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A limiting property of the inverse of sampled-data systems on a finite-time interval
Sogo, T; Adachi, N
IEEE TRANSACTIONS ON AUTOMATIC CONTROL (2001), 46(5): 761-765
2001-05
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http://hdl.handle.net/2433/50315
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(c)2001 IEEE. Personal use of this material is permitted. However, permission to reprint/republish this material for advertising or promotional purposes or for creating new collective works for resale or redistribution to servers or lists, or to reuse any copyrighted component of this work in other works must be obtained from the IEEE.
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Journal Article
Textversion
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Kyoto University
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
A Limiting Property of the Inverse of Sampled-Data Systems on a Finite-Time Interval Takuya Sogo and Norihiko Adachi
Abstract—If one considers a sampled-data system derived from a continuous-time system with a relative degree of one or two on a finite-time interval, it is not simple to predict the behavior of the output of the inverse of the sampled-data system as the sampling period goes to zero. This is because the number of sample points increases while the zeros of the pulse-transfer function tend to the boundary between the stable and unstable areas. This paper shows that the output of the sampled-data inverse systems converges to the output of the continuous-time inverse systems independently of the stability of zeros.
761
It should be noted that this property of zeros is common for systems whose relative degree is one or two [2], [8], [9]. In this paper, we will discuss such a limiting problem on the fixed time interval for systems with a relative degree of one or two. We will demonstrate that the output of the sampled-data inverse systems converges to the output of the continuous time inverse systems independently of the stability of the zeros when 1 goes to zero. II. MATHEMATICAL PRELIMINARIES Consider a linear continuous-time single-input–single-output (SISO) system
d x(t) =Ac x(t) + bc u(t) dt y (t) =cx(t)
Index Terms—Inverse systems, iterative learning control, limiting zeros, nonminimum phase systems, sampled-data systems.
I. INTRODUCTION Stable inverse systems or stable zeros of transfer functions are often required in many kinds of control problems defined on the infinite time horizon. However, since there is no simple relation between zeros of the pulse-transfer function of sampled-data systems and zeros of the transfer function of continuous-time systems, the behavior of the zeros of sampled-data systems has drawn much attention from researchers. Several approaches to determine the stability of zeros or avoid unstable zeros have been presented [1]–[10]. On the other hand, when control problems are defined on a fixed finite-time interval [0; tf ], one can admit unstable systems unless signals become too large inside [0; tf ]. One such finite-time control problems is, for example, iterative learning control which is a trial-based iterative method to improve the transient response on a short time interval [11], [12]. When continuous time systems or their inverse systems are considered on [0; tf ], the peak of the signals is simply determined by the distance between the imaginary axis and poles or zeros, respectively. However, when sampled-data systems with a sampling period 1 are considered on [0; tf ], the peak of the signals depends on the variable 1. This relationship is not simple because 1 changes both the number of sample points inside [0; tf ] and the location of poles and zeros. Furthermore, the zeros as a function of 1 are much more complicated than the poles. For example, consider the continuous time systems G1 (s) = (s 0 1)=s2 and G2 (s) = (2s 0 3)=f(s + 1)(s + 2)(s + 3)g. Then, the sampled-data systems derived from each with a sampler and a zero-order hold are H1 (z ) = f(2 0 1)1(z 0 (2 + 1)=(2 0 1))g=f2(z 0 1)2 g and H2 (z) = f (1)(z 0 q1 (1))(z 0 q2 (1))=f(z 0 exp(01))(z 0 exp(021))(z 0 exp(031))g, respectively, where q1 (1) = 1 + 31=2 + O(12 ) and q2 (1) = 01 + 51=2 + O(12 ) [8], [9]. Consider the inverse systems H101 (z ) and H201 (z ) on f0; 1; . . . ; tf =1g where tf =1 is assumed to be a natural number and give fyd (0); yd (1); . . . ; yd (tf )g that is the sampled-data of a function yd (t) to H101 (z ) and H201 (z ) as their inputs. Then, the unstable zeros for a small 1 make the output of the inverse systems increase exponentially. However, it is not easy to determine whether the output of the inverse systems diverges or converges inside [0; tf ] when 1 goes to zero, because all the zeros tend to points on the unit circle, i.e., the boundary between the stable and unstable areas, while the number of sample points increases inside [0; tf ].
where x 2 Rn , u 2 R, and y 2 R and a sampled-data system derived from (1) with a zero-order hold and a sampler with a sampling period 1. Then, we have
x((k + 1)1) =A1 x(k1) + b1 u(k1) y (k1) =cx(k1)
(2)
1
where A1 = exp Ac 1 and b1 = 0 exp(Ac )bc d . Assume that the transfer function G(s) = c(sI 0 Ac )01 bc is expressed as
G(s) =
K (s 0 1 )(s 0 2 ) 1 1 1 (s 0 m ) : (s 0 p1 )(s 0 p2 ) 1 1 1 (s 0 pn )
(3)
Then, since there exists a positive constant 0 such that cb1 6= 0 for all 1 2 (0; 0 ), the pulse-transfer function H (z ) = c(zI 0 A1 )01 b1 can be expressed as
H (z ) =
cb1 (z 0 q1 (1)) 1 1 1 (z 0 qn01 (1)) : (z 0 exp(p1 1)) 1 1 1 (z 0 exp(pn 1))
(4)
Next, consider a system (1) with the initial condition x(0) = 0 on a finite-time interval [0; tf ]. Then, the input–output mapping defined by (1) is expressed as y = Su (u; y 2 L2 [0; tf ]) where
Su =
t
0
ceA
(t0 ) bc u( )d :
(5)
Moreover, assume that the sampling period 1 satisfies 1 = tf =N , where N is a natural number. Then, the input–output relationship on the sample points f0; 1; 21; . . . ; tf 0 1; tf g is
w1
= 01 v1
(6)
where
= [ u(0) u(1) = [ y(0) y(1) 0 m1 (1) 01 = .. ..
1 1 1 u(tf ) ]T 1 1 1 y(tf ) ]T
v1 w1
.
Manuscript received November 29, 1999; revised August 3, 2000 and October 31, 2000. Recommended by Associate Editor T. Chen. The authors are with the Graduate School of Informatics, Kyoto University, Kyoto 606-8501, Japan (e-mail: [email protected]). Publisher Item Identifier S 0018-9286(01)03616-9.
(1)
m1 (N )
0
.
111
where the Markov parameter m1 (i) (i m1 (i) = cAi101 b1 .
0018–9286/01$10.00 ©2001 IEEE
(7) (8)
(9)
m1 (1)
0
= 1; . . . ; N ) is
defined as
762
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
Note that cb1 system x((k
6= 0 for almost all 1 > 0. Then, we have the inverse
+ 1)1) =fA1 0 b1 (cb1 )01 cA1 gx(k1) + b1 (cb1 )01y((k + 1)1) 01 u(k 1) = 0 (cb1 ) cA1 x(k 1) + (cb1 )01 y((k + 1)1) (k =0; . . . ; N 0 1):
k1 01+ 1 y3 0 u3 k1 j01+ 1 y3 0 1u3 jN1 + k1 1 u3 0 u3 k1 + (1 Su3 0 1 S1 1 u3 )j1 N =j01 3 3 + k1 1 u 0 u k1 :
(10)
By letting u(N 1) = 0, (10) defines a mapping from w1 to v1 . We can denote the mapping as 0+ 1 , the Moore-Penrose generalized inverse + of 01 , which is the mapping 01 from w1 to the minimizer of (w1 0 01 v1 )T (w1 0 01 v1 ) with the minimum norm. In the following discussions, we will use the sampling operator 1 : N +1 and the zero-order hold operator 1 : RN +1 ! L2 [0; tf ] ! R L2 [0; tf ], defined as follows:
1 [u] = [u(0)u(1) 1 1 1 u((N 0 1)1) u(tf )]T v (k ) if t 2 [(k 0 1)1; k1] t k = 1; 2; . . . ; 1 [1 v](t) = : t v 1 +1 if t = tf
(11)
= supfju(t)j; t 2 [0; tf )g, = 1; 2; . . . ; N (= tf =1)g, C k [0; tf ] denotes set of k -th continuously differentiable functions on [0; tf ]. Note that 01 v = 1 S1 v and if u3 2 C k [0; tf ] (k 1) then lim1!0 k1 1 u3 0 u3 k1 = 0. i
N !0 j0+1(1 Su3 0 1 S1 1 u3 )j1
k101+ 1 y3 0 u3 k1 ! 0 as
1
..
0
..
.
0
01 1
n0m
=fAc + bc F 01 cAc gx(t) + bc F 01 d dt dt n0m d 01 cAn0m x(t) + F 01 y (t) u(t) =F c x(t)
dt
1(z 0 exp(p 1)) 1 1 1 (z 0 exp(p 1)) 1 (z)01 = cb (z 0 1)(z 01 q (1)) 1 1 1 (z 0 q n (1)) 1 1 n01 (18 )
01 = (z 0 1) : H2 (z ) 1 Lemma 1: Assume n
311 defined as
(19)
0 m = 1 and consider the N 2 N matrix 11 (0)
0
m
311 =
.. .
..
(20)
.
11 (N 0 1) 1 1 1
11 (0)
0m01 bc and x(0) = 0. Let S 01 be the input–output where F = cAn c mapping of (14) on [0; tf ]. Then, u3 = S 01 y 3 ; (13) is equivalent to k1 01+ 1 y3 0 S 01 y3 k1 ! 0. This means that the discrete-time inverse system (10) converges to the continuous-time inverse system (14) as 1 ! 0.
m
= 0; . . . ; N 0 1) are defined as 11 (i) = 1=2j C H1 (z)01 zi01 dz (C : a simple path enclosing all poles) Then, sup j311 v jN 1 =jvjN1 ; v 2 RN < +1 for 1 2 (0; 0 ). where Markov parameters m11 (i) (i
m
Proof: See Appendix. Lemma 2: Assume n 0 m = 1. Then, j321 (1 Su3 0 3 N 1 S1 1 u )j1 ! 0 as 1 ! 0. Proof: See Appendix. Next, we consider the case of n 0 m = 2 and N 2 N matrix 331 + (N; N +1) = 331 341 351 where 341 is an N 2(N 01) such that 01 matrix defined as
y (t)
(14)
(17)
H
m
Remark 1: The convergence (13) is presented only on [0; tf ), while + 1 y3 0 u3 is defined on [0; tf ]. However, this is the the function 1 01 + we have [1 0+ 1 y3 0 best result because from the definition of 01 1 3 3 u ](tf ) u (tf ), which is independent of 1. It should be noted that the result of Theorem 1 is independent of the + 1 y3 converges even if some zeros stability of the zeros of H (z ); 01 go to the unit circle from the outside. If a function y 3 (t) (t 2 [0; tf ]) satisfies the assumption in Theorem 1, u3 can be obtained such that y 3 = Su3 by using the following continuous-time inverse system of (1)
:
.
Consider the inverse system (10) on the infinite interval and the pulse-transfer function H (z )01 . Then, we can see that the + (N; N + 1) given above corresponds to decomposition of 01 0 1 0 1 H (z ) = H1 (z ) H2 (z )01 where
(13)
! 0.
d
01 1
321 = 1 1
III. THE MAIN RESULTS
2
(16)
as 1 ! 0. First, we consider the case of n 0 m = 1 and N 2 N matrix 311 + (N; N + 1) = 311 321 where 0+ (N; N + 1) indicates such that 01 + without its N +1-th row; 321 is1an N 2 (N +1) matrix the matrix 01 defined as
In the following sections, we consider the inverse discrete-time system with sampled data of a fixed function y 3 defined in [0; tf ] as + 1 y3 or (10) whose y((k + 1)1) is substituted the input, i.e., 01 3 with y ((k + 1)1).
+ 1 y3 as 1 ! 0. In this section, we present the limit of 01 Theorem 1: Assume that n 0 m = 1 or 2 and there exists u3 n 0m01 [0; tf ] such that y3 = Su3 . Then C
(15)
What we have to show is
(12)
We define the following notations: kuk1
jvjN1 = supfjvi j;
The proof of Theorem 1 will be established in the following sequence of lemmas1 ; all assumptions in Theorem 1 will be preserved. Since we have
341 =
0 1 01 .. .
(01)N
0m
0 1
..
.
0 ..
.
..
.
(21)
0
1 1 1 01 1
1Theorem for the case of n = 1 has been proved in [13], [14]; the approach was based only on a state-space representation. In this paper, we present a refined proof which uses an asymptotic property of zeros of the pulse transfer function.
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
Fig. 1. Left: u
= 0
and 351 is an (N
y with u . Right: y
0y
= S 0
y
763
0y
0 1) 2 (N + 1) matrix defined as 1 02 1 0
351 = 12 1
..
.
..
.
..
.
:
(22)
1 02 1 We can see that the decomposition of 0+ 1 (N; N + 1) given above corresponds to H (z )01 = H3 (z )01 H4 (z )01 H5 (z )01 where 01 = 12 (z + 1)(z 0 exp(p1 1)) 1 1 1 (z 0 exp(pn 1)) ; (23) H3 (z ) cb1 (z 0 1)2 (z 0 q1 (1)) 1 1 1 (z 0 qn01 (1)) 01 = 1=(z + 1) and H5 (z )01 = (z 0 1)2 =12 . H4 (z ) Lemma 3: Assume n 0 m = 2 and consider the N 2 N matrix 331 defined as m31 (0) 0 .. . . 331 = (24) . . m31 (N 0 1) 1 1 1 m31 (0) 0
= 0; . . . ; N 0 1) are defined as 31 (i) = (1=2j ) C H3 (z)01 zi01 dz (C : a simple path enclosing all poles). Then, sup j331 v jN 1 =jvjN1 ; v 2 RN < +1 for 1 2 (0; 0 ). where Markov parameters m31 (i) (i
m
Proof: See Appendix. Lemma 4: Assume n 0 m = 2. Then, j341 351 (1 Su3 0 3 N 1 S1 1 u )j1 ! 0 as 1 ! 0. Proof: See Appendix. We are now ready to establish Theorem 1. We have seen that + (N; N + 1) = 311 321 or 331 341 351 . If n 0 m = 1 or 2, 01 we can obtain (16) from Lemma 2 with Lemma 1, or Lemma 4 with Lemma 3, respectively. IV. NUMERICAL EXAMPLES AND INTERPRETATION OF THE MAIN RESULT Consider G2 (s) given in Section I on the finite-time interval [0; 5] with u3 (t) = 5t + 1 and y 3 = Su3 . Then, the pulse transfer function H2 (z ) has the unstable zero q1 (1) for a small sampling period 3 3 as shown in Section I. Fig. 1 shows u = 1 0+ 1 1 y with u and + 3 3 3 y 0y = S1 01 1 y 0y for 1 = 0:5 and 0.25. When 1 = 0:5, the unstable zero makes the signal u very large at the right end of the time interval; this causes intersample ripples as shown in the right figure.
.
However, when 1 is shrunk to 0.25, we can observe that the signal u is near the ideal input u3 and therefore the intersample ripples are eliminated. The main result shown in this paper implies that the solution of an t optimal control problem minimizing 0 fy (t) 0 y 3 (t)g2 dt or the inverse problem finding u(t) such that y 3 = Su can be approximated by the solution of the finite-dimensional optimal control problem mini+ 1 y3 when the relative degree is 1 mizing j01 v 0 1 y 3 j2 , namely 01 or 2. This property is favorable for iterative learning control [11], which is a method to realize precise output tracking on [0; tf ] by repetitive improvement based on experimental input–output data; inter-sample residuals of the output can be reduced simply by shrinking the sampling period 1 as far as precise output tracking is achieved only on the sample points. V. CONCLUDING REMARKS We demonstrated that when the relative degree is one or two, the inverse of sampled-data systems approximates to the inverse of continuous-time systems independently of the stability of the zeros. It should be noted that such a property is uncommon for a relative degree greater than two, because there is at least one zero that converges to a point exterior to the unit circle [2]. APPENDIX I PROOF OF LEMMA 1 Since the first equation shown at the bottom of the next page holds true, the system (A1 ; b1 ; c1 ; 1=cb1 ) is converted to the controllable canonical form. r0 (1) 1 .. ~ ; [ 0 1 1 1 0 cb1 ] ; (25) A1 ; . cb1 rn (1)
where jA~1 j = maxf1; jq1 (1)j; . . . ; jqn01 (1)jg. Since we can see jrj (1)j=1 +1 (j = 0; . . . ; n), we have
r0 (1) .. . rn (1)
M 1 1
(26)
1 is a positive constant; j1j indicates the Euclidean norm and its where M induced norm. From the Taylor expansion of the intrinsic zero, namely 2 1 eM 1 qi (1) = 1 + i 1 + O(12 ) [8], we have jA~1 j 1 + M
764
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
1 (0 ) 1 0
for 2 ; tf , where M2 is a positive constant. Note that there exists a positive constant M3 > such that j =cb1 j < M3 for almost all > . Then, we have 11 v 1 M3 jv1 j and
0
1
[3 ] [311 v]k MM1M2 3 e(k01)M 1 0 1 2 maxfjvj j; j = 1; . . . ; k 0 1g + M 3 jvk j for k = 2; . . . ; N . Those inequalities imply that sup j3 11vjN1 =jvjN1; v 2 R N < +1 because e(k01)M 1 0 1 e(N 01)M 1 0 1 < eT M 0 1. PROOF OF LEMMA 2 Note
that cAc 0 eA ( 0) bc u d cbc u d . Then, we have the second equation shown at the bottom of the page, for . Moreover, we have k ; ; ; N 0 , where k 2 k ; k
() +
t
0
= 0 1 ...
Since we have qi
i O 2 (i ; ; n0 ) [15] for m n the intrinsic zeros and qn01 0
0 = i i=1 i=1 pi 2 O for the discretization zero [9], there exists a constant M4 such that f ; jq1 j; ; jqn01 jg N eMN 1 for N 2 ; tf . Therefore, we can see that j 31vj1=jvj1 v 2 R < 1 in the same manner as the proof of Lemma 1.
(1) = 1+ 1+ (1 ) = 1 . . . 2 (1) = 1+ 1 3+ (1 ) max 1 (1) . . . 1 (0 ) (1) ; + sup 3 APPENDIX IV PROOF OF LEMMA 4
1( 1 1) + ( (2) 1 0
Consider an N 0 tf = 0 -dimensional vector w1 which satisfies jw1 jN 101 < 1 and jv1 jN102 ! as ! T , where v1 w1 0 w1 1 1 1 w1 N 0 0 w1 N 0 . Then, j 41w1 jN1 tends to 0 as ! . This is because we have the third equation, shown at the bottom of the page, which implies the fourth equation shown at the bottom of the previous page. To establish the lemma, we will show that 51 1 S1 , where 1 u3 0 1 1 u3 has the same limiting property as w1 given above. Let 1 S1 . Then
= [ (1) 3 1
APPENDIX II
[Su](t) =
APPENDIX III PROOF OF LEMMA 3
()
[ 1 ( +1)1] 1 j[321(1 S (u3 0 11 u3 ))](k)j t sup cAc eA (t0) bc d + cbc 2[0;t ] 0 3 2 ku 0 1 1 u3 k01
0 0 where kuk1 fju t j t 2 ;3tf g. Since ku3 03 N11 u3 k1 as ! , we have j 21 1 Su 0 1 S1 1 u j1 ! .
= sup ( ) ; [0 ] 3 ( 0 1 0
)
3
=
(2) t 1
t
0
!
2
=
1
t
0 + cA2c bc1 (t) + cAc bc d u3 (t): dt
z n01 + r1 (1)z n02 + 1 1 1 + rn (1) ( ) = cb11 1 + r0(z(1) 0 1)(z 0 q1 (1)) 1 1 1 (z 0 qn01 (1))
[321 (1S (u3 0 1 1 u3 ))](k) 3 3 0 [S (u3 0 1 1u3 )](k1) = [S (u 0 1 1 u )]((k + 1)1) 1 A ( 0 ) 3 bc (u 0 1 1 u3 )d + cbc (u3 (k ) 0 u3 (k1)) = cAc e 0
(1)1 k01)=2 1 fw1 (i) 0 w1 (i + 1)g [341w1 1](k) = (01)k (i=1 w1 (k 0 1)1 2 +(01)k k= i=1 1 fw1 (i) 0 w1 (i + 1)g w1
0 1 0 ( 1)]
( ) = dtd 2 S1 =cA2c eA (t0 )bc 1 ( )d 0 + cAc bc 1 (t) (3) (t) = d3 S1 =cA3 t eA (t0 ) bc 1 ( )d 1 c dt3
H101 z
0
2)
k k k
=1 =2 = 3; 5; . . .
k
= 4; 6; . . .
j[341w1 1](k)j max jw1(1)j1; (tf =12 0 1) 1 jv1 jN102 ; jw1 (k 0 1)j1 + t1f 1 jv1 jN102
:
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
765
[351 1S1](k) = 1 ((k 0 1)1) 0 211(k2 1) + 1 ((k + 1)1) (3) (3) 1 1 =1(2) (k1) + 1 ((k +6 k )1)1 0 1 ((k 06 k )1)1 ((
1)1) [0 1] ( )
(( + 1)1) ( 1) =
By using the Taylor expansion of 1 k 0 and 1 k , we have the equation shown at the top of the page, where and k1 ; k1 2 ; . Since 1 k k ; ; ;N 0 0 , we have 1 k and k1 k01 = k d=dt u3 k1
= 1 2 ... 1 (( + 1)1) = 0 (2) 1 (k1)
1
t
1
2
sup
cAc
2[0;t ]
t
0
( 0 ) bc d
A t
e
d 3 u dt
Similarly,
we
can
also
see
(3) ((1 0 1 )1)1=6 1 1 1 (3) ((N 0 1 0 1 )1)1=6 1 N 01 1
that
1
does likewise.
ACKNOWLEDGMENT
0
The authors would like to thank T. Hagiwara and the anonymous reviewers for their comments and suggestions.
1
REFERENCES and by the mean-value theorem
(2) ((k + 1)1)
1
1
(2)
0 1 1(k1)
3 (k+ )1 eA ((k+ )10 ) bc 1 ( )d 0
=
cAc
+ cA2c bc 1 ((k + k1 )1) sup
2[0;t ]
t
t
3
cAc
0
( 0 ) bc d + cA2c bc k1 k0 1
A t
e
(27)
0 is bounded and k1 k1 0 tends where k1 2 ; . Since k d=dt u3 k1 (2) (2) to 0, we can see that 1 1 1 1 1 N 0 has the same limiting property as w1 . Moreover, we have
[0 1]
( ) (1) ((
1
1)1)
(3) ((k + 1 )1) k
1
sup
2[0;t ]
t
+ jcAc bc j
3
cAc
t
0
d 3 u dt
( 0 ) bc d + cA2c bc k1 k0 1
A t
e
0 1
(28)
and
(3) ((k + 1 + 1 )1) 0 (3) ((k + 1 )1) k+1 k 1 t 0 2 sup cA3c eA (t0 )bc d + cA2c bc k1 k1 t2[0;t ] 0 + jcAc bc j sup dtd u3 (t1 ) 0 dtd u3 (t2 ) : jt 0t j1
1
(d=dt)u3 2 C 0 [0; tf ], we can see (3) (3) 1 101 )1)1=6 has 1 ((1 + 1 )1)1=6 1 1 1 1 ((N 0 1 + N same limiting property as w1 1. Since
(29)
that the
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Author(s)
Citation
Issue Date
A limiting property of the inverse of sampled-data systems on a finite-time interval
Sogo, T; Adachi, N
IEEE TRANSACTIONS ON AUTOMATIC CONTROL (2001), 46(5): 761-765
2001-05
URL
http://hdl.handle.net/2433/50315
Right
(c)2001 IEEE. Personal use of this material is permitted. However, permission to reprint/republish this material for advertising or promotional purposes or for creating new collective works for resale or redistribution to servers or lists, or to reuse any copyrighted component of this work in other works must be obtained from the IEEE.
Type
Journal Article
Textversion
publisher
Kyoto University
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
A Limiting Property of the Inverse of Sampled-Data Systems on a Finite-Time Interval Takuya Sogo and Norihiko Adachi
Abstract—If one considers a sampled-data system derived from a continuous-time system with a relative degree of one or two on a finite-time interval, it is not simple to predict the behavior of the output of the inverse of the sampled-data system as the sampling period goes to zero. This is because the number of sample points increases while the zeros of the pulse-transfer function tend to the boundary between the stable and unstable areas. This paper shows that the output of the sampled-data inverse systems converges to the output of the continuous-time inverse systems independently of the stability of zeros.
761
It should be noted that this property of zeros is common for systems whose relative degree is one or two [2], [8], [9]. In this paper, we will discuss such a limiting problem on the fixed time interval for systems with a relative degree of one or two. We will demonstrate that the output of the sampled-data inverse systems converges to the output of the continuous time inverse systems independently of the stability of the zeros when 1 goes to zero. II. MATHEMATICAL PRELIMINARIES Consider a linear continuous-time single-input–single-output (SISO) system
d x(t) =Ac x(t) + bc u(t) dt y (t) =cx(t)
Index Terms—Inverse systems, iterative learning control, limiting zeros, nonminimum phase systems, sampled-data systems.
I. INTRODUCTION Stable inverse systems or stable zeros of transfer functions are often required in many kinds of control problems defined on the infinite time horizon. However, since there is no simple relation between zeros of the pulse-transfer function of sampled-data systems and zeros of the transfer function of continuous-time systems, the behavior of the zeros of sampled-data systems has drawn much attention from researchers. Several approaches to determine the stability of zeros or avoid unstable zeros have been presented [1]–[10]. On the other hand, when control problems are defined on a fixed finite-time interval [0; tf ], one can admit unstable systems unless signals become too large inside [0; tf ]. One such finite-time control problems is, for example, iterative learning control which is a trial-based iterative method to improve the transient response on a short time interval [11], [12]. When continuous time systems or their inverse systems are considered on [0; tf ], the peak of the signals is simply determined by the distance between the imaginary axis and poles or zeros, respectively. However, when sampled-data systems with a sampling period 1 are considered on [0; tf ], the peak of the signals depends on the variable 1. This relationship is not simple because 1 changes both the number of sample points inside [0; tf ] and the location of poles and zeros. Furthermore, the zeros as a function of 1 are much more complicated than the poles. For example, consider the continuous time systems G1 (s) = (s 0 1)=s2 and G2 (s) = (2s 0 3)=f(s + 1)(s + 2)(s + 3)g. Then, the sampled-data systems derived from each with a sampler and a zero-order hold are H1 (z ) = f(2 0 1)1(z 0 (2 + 1)=(2 0 1))g=f2(z 0 1)2 g and H2 (z) = f (1)(z 0 q1 (1))(z 0 q2 (1))=f(z 0 exp(01))(z 0 exp(021))(z 0 exp(031))g, respectively, where q1 (1) = 1 + 31=2 + O(12 ) and q2 (1) = 01 + 51=2 + O(12 ) [8], [9]. Consider the inverse systems H101 (z ) and H201 (z ) on f0; 1; . . . ; tf =1g where tf =1 is assumed to be a natural number and give fyd (0); yd (1); . . . ; yd (tf )g that is the sampled-data of a function yd (t) to H101 (z ) and H201 (z ) as their inputs. Then, the unstable zeros for a small 1 make the output of the inverse systems increase exponentially. However, it is not easy to determine whether the output of the inverse systems diverges or converges inside [0; tf ] when 1 goes to zero, because all the zeros tend to points on the unit circle, i.e., the boundary between the stable and unstable areas, while the number of sample points increases inside [0; tf ].
where x 2 Rn , u 2 R, and y 2 R and a sampled-data system derived from (1) with a zero-order hold and a sampler with a sampling period 1. Then, we have
x((k + 1)1) =A1 x(k1) + b1 u(k1) y (k1) =cx(k1)
(2)
1
where A1 = exp Ac 1 and b1 = 0 exp(Ac )bc d . Assume that the transfer function G(s) = c(sI 0 Ac )01 bc is expressed as
G(s) =
K (s 0 1 )(s 0 2 ) 1 1 1 (s 0 m ) : (s 0 p1 )(s 0 p2 ) 1 1 1 (s 0 pn )
(3)
Then, since there exists a positive constant 0 such that cb1 6= 0 for all 1 2 (0; 0 ), the pulse-transfer function H (z ) = c(zI 0 A1 )01 b1 can be expressed as
H (z ) =
cb1 (z 0 q1 (1)) 1 1 1 (z 0 qn01 (1)) : (z 0 exp(p1 1)) 1 1 1 (z 0 exp(pn 1))
(4)
Next, consider a system (1) with the initial condition x(0) = 0 on a finite-time interval [0; tf ]. Then, the input–output mapping defined by (1) is expressed as y = Su (u; y 2 L2 [0; tf ]) where
Su =
t
0
ceA
(t0 ) bc u( )d :
(5)
Moreover, assume that the sampling period 1 satisfies 1 = tf =N , where N is a natural number. Then, the input–output relationship on the sample points f0; 1; 21; . . . ; tf 0 1; tf g is
w1
= 01 v1
(6)
where
= [ u(0) u(1) = [ y(0) y(1) 0 m1 (1) 01 = .. ..
1 1 1 u(tf ) ]T 1 1 1 y(tf ) ]T
v1 w1
.
Manuscript received November 29, 1999; revised August 3, 2000 and October 31, 2000. Recommended by Associate Editor T. Chen. The authors are with the Graduate School of Informatics, Kyoto University, Kyoto 606-8501, Japan (e-mail: [email protected]). Publisher Item Identifier S 0018-9286(01)03616-9.
(1)
m1 (N )
0
.
111
where the Markov parameter m1 (i) (i m1 (i) = cAi101 b1 .
0018–9286/01$10.00 ©2001 IEEE
(7) (8)
(9)
m1 (1)
0
= 1; . . . ; N ) is
defined as
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IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
Note that cb1 system x((k
6= 0 for almost all 1 > 0. Then, we have the inverse
+ 1)1) =fA1 0 b1 (cb1 )01 cA1 gx(k1) + b1 (cb1 )01y((k + 1)1) 01 u(k 1) = 0 (cb1 ) cA1 x(k 1) + (cb1 )01 y((k + 1)1) (k =0; . . . ; N 0 1):
k1 01+ 1 y3 0 u3 k1 j01+ 1 y3 0 1u3 jN1 + k1 1 u3 0 u3 k1 + (1 Su3 0 1 S1 1 u3 )j1 N =j01 3 3 + k1 1 u 0 u k1 :
(10)
By letting u(N 1) = 0, (10) defines a mapping from w1 to v1 . We can denote the mapping as 0+ 1 , the Moore-Penrose generalized inverse + of 01 , which is the mapping 01 from w1 to the minimizer of (w1 0 01 v1 )T (w1 0 01 v1 ) with the minimum norm. In the following discussions, we will use the sampling operator 1 : N +1 and the zero-order hold operator 1 : RN +1 ! L2 [0; tf ] ! R L2 [0; tf ], defined as follows:
1 [u] = [u(0)u(1) 1 1 1 u((N 0 1)1) u(tf )]T v (k ) if t 2 [(k 0 1)1; k1] t k = 1; 2; . . . ; 1 [1 v](t) = : t v 1 +1 if t = tf
(11)
= supfju(t)j; t 2 [0; tf )g, = 1; 2; . . . ; N (= tf =1)g, C k [0; tf ] denotes set of k -th continuously differentiable functions on [0; tf ]. Note that 01 v = 1 S1 v and if u3 2 C k [0; tf ] (k 1) then lim1!0 k1 1 u3 0 u3 k1 = 0. i
N !0 j0+1(1 Su3 0 1 S1 1 u3 )j1
k101+ 1 y3 0 u3 k1 ! 0 as
1
..
0
..
.
0
01 1
n0m
=fAc + bc F 01 cAc gx(t) + bc F 01 d dt dt n0m d 01 cAn0m x(t) + F 01 y (t) u(t) =F c x(t)
dt
1(z 0 exp(p 1)) 1 1 1 (z 0 exp(p 1)) 1 (z)01 = cb (z 0 1)(z 01 q (1)) 1 1 1 (z 0 q n (1)) 1 1 n01 (18 )
01 = (z 0 1) : H2 (z ) 1 Lemma 1: Assume n
311 defined as
(19)
0 m = 1 and consider the N 2 N matrix 11 (0)
0
m
311 =
.. .
..
(20)
.
11 (N 0 1) 1 1 1
11 (0)
0m01 bc and x(0) = 0. Let S 01 be the input–output where F = cAn c mapping of (14) on [0; tf ]. Then, u3 = S 01 y 3 ; (13) is equivalent to k1 01+ 1 y3 0 S 01 y3 k1 ! 0. This means that the discrete-time inverse system (10) converges to the continuous-time inverse system (14) as 1 ! 0.
m
= 0; . . . ; N 0 1) are defined as 11 (i) = 1=2j C H1 (z)01 zi01 dz (C : a simple path enclosing all poles) Then, sup j311 v jN 1 =jvjN1 ; v 2 RN < +1 for 1 2 (0; 0 ). where Markov parameters m11 (i) (i
m
Proof: See Appendix. Lemma 2: Assume n 0 m = 1. Then, j321 (1 Su3 0 3 N 1 S1 1 u )j1 ! 0 as 1 ! 0. Proof: See Appendix. Next, we consider the case of n 0 m = 2 and N 2 N matrix 331 + (N; N +1) = 331 341 351 where 341 is an N 2(N 01) such that 01 matrix defined as
y (t)
(14)
(17)
H
m
Remark 1: The convergence (13) is presented only on [0; tf ), while + 1 y3 0 u3 is defined on [0; tf ]. However, this is the the function 1 01 + we have [1 0+ 1 y3 0 best result because from the definition of 01 1 3 3 u ](tf ) u (tf ), which is independent of 1. It should be noted that the result of Theorem 1 is independent of the + 1 y3 converges even if some zeros stability of the zeros of H (z ); 01 go to the unit circle from the outside. If a function y 3 (t) (t 2 [0; tf ]) satisfies the assumption in Theorem 1, u3 can be obtained such that y 3 = Su3 by using the following continuous-time inverse system of (1)
:
.
Consider the inverse system (10) on the infinite interval and the pulse-transfer function H (z )01 . Then, we can see that the + (N; N + 1) given above corresponds to decomposition of 01 0 1 0 1 H (z ) = H1 (z ) H2 (z )01 where
(13)
! 0.
d
01 1
321 = 1 1
III. THE MAIN RESULTS
2
(16)
as 1 ! 0. First, we consider the case of n 0 m = 1 and N 2 N matrix 311 + (N; N + 1) = 311 321 where 0+ (N; N + 1) indicates such that 01 + without its N +1-th row; 321 is1an N 2 (N +1) matrix the matrix 01 defined as
In the following sections, we consider the inverse discrete-time system with sampled data of a fixed function y 3 defined in [0; tf ] as + 1 y3 or (10) whose y((k + 1)1) is substituted the input, i.e., 01 3 with y ((k + 1)1).
+ 1 y3 as 1 ! 0. In this section, we present the limit of 01 Theorem 1: Assume that n 0 m = 1 or 2 and there exists u3 n 0m01 [0; tf ] such that y3 = Su3 . Then C
(15)
What we have to show is
(12)
We define the following notations: kuk1
jvjN1 = supfjvi j;
The proof of Theorem 1 will be established in the following sequence of lemmas1 ; all assumptions in Theorem 1 will be preserved. Since we have
341 =
0 1 01 .. .
(01)N
0m
0 1
..
.
0 ..
.
..
.
(21)
0
1 1 1 01 1
1Theorem for the case of n = 1 has been proved in [13], [14]; the approach was based only on a state-space representation. In this paper, we present a refined proof which uses an asymptotic property of zeros of the pulse transfer function.
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
Fig. 1. Left: u
= 0
and 351 is an (N
y with u . Right: y
0y
= S 0
y
763
0y
0 1) 2 (N + 1) matrix defined as 1 02 1 0
351 = 12 1
..
.
..
.
..
.
:
(22)
1 02 1 We can see that the decomposition of 0+ 1 (N; N + 1) given above corresponds to H (z )01 = H3 (z )01 H4 (z )01 H5 (z )01 where 01 = 12 (z + 1)(z 0 exp(p1 1)) 1 1 1 (z 0 exp(pn 1)) ; (23) H3 (z ) cb1 (z 0 1)2 (z 0 q1 (1)) 1 1 1 (z 0 qn01 (1)) 01 = 1=(z + 1) and H5 (z )01 = (z 0 1)2 =12 . H4 (z ) Lemma 3: Assume n 0 m = 2 and consider the N 2 N matrix 331 defined as m31 (0) 0 .. . . 331 = (24) . . m31 (N 0 1) 1 1 1 m31 (0) 0
= 0; . . . ; N 0 1) are defined as 31 (i) = (1=2j ) C H3 (z)01 zi01 dz (C : a simple path enclosing all poles). Then, sup j331 v jN 1 =jvjN1 ; v 2 RN < +1 for 1 2 (0; 0 ). where Markov parameters m31 (i) (i
m
Proof: See Appendix. Lemma 4: Assume n 0 m = 2. Then, j341 351 (1 Su3 0 3 N 1 S1 1 u )j1 ! 0 as 1 ! 0. Proof: See Appendix. We are now ready to establish Theorem 1. We have seen that + (N; N + 1) = 311 321 or 331 341 351 . If n 0 m = 1 or 2, 01 we can obtain (16) from Lemma 2 with Lemma 1, or Lemma 4 with Lemma 3, respectively. IV. NUMERICAL EXAMPLES AND INTERPRETATION OF THE MAIN RESULT Consider G2 (s) given in Section I on the finite-time interval [0; 5] with u3 (t) = 5t + 1 and y 3 = Su3 . Then, the pulse transfer function H2 (z ) has the unstable zero q1 (1) for a small sampling period 3 3 as shown in Section I. Fig. 1 shows u = 1 0+ 1 1 y with u and + 3 3 3 y 0y = S1 01 1 y 0y for 1 = 0:5 and 0.25. When 1 = 0:5, the unstable zero makes the signal u very large at the right end of the time interval; this causes intersample ripples as shown in the right figure.
.
However, when 1 is shrunk to 0.25, we can observe that the signal u is near the ideal input u3 and therefore the intersample ripples are eliminated. The main result shown in this paper implies that the solution of an t optimal control problem minimizing 0 fy (t) 0 y 3 (t)g2 dt or the inverse problem finding u(t) such that y 3 = Su can be approximated by the solution of the finite-dimensional optimal control problem mini+ 1 y3 when the relative degree is 1 mizing j01 v 0 1 y 3 j2 , namely 01 or 2. This property is favorable for iterative learning control [11], which is a method to realize precise output tracking on [0; tf ] by repetitive improvement based on experimental input–output data; inter-sample residuals of the output can be reduced simply by shrinking the sampling period 1 as far as precise output tracking is achieved only on the sample points. V. CONCLUDING REMARKS We demonstrated that when the relative degree is one or two, the inverse of sampled-data systems approximates to the inverse of continuous-time systems independently of the stability of the zeros. It should be noted that such a property is uncommon for a relative degree greater than two, because there is at least one zero that converges to a point exterior to the unit circle [2]. APPENDIX I PROOF OF LEMMA 1 Since the first equation shown at the bottom of the next page holds true, the system (A1 ; b1 ; c1 ; 1=cb1 ) is converted to the controllable canonical form. r0 (1) 1 .. ~ ; [ 0 1 1 1 0 cb1 ] ; (25) A1 ; . cb1 rn (1)
where jA~1 j = maxf1; jq1 (1)j; . . . ; jqn01 (1)jg. Since we can see jrj (1)j=1 +1 (j = 0; . . . ; n), we have
r0 (1) .. . rn (1)
M 1 1
(26)
1 is a positive constant; j1j indicates the Euclidean norm and its where M induced norm. From the Taylor expansion of the intrinsic zero, namely 2 1 eM 1 qi (1) = 1 + i 1 + O(12 ) [8], we have jA~1 j 1 + M
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IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
1 (0 ) 1 0
for 2 ; tf , where M2 is a positive constant. Note that there exists a positive constant M3 > such that j =cb1 j < M3 for almost all > . Then, we have 11 v 1 M3 jv1 j and
0
1
[3 ] [311 v]k MM1M2 3 e(k01)M 1 0 1 2 maxfjvj j; j = 1; . . . ; k 0 1g + M 3 jvk j for k = 2; . . . ; N . Those inequalities imply that sup j3 11vjN1 =jvjN1; v 2 R N < +1 because e(k01)M 1 0 1 e(N 01)M 1 0 1 < eT M 0 1. PROOF OF LEMMA 2 Note
that cAc 0 eA ( 0) bc u d cbc u d . Then, we have the second equation shown at the bottom of the page, for . Moreover, we have k ; ; ; N 0 , where k 2 k ; k
() +
t
0
= 0 1 ...
Since we have qi
i O 2 (i ; ; n0 ) [15] for m n the intrinsic zeros and qn01 0
0 = i i=1 i=1 pi 2 O for the discretization zero [9], there exists a constant M4 such that f ; jq1 j; ; jqn01 jg N eMN 1 for N 2 ; tf . Therefore, we can see that j 31vj1=jvj1 v 2 R < 1 in the same manner as the proof of Lemma 1.
(1) = 1+ 1+ (1 ) = 1 . . . 2 (1) = 1+ 1 3+ (1 ) max 1 (1) . . . 1 (0 ) (1) ; + sup 3 APPENDIX IV PROOF OF LEMMA 4
1( 1 1) + ( (2) 1 0
Consider an N 0 tf = 0 -dimensional vector w1 which satisfies jw1 jN 101 < 1 and jv1 jN102 ! as ! T , where v1 w1 0 w1 1 1 1 w1 N 0 0 w1 N 0 . Then, j 41w1 jN1 tends to 0 as ! . This is because we have the third equation, shown at the bottom of the page, which implies the fourth equation shown at the bottom of the previous page. To establish the lemma, we will show that 51 1 S1 , where 1 u3 0 1 1 u3 has the same limiting property as w1 given above. Let 1 S1 . Then
= [ (1) 3 1
APPENDIX II
[Su](t) =
APPENDIX III PROOF OF LEMMA 3
()
[ 1 ( +1)1] 1 j[321(1 S (u3 0 11 u3 ))](k)j t sup cAc eA (t0) bc d + cbc 2[0;t ] 0 3 2 ku 0 1 1 u3 k01
0 0 where kuk1 fju t j t 2 ;3tf g. Since ku3 03 N11 u3 k1 as ! , we have j 21 1 Su 0 1 S1 1 u j1 ! .
= sup ( ) ; [0 ] 3 ( 0 1 0
)
3
=
(2) t 1
t
0
!
2
=
1
t
0 + cA2c bc1 (t) + cAc bc d u3 (t): dt
z n01 + r1 (1)z n02 + 1 1 1 + rn (1) ( ) = cb11 1 + r0(z(1) 0 1)(z 0 q1 (1)) 1 1 1 (z 0 qn01 (1))
[321 (1S (u3 0 1 1 u3 ))](k) 3 3 0 [S (u3 0 1 1u3 )](k1) = [S (u 0 1 1 u )]((k + 1)1) 1 A ( 0 ) 3 bc (u 0 1 1 u3 )d + cbc (u3 (k ) 0 u3 (k1)) = cAc e 0
(1)1 k01)=2 1 fw1 (i) 0 w1 (i + 1)g [341w1 1](k) = (01)k (i=1 w1 (k 0 1)1 2 +(01)k k= i=1 1 fw1 (i) 0 w1 (i + 1)g w1
0 1 0 ( 1)]
( ) = dtd 2 S1 =cA2c eA (t0 )bc 1 ( )d 0 + cAc bc 1 (t) (3) (t) = d3 S1 =cA3 t eA (t0 ) bc 1 ( )d 1 c dt3
H101 z
0
2)
k k k
=1 =2 = 3; 5; . . .
k
= 4; 6; . . .
j[341w1 1](k)j max jw1(1)j1; (tf =12 0 1) 1 jv1 jN102 ; jw1 (k 0 1)j1 + t1f 1 jv1 jN102
:
IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL. 46, NO. 5, MAY 2001
765
[351 1S1](k) = 1 ((k 0 1)1) 0 211(k2 1) + 1 ((k + 1)1) (3) (3) 1 1 =1(2) (k1) + 1 ((k +6 k )1)1 0 1 ((k 06 k )1)1 ((
1)1) [0 1] ( )
(( + 1)1) ( 1) =
By using the Taylor expansion of 1 k 0 and 1 k , we have the equation shown at the top of the page, where and k1 ; k1 2 ; . Since 1 k k ; ; ;N 0 0 , we have 1 k and k1 k01 = k d=dt u3 k1
= 1 2 ... 1 (( + 1)1) = 0 (2) 1 (k1)
1
t
1
2
sup
cAc
2[0;t ]
t
0
( 0 ) bc d
A t
e
d 3 u dt
Similarly,
we
can
also
see
(3) ((1 0 1 )1)1=6 1 1 1 (3) ((N 0 1 0 1 )1)1=6 1 N 01 1
that
1
does likewise.
ACKNOWLEDGMENT
0
The authors would like to thank T. Hagiwara and the anonymous reviewers for their comments and suggestions.
1
REFERENCES and by the mean-value theorem
(2) ((k + 1)1)
1
1
(2)
0 1 1(k1)
3 (k+ )1 eA ((k+ )10 ) bc 1 ( )d 0
=
cAc
+ cA2c bc 1 ((k + k1 )1) sup
2[0;t ]
t
t
3
cAc
0
( 0 ) bc d + cA2c bc k1 k0 1
A t
e
(27)
0 is bounded and k1 k1 0 tends where k1 2 ; . Since k d=dt u3 k1 (2) (2) to 0, we can see that 1 1 1 1 1 N 0 has the same limiting property as w1 . Moreover, we have
[0 1]
( ) (1) ((
1
1)1)
(3) ((k + 1 )1) k
1
sup
2[0;t ]
t
+ jcAc bc j
3
cAc
t
0
d 3 u dt
( 0 ) bc d + cA2c bc k1 k0 1
A t
e
0 1
(28)
and
(3) ((k + 1 + 1 )1) 0 (3) ((k + 1 )1) k+1 k 1 t 0 2 sup cA3c eA (t0 )bc d + cA2c bc k1 k1 t2[0;t ] 0 + jcAc bc j sup dtd u3 (t1 ) 0 dtd u3 (t2 ) : jt 0t j1
1
(d=dt)u3 2 C 0 [0; tf ], we can see (3) (3) 1 101 )1)1=6 has 1 ((1 + 1 )1)1=6 1 1 1 1 ((N 0 1 + N same limiting property as w1 1. Since
(29)
that the
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