Transfer Functions. Olof Staffans

Transcription

1 Transfer Functions Olof Staffans March 10, 2009

2 Contents 1 Introduction State Space Model of a BB An Extended BB A Distributed Parameter Example A General Black Box Model A Discrete Time BB Model Transfer Functions Realizations of Rational Transfer Functions Different Stability Notions Exponential Stability Power Stability Input/Output Stability The Cayley Transform Time Discretization The Z-transform Versus the Fourier Series The Frequency Domain Notations L p -Spaces The Discrete Time Case The Continuous Time Case Plancherel s Theorem H p -Spaces Frequency Domain Stability The Standard Problem The General Setting H -minimization Problem The Model Matching Problem Tracking Problem

3 5 Coprime Factorizations The SISO Case Discrete Time Continuous Time The MIMO Case Computation of Doubly Coprime Factorizations Stabilization Factorization of the Standard Problem Stabilization of the Standard Problem The Youla Parameterization

4 Chapter 1 Introduction 1.1 State Space Model of a BB Let s consider a BB = Black Box. input u Black box output y A standard engineering assumption is that the dynamics of the black box can be described by { ẋ(t) = Ax(t) + Bu(t), x(0) = x0 (1.1) y(t) = Cx(t) + Du(t), where x u y is an n-dimensional vector, the state ( tillstånd eller fas ), is a p-dimensional vector, the input ( insignal ), is a q-dimensional vector, the output ( utsignal ), and A R n n, B R n p, C R q n, D R q p are matrices. Often p = 1 ( single input ) and q = 1 ( single output ). We denote 3

5 SISO = single input, single output, MIMO = multiple input, multiple output etc. If n =, then we have a Distributed Parameter System (treated in the previous course). The solution of (1.1) is obtained (from standard theory of ODE:s) x(t) = e At x 0 + y(t) = Ce At x 0 + t 0 t e A(t s) Bu(s) ds, 0 Ce A(t s) Bu(s) ds + Du(t). The system is exponentially stable if all the eigenvalues of A have strictly negative real parts. In this case there exist constants M < and ε > 0 such that e At Me εt. (The constant ε is called the stability margin.) The initial state corresponds to excitations of the system due to past inputs or initialization terms (which pop up when you switch on the current to the box). If the system is at rest at the time t = 0, then x 0 = 0. In that case y(t) = t 0 Ce A(t s) Bu(s) ds + Du(t) = Du(t) + where K(t) = Ce At B. We call the distribution Dδ + K t 0 K(t s)u(s) ds, (where δ is the Delta-function ) the impulse response of the box. Here D is the feedthrough or feedforward and K is the strictly proper part of the impulse response. 1.2 An Extended BB We group the input and output vectors into two parts. 4

6 u 1 u 2 Plant = P y 1 y 2 Controller = Q Both P and Q are black boxes, u 1 R p 1 u 2 R p 2 y 1 R q 1 y 2 R q 2 = external input = control input = external output = measured output and p 1 + p 2 = p = total input dimension q 1 + q 2 = q = total output dimension. Task (uppgift): Design a controller Q in such a way that y 1 behaves in the desired way as u 1 varies. The desired way may include tracking, the output follows the input (e.g. a power amplifier) disturbance rejection, y 1 0 independently of how u 1 behaves L 2 -optimal control, minimize the effect of initial transients L 1 -optimal control, minimize the maximal value of the output H -optimal control, minimize the amount of energy which passes from u 1 to y 1. Classes of permitted controllers: the same type as P (black box) digital implementation (very common). 5

7 1.3 A Distributed Parameter Example Consider a bar with insulation on the sides. The end point is heated. Heat evaporates through the end point by radiation. A sensor measures the radiation, and it gives us the temperature at the end point. Let T 0 + T(t, x) be the temperature of the bar at the point x [0, ) at time t [0, ). The constant T 0 is the surrounding temperature. Standard PDE-theory gives T t (t, x) = T xx (t, x), x 0, t 0, T x (t, 0) = αt(t, 0) }{{} u(t) }{{}, t 0, radiation heat source T(0, x) = 0 The solution is given by Green s formula, T(t, x) = 1 π t 0 (initial rest temperature). 1 e x2 4(t s) (u(s) αt(s, 0)) ds. t s Thus, if we know T(s, 0), then we can compute the rest of the temperature distribution from this formula. Substitute x = 0 above, and denote y(t) = T(t, 0), k(t) = 1 πt. Then y(t) = t 0 k(t s)(u(s) y(s)) ds. u u y y k + 6

8 This can be interpreted as a feedback connection of the BB which maps u into (k u)(t) = t 0 k(t s)u(s) ds. This function k cannot be written in the form k(t) = Ce At B for any matrices C, A, B (of finite size). General questions: What is a reasonable requirement on the impulse response of a BB? How do we deal with unstable systems (like an atomic reactor or a modern combat air plane like JAS Gripen)? What do we do when we have many inputs and outputs? (MIMO) Answers later. 1.4 A General Black Box Model Let us make a number of natural assumptions, and use these to derive a model of a general BB: u P y = Pu (dimension p) (dimension q) Hyp. 1 Causality: Future inputs have no influence on past outputs, if u 1 (s) = u 2 (s) for s T, then (Pu 1 )(s) = (Pu 2 )(s) for s T. Hyp. 2 Time independence: The response of P is the same today as tomorrow. If the input signal arrives h seconds later, then the response is the same, except that it also is delayed by h seconds. If u 2 (s) = u 1 (s + h) for some constant h, then (Pu 2 )(s) = (Pu 1 )(s + h), s R (this hypothesis can be removed, the model then becomes more complicated). 7

9 Hyp. 3 Continuity or well-posedness: Small changes in the input causes small changes in the output. For all ε > 0 there is a δ > 0 such that if (i) u 1 (s) = u 2 (s) for s S (ii) u 1 (s) u 2 (s) δ for S s T, then (iii) (Pu 1 )(s) (Pu 2 )(s) ε for S s T. (This hypothesis can be slightly weakened.) Hyp. 4 Linearity: Two signals do not interfere with each other, and if we increase the amplitude of the input, then the amplitude of the output increases by the same scale. If u(t) = λ 1 u 1 (t) + λ 2 u 2 (t), then (Pu)(t) = λ 1 (Pu 1 )(t) + λ 2 (Pu 2 )(t). (Standard requirement for all amplifiers: We want to avoid distortion of the original signals, and different frequency components should not interact.) (True in a limited amplitude range, typically.) Hyp. 5 Now pure dealys: If we feed in a step signal { 1, t 0 u(t) = 0, t < 0, then the output (P u)(t) is an (absolutely) continuous function (we allow a jump discontinuity at zero). (This hypothesis can be removed, at the expense of a more complex model.) Theorem Assume Hyp Then there is a matrix-valued function k L 1 loc (R+ ; R q p ) which represents P in the following sense. For every continuous input signal u which satisfies u(t) = 0 for all t T (for some T R) we have (Pu)(t) = Du(t) + ( = Du(t) + t t T k(t s)u(s) ds, t R ) k(t s)u(s) ds. Proof. Bypassed because of lack of time. Not terribly difficult, but one needs some knowledge of integration theory. Instead we prove the discrete time version of this result. 8

10 1.5 A Discrete Time BB Model Consider a discrete time variable t {..., 2, 1, 0, 1, 2,... } = Z. u(k) BB y(k) = (P u)(k) We still have p inputs and q outputs, so u(k) R p, y(k) R q, k Z, and u and y are vector-valued sequences, u = {u(k)} k=, y = {y(k)} k=. We still make the same hypothesis as in the last section. Hyp. 1 Causality: If u 1 (k) = u 2 (k) for k K, then (Pu 1 )(k) = (Pu 2 )(k) for k K. Hyp. 2 Time independence: If u 2 (k) = u 1 (k + h) for some h Z and all k Z, then (Pu 2 )(k) = (Pu 1 )(k + h), k Z. Hyp. 3 Continuity or well-posedness: To every given input u which vanishes for all k K (for some K) there is a unique output y. Hyp. 4 Linearity: If u = λ 1 u 1 + λ 2 u 2, then Hyp. 5 Properness: Is not needed. Pu = λ 1 Pu 1 + λ 2 Pu 2. Theorem If Hyp. 1-4 hold, then there exists a matrix-valued sequence {K(k)} k=0 (each K(k) Rq p ) which represents P in the following sense. If u(k) = 0 for all k < M (for some M Z), then ( ) k k (Pu)(k) = K(k l)u(l) = K(k l)u(l). (1.2) l=m 9 l=

11 Note If (1.2) holds, then it is easy to show that Hyp. 1-4 are satisfied. Thus, Hyp. 1-4 are necessary and sufficient for the representation (1.2). Definition We call {K(k)} k=0 the (discrete time) impulse response of P. Proof of Theorem Step 1. Find K(0): Suppose that u(k) = 0 for k < 0, and look at (Pu)(0). By Hyp. 1, 3 and 4, this is a linear function of u(0) only. But every linear map R p R q has a matrix representation, there is a matrix K(0) R q p such that (Pu)(0) = K(0)u(0), if u(k) = 0 for k < 0. Step 2. Find K(1): Suppose that u(k) = 0 for all k 0, k 1. Consider the value (Pu)(0). As above we conclude that there is a matrix in R q p, which we denote by K(1), such that (Pu)(0) = K(1)u( 1), if u(k) = 0 for all k 0, except k = 1. Step 3. Find K(m) m 2: Take u(k) = 0 for all k 0 except for k = m. As above we find a matrix K(m) R q p such that if u(k) = 0 for k 0, except k = m. (Pu)(0) = K(m)u( m), Step 4. Find (Pu)(0) for general u: Suppose that u(k) = 0 for all k M, where M Z, M 0. We can then write {u(k)} 0 k=m in the form {u(m), u(m + 1),...,u( 1), u(0)} = {u(m), 0,..., 0, 0} + {0, u(m + 1), 0,..., 0}. + {0,..., 0, u( 1), 0} + {0,..., 0, 0, u(0)}. 10

12 Each piece is of the type considered in Step 3. By Hyp. 4 (linearity), (Pu)(0) = (P(first seq.))(0) + (P(second seq.))(0) +... = K( M)u(M) + K(1 M)u(M + 1) +... = = + K(1)u( 1) + K(0)u(0) 0 K( l)u(l) l=m 0 l= K( l)u(l) (where all terms are zero if l < M). Note. So far we have not used the time invariance, Hyp. 2. Step 5. Find (Pu)(k) for all K: Define v(l) = u(k + l), l Z (a shifted sequence). The time invariance gives which proves (1.2). (Pu)(k) = (Pv)(0) 0 = K( l)v(l) = = l= 0 l= k m= K( l)u(k + l) (put k + l = m) K(k m)u(m) Definition We call the sum above the convolution of K and u, and denote it by K u. Thus, k (Pu)(k) = (K u)(k) = K(k l)u(l) l= (where all but finitely many terms are zero). Comment Without any further knowledge of K, we may compute (1.2) by using the FFT, so it is easy to implement a Discrete BB of this type on a computer. Note that K(1), K(2),... represent delayed terms (we have time to compute them), but K(0) is needed immediately (implement by a direct coupling). 11

13 1.6 Transfer Functions Definition The transfer function of the discrete time system described in Theorem is the Z-transform of the impulse response, i.e., ˆK(z) = K(k)z k k=0 (defined for those z C for which the sum converges, and analytically extended to a domain which is as big as possible). Definition The transfer function of the continuous time system in Theorem is given by G(s) = D + ˆk(s), where ˆk is the Laplace transform of k, ˆk(s) = 0 e st k(t) dt (defined for those s C for which the integral converges, and extended by analyticity). Note Typically ˆK(z) is defined and analytic for z large enough (close to complex infinity). Typically G(s) is defined for Re(s) large enough (in some right half-plane). In the rational case they can be extended to all but finitely many points in C. Note Some people replace z k by z k in Definition Realizations of Rational Transfer Functions Recall the engineering model of Section 1.1: ẋ(t) = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t), x(0) = x 0, (1.3) 12

14 whose solution is (if x 0 = 0 and t 0) x(t) = The impulse response is t y(t) = Du(t) + 0 e A(t s) Bu(s) ds, t 0 t 0 Ce A(t s) Bu(s) ds. k(t) = D δ(t) +Ce At B. }{{} δ-function Problem Find the transfer function. Solution: Take Laplace transforms in (1.3) to get { sˆx(s) = x(0) + Aˆx(s) + Bû(s), ŷ(s) = Cˆx(s) + Dû(s), (Re(s) large enough). This implies (s A)ˆx(s) = x(0) + Bû(s) and so If x(0) = 0, then ˆx(s) = (s A) 1 (x(0) + Bû(s)), ŷ(s) = C(s A) 1 (x(0) + Bû(s))Dû(s). ŷ(s) = C(s A) 1 Bû(s) + Dû(s). On the other hand, by standard Laplace transform theory, if then y(t) = Du(t) + t 0 k(t s)u(s) ds, ŷ(s) = Dû(s) + ˆk(s)û(s). Comparing these two to each other we get the following Theorem. Theorem The transfer function of the continuous time system (1.3) is G(s) = D + C(s A) 1 B. (1.4) In particular, each component of this function is rational. 13

15 Proof. We derived (1.4) above. This function is rational, because we may write (by Cramer s rule) (s A) 1 = adj(s A) det(s A), where det(s A) is a polynomial of degree n (dimension of state space), and each element of adj(s A) is a polynomial of degree n 1 (it is ± the sub determinant which you get by dropping one row and one column of (s A)). Corollary If the dimension of the state space is n, then the transfer function has at most n poles (counted according to their multiplicity). Proof. det(s A) is a polynomial of degree n in s. A similar result is true in discrete time. A standard engineering assumption is that we can describe the input/output relation by using a system of difference equations, x(k + 1) = Ax(k) + Bu(k), k Z + y(k) = Cx(k) + Du(k), (1.5) x(0) = x 0, where Z + = {0, 1, 2,... }. We multiply this by z k and add, z k x(k + 1) = A z k x(k) + B k=0 k=0 z k Bu(k), where the left hand side is [ ] [ ] z z k 1 x(k + 1) + x(0) x(0) = z z k x(k) x 0 = z[ˆx(z) x 0 ]. k=0 k=0 k=0 Thus, so which implies z(ˆx(z) x 0 ) = Aˆx(z) + Bû(z), (z A)ˆx(z) = zx 0 + Bû(z) ˆx(z) = (z A) 1 (zx 0 + Bû(z)), ŷ(z) = C(z A) 1 (zx 0 + Bû(z)) + Dû(z). 14

16 If again x 0 = 0, then we get ŷ(z) = [D + C(z A) 1 B]û(z). On the other hand, by taking z-transforms (multiply by z k and add) in formula (1.2) we get ŷ(z) = ˆK(z)u(z), so we arrive at the following theorem. Theorem The transfer function of the discrete time system (1.5) is In particular, it is rational. ˆK(z) = D + C(z A) 1 B. There is also a converse to Theorems and Theorem Every proper rational matrix-valued function with rational elements can be interpreted as the transfer function of a continuous time system of the type described in Theorem 1.7.2, and also as a transfer function of a discrete time system of the type described in Theorem In other words it has a representation of the form ˆK(s) = D + C(s A) 1 B for suitably chosen matrices A, B, C and D (and a state space R n of suitable dimension). Note By Corollary 1.7.3, the dimension of the state space is at least as large as the number of poles of the given rational transfer function (counted according to their multiplicity). Definition The minimal dimension of the state space in Theorem is called the Mac-Millan degree of the given rational matrix-valued function. 15

17 Chapter 2 Different Stability Notions 2.1 Exponential Stability Definition The continuous time system ẋ(t) = Ax(t) + Bu(t), t 0 y(t) = Cx(t) + Du(t), t 0 (2.1) x(0) = x 0 is exponentially stable if e At 0 exponentially, i.e., there exist constants ε > 0 and M < such that e At Me εt for all t 0. Theorem The system (2.1) is exponentially stable if and only if all the eigenvalues of A lie in the open left half-plane. In particular, (s A) 1 then has no poles in the closed right half-plane. Proof (outline). Based on Jordan s canonical form. After a similarity transform in the state space we get a number of independent Jordan blocks, and it suffices to prove this for one single block. Corollary If (2.1) is exponentially stable, then its transfer function D + C(s A) 1 B has no poles in the closed right half-plane. Proof. The function (s A) 1 has no poles there. Comment The converse is not true. Even if the transfer function has no poles in the right half-plane, the system need not be exponentially stable (unless it is minimal, i.e., both controllable and observable). 16

18 2.2 Power Stability Definition The discrete time system x(k + 1) = Ax(k) + Bu(k), k Z + y(k) = Cx(k) + Du(k), x(0) = x 0, (2.2) is power stable if the solution with u(k) 0 tends to zero as k with a geometrical rate, x(k) Mη k for some M < and 0 η < 1. Theorem The system (2.1) is power stable if and only if the eigenvalues of A lie strictly inside the unit disc, λ < 1 for every eigenvalue λ of A. Proof. Based on the Jordan decomposition of A, and the fact that x(k) = A k x 0, k Z +, if u(k) = 0 for all k. Corollary If (2.1) is power stable, then the transfer function of (2.1) has no poles in the closed exterior of the unit disc, i.e. in the region {z C z 1}. Proof. The function (z A) 1 has no poles there. Comment Comment applies also to the discrete case. 2.3 Input/Output Stability If we have no state space representation of an impulse response, then the preceding stability definitions are not valid. Instead we must look directly at the input/output map. Several different stability notions exist. 17

19 Definition Consider the discrete time input/output map (assuming that u(k) = 0 for k < 0) y(n) = n K(n k)u(k), (n 0). k=0 This map is (i) l - or BIBO-stable (Bounded Input Bounded Output) if there is a constant M < such that for all u, sup n y(n) M sup u(n). n (ii) energy stable, or l 2 stable, if there is a constant M so that ( ) 1/2 ( ) 1/2 y(n) 2 M u(n) 2. n } {{ } l 2 -norm of y n } {{ } l 2 -norm of u (iii) l 1 stable if there is a constant M so that y(n) M u(n). n n In this course we only study energy stability, which we simply call input/output stability. The same notions are also used in continuous time, with the sums replaced by integrals. 2.4 The Cayley Transform We will now review a simple method for solving a given continuous time system. We transform the system into a discrete system in such a way that the discrete trajectory can be considered a numerical approximation of the continuous trajectory. Let h > 0 to make {t n } n=0, t n = nh, an equidistant discretisation of the interval [0, ). By integrating the standard input equation ẋ(t) = Ax(t) + Bu(t) 18

20 (h,x(h)) (h, x(1)) (nh, x(n)) ε n (nh,x(nh)) (0,x(0)= x(0)) (t,x(t)) h nh t Figure 2.1: The internal Cayley transform interpreted as a discrete approximation of the original system. The points ( nh, x n ) do not lie on the curve ( t, x(t) ), but the error ε[t/h] is proportional to h 2. from t n to t n+1 we get x(t n+1 ) x(t n ) = tn+1 t n ( Ax(t) + Bu(t) ) dt. If the continuous integrand is approximated by a straight line, this becomes x(t n+1 ) x(t n ) h 2( Ax(tn+1 ) + Ax(t n ) + Bu(t n+1 ) + Bu(t n ) ). (2.3) Introduce α := 2/h > 0 and u n = ( u(t n+1 ) + u(t n ) ) / 2α. If α ρ(a), then (2.3) can be solved for x(t n+1 ): x(t n+1 ) (α A) 1 (α + A)x(t n ) + 2α(α A) 1 Bu n, Thus define x n by x 0 = x(0), x n+1 = (α A) 1 (α + A)x n + 2α(α A) 1 Bu n (2.4) in order to make x n an approximation of x(t n ). If d 2 u/dt 2 exists for all t 0, then x [t/h] x(t) C(t)h 2. See Figure 2.1 for an illustration. The scaling of the input has been chosen so that the l 2 -norm of the input sequence u k is approximately equal to the L 2 -norm of the original input function u. 19

21 The above transform (together with a similar discretization of the output) is called the Cayley transform. The numerical properties of this transform has also been studied recently in [MH04]. The above discussion is taken from [KS05]. The Cayley transform is also used is used a lot in theoretical proofs, to convert a theorem concerning continuous time to discrete time and back. However, in that connection one uses a different correspondence between the continuous time functions x, u, and y and their discrete counterparts. Instead of sampling the continuous time functions one defines the coefficients of the discrete time functions to be the Fourier exponents of the continuous time functions with respect to a Laguerre basis. In this version the correspondence between a continuous time signal and a discrete time signal is not causal, and it cannot be used in on-line computations. The place to read more about this is Chapter 11 of [Sta05]. Definition We consider the continuous time system ẋ(t) = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t), x(0) = x 0. (2.5) Let Re(α) > 0 and suppose that (α A) is invertible (α is not an eigenvalue). Then the discrete time system x n+1 = A(α)x n + B(α)u n, y n = C(α)x n + D(α)u n, x 0 = x 0 (2.6) is called the Cayley transform of (2.5) with parameter α. Here A(α) = (α + A)(α A) 1 B(α) = 2Re(α)(α A) 1 B C(α) = 2Re(α)C(α A) 1 D(α) = G(α), and G(s) is the transfer function of (2.5). Note. The most popular values of α are α = 1 and α = 1/2 (in the latter case 2Re(α) = 1). 20

22 Lemma There are many interesting connections between the original system and it s Cayley transform (2.6). The most interesting ones in this course are (i) (2.5) is exponentially stable if and only if (2.6) is power stable (here it is important that the dimension of the state space is finite). (ii) The transfer functions of the two systems are related as follows: If G is the transfer function of (2.5), and Q is the transfer function of (2.6), then G(s) = Q(z) if we choose s and z to satisfy z = α + s α s s = αz α z + 1. (Thus, G(s) = Q( α+s ) and Q(z) = G(αz α ).) α s z+1 Proof. A lengthy computation (manipulation of matrices and their inverses). Comment The same transform is also used in the case of the general models described in Theorem and Theorem We say that the discrete time impulse response {K(j)} j=0 is the Cayley transform with parameter α of the continuous time impulse response Dδ + k if their respective transfer functions satisfy G(s) = ˆK(z), s and z as above. (G is the Laplace transform of Dδ + k and ˆK is the Z-transform of {K(n)} n=0.) 2.5 Time Discretization The Cayley transform behaves mathematically in a nice way, but in its mathematical Laguerre version it cannot be used for online control purposes (because of the lack of causality). In digital control one often uses direct sampling of the input/output map instead. This method has the advantage that it does not need a state space representation of the system. 21

23 If we in the continuous time formula y(t) = t 0 k(t s)y(s) ds take t to be a multiple of h, t = nh (h > 0 is the sampling interval), and approximate the integral by y(nh) h n k(nh jh)u(jh), j=0 then we get a discrete time impulse response. u(jh) u j y(jh) y j hk(jh) K j y n = n K n j u j j=0 The discrete time transfer function ˆK is ˆK(z) = hk(nj)z j. j=0 This can also be interpreted as an approximation of the original continuous time kernel. We replace K(t) by a sum of δ-functions located at the points 0, h, 2h,.... k(t) h[k(0)δ 0 + k(h)δ h + k(2h)δ 2h +... Obviously, this transformation cannot be inverted exactly (since we loose information). The Laplace transform of the sum of δ-functions is ˆk(s) = h(k(0) + k(h)e hs + k(2h)e 2hs +... ) = h(k 0 + K 1 e hs + K 2 (e hs ) 2 + K 3 (e hs ) ) = h K j (e hs ) j j=0 = h ˆK(e hs ) = h ˆK(z), where z = e hs. We have for all integers n that e 2πn = 1, so e hs = e h(s+2πn/h) 22

24 for all n. That is, the new continuous transfer function is periodic in the direction of the imaginary axis, with period 2π/h. If we are only interested in the frequency band π h < Im(s) π s, then we can map this part of the plane one-to-one onto the discrete time frequency variable z = e hs. Note that Re(s) 0 z > 1, so the appropriate part of the unstable part of the continuous time frequency domain is mapped into the unstable part of the discrete time frequency domain. Note. If the original transfer function is rational, then the transformed function is not. We loose the state space representation (the function s e hs is not rational). More on time discretization in the course on wavelets. 2.6 The Z-transform Versus the Fourier Series The Z-transform is very closely related to the Fourier series. The Z-transform of the sequence {u(k)} k=0 is û(z) = u(k)z k. k=0 Put z = re 2πiφ, where r = z and 2πφ, 0 φ < 1 is the argument of z. Then û(z) = u(k)r k e 2πikφ k=0 is the Fourier series (a function defined for 0 φ < 1) of the sequence u(k)r k. In particular, taking r = 1 we find that we get the Fourier series of a sequence {u(k)} k=0 by evaluating it s Z-transform on the unit circle z = 1. Therefore, you can use any book on Fourier series (the inverse of the finite Fourier transform) to learn more about the Z-transform. 23

25 Chapter 3 The Frequency Domain Terminology: We work in the frequency domain when we study either how the Laplace transforms (continuous time) or Z-transforms (discrete time) of the inputs and outputs behave. 3.1 Notations R = (, ) R + = [0, ) Z = {0, ±1, ±2,... } Z + = N 0 = {0, 1, 2,... } N = {1, 2, 3,... } C = {all complex numbers} C + = {z C Re(z) 0} C + = {z C Re(z) > 0} D = {z C z 1} D= {z C z < 1} E= {z C z > 1} E = {z C z 1} R n m = n m-dimensional matrix with real elements C n m = n m-dimensional matrix with complex elements R n = R n 1 C n = C n 1 Note. Most mathematicians like to work in the interior D of the unit disc in discrete time. Most control engineers like to work in the exterior E of the unit disc. The mathematical Z-transform is u(k)z k, k=0 24

26 and the control Z-transform is u(k)z k. k=0 Here we use the control version throughout the course. (Replace z by 1/z to get the mathematical version.) 3.2 L p -Spaces The Discrete Time Case Definition l p (Z; C m n ) = The set of sequences K(k), k Z, that are (m n)-matrix valued, and satisfy K(k) p <. k= Here 1 p <, and K(k) is some norm in C m n. The norm in l p is ( ) 1/p K p = K(k) p. k= l (Z; C m n ) = The set of bounded C m n -valued sequences with norm K = sup K(k). k Z l p (Z + ; C m n ) = As above, but k 0. l p (N; C m n ) = As above, but k 1. Note. In the last two cases we typically define K(k) = 0 for k 1 respectively k 0. Note. We typically require that the norm in C m n or R m n is adapted to the norm in C m and C n etc. If we use the Euclidean norms in R n and C n, then K(k) = sup x R n x =0 (This is the operator norm of K(k).) 25 K(k)x R m. x R n

27 3.2.2 The Continuous Time Case Definition L p (R; C m n ) = The set of (m n)-matrix valued functions on R which are measurable and satisfy ( 1/p K p = K(t) dt) p <. Here 1 p <. L (R; C m n ) = The set of (m n)-matrix valued measurable functions which are essentially bounded, with norm K = (ess)sup t R K(t). L p (R + ; C m n ) and L (R + ; C m n ) same as above, but t R + instead. The variables k (discrete time) and t (continuous time) live in the time domain (which is R or Z). The appropriate transforms (Laplace or Z) live in the frequency domain. Interpretation: L 2 and l 2 : L and l : L 1 and l 1 : energy maximal amplitude total mass. 3.3 Plancherel s Theorem To move back and forth between the time domain and the frequency domain we need two versions of Plancherel s theorem (or Parseval ), one for continuous time and one for discrete time. Lemma (Continuous time Plancherel.) Let u L 1 (R; C n ) L 2 (R; C n ). Then the Fourier transform of u, ũ(ω) = e iωt u(t) dt belongs to L 2 (R; C n ), and if we use the Euclidean norm in C n, that is ( ) 1/2 u(t) C n = ui (t) 2 26

28 where u(t) = u 1 (t) u 2 (t). u n (t) u 2 2 =, then u(t) 2 dt = 1 ũ(ω) 2 dω = 1 2π 2π ũ 2 2. Note Throughout the rest of this course we use the Euclidean norm in R n and C n. Lemma (Discrete time Plancherel.) Let u l 1 (Z; C n ) (then also u l 2 (Z; C n )). Then the (discrete) Fourier transform ũ(φ) = satisfies ũ C[0, 2π], and u 2 2 = k= k= e ikφ u(k), 0 φ 2π, u(k) 2 = 1 2π ũ(φ) 2 dφ = 1 2π 0 2π ũ 2 2. Thus, the Fourier transform maps L 2 (R) L 2 (R) in continuous time, l 2 (Z) L 2 ([0, 2π]) in discrete time. 3.4 H p -Spaces From the preceding two lemmas we can derive certain results for Laplace and Z-transforms. We again begin with continuous time. û(s) = 0 e st u(t) dt is the Laplace transform of u. Define u(t) = 0 for t < 0, and write s = α+iω. Then û(α + iω) = 0 27 e iωt e αt u(t) dt

29 which is the Fourier transform of the function { e αt u(t), t 0 0, t < 0. By Lemma 3.3.1, we have for all α > 0, 1 2π û(α + iω) 2 dω = 0 0 e αt u(t) 2 dt = u(t) 2 dt = u e 2αt u(t) 2 dt Thus, there is a constant (= 2π u 2 2) such that û(α + iω) 2 dω K for all α > 0 (with equality as α 0+). In addition û(s) is analytic in C + = {s C Re(s) > 0}. This is true whenever u L 2 (R + ; C n ). Definition Let 1 p <. The space H p (C + ; C m n ) consists of all C m n -valued functions f which are analytic in C + and satisfy the condition ( 1/p f H p (C + ) = f(α + iω) dω) p. sup α>0 The space H (C + ; C m n ) consists of all C m n -valued bounded and analytic functions in C +, with norm f H (C + ) = sup f(s). Re(s)>0 The computation on the preceding page proves Theorem If u L 2 (R + ; C n ), then the Laplace transform û of u satisfies û H 2 (C + ; C n ) (with equality of norms). 28

30 Let us also mention without proof the following theorem. Theorem Every function in H 2 (R + ; C n ) is the Laplace transform of some function u L 2 (R + ; C n ). Thus, we get a one-to-one unitary (i.e. norm preserving) map between L 2 (R + ; C n ) and H 2 (C + ; C n ): 1 2π Laplace L 2 (R + ; C n ) H 2 (C + ; C n ) 2π Inverse Laplace We continue with discrete time. Let {u(k)} k=0 l2 (Z + ; C n ). Then û(z) = z k u(k). k=0 By writing z = re iφ we get û(re iφ ) = e ikφ r k u(k) k=0 which is the discrete Fourier transform of the sequence {r k z(k)} k=0 (we define u(k) = 0 for k < 0). By Lemma 3.3.2, if r 1, then 1 2π û(re iφ ) 2 dφ = 2π 0 r 2k u(k) 2 k=0 u(k) 2 = u 2 2. k=0 Thus, there is a constant (= u 2 2 ) such that for all r > 1, the total energy of û on a circle of radius r is bounded, i.e., 1 2π û(re iφ ) 2 dφ K 2π 0 (with equality as r 1+). In addition û is analytic in E= {z C z > 1}, and also at infinity. 29

31 Analytic at infinity means that the function û(1/z) is analytic at zero. Thus, û(1/z) is analytic in D. Definition Let 1 p <. The space H p (D; C m n ) consists of all C m n -valued analytic functions in D= { z < 1} which satisfy ( 1 2π 1/p f H p (D) = sup f(re iφ ) dφ) p <. 0 r<1 2π 0 The space H (D; C m n ) consists of all bounded analytic C m n -valued functions, with norm f H (D) = sup f(z). z D H (E; C m n ) as above but D is replaced by E= {z C z > 1}. (Here f must also be analytic at infinity.) H p (E; C m n ) as above, but r > 1 instead of r < 1 (and f is analytic also at infinity). Thus, the argument on the preceding page proves one half of the following theorem. Theorem If u l 2 (Z + ; C n ), then û H 2 (E ; C n ), and conversely, every f H 2 (E ; C n ) is the Z-transform of some u l 2 (Z ; C n ). The Z- transform is a unitary map of l 2 (Z + ; C n ) onto H 2 (E ; C n ) (unitary means that it is one-to-one onto, and preserves norms and inner products). 3.5 Frequency Domain Stability We discussed different versions of input/output stability in Section 2.3. Only one of them has a simple frequency domain interpretation, namely the energy version (L 2 or l 2 ). 30

32 Theorem A continuous time input/output map u y(t) = Du(t) + t 0 k(t s)u(s) ds is (energy) input/output stable if and only if its transfer function G satisfies G H (C + ; C q p ). Theorem A discrete time input/output map u y(n) = n K(n k)u(k) k=0 is (energy) input/output stable if and only if its transfer function Q satisfies Q H (E ; C q p ). Continuous time: No poles in the right half plane. Discrete time: No poles outside the unit disc. Proof (partial) of Theorem Suppose that Q is analytic and bounded in E. Let u l 2 (Z + ; C p ). Then û H 2 (E ; C p ). The Z-transform of the output y is ŷ(z) = Q(z)û(z) ( z > 1) (the Z-transform maps convolution into pointwise multiplication like the Laplace transform). Therefore we find that ŷ is analytic in E, and that for all r 1, 1 2π 2π 0 ŷ(re iφ ) 2 dφ = 1 2π 1 2π M2 2π 2π 0 2π 0 2π 0 Q(re iφ )û(re iφ ) 2 dφ Q(re iφ ) 2 û(re iφ ) 2 dφ }{{} M 2 û(re iφ ) 2 dφ M2 2π u 2 2 (by the computation on p. 28). Therefore ŷ H 2 (E ; C n ), and by Theorem 3.4.5, the inverse Z-transform of ŷ, i.e., y itself, must therefore belong to l 2 (Z + ; C n ). 31

33 Thus, input/output (energy) stable transfer function is in H. Proof of converse: Take u(t) = e st u 0, t 0. 32

34 Chapter 4 The Standard Problem The results presented in this section are also found in [Fra87], Chapter The General Setting w G z u y K All signals are vectors, A more detailed picture w = external input (disturbance?) R p 1 u = control input R p 2 z = external output R q 1 y = measured output R q 2 33

35 w G 11 G 12 + z + G 21 G 22 u K z = G 11 w + G 12 u y = G 21 w + G 22 u u = Ky Think about these as frequency domain signals. G 11, G 12, G 21 and G 22 are rational transfer functions. To study the stability of this system we add two more disturbance signals v 1 = actuator noise (reglerarens brus) v 2 = sensor noise (mätfel) y u w G z + + v 2 v K y This is identical to the preceding picture if v 1 = v 2 = 0, but in the new setting we have (external) inputs: w, v 1, v 2 (internal input: u) (external) outputs: u, z, y (out of which the output u is fed back into the system). This gives us the system of equations: z = G 11 w + G 12 u y = G 21 w + G 22 u + v 2 u = Ky + v 1. 34

36 This system defines (implicitly) u, z, y as functions of w, v 1, v 2. We can solve it, e.g., as follows. u = Ky + v 1 = K(G 21 w + G 22 u + v 2 ) + v 1 (1 KG 22 )u = KG 21 w + Kv 2 + v 1 u = (1 KG 22 ) 1 (KG 21 w + Kv 2 + v 1 ). (4.1) Substitute into the equations above to get y and z. If this is to be used as an industrial process, then it must be stable in some sense. To simplify the mathematics we use energy stability : For all inputs w, v 1, v 2 L 2 (R + ) (with appropriate dimensions), the signals produced by the system, i.e., u, y, z should also belong to L 2 (R + ) (and the energy amplification should be finite). We may solve u, z, y explicitly by using the method described above, and get a formula of the type z F 11 F 12 F 13 w u = F 21 F 22 F 23 v 1. y F 31 F 32 F 33 v 2 By the discussion in section 3.5, this means that we need Every F ij must belong to H ( C + ) (if continuous time) or H ( E) (if discrete time). One especially important part of F is F 22 = (1 KG 22 ) 1 (see (4.1)). If we denote the feedthrough operators by feedthrough of K = K( ), feedthrough of G 22 = G 22 ( ), then F 22 ( ) = (1 K( )G 22 ( )), so a necessary condition is that 1 K( )G 22 ( ) is invertible. Let us suppose this invertibility condition in the sequel. True, e.g., if G 22 ( ) = 0 (which is the standard assumption made in many text books), or if K( ) = 0. 35

37 4.2 H -minimization Problem Find the compensator K which makes the closed loop system stable, and minimizes the H -norm of F 11 (this is the transfer function from w to z). By checking the computations leading to Theorem and Theorem we find that we will minimize the energy amplification of the input/output map of the time domain u to the time domain z (which is the norm of this operator as a mapping L 2 (R + ) L 2 (R + ) or l 2 (Z + ) l 2 (Z + )). Let s compute this transfer function. z = G 11 w + G 12 u = [G 11 + G 12 (1 KG 22 ) 1 KG 21 ]w + terms which depend on v 1 and v 2 F 11 = G 11 + G 12 (1 KG 22 ) 1 KG 21 Theorem. In the case of a rational matrix-valued transfer function this problem has a solution, which can be computed (although the solution is rather complicated). There are MATLAB toolboxes that solve this: Robust Control Toolbox µ-systems toolbox LMI-Toolbox 4.3 The Model Matching Problem Find the compensator Q which minimizes the H -norm of the following system. w T 1 y u T 3 Q T 2 + z 36

38 ( z y Here T 1, T 2, T 3, Q H ( stable ). y = T 3 w z = T 1 w T 2 u u = Qy ) ( T1 T = 2 T 3 0 u = Qy ) ( w u Homework? Reformulate this so that it becomes a special case of the standard setup presented on p. 35. (I.e., find G 11, G 12, G 21, G 22.) ) 4.4 Tracking Problem Find the compensators C 1 and C 2 which solve the following minimization problem. w + u v C 1 P + C 2 Here we use a two degree of freedom compensator pair C 1 and C 2. The cost of the control is ( ) w(t) v(t) 2 + u(t) 2 dt 0 }{{}}{{} quality of tracking, cost of control want v(t) w(t) To turn this into a minimization problem we replace the original output v by a new output z, and interpret also u as an output. w + C 1 + u P C 2 + z v 37

39 Write the compensator in the form K = [C 1, C 2 ] (these have the same output dimension, can be combined into a larger matrix, also z, w and v have the same dimension). Summary: Controller K = [C 1 C 2 ] External input: w Control input: u [ ] z External output: = z [ u ] w Control output: = y v The equations become [ ] z = y z u w v = = 1 P P }{{} 2 G 4 11 G 12 G 21 G [ w u u = Ky = [ [ ] w C 1 C 2 }{{} v K ] = ] w Pu u w Pu We realize [ that this ] is a special case of the H minimization problem with G11 G G = 12 given above. G 21 G 22 These are minimax problems. For example, model matching is min Q max T 1(s) T 2 (s)q(s)t 3 (s). Re s>0 38

40 Chapter 5 Coprime Factorizations Idea: Factor an unstable transfer function into a numerator and a denominator with no common factors (a minimal factorization in some sense). 5.1 The SISO Case Discrete Time In the SISO case the transfer function is scalar (matrix of dimension 1 1). Let us suppose that k(z) is a rational transfer function. It is analytic at infinity, so it can be written as a quotient k(z) = p(z) q(z) = a 0 + a 1 z + + a m z m b 0 + b 1 z + + b n z n, where b n 0 and n m (since k is analytic at infinity). From a stability point of view, this is not a good representation. We can without loss of generality take m = n (define a k = 0 for k > m). Neither p(z) = a 0 + a 1 z + +a n z n nor q(z) = b 0 +b 1 z+ +b n z n is a proper transfer function (proper = bounded at infinity). We get a better representation by writing (divide both p(z) and q(z) by z n ) k(z) = p(1/z) q(1/z) = a 0 + a 1 z a n z n b 0 + b 1 z b n z n, where b 0 0. (I have changed b k to b n k and a k to a n k and denoted z n p(z) by p(1/z) and z n q(z) by q(1/z). Note that p and q are polynomials.) 39

41 Basic stability requirement: We cannot allow p(1/z) and q(1/z) to have any common zeros in E (outside the unit disk). Equivalently, p(z) and q(z) must not have common zeros in D (inside unit disk). Motivation: When the system gets older the parameters change slightly, and the zeros move, and we suddenly get an unstable pole (which is very close to an unstable zero) from nowhere, and the system goes unstable. Principle: We do not allow unstable zero-pole cancellations. This is not a problem with stable zero-pole cancellations. Even if these change slightly, the input/output behavior does not change much. Near zero-pole cancellations: An unstable zero of p(1/z) which almost cancels an unstable zero of q(1/z) causes the system to be badly behaved (difficult to control). Note. The function k is (energy) stable if and only if q(1/z) has no zeros in E (assuming no unstable zero-pole cancellations). Theorem If the polynomials p and q have no common zeros, then there exist unique polynomials x and y such that p(z)x(z) + q(z)y(z) 1 (z C). (5.1) Proof. Standard division algorithm for polynomials taught in school. Note that (5.1) also can be written in transfer function form as p(1/z)x(1/z) + q(1/z)y(1/z) 1 (z 0). (5.2) Definition (i) Two polynomials p and q are coprime (relativa primpolynom, relativt odelbara) if they have no common zeros (anywhere in C). (ii) Equation (5.1) is called the Bezout identity. Comment The terminology comes from prime number theory. Two positive integers are coprime (relativa primtal) if they have no common factors. Two polynomials have a common polynomial as a factor if and only if they have a common zero. Definition The polynomials p(z) and q(z) are coprime with respect to D (the unit disk) if they have no common zeros in D. (Then p(1/z) and q(1/z) have no common zeros in E.) 40

42 5.1.2 Continuous Time We copy the same idea to the continuous time setting, and write the rational (transfer) function k(s) in the form k(s) = p(s) q(s), where p and q are no longer polynomials, but they are rational transfer functions in RH ( C + ). Here RH = rational H -function. Definition Two functions in RH ( C + ) are coprime (relativt odelbara) (with respect to RH ( C + )) if they have no common zeros in the closed right half-plane C +. Theorem Two functions p and q in RH ( C + ) are coprime if and only if there exist two functions x and y in RH ( C + ) such that p(s)x(s) + q(s)y(s) 1, Re(s) 0. (5.3) Proof. (Outline.) If (5.3) holds, then p and q cannot have any common zeros in Re(s) 0 (including the imaginary axis). The converse direction is more complicated, straightforward but long. It is based on Theorem (polynomial division). 5.2 The MIMO Case What does a zero of a matrix-valued function mean? The answer does exist, but it is too complicated to be presented here (the Smith-MacMillan canonical form). Instead we use the Bezout identity as a definition of coprimeness. Definition (i) Assume that F and G are two matrix-valued functions in RH ( C + ), with the same input dimension (number of columns). Then F and G are right coprime (with respect to RH ( C + )) if 41

43 there exist matrix-valued RH ( C + )-functions X and Ỹ (not necessarily unique) such that or in block matrix form [ ] [ F X Ỹ G X(s)F(s) + Ỹ (s)g(s) 1 (Re (s) 0), (5.4) ] = 1 (identity matrix, square) where we have the same number of rows in X and Ỹ, and the same number of columns in F and G. (På svenska relativt odelbara från höger.) (ii) If F and G in RH have the same output dimension (same number of rows), then they are left coprime (over RH ( C + )) if there exist matrix-valued RH ( C + )-functions X and Y such that [ ] [ ] X F G = Y FX + GY = 1. Lemma If (and only if) F and G are right coprime, then every common right factor (with the same number of rows and columns) which belongs to RH ( C + ) is invertible in RH ( C + ). Right coprime no nontrivial right factors. Proof. Suppose that F = F 1 Z and G = G 1 Z, where Z RH ( C + ). Then by the Bezout identity (5.1), XF + Ỹ G = XF 1 Z + Ỹ G 1Z = ( XF 1 + Ỹ G 1)Z 1. Thus, XF1 + Ỹ G 1 is a left inverse of Z. Both XF 1 + Ỹ G 1 and Z are square matrices, so the left inverse is also a right inverse. Therefore, Z 1 = ( XF 1 + Ỹ G 1) RH ( C + ). Analogous result with right replaced by left is also true. 42

44 Comment (i) The functions F and G are right coprime in RH ( C [ ] + ) [ ] F if and only if the combined matrix has a left inverse X Ỹ G RH ( C + ). (ii) The functions F and G are left coprime in RH ( C + ) if and only if the [ ] [ ] X combined matrix F G has a right inverse RH Y ( C + ). This is obvious from Definition Definition Let G be an m n-dimensional rational function which is bounded at infinity (it has no pole there). Then (i) A right coprime factorization (relativt odelbar högerfaktorisering) of G is of the form G = NM 1 (G is m n, N is m n and M is n n), where det(m) is not identically zero and N and M are right coprime in RH ( C + ). (ii) A left coprime factorization is of the form G = M 1 Ñ (here M is n n and Ñ is n m), where det( M) is not identically zero and M and Ñ are left coprime in RH ( C + ). (iii) A doubly coprime factorization of G is of the form G = NM 1 = M 1 Ñ, (det( M) and det(m) are not identically zero) where we in addition have four more functions in RH ( C + ): X, X, Y, Ỹ such that [ ] [ ] [ ] X Ỹ M Y 1 0 Ñ M = N X 0 1 and also the other way around: [ ] [ M Y X Ỹ N X Ñ M (square matrices (n + m) (n + m)). 43 ] = [ ] (5.5) (5.6)

45 Note. We know from standard matrix theory that a left inverse of a square matrix is also a right inverse (= a full inverse), and conversely. Thus, (5.5) and (5.6) are equivalent (when the input and output dimensions are finite). Note. All the above functions belong to RH ( C + ) (except M 1, M 1 and G which can be unstable). We can write out the four equations in (5.5): XM Ỹ N = 1 ( M, N are right coprime and X, Ỹ left coprime) XY Ỹ X = 0 ÑM + MN = 0 ÑY + MX = 1 ( Ñ, M are left coprime and X, Y right coprime). In particular, NM 1 is a right coprime factorization of G and M 1 Ñ is a left coprime factorization of G. From (5.6) we get four more equations. Problem: How to compute suitable X, Y, X, Ỹ, M, N, M, Ñ? Solution 1: Polynomial factorization. This is possible, but not much used today. Solution 2: Use state space formulas. (Can be computerized.) 5.3 Computation of Doubly Coprime Factorizations Step 1: Write the given transfer function G in state space form G(s) = D + C(s A) 1 B. For this you can use a matrix analogue of the homeworks (the canonical controllable or observable realization, although these are not the best possible ones). Step 2: Choose two arbitrary matrices F (state feedback, tillståndsåterkoppling) and H (output injection, utgångsinjicering) such that A F = A + BF is stable 44

46 H H = A + HC is stable. This is equivalent to (s A BF) 1 RH ( C + ) (s A HC) 1 RH ( C + ), that is, no poles in the closed r.h.p. Then check the following connections: w H + u x + y B (s A) 1 + C + D v + + F Figure 5.1: Right coprime Factor (here the state feedback loop is closed) w H u x + y B (s A) 1 + C D v + F Figure 5.2: Left Coprime Factor (here the output injection loop is closed) In Figure 5.1 we have w and v as inputs, and define M, N, Y, X to be the transfer functions from these variables to u and y, [ ] [ ][ ] u M Y v =. y N X w Note that { y = w + Cx + Du u = v + Fx. 45

47 In Figure 5.2 we have used the same trick as in one of the homeworks and turned u and y into inputs and w and v into outputs. We define M, Ñ, Ỹ, X to be the transfer functions [ ] [ ] [ ] v X Ỹ u = w Ñ M. y The relationships between all the signals in Figure 5.1 and Figure 5.2 are the same, so we must have [ ] [ ] 1 X Ỹ M Y Ñ M =. (5.7) N X Homework: Compute all these transfer functions. (The result is the same as in [Fra87], but there it is not explained why (5.7) holds. It is pulled out of the hat.) Above we inverted both lines w y and v u. We can also invert only one of the two lines. This operation is called a scattering transformation. w H + u x + y B (s A) 1 + C + v + D F Figure 5.3: Extended Original System (both loops open) 46

48 w H u x + y B (s A) 1 + C D v + + F Figure 5.4: Luenberger Observer (both loops closed) The Luenberger observer will appear later in the course (by connecting it in series with the original system we get a stable system; this is called dynamic stabilization). 47

49 Chapter 6 Stabilization 6.1 Factorization of the Standard Problem Repetition: The Standard Problem is the following. u w G z + + v 2 v K y Figure 6.1: [ G11 G Let us factor both G = 12 G 21 G 22 ] and K into right coprime factors K = UV 1, G = NM 1 [ ] [ ][ G11 G G = 12 N11 N = 12 M11 M 12 G 21 G 22 N 21 N 22 M 21 M 22 ] 1. 48

50 u w M 1 ξ N z + + v 2 v U η V 1 y Figure 6.2: Here we have given name to the signals ξ between M 1 and N η between V 1 and U. We can, if we want, turn ξ and η into inputs as follows. u w M ξ N z + v 2 v 1 + U η V y Figure 6.3: Note that obviously all the transfer functions (ξ, η) (w, v, v 1, z, v 2, y) are stable. More precisely, we split M and N into parts, [ ] M1 dimension of w M = dimension of ξ M 2 dimension of u }{{} dim. of ξ N = [ N1 N 2 ] }{{} dim. of ξ dimension of z dimension of y 49

51 From either Figure 6.2 or Figure 6.3 we get the following set of equations. w = M 1 ξ (6.1) u = M 2 ξ (6.2) z = N 1 ξ (6.3) y = N 2 ξ + v 2 (6.4) y = V η (6.5) u = Uη + v 1 (6.6) [ ] w where (1) and (2) are equivalent to = Mξ and (3) and (4) are equivalent to = Nξ +. From this set of equations we may eliminate [ ] [ ] u z 0 y v 2 either (ξ, η) or (z, u, y). Lemma (Improved version of [Fra87], p. 27.) The following conditions are equivalent. (i) All nine transfer functions (w, v 1, v 2 ) (z, u, y) are stable (Figure 6.1). (ii) All six transfer functions (w, v 1, v 2 ) (ξ, η) are stable (Figure 6.2). (iii) All three transfer functions (w, v 1, v 2 ) ξ are stable (delete η in Figure 6.2). Note. Dimensions are (i) (n w + n u + n y ) (n z + n u + n y ) (ii) (n w + n u + n y ) (n w + n u + n y ) (iii) (n w + n u + n y ) (n w + n u ) Proof. (iii) = (i): Use equations (6.2), (6.3) and (6.4). (ii) = (iii): Obvious. (i) = (ii): Since M and N are right coprime, there exist X 1 and Ỹ1 so that X 1 M + Ỹ1N = 1 = X 1 Mξ + Ỹ1Nξ = ξ = 50

52 ξ = X 1 [ w u ] + Ỹ1 [ z y v 2 Thus, the transfer functions (w, v 1, v 2 ) ξ are stable, since the transfer functions (w, v 1, v 2 ) (z, u, y) are stable and both X 1 and Ỹ1 are stable (they belong to RH ( C + )). Same proof for η: There exist X 2 and Ỹ2 so that X 2 V + Ỹ2U = 1 = ]. η = X 2 V η + Ỹ2Uη = X 2 y + Ỹ2(u v 1 ), so the transfer function (w, v 1, v 2 ) η is stable. ξ X 1 Ỹ w u M 1 ξ N + + z v 2 v U η V 1 Ỹ η X2 y Figure 6.4: Picture of the proof: ξ = x and η = η. Definition K stabilizes G if (1 KG 22 ) 1 exists (determinant not 0), and all the transfer functions (w, v 1, v 2 ) (z, u, y) are stable. There is also a left version of Lemma See the homeworks. 51

53 Theorem ([Fra87], p. 30, improved version.) Factor G = NM 1 = M 1 Ñ, and split N, M, Ñ, M into parts of appropriate dimensions: ] ] M = M = [ M1 M 2 ξ [ M1 M2 ] z y w u ξ,, N = Ñ = [ N1 N 2 ξ [Ñ1 Ñ 2 ] w u Also factorize K = UV 1 = Ṽ 1 Ũ. Then the following conditions are equivalent. (i) K stabilizes G M 1 0 (ii) M 2 U V N 2 1 RH ( C + ) (i.e. stable) z y (ii) the top row (corresponding to ξ) of the above inverse is stable ξ, (iii) ( M1 M2 Ñ 2 0 Ũ Ṽ ) 1 RH ( C + ) (stable) (iii) the left column of the above inverse (corresponding to ξ) is stable. Interpretation: (i) Only those parts of N and Ñ which are part of the feedback loops are important. (ii) M must have full row rank and M 1 must have full column rank. Proof. From Figure 6.2 (or the corresponding equations) [ ] [ ] w M1 = ξ, u M 2 [ ] [ ] z N1 = ξ + v y N 2, 2 [ ] [ ] [ ] u u v1 = η +. y v 0 52