Unformatted text preview: SteadyState Error:
r
 e L(s) = DG(s) y The transfer function from reference to error is: E s = 1 + 1 s Rs; Ls = DGs: L If the feedback system is stable, then the nalvalue theorem e = slim sE s = slim 1 + s s Rs: ! ! L
ss 0 0 For unit step input, Rs = 1 , and thus s ess = slim0 1 + 1 s = 1 + lim1 Ls = 1 +1K ; ! L s!0 p where Kp = lims!0 Ls is called position constant. For unit ramp input, Rs = s12 , and thus 1 1 ess = slim0 1 + 1 s s = lim 1sLs = K ; ! L s!0 v where Kv = lims!0 sLs is called velocity constant. For an input of the form Rs = sk1+1 , ess = slim0 1 + 1 s s1k = lim 1sk Ls : ! L s!0
1 Signal Type:
A signal is called type k, if it has the form tk . A unit kth order reference signal has the form tk ; Rs = 1 : rt = k! sk+1 System Type: A feedback system is called type 0, if Kp = slim Ls = nonzero constant: !
0 A feedback system is called type 1, if Kv = slim sLs = nonzero constant: !
0 Ls is called type 2, if Ka = slim s Ls = nonzero constant: !
2 0 A feedback system is called type k, if Equivalently, k = number of poles of Ls at the origin. lim sk Ls = nonzero constant: s!0 2 Fact: Type k systems can track kth order input with nonzero nite error, and k , ith order input with zero error, but unable to track k + lth order input where i and l are positive integers. For type 0 systems, Kp is a nonzero constant, and Kv = Ka = 0. Thus 8 1 ; if input = unit step; 1 + Kp ess = : 1; if input = ramp or higher order. For type 1 systems, Kv is a nonzero constant, and Kp = 1, Ka = 0. Thus if input = step; ess = K ; if input = unit ramp; v : 1; if input = parabola or higher order. See Figure 4.34 on page 204 for the physical meaning of ess. For type 2 systems, Kp = Kv = 1, and Ka = nonzero constant. Thus 8 0; if input = step; ess = 0; if input = ramp; 1 ; if input = unit parabola. :
8 0; 1 Ka 3 Design Example: Suppose a D.C. motor is given by
Design PI compensator 800 Gs = s + 4s + 15 :
0 Ds = K @1 + such that ess = 0 for step input, ess 1 for unit ramp, P.O. 5, and ts 1 sec. Answer: For PI compensator, choose TI = 0:25. Hence Ls = DGs = s800K : s + 15 This is type 1 system. Since ess 1 is equivalent to Kv 1, we have that 15 Kv = slim0 sLs = 800K 1 K 800 = 0:0187: ! 15 For P.O. requirement, we have that 0 = r 2j log0:05j 2 = 0:69: + j log0:05j The closedloop transfer function is 2 !n 800K T s = s2 + 15s + 800K = s2 + 2
! s + !2 : n n sTI 1 1 A 4 Therefore !n = 800K , and 2
!n = 15
0 @ p The above is equivalent to 15 = 2p800K 0:69: 15 12 K 20 0:69 A =8 = 0:1477: Finally we verify the settling time: 3: ts = 3:22 = 3:22 = 1522 = 0:4293 1s: !n =2 That is, settling time is independent of K . Now we conclude 0:0187 K 0:1477 will ensure ess 1 for unit ramp input, and P.O. no larger than 5. We note that the cancellation of the pole at ,15 by choosing TI = 1=15 does not work. In this case, 800K 800K Ls = DGs = ss + 4 ; T s = s2 + 4s + 800K : Hence 2
!n = 4 that yields = !n = 2. We then have 22 ts = 3:22 = 3:2 = 1:61 1; violating the requirement. 5 Another Design Example:
We continue our example in the previous note for the plant 25 104 Gs = s + 62 + 144 s + 202 + 100 : The design requirements are now: i P.O. 10, and ii Steady state error 0:1 for unit ramp input. s Ds = Kpss+ as+ b ; p = 100; a = 12; b = 180; +p achieves the 10 P.O. for K = 0:0132. Its requirement on steady2 Answer: As in the previous note, state error for unit ramp input is: ess = 1 0:1; Kv 10: For this example, Hence Kv 6 103Kp Ls = ss + ps + 12 6 103K Kv = slim0 sLs = 12 = 6:6 10 ! violating the requirement of ess = 0:1. The actual tracking error in steadystate is ess = 1=6:6 = 0:1515 0:1:
6 Disturbanace Rejection:
w r
 e D u G y The transfer function from disturbance to output is given by Twy s = 1 +Gs s : DG Assuming that G0 6= 0. If the disturbance input is Then the steadystate response of the output due to disturbance is yss = slim0 sTwy sW s = slim0 1 + G0w 0 : ! ! D s G If Ds is type 0, then Dp = lims!0 Ds 6= 0, and yss = 1 +G0 0 : DpG If Ds is type 1, then W s = w ; w 0: s Dv = slim sDs 6= 0: !
0 7 Furthermore, Dp = 1 which leads to yss = 0. For ramp disturbance, w W s = s2 ; where w is unknown. Then type 1 assumption for Ds yields G0 w yss = slim0 s + sDswG0 = lim wsDs = D : ! s!0 v given by A Similar Design Example: Suppose that a D.C. motor is
800 Gs = s + 4s + 15 :
0 Ds = K @1 + Design PI compensator such that ess = 0 for step input, ess 5 for unit ramp disturbance input, P.O. 5, and ts 1 sec. sTI 1 1 A Answer: For PI compensator, choose TI = 0:25. Hence
Since Ds is type 1, ess 5 is equivalent to Dv 0:2, we have that Ls = DGs = s800K : s + 15 K Dv = slim sDs = T = 4K 0:2 = K 0:05: !
0 I 8 For P.O. requirement, we have a similar situation: 0 = r 2j log0:05j 2 = 0:69: + j log0:05j The closedloop transfer function is 2 800K !n T s = s2 + 15s + 800K = s2 + 2
! s + !2 : n n Therefore !n = 800K , and 2
!n = 15
0 @ p The above is equivalent to 15 = 2p800K 0:69: 15 12 K 20 0:69 A =8 = 0:1477: Finally we verify the settling time: 3: ts = 3:22 = 3:22 = 1522 = 0:4293 1s: !n =2 That is, settling time is independent of K . Now we conclude 0:05 K 0:1477 will ensure ess 5 for unit ramp disturbance input, and P.O. no larger than 5. 9 Nonunity Feedback:
r  u G(s) H (s) y There holds 2 4 E s = Rs , Y s = 61 , For order k input, Rs = e ss A feedback system capable of tracking order k input with nite error is called type k system. Note that type k systems with nonunity feedback are di erent from those with unity feedback. 2 1 = s!0 6 k lim 4 s sk , and
+1 1 Gs 7 Rs: 5 1 + GH s
3 3 5 , Tsks 7 : 10 For unit step input, k = 0, and e = 1 , T 0:
ss For unit ramp input, k = 1, and ess = slim0 1 , T s : ! Similar properties hold for nonunity feedback systems. s Example: Consider feedback system given by
Discuss steadystate errors. Gs = s s1+ 1 ; H s = h = constant: Answer:: The closedloop transfer function is
2 For unit step input,
ss T s = 1 +Gs s = s +1s + h : GH
8 1 e = 1 , T 0 = 1 , h = : 0; For unit ramp input,
2 ss 0 0 const.;
8 if h = 1; if h 6= 1: T e = slim 1 , s s = slim s + ss+ h , 1 : 1; if h = 1; ! ! 1; if h 6= 1: 11 Truxal's Formula:
Suppose that T s = K s , z s , z s , zm : s , p s , p s,p
1 2 1 2 Assume that T 0 = 1. Then the tracking error to the unit ramp input is: T ess = slim0 1 , s s = , slim0 dT s ; ! ds ! by L'Hopital's rule. Since T 0 = 1, 1 ess = , slim0 dT s = , slim0 T s dT s = , slim0 d logT s : ! ds ! ! ds ds Substituting the expression of T s yields Hence by taking the derivative for each individual term, and taking the limit s ! 0, we obtain m n ess = 1 = X 1 , X 1 :
m n d 2K + X logs , z , X logs , p 3 : 4 e = , slim ds i i5 ! i i
ss 0 =1 =1 n Kv i=1 zi i=1 pi 12 Stability Testing
Suppose that a system admits rational transfer function: bs = b0sm + b1sm,1 + bm : T s = as sn + a sn,1 + a 1 1 n If bs and as have no common roots, then its stability is determined by roots distribution of as. Routh array is the tool to determine roots distribution of a given polynomial. Routh's Stability Criterion Given a polynomial as as in 1. Routh array is given by Row n Row n , 1 Row n , 2 Row n , 3 .. Row 1 Row 0 sn: sn, : sn, : sn, :
1 2 3 a 1
1 1 a a 2 3 2 a a 4 5 3 s s
1 0 1 0 2 0 0 3 0 0 The formula to compute k 's, k 's, and all others can be found in book, and will be given in the next page. 13 The computation of the element at n , ith row and kth column use the 2 2 elements in the previous two rows, one at the rst column, and the other at the next column. Speci cally they are given by 1 2 3 1 2 a a a a ,a ; = a 3 a 2 6 1 a 7 7 det 6 4 a a 5 a a ,a = , a 3= a ; 2 6 1 a 7 7 det 6 4 a a 5 a a ,a = = , a a ; 2 3 a a 7 6 7 det 6 4 5 a ,a ; = , = 2 3 6a a 7 7 det 6 4 5 a ,a ; = , =
= , 2 6 det 6 4 1 a2
1 3 1 3 7 7 5 1 2 3 1 4 5 1 1 4 5 1 1 6 7 1 1 6 7 1 1 1 3 1 2 1 3 1 2 1 1 1 5 1 3 1 5 1 3 3 = , 2 6 det 6 4 a a
1 1 1 1 7 4 3 7 7 5 1 = 1 a ,a :
7 1 4 1 14 Example: Consider polynomial
as = s + 4s + 3s + 2s + s + 4s + 4:
6 5 4 3 2 The Routh array is then given by s s s s s s s 6 5 4 3 2 ,
0 1 4 2:5 2 3 4
76 15 , 3 2 0 4 0 0
12 5 1 4 4 0 0 0 0 4 0 0 0 0 0 0 The row of s4 is computed as 43,12 41,41 44,10 = 2:5; = 0; = 4: 4 4 4 The row of s3 is computed as 2:5 2 , 4 0 2:5 4 , 4 4 12 = 2; =, : 2:5 2:5 5 The row of s2 is computed as 2 0 , 2:5,12=5 24,20 = 3; = 4: 2 2 The rows of s and s0 are computed as 76 4 ,76=15 , 0 3,12=5 , 8 =, ; 3 15 ,76=15 = 4:
15 Roots Location: The number of sign change at the rst column of the Routh array is the same as the number of roots on open right half plane. Hence for the example in the previous page, two roots of as lie on open right half plane, and thus as is unstable. Special Case 1: Zero element in the rst column. In this case, replace 0 by 0, and proceed as before. Limit ! 0 will be taken to determine stability. Example: Consider polynomial as = s + 3s + 2s + 6s + 6s + 9:
5 4 3 2 Determine number of roots on ORHP. Answer: The Routh array is: s s s s s s 5 4 3 2 0!
6 1 3 ,9
2 0 3 + 99 6 , 3 2 6 9 3 0 0 6 9 0 0 0 0 Thus there are two poles on ORHP two sign changes as ! 0.
16 Special Case 2: Zero row. In this case, set auxiliary polynomial a1s using the previous row as coe cients, and then compute a01s or derivative with respect to s. Replace the zero row by coe cients of a01s, and then proceed. Example: Consider polynomial as = s + 5s + 11s + 23s + 28s + 12:
5 4 3 2 Determine number of roots on ORHP. Answer: The Routh array is: s s s s s New s s
5 4 3 2 0 1 5 6:4 3 0 6 12 11 23 25:6 12 0 0 0 28 12 0 0 0 0 0 The auxiliary polynomial is: a s = 3s + 12; a0 s = 6s
1 2 1 which replaces the zero row of s. No sign change in the 1st column implies no root on ORHP. 17 Fact: The roots of as on the imaginary axis are also the
roots of auxiliary polynomial a1s. Reexamine the previous example, the roots of the auxiliary polynomial are a s = s + 4 = 0; or s ; = 2j; j = ,1:
1 2 12 p Substitute s = s1;2 = 2j into as yields a2j = j 2 + 5j 2 + 11j 2 + 23j 2 +28j 2 + 12 = 32j + 80 , 88j , 92 + 56j + 12 = 0:
5 4 3 2 Hence s = 2j are also roots of as. Roots distribution: An nth order real coe cient polynomial has n roots. The number of roots on open right half plane nr is the same as the number of sign change in the 1st column of the Routh array. The number of roots on the imaginary axis nc can be determined by the roots of auxiliary polynomial. Finally the number of roots on open left half plane rl is given by rl = n , nr , nc: due to the fact that nl + nr + nc = n. 18 Example: Determine roots distribution of
as = s + 2s + 7s + 14s + 16s + 32:
5 4 3 2 Answer: The Routh array is:
s s s New s s s s
5 4 3 3 2 0 0 32
2 1 2 0 8 7 7 14 0 28 32 0 0 16 32 0 0 0 0 0
3 The auxiliary polynomial is: a s = 2s + 14s + 32; a0 s = 8s + 28s:
1 4 1 The 1st elements for the row of s2 and s are 7 28 , 8 32 8 14 , 2 28 = 7; = ,8:5714: 8 7 Hence two sign changes imply nr = 2: Two roots on ORHP. For roots on imaginary axis, set x = s2 yields px = 2x + 14x + 32 = 0 x = ,3:5 j p3:75: a ;
1 2 12 19 Note that in polar form, x ; = 4ej
12 180o 28:955o : 14:4775o that gives the four roots of a1s: s ; = 2ej
12 90o 14:4775o; s ; = 2ej
34 270o We thus conclude nc = 0: No root on the imaginary axis. The number of roots on OLHP is now given by nl = n , nr , nc = 5 , 2 , 0 = 3:
MATLAB gives the exact roots as follows: s ; = ,0:5000 1:9365j; s ; = 0:5000 1:9365j;
12 34 and s5 = ,2:0000. Note that the rst four roots are in mirror pattern in the sense that if s is one root, then s; s are also roots. This holds for the roots of auxiliary roots. Stability Versus Parameter Range:
Stability of feedback system depends on parameters of the plant and the compensator. It is some time crucial to determine range of the underlying parameter which ensure stability. 20 system: Example: Find stability range of K for the following feedback
r
 e K u s+1 s(s1)(s+6) y Answer: The closedloop trasnfer function is:
3 2 Hence the poles are the roots of denominator polynomial. Using RouthHurwitz method we obtain + 1 T s = s + 5s K sK , 6s + K : + s s s s
3 2 0 1 5 5K , 30 1 5 K,6 K
0 0 K Stability of the feedback system is equivalent to 5K , 30 0; and K 0; 5 which is in turn equivalent to K 6 and K 0. Hence K ensures stability. 6 21 Example: Find stability range of K for
4 3 2 Answer: The Routh array is:
s s s s s
4 3 2 0 K Gs = s + s + 5Ks + 2Ks + 12K + 18 :
1 1 3K xK 12K + 18 5K 2K 12K + 18 0 0 12K + 18 0 0 0 0 where by stability 6K 2 , 12K , 18 xK = 0 3K i 6K 2 , 12K , 18 0; and ii K
r
2 0:
8 For i, its roots are: Hence i is true if K 3, or K ,1. But K 0 in ii rules out the possibility of K ,1. We thus conclude the stability range of K as: K 3. Note that 12K + 8 0 is true if K 0. 12 ,12 + 4 6 18 3 K1;2 = =12=: 6 ,1 22 Ultimate Gain:
The ultimate gain method requires computation of the critical gain Kc. The Routh array provides a way to determine Kc. Let the plant be bs = b0sm + b1sm,1 + + bm : Gs = as sn + a sn,1 + + a 1 n Then the closedloop denominator is a function of Kc also. The oscillation of the closedloop system is equivalent to the existence of poles on j!axis, which is equivalent to the existence of zero row in the Routh array. Hence Kc can be obtained accordingly. How to calculate the ultimate period? Note that the roots of the auxiliary polynomial 1s = 0 are also the roots of s = 0. They are on the imaginary axis, and thus roots = j!o: The corresponding inverse Laplace transform has the form A cos!ot + which has the ultimate period Pu = 2 : !
o 23 Example: Consider plant model
4 3 Determine its associated ultimate gain. Answer: We rst note that Gs is stable that has poles at 1000 Gs = s + 27s + 160s + 450s + 1000 :
2 s ; = ,1 3j; s = ,5; s = ,20:
12 3 4 The characteristic polynomial for the closedloop is: s = s4 + 27s3 + 160s2 + 450s + 1000 + 1000K which is stable for K = 0. For K 0, its stability needs to be determined using RouthHurwitz method. Routh array for s is: s s s s s
4 3 2 0 1 27 1000 + 1000K xK 430 3 160 450 1000 + 1000K 0 0 1000 + 1000K 0 0 0 0 24 The expression of xK is given by xK = 150 430 , 27 1000 + 1000K :
Hence at K = 0, xK = 0 = 150 430 , 27 1000 = 37500 0:
That is, s, or Gs is stable at K = 0. As K increases to K = Kc, xKc = 0 eventually that gives zero row. Since no pole on open right half plane at K = Kc, Kc is the ultimate gain which can be solved by setting xKc = 0: To compute the ultimate period, set auxiliary polynomial 430 s = s2 + 1000 + 1000Kc = 0 1 3 which gives
v u u 30001 + Kc t s = j u xKc = 150 430 , 27 1000 + 1000Kc = 0 Kc = 5 43 , 1 1:38889: 90 Hence the ultimate gain is: 430 = j 4:0833: Pu = 2=4:0833 = 1:5388: 25 Stability Versus Two Parameter Ranges:
Find stability ranges of K; KI for the PI control system:
r
 e K K+ s I u 1 (s+1)(s+2) y is: Answer: The characteristic equation of the closedloop system
as = s + 3s + 2 + K s + KI = 0:
3 2 Its Routh array is: s s s s
3 2 0 1 3 6 + 3K , KI 1 3 2+K KI
0 0 KI For stability, we must have KI 0; KI 3K + 6:
The shaded region of the following picture shows the stability range of K; KI see next page: 26 KI 6 KI =3K+6 K 2 Stability of the feedback system is insured only if the parameters of KI ; K fall into the above shaded region. It is note that the textbook draws it di erently reverse the axis of K and KI . not only to be stable, but also to admit some stability margin. For instance, we would like to place all the closedloop poles on the left of s = ,r, r 0. Routh Hurwitz method can be modi ed to test if all roots on the left of s = ,r. Stability Margin: In some applications, the system is required Since as 6= 0 for all s such that Re s ,r is equivalent to a~ = a~ , r 6= 0 for all s = s + r on closedloop half plane, ~s s ~ we can applied Routh array to a~. ~s 27 Example: Determine roots distribution with respect to s = ,1
for the following polynomial:
4 as = s + 5s + 11s + 13s + 8:
3 2 ~ Answer: Replacing s by s , 1 gives a~ = ~ , 1 + 5~ , 1 + 11~ , 1 + 13~ , 1 + 8 ~s s s s s
4 3 2 = s4 + s3 + 2~2 + 2~ + 2: ~ ~ s s The Routh array is: s ~ s ~ s ~ s ~ s ~
4 3 2 0 0!
2 1 1 2 ,2 2 2 2 0 0 2 0 0 0 0 As ! 0, the rst column has two sign changes. Hence two roots of a~ on open RHP which implies that the other two are on open ~s LHP, because there is no root on j!axis. Equivalently, as has two roots on the right of s = ,1, and two on the left of s = ,1. 28 Fact 1: If any coe cient of as is zero or negative, as has
some roots on the CRHP.
1 Proof: Let rk be the kth root of as. Then as = sn + a sn, + + an = s , r s , r s , rn:
1 1 2 It follows that Stability implies that Re ,rk and nonzero.
n X a1 = , rk ; k=1 a = X rirk ; ; an = ,1nr r rn:
2 i6=k 1 2 0, and thus ak are real positive zero, as has some roots on the CRHP . element for the next row. Fact 2: If any number in the rst column of the Routh array is Proof: In this case, 0 is replaced by which gives a negative
Actually Fact 2 can be modi ed to include the case where zero or negative element appear in the middle of the Routh array not include the last a few zero elements in the same row. Therefore, stability test can terminate whenever we see a zero or negative element in the middle of the Routh array. 29 Example: Determine stability for
as = s + 2s + 6s + 3s + s + 2s + 2:25:
6 5 4 3 2 Answer: Its Routh array is:
s s s s s s s
6 5 4 3 2 0 1 2 4:5 3 ,1:5 5:5 2:25 6 3 0 1 2:25 0 0 1 2 2:25 0 0 0 0 2:25 0 0 0 0 0 0 For stability test, we can stop at the row of s4 because the second element is zero which will eventually bring a negative element to the rst column. However if we stop at the row of s4, we will not be able to determine roots distribution of as. 30 ...
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 Fall '07
 Chen
 Trigraph, Routh Hurwitz, Routh array

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