is positive, e(oo)=0 (b) To compute the error signal to the ramp input, we can use Eqn(4)with now X(s= E s2+10s+Ks2 (s2+10s+K 2#s+1-
E(s) X(s) = 1 − Y (s) X(s) � Eqn (3) K = 1 − s(s + 1) � Eqn (2) + K s(s + 10) = s(s + 10) + K E(s) s(s + 10) � = . (4) X(s) s2 + 10s + K Thus, using Eqn (4), E(s) for the unit step input x(t) = u(t) is: s(s + 10) E(s) = X(s) s2 + 10s + K s(s + 10) 1 = s2 + 10s + K s s + 10 = s2 + 10s + K , so if K → 0, we can see from Routh-Hurwitz criteria that E(s) has at least one of the poles in the right half plane, i.e, the real part of the pole is positive. Hence, in this case, the steady state tracking error diverges to ∗. Now, consider the case when K > 0. Then, both poles of E(s) are in the left half plane and the order of the numerator is strictly less than that of the denominator (the latter condition is usually referred to as being strictly proper) . Thus we can apply the final value theorem to compute the steady state tracking error: e(∗) = lim e(t) t�� = lim sE(s) s�0 s + 10 = lim s s�0 s2 + 10s + K = 0. Thus, regardless of the value of K, as far as it is positive, e(∗) = 0. (b) To compute the error signal to the ramp input, we can use Eqn(4) with now X(s) = s 1 2 : s(s + 10) 1 E(s) = 2 s2 + 10s + K s s + 10 = s(s2 + 10s + K) 10 10 s + 1 − 100 K + = − K K . s s ���� 2 + 10s + K � �� � E1(s) E2(s) 7
el(t), the corresponding time signal to E1(s)is a scaled step. E2(s) has two poles and they remain in the left-half plane as long as K>0. In this case e2(t), the corresponding time signal to E2(s) will decay to 0 as t-00. Thus, e(oo)=k, i. e, the steady state error will not be On the other hand, if K <0. Then, E2(s)now has at least one of the poles in the right-half plane.Thus,e2(∞)→∞ast→∞; therefore e(o)→∞. (c)For this part, first we would like to get the transfer function from a (t) to y(t). Let's denote the output of Kf, af(t), then X(s)=KyX(s) 1 Y(s (s+10) r (s)+KsX(s-y(s) s(s+10) Y(s) (s+10 (K+Ks)X(s) (Kf+Ks) X 1 X(s) (s+10)+1 Since e(t)=r(t)-y(t), we can compute the Laplace transform of e(t), E(s)as follows E(s)=X(s-Y(s) E(s The input a(t) to the system is a ramp, so X(s=. Thus, E(s) is Kf+K 10s+1-(Kf+K) 82(s2+10s+1 +10s+1-(Kf+K) s(s2+10s+1) To be able to apply the final value theorem, we want to choose K,(s)and Ks(s)such that the poles of sE() are in the left-half plane. In addition, we need to choose K(s)and Ks() such lim SE(s)=0 o ensure that the stead state error to the ramp input is zero. Combining these two conditions we see that we need to make the numerator of sE(s)equal to s. From this conclusion alone, we only know that K(s)+Ks(s)=10s+ 1. Assuming that neither is 0, then we can choose
� = . e1(t), the corresponding time signal to E1(s) is a scaled step. E2(s) has two poles and they remain in the left-half plane as long as K > 0. In this case e2(t), the corresponding time signal to E2(s) will decay to 0 as t ∩ ∗. Thus, e(∗) = 10 K , i.e., the steady state error will not be 0. On the other hand, if K → 0. Then, E2(s) now has at least one of the poles in the right-half plane. Thus, e2(∗) ∩ ∗ as t ∩ ∗; therefore e(∗) ∩ ∗. (c) For this part, first we would like to get the transfer function from x(t) to y(t). Let’s denote the output of Kf , xf (t), then Xf (s) = KfX(s). 1 Y (s) = s(s + 10)(Xf (s) + KsX(s) − Y (s)) � 1 � 1 1 + s(s + 10) Y (s) = s(s + 10)(Kf + Ks)X(s) 1 s(s+10) Y (Kf + Ks) (s) = X(s) 1 + 1 s(s+10) Y (s) Kf + Ks X(s) s(s + 10) + 1 Since e(t) = x(t) − y(t), we can compute the Laplace transform of e(t), E(s) as follows: E(s) = X(s) − Y (s) � Y (s) � E(s) = 1 − X(s) X(s). The input x(t) to the system is a ramp, so X(s) = s 1 2 . Thus, E(s) is: � Kf + Ks � 1 2 E(s) = 1 − s(s + 10) + 1 s s2 + 10s + 1 − (Kf + Ks) = s2(s2 + 10s + 1) ∩ sE(s) = s2 + 10s + 1 − (Kf + Ks) . s(s2 + 10s + 1) To be able to apply the final value theorem, we want to choose Kf (s) and Ks(s) such that the poles of sE(s) are in the left-half plane. In addition, we need to choose Kf (s) and Ks(s) such that lim sE(s) = 0 s�0 to ensure that the stead state error to the ramp input is zero. Combining these two conditions, we see that we need to make the numerator of sE(s) equal to s2. From this conclusion alone, we only know that Kf (s) + Ks(s) = 10s + 1. Assuming that neither is 0, then we can choose 8
Kf(s)=10s and Ks =1. Then, SE(s +10s+1-(10s+1) s(s2+10s+1) (s2+10s+1) 2+10s+ Now, we can safely apply the final value theorem to obtain the steady state error and we indeed obtain o as lim SE(s= lim Problem 3(O&W 11.27) Given H()=2+25+4Cs)=K First, let's identify the poles and zeros of H(s). Since the system H(s) is a second order system there are two poles and two zeros. It is clear that one of the zeros is of finite at -2 and the other is at oo. The poles are the solutions of s+2s+4=0, s0-1++V3. Thus, the pole-zero diagram is as shown below Sm Since H(s)is causal, the ROC for H(s)is Res)>-1 The closed loop poles are the solution to the following equation 1+KH(s)=1+K
Kf (s) = 10s and Ks = 1. Then, s2 + 10s + 1 − (10s + 1) sE(s) = s(s2 + 10s + 1) 2s = s(s2 + 10s + 1) s � sE(s) = . s2 + 10s + 1 Now, we can safely apply the final value theorem to obtain the steady state error and we indeed obtain 0 as expected: s lim sE(s) = lim = 0. s�0 s�0 s2 + 10s + 1 Problem 3 (O&W 11.27) Given: s + 2 H(s) = s2 + 2s + 4 , G(s) = K. First, let’s identify the poles and zeros of H(s). Since the system H(s) is a second order system, is at ∗. The poles are the solutions of s is as shown below: − 2 s − ±⇒ there are two poles and two zeros. It is clear that one of the zeros is of finite at 2 and the other + 2 + 4 = 0, so 1 + 3. Thus, the pole-zero diagram ≤m ↓e × × ⇒3 − ⇒ 3 −2 −1 Since H(s) is causal, the ROC for H(s) is ↓e{s} > −1. The closed loop poles are the solution to the following equation: s + 2 1 + KH(s) = 1 + K = 0, s2 + 2s + 4 9
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= 6
or 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 closedloop impulse response does not exhibit any oscillatory behavior simply means that the closedloop 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
