the pole and zeros.pdf

8/16/2019 the pole and zeros.pdf

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REND. SEM. MAX.UNIV. POLITEC. TORINO

FASC. SPEC. CONTROL THEORY

Edward W. Kamen

THE POLES AND ZEROS OF A LINEAR TIME-VARYING SYSTEM*

Summary. For linear time-varying discrete-time and continuous-time systems, a notion of poles and zeros is developed in terms of factorizations of operator polynomials withtime-varying coefficients. In the discrete-time case, it is shown that factorizations canhe computed hy solving a nonlinear recursion with time-varying coefficients. In thecontinuous-time case, one must solve a nonlinear differential equatìon with time-varying coefficients. The theory is applied to the study of the zero-input responseand asymptotic stability. It is shown that if a time-varying analogue of the Vander-monde matrix is invertihle, the zero-input response can he decomposed into a sum ofmodes associate d with the poles. Stability is then studied in terms of the componentsof the modal decomposition.

1. Introduction

In this paper we develop a notion of poles and zeros for the class of lineartime-varying systems. Our notion is defined in terms of the factorization ofpolynomials with time-varying coefficients. Although the idea of factoringpolynomials with time-varying coefficients is not new (e.g., see Ore [1933],

* A prelimrnary version of this paper was presented at the Twentieth Annual Conferenceon Information Sciences and Systems, Princeton, NJ, March 1986. This work was suppor-ted in part by the National Science Foundation under Grant No. ECS-8400832 and bythe U.S. Army Research Office under Contract No. DAAG29-84-K-0081.


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Amitsur [1954], Newcomb [1970]), we present a new approach to construc-ting factorizàtions given in terms of the solutions to a nonlinear equation.

We begin in the next section with the study of a linear time-varying

discrete-time system specified by a second-order input/output differenceequation. Various properties of factorizàtions are studied in Section 3, andthen in Section 4, we consider a general class of linear time-varying dicrete-time systems specified by an nth-order input/output difference equation. InSection 5 the theory is applied to the study of the zero-input response andasymptotic stability. The continuous-time case is considered in the last sectionof the paper.

2. Poles of a Second-Order System

With Z equal to the set of integers and R equal to the set of realnumbers, let A denote the set of ali functions defined on Z with values in

R . With C equal to the set of complex numbers, let AQ denote theset of ali complex-valued functions defined on Z . Clearly, the set A canbe viewed as a subset of Ac . Both A and Ac are commutative ringswith pointwise addition and multiplication defined by

(a + b)(k) = a(k) + b(k)

(ab)(k) = a(k)b(k) ,

where a(k),b(k) are arbitrary elements of A (or i4c) and k is thediscrete-time index.

Now consider the linear discrete-time system given by the second-orderinput/output difference equation

(1) y (k + 2) + « t (k)y (k + 1) +a0 (k)y (k) = b(k) u (k)

where y (k) is the output at time k , u (k) is the input at time k , and«o (k),ax (k) are elements of A . We shall write the left side of (1) inoperator form as follows.

For any positive integer i, let z1 denote the z'-step left shift operatoron A defined by

z'7(*)=/(*+/) , feA .

s


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For any a(k)eA,'.let a (k)z* denote the operator on A definedby

fe -«(*)/(*+0 •

Then we can write (1) in the operator form

(2) (z2 + a x (*)*- + a0 ih))y (*) = b(k)u (k).

Suppose that there exist functioris p x (k) , p 2 (k) belonging to A csuch that

(3) (z 2+a x (k)z+a0(k))y(k) = (z-p x (k))[(z-p 2 (k))y(k)].

It follows from (3) that the given system can be viewed as a cascade connec

tion of two first-order subsystems. To see this, let

(4) v(k) = (z-p 2(k))y(k) ,

so that

(5) y(k + l)-p 2(k)y(k) = v(k) .

Inserting the expression (4) for v (k) into (3) and using (2), we have that

(z- Pl (k))v(k) = b(k)u(k) ,

or

(6) v (k + 1) -p x (k)v {k) = b(k)u (k) .

From (5) and (6) , we have the cascade realization of the system shown in

Figure 1 below.

Given the decomposition of the system into the first-order subsystems

shown in Figure 1, it is tempting to cali p x (k) and p 2 (k) the poles of

the system. We shall give a formai definition later. First, we want to characte-

rize the cascade decomposition in terms of a polynomial factorization, and

then we consider the existence and construction of factorizations.

Again suppose that there exist p x {k),p 2 (k) e AQ such that (3) is

satisfied. We want to define a product


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so that

(7) [(2 - P ! (*)) o (2 - p 2 (k))]y (k) = (2 -/>, (*)) [(« ~P2 (ft))^ .

Equation (8) shows that the multiplication ° is the usuai polynomial multiplication except that

(9) 2 0p2 (fc)=p2 (£+ 1)2

As a result of the property (9), the multiplication ° is called a skew polynomial multiplication. In Section 4 we define a ring of polynomials withthe property (9).

We now consider the existence of a skew polynomial factorization ofthe form (8). Combining (3), (7), and (8), we have that

(10) z 2-'(pl (k) + p 2 (fc + DJz+p! (k)p 2 (*) = z2 +*! (k)z+a0(k).


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Equating coefficients of 2 in (10), we see that there is a factorization ofthe form (8) if and only if

(11) Pi (* )+p 2 (* + D = --*i (*)

(12) Pi(k)p 2(k) = a0(k) .

Multiplying both sides of (ll)by p2 (k) and using (12), we obtain

(13) p 2 (* + Dp2 (*)+«i (k)p2 (*) + *0 (*) = 0 .

Note that (13) is a nonlinear first-order difference equation with time-

varying coefficients. It is also interesting to note that the left side of (13)looks like the polynomial 22 + a1 (k)z + a0 (k) evaluated at 2=p2 (k)with

Zk-p,( ik) =?»(*) a n d z2 \z=p 2(k)= P2^^^p2(k).

Given the initial value p2 (k 0) at initial time k 0 , we can compute p2 (k) for k>k 0 by solving (11) and (12) recursively, or by solving

(13) recursively. If p2{k)^0 for ali k>k 0 , the solution is unique.Given p2 (k 0) selected at random, the probability is zero that p2 (k)will be zero for some value of k > k 0 . In other words, for almost ali initialvalues p2 (k 0)e C , (13) has a unique solution p2 (k) with p2 (^)=^0for ali k > k 0 , Futhermore, p x (k) can be computed from (11). We there-fore have the following result.

Proposition 1: For amost ali yeC, the operator polynomial 22

4-al(k)z+a0(k) has a unique skew polynomial factorization (2-pt '(k)) o(2 - p 2 (k)) for k >k0 with p 2 (k0 )-y '.

^-^it should be stressed that we are asserting the (generic) existence of a unique factorization over the time interval k >k 0 with the given initial valuePi (^o) • If w e allow the initial value p2 (fc0) to range over ali of C, in general we will obtain an infinite collection of skew polynomial factorizations.

If the given system is time invariant for k


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k=k 0 , we could take p2 (k 0) to be one of the zeros of the polynomial z2 +a x (k 0)z +a0 (k 0).

If p x (k), p2 (k) are the solutions to (11) and (12) with p2(k 0) =

7 e C , then the complex conjugates p x (k), p2 (k) are the solutions to(11) and (12) with initial value 7 . This can be seen by simply taking thecomplex conjugate of both sidesof (11) and (12). Unlike the time-invariantcase, in general ~p x (k)i=p2 (k) and /?2 (h)ìp x (k). Thus, in generalthere are two different solutions to (11) and(12) with initial value p2 (k 0) =7 or p2 (^o)

= 7 • We cali the ordered sets (pi (k), p2 , (&)) and (pt (k),Pi (£)) P°le sets on k >k 0 with respect to the initial values 7 , 7 - Theelements p2 (k) and p2 (k) are called the right poles on k>k 0 .

In the following two examples, the pole sets were computed by solving(11) and (12) recursively. In particular, given p2 (k 0) we used (12) tocompute p x {k Q), then (11) to compute p2 (k 0 + 1) , then (12) tocompute pi (k 0 + 1), and so on.

Example 1: The coefficients of the input/output difference equation (1) are

tf 0 (£) = 0.5 forali keZ,a1(k)= < - 1 + (l5fc/200), 0


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As in Example 1, the initial poles are 0.5 '+ j0.5., 0.5 - jO.5 , and as k -•


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3. Properties of Factorizations

Again consider the discrete-time system given by the second-order input/

output difference equation (1). Let p x (k) and p2 (k) denote the zerosof the polynomial z2 + h x (k) z + a0 (k) ; that is,

(z-p x(k))(z~p2 (&)) = %2 +a x (k)z+a0 (k) ,

where (z -p x (&)) (z -p2 (k)) is the ordinary product of two polynomials.We can write this product in the form

{z-p, (k))(z~p 2 (*))=**- k Q , from (14) we have that

(15) (z -p\ (*)) (z -p\ (*)) - (z - p \ (*)) o (2 - p 2 (*)).

Here = means that the magnitude of the difference of the coefficients ofthe polynomials is small for k>k 0 . The relationship (15) corresponds tothe well-known result that slowly varying systems can be studied using the"frozen time appróach".

Now suppose that a0(k)-*c0 a n d #i (&) ~*c x as k-+°°. Let r x

and r 2 denote the zeros of z2 + c x z + c0 . If (p x (k), p2 (k)) is a

pole set on k > k 0 , it turns out that p2 (k) will not converge to r x orr 2 in general. In particular, if r x and r 2 are complex numbers andP2 (&o) is a r ea l number, by (13) p 2 (&) is real for ali k>k 0 , andthus p2 (k) cannot converge to f x or r 2 .

As a special case, suppose that

a0 (k)~c0 and #! (k):=c x forali k>k x .

Given the pole set (pj (fc), p 2 (&)) , let A (fc) = p 2 (*) - r2 ' so that

(16) A(* + l ) = p 2 (* + l ) - r 2

s


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By(13),

#n (k)

p2Ìh)

and thus for k > k ì ,

(17) p (* + 1 ) = _ , - - £ S _ .

Inserting (17) into (16) gives

A f t + 1)=Z(£L±a)£L»)Z£oPi (*)'

Now /?2 (&) = A (&) + r2 , and therefore

, x A . ~ (r\ •+• Ci r2 + £'o) "" fai + r2) A (&) •(18) A (* + 1.) = —1 -Ì —1 -2 -— Li —i -L 2 ; .

A (k) + r2

But since r x and r2 a r e t n e zeros of z2 + Cj z + c0 , we have that

(19) r 2 4- c x r 2 + cQ = 0 and ---(^ 4- r 2)-r x .

Using (19) in (18), we obtain

r, A (k)(20) A (* + 1) = T / •; *. ->*i. .

A (&) + r2

We can convert (20) into a linear difference equation by using the transforma-tion

(21) g &)=-*—^-,.


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Combining (20) and (21) gives

(22) £ (* + i) = I l [£(*) + l ] .

•ri

Now (22) can be solved using standard techniques, and once we have deter-mined g (k), we can compute A (k) using(21).If r 1¥

:r 2 and k x = 0,

the resùlt is

(r 1-r 2) A(0)(23) W'Moy+m-rs-uom,»*-'*

0-

From (23), we see that if A (0)i=r y -r 2 (i.e., p2 (0)#rj ) and I r 2 /r l l> 1,then A (&) converges to zero as &-» 1 requires that rj andr 2 be real (not complex conjugates).

The derivation given above can be generalized to include the case when#o ( ^ ) - ^0 a n d and

let f^, r2 be the zeros of • z2 + Cj z 4- c0 . If l^/rj l > 1 , then for anypole set (pt (k), p2 (k)) with p2 (k 0)=£r 1 , p2 (k) converges to r 2 ask-+.

The hypothesis of Proposition 2 is satisfied in Examples 1 and 2 given inthe previous section, and thus in these cases p2 (k) does converge tor 2 ( = - 1 ) . In the following example, we modify Example 2 so that r t andr 2 are complex.

Example 3 : Suppose that

( - 1 ,k


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In this example, r x =- 0. 7- j0 .1 and r 2 =7i . The poles with p2 (0) =0.5 -jO.5 are plotted in Figure 4. In this case, the poles pi(k) and p2 {k)continuously encircle points ih the complex piane as k -* °° .

Imag

Pi(k)

P2(k)

• \• • a « a

r '

01

O-0.5

V .• '• ' • •

• • s

'. 0.5+J0.5

0 0.5

« 0.5 -i0.5

Re

Figure 4. Poles with p2 (0) = 0.5 - j0.5 .

To avoid the type of behavior displayed in Example 3, we can reinitializethe recursion for computing the poles. For example, again letting p x {k)and p2 (k) denote the zeros of z

2 +at {k)z + a0 {k) , if \p2 {k + 1) — p\ {k)\ is small for some range of k


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we can write (26) in the form

(27) Pi(k)w{k) = q2(k)w(k + 1).

If w (k) =£ 0 for ali k > k 0 , from (27) we hàve that

w (k 4- 1)(28) pi (k) = ?2 (*) , k > k 0 .

w (&)

The relationship (28) shows that it is possible to compute the element p x (k)of the pole set (pi (k), p 2 (k)) from the right pòle q2 (&) of the pole set(


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and

b(z,k)= S b.(k)zl eA [z],

consider the linear discrete-time system defined by the input/output equation

a(z,k)y(k) = b(z,k)u(k) ,

where as before, y (k) is the output and u (k) is the input.

We cali p n (k) e AQ a right pole of the system if there exists a e (k, z) e

A c [:z.] such that

(29) a(z,k) = e(z,k)o(z-pu(k))'.

We cali q (k)eAQ a right zero of the system if there exists a h (z, k) e

A c [z] such that

b(z, k) = b (z, k)o(z-q(k)).

Suppose that there exist p n (k) and e (k, z) such that (29) holds. Then

writing

e(k,z) = z n~1+ n 2 2 e.{k)zi ,

we have that (29) is equivalent to the equations

(30) e n_ 2 ik)=p n tt.+ i f - D + tf^ (*)

(31) e._1 (k) = e. (k)p n (k H ) + a. (k) , i = »- 2, »-3, . . . , 1

(32). 0 = ^ 0 ( * ) P . ( * ) + « o ( * ) .

Given the initial values p n (k0 - n + 1) , pw (^0-» + 2), ..., pB (^0) » we can

solve (30)-(32) recursively to compute p n (k) and the e{ (k) for k>k0 .

As in the case n = 2 , a unique solution exists for almost ali possible values of

the initial data.

Once we have determined the factorization (29), we can then pulì out a

right factor from e (k, z), and so on, until we obtain the pole set (pij (fe),


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p2 (&)> •••>/>„ (^)) • I* is important to note that this is an ordered set. Due tothe noncommutativity of multiplication in the ring AQ [2], a permutation ofthe elements in a pole set would not result in anotherpole set (ùnless of coursethe pf(k) are Constant).

As we now show, Equations (30)-(32) can be combined to yield a singleequation for pn (k). First, multiplying both sides of (30) by p (k + n - 2)and using (31) with i = n - 2 , we obtain

(33) V 3 ( ^ , ( ^ + ^ ^ + ^ 2 ) ¾ ( * ) p f » + « - 2 ) + « ^ » ) .

Multiplying both sides of (33) by pn(k+n~3) and using (31) withi = n — 3 gives

*„_4 * >?n •

Continuing, we obtain

0^n{k+n-\)pn(k+n-2)...pn(k)+nì ai(k)pn(k+i-l)pn(k^i-2)...pn(k)+a0(k)

(34)

If pn (k) ¥= 0 for ^ > f e 0 - « + l , we can rewrite (34) in the form (assu-ming n > 3)

(35)flp (fe) _

pn(k+n-2)pn(k+n-ì)...pn{k) '

Equation (35) can be solved recursively to compute the right pole pn (k) .We can compute a collection of right poles by solving (35) for different

initial conditions. The calculation of right poles using (35) can be carried outin parallel. In the next section we will show that the computation of a set ofright poles arises in the derivation of a modal decomposition of the zero-inputresponse.

We conclude this section by showing that our notion of a zero can be

•»


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interpreted to be a transmissión blocking zero as .in the time-invariant case.Let q (k) be a right zero of the system defined above so that

b(z,k) = h{z,k)o(z-q(k)).

Define /

é (k , k0 )=h (k-l)q (k-2) ...q(k0 ), k>k0i

/o ,k k 0 . First, we have that

b(z,k)u(k) = h(z,k)[(z~q(k))u(k)]

= b{z,k)[4>q(k + l t k 0)-q (k)(j>q (k , *0)]

By definition of 0 (k, k 0) ,

. ^ ( * + l , *o ) = * (* )0 , » , Aio) , * > * o ,

and thus

b(z,.k)u(k) = 0 for &>&0 •

Returning to the input/output difference equation, we then have that

(36) y(k+n) = - "Ì a.(k)y(k+i), k>k 0 .

If y (&0 + /) = 0 for i = 0, 1, ..., n - 1 , it follows from (36) that y (&)= 0

forali k>k 0 as desired . If d 0 (&)=£() for & = &0 - 1 ,&0 - 2 , ..., &0 - » ,


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it Is a standard aonstruction to show that the initial values y (k 0 - 1 ) , y (k 0 - 2 ) , ...,y (k 0 -n) can be chosen so that y (k 0 + i) = 0 for / = 0 , 1 ,. . . , « - ! • . Thus we have the following result.

Propositìon 3: Suppose that a0 (k) =£ 0 for k = k 0 - 1 , k 0 - 2, ..., k 0 - n .Then with u fk) == 0 (&, fc0) , for some initial values y (k 0 - 1), y (k 0 - 2),•••» y (&o ""n) t n e output y (k) is identically zero for ali k > &0 .

5. The Zero-Input Response and Stability

In this section we assume that the input u (k) to the system is zero forali k > k 0 , so that the input/output difference equation reduces to

(37)»—i

y{k+rì)+ X a.(k)y(k+i) = 0

Using our operator notation, (37) can be written in the form

(38) a (z, k)y (&) = 0

The initial conditions for (37) (or (38)) are the values y (k 0), y (k 0 + 1),..., y(k 0 + » - l ) .

Let pnl (k), pn2 (k),...,pnn (k) be n right poles; that is, supposethat there exist polynomials ei (z , k) e Ac [z] such that

a(z,k) = e.(z;k)o(z-pn.(k)) for i = l, 2, .. .,». .

Let V (k) denote the time-varying version of the n X n Vandermondematrix defined by

K


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Finally, Jet {k', k 0) denote the mode associated with the pole p^-W- ni

Recali that

\ 1, k ~ k Q

P ..(¾. *o) = „,(* - DP„,- (* - 2 ) .../»., (ho), k>k 0

/o,kk 0.r= 1 * Pni

The constants ci in (39) depend on the initial values y (k 0), y (k 0 -f- 1),...,3/(^0 + » - l ) .

Proof: We will first show that y{k)i-c^ (k, k 0) is a solution to "a (z,k)y(k)=0.We Have that

a (2, k) ce '(k, k Q) = cei


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from (39) we also have

"> k 0 , then (39) simplifies to the modal decomposition in the time-invariantcase given by

y(k)= jt t c.p***, k>k0 .

The repeated root case is left to another paper (see Kamen [1987]).We conclude this section by applying the modal decomposition (39) tothe study of asymptotic stability. Recali that the given system is asymptotically stable (a.s.) on k>k 0 if the zero-input response y ih) convergesto zero as k-+°° for any initial values y (k 0 ) , y (k 0 + 1), ..., y (k 0 + n — 1).The system is uniformly asymptotically stable (u.a.s.) on k > k 0 if for anyreal number e > 0 , there is a positive integer N such that

\y(k+N)\k 0.

Theorem 2: Let pnl (k),pn2 (k) y...,pnn (k) be n right poles with asso-ciated Vandermonde matrix V (k), and suppose that V (k 0) is invertible.Then the system is a.s. on k>k 0 if and only if

(40) lp^- l )p M t ^ - 2 ) . . . p w ^ 0 ) l^0 as *->oof or i = 1, 2, .... » .

The system is u.a.s. on k > k 0 if and only if for any real number e > 0 ,


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there is a positive integer N such that

\pni(k+N e-l)pni(k+N r 2)...pJk)\ k 0 and i = 1, 2, ..., n.

The proóf of Theorem 2 follows easily from the modal decomposition(39). The details are omitted.

By Theorem 2, we see that testing for a.s. or u.a.s. can be reduced totesting the stability of the first-order systems corresponding to the componentsof the modal decomposition (39).

It follows from (40) that a sufficient condition for asymptotic stability onk>k 0 is

(41) \pni (k)\k Q

with period two, the system is a.s. if and only if

\pni(k 0 + Dpni(k Q)\


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We assume that the coefficients a: (t) and b. (t) can be differentiated asuitable number of times.

We begin by considering the second-order case (n = 2 ) , so that

a (D, t)~D z +at (t)D+a0 (t) .

Suppose that there exist functions pi (t) and p2 (t) (both of which maybe complex valued) such that

(43) (D2 +* j (t)-D+a0 (t))y (t)~(D~p x (?)) [(D ~p2 (;t))y (t)} .

Expanding the right side of (43), we obtain

(D-Pl (t))[(D~p2 (t))y(t))~[D2~(pl (t) + p2 (t))D +

(44)Pi(t)p2 (t)~p2 (t)]y(t) ,

where p2 (t) is the derivative of p2 (t).

Now we want to define a polynomial multiplication o such that

(45). [ ( D - Pl (t))o (D-p2 (t))]y(t) = (p-Pl (t))[(D-p2 (t))y(t)] .

Comparing (44) and (45), we see that

(D~Pl (t))o(D-p2 (t))~D2-(Pì « + p 2 (t))D +

( 4 6 ) Pi (t)p2 (t)~p2 (t) .

Note that

D o p2(t)=p2 (t)D-p2 (t) ,

and thus the multiplication o is noncommutative. Also note that the non-commutativity is different from that in the discrete-time case (see (9)).

If there exist p x (t) and p2 (t) such that (43) holds, we cali the or-dered set (pi (t), p2 (t)) a pole set, and we cali p2 (t) a right pole.It follows from (43)-(46) that (pi (t), p2 (t)) is a pole set if and only if

(47) pi W +P 2 ( 0 = -^1 (t)

\


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(4.8) . p x (0 p2 (t)~p2 \t) = a0 (t) .

Multiplying both sides of (47) by p2 (t) and using (48), we have that

(49) pi (0 4- p2 (t) + a ! (t) p2 (0 4- a0 (0 = 0 .

Note that (49) is a nonlinear first-order differential equation with time-varying coefficients. In fact, the left side of (49) looks like the polynomial

D2 4-0, (t)D+aQ (t) evaluatedat D~p2 (t) with

D |D=MO ^ 2 {t) a n d D " W,(0 = ^ ( t ) + ^2 ( t ) '

We conjecture that (49) has a unique solution p2 (t) for almost aliinitial conditions p2 (t 0 ) e C . We will show that this is the case whentfo (0 = co a n d ^ i ( 0 ~ c i forali t>t 0 .

Let r t and r2 denote the zeros of D2 + c t D f c0 and define

A(t ) = p2 ( 0 - r 2 .

Then

A(f)=p2 (0 = ~p2 (*)-Ci P2 (O-^o f o r *>'o

Inserting p2 (t) = A (t) + r2 into this expression gives

(50) À( 0 = - A 2 ( 0 - ( r 2 - r t ) A (0 for t > t 0 .

Defining g (t) - I/A (t) , we can convert (50) into the linear differentialequation

(51) £(£) = 0 - 2 - ^ ( 0 + 1 •

Solving (51) and using the relationship A (?) = \lg (t) , we obtain

A , , (r' i-r a)A(0) ^^(52) A (t)= — , t>0.

A(0) + (r , - r 3 ) - A ( 0 ) exp (-(r t - r2 ) 0

In deriving (52), we have assumed that r x # r 2 and t 0 = 0 .


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It is interesting that (52) has exactly the same form as in the discrete-timecase (see (23)), except that the exponential terni is replaced by (r 2 /ri)

k inthe discrete-time case. From (52), we see that there is a unique solution

p2 (t) = A (t ) + r 2 for any initial condition p2 (0).Returning to the nth-order case defined by (42) , we cali pn (t) a rightpole of the system if there is a polynomial e (D, t) such that

a(D,t) = e(P,t)o(p-pn(t))t

and we cali q (t) a right zero of the system if there exists a polynomialh (D, t) such that

b(D,t) = b(D,t)o(D-q(t)).

It is assumed that the coefficients of e (D, t) and h (D, t) belong to somespace of functions that can be differentiated an appropriate number of times.The precise specification of the space of coefficients will not be considered.

If p (t) is a right pole or a right zero, we define the mode associatedwith p (t) by

v 0 , t < t0

exp i>(X)

dX , t>t0

We then have the following continuous-time counterpart to Proposition 3.

Proposition 4: Let q (t) be a right zero of the system. Then with the input, t Qu (t) = (j) (t, t 0) , there exist initial conditions y

(l ) (t 0), i = 0, 1, ..., n - 1

The proof of this proposition is very similar to the proof of Proposition 3 andis therefore ornitted.

Now suppose that pnl (t), pn2 (t),...,/?Wf l (t) are n right poles ofthe system. Define the operator

S p .(f(t)) = P ni(t)f(t)+f(t),* ut

•t


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and let V (t) denote the generalized Vandermonde matrix defined by

V(t) =

1

Pnl

S (p At))

1

Pnl - ( t ) )

p (t)

s„ (p (*))WK

S"~2(/> (0

« K

We then have the following counterpart to Theorem 1

Theorem 3: Suppose that the determinant of V (t0) is nonzero. Then forany initial conditions y^ (t 0), i — 1, 2, ...,'w — 1 , there exist constantsclf c2 , ..., cn such that the solution to a (D, t) y (t)~0 can be written inthe form

(53) y ( 0 = S e (j> (t, t 0) for t>t 0 .

Again the proof is very similar to the one given in the discrete-time case, so weshall omit the details.

If the right poles pni (t) are Constant, so that pni (t)—p. for t> t 0 ,the modal decomposition (53) simplifies to the well-known form in the time-invariant case given by

y (t) = 2 c.exp (p. t) , t>t 0 .

It follows directly from (53) that the system is asymptotically stable(a.s.) on t > t 0 if and only if

(54) li (t,t 0)\-+0 as t->oo for « = 1,2, . . . , » ."ni

A condition which imples (54) is

(55) Rcp

ni(t)


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If the p . (t) in (55) are replaced by the ordinary zeros of a (D, t) , thenthe condition is no longer sufficient for stabili ty.

REFERENCES

[1] O. Ore, Theory of noncommutative polynomiah, Ann. Math., Voi. 36, pp. 480-508,1933.

[2] A.C. Amitsur, Differential polynomiah and division algebras, Ann Math., Voi. 59,

pp. 245-278, 1954.

[3] R.W. Newcomb, Locai time-variable synthesis, in Proc. Fourth Colloquim on Micro-ware Communications, Budapest, 1970.

[4] K.M. Hafez, New results on discrete-time time-varying linear systems, Ph. D. Disserta-tion, G'eorgia:Institu*te of Technology, Atlanta, 1975.

[5] E. W. Kamen and K. M. Hafez, Algebraic theory of linear time-varying systems, SIAMJ. Control and Optimization, Voi. 17, pp. 500-510, 1975.

[6] E.W. Kamen, P. P. Khargonekar, and K. Poolla, A transfer fimction approach to lineartime-varying discrete-time systems, SIAM J. Control and Optimization, Voi. 23, pp.550-565, 1985.

[7] E.W. Kamen, On the factorization of difference and differential polynimials, in prepa-ration.

EDWARD W. KAMENPittsburgh, PA 15261

- Department of Electrical Engineering University of Pittsburgh

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