Lectures on Dynamic Systems and Control Mohammed Dahleh Munther A. Dahleh George Verghese Department of Electrical Engineering and Computer Science Massachuasetts Institute of Technology1 1�cChapter 10 Discrete-Time Linear State-Space Mo dels 10.1 Intro duction In the previous chapters we showed how dynamic mo dels arise, and studied some sp ecial characteristics that they may p ossess. We fo cused on state-space mo dels and their prop erties, presenting several examples. In this chapter we will continue the study of state-space mo dels, concentrating on solutions and prop erties of DT linear state-space mo dels, b oth time-varying and time-invariant. 10.2 Time-Varying Linear Mo dels A general nth-order discrete-time linear state-space description takes the following form: x(k + 1) � A(k)x(k) + B(k)u(k) y(k) � C(k)x(k) + D(k)u(k) � (10.1) where x(k) 2 Rn . Given the initial condition x(0) and the input sequence u(k), we would like to �nd the state sequence or state trajectory x(k) as well as the output sequence y(k). Undriven Resp onse First let us consider the undriven response, that is the resp onse when u(k) � 0 for all k 2 Z. The state evolution equation then reduces to x(k + 1) � A(k)x(k) : (10.2)The resp onse can b e derived directly from (10.2) by simply iterating forward: x(1) � A(0)x(0) x(2) � A(1)x(1) � A(1)A(0)x(0) x(k) � A(k ; 1)A(k ; 2) : : : A(1)A(0)x(0) (10.3) Motivated by (10.3), we de�ne the state transition matrix, which relates the state of the undriven system at time k to the state at an earlier time `: x(k) � �(k � `)x(`) k � ` : (10.4) The form of the matrix follows directly from (10.3): (�(k � `) � A(k ; 1)A(k ; 2) � � � A(`) � k � ` � 0 : (10.5)I � k � ` If A(k ;1), A(k ;2),. . . , A(`) are all invertible, then one could use the state transition matrix to obtain x(k) from x(`) even when k � `, but we shall typically assume k � ` when writing �(k � `). The following prop erties of the discrete-time state transition matrix are worth highlight-ing: �(k � k) � I x(k) � �(k � 0)x(0) �(k + 1� `) � A(k)�(k � `): (10.6) Example 10.1 (A Su�cient Condition for Asymptotic Stability) The linear system (10.1) is termed asymptotical ly stable if, with u(k) � 0, and for all x(0), we have x(n) ! 0 (by which we mean kx(n)k ! 0) as n ! 1. Since u(k) � 0, we are in e�ect dealing with (10.2). Supp ose kA(k)k � � � 1 (10.7) for all k, where the norm is any submultiplicative norm and � is a constant (inde-p endent of k) that is less than 1. Then k�(n� 0)k � �n and hence kx(n)k � �nkx(0)k so x(n) ! 0 as n ! 1, no matter what x(0) is. Hence (10.7) constitutes a su�cient condition (though a weak one, as we'll see) for asymptotic stability of (10.1).Example 10.2 (\Lifting" a Perio dic Mo del to an LTI Mo del) Consider an undriven linear, periodical ly varying (LPV) mo del in state-space form. This is a system of the form (10.2) for which there is a smallest p ositive integer N such that A(k + N) � A(k) for all k� thus N is the period of the system. (If N � 1, the system is actually LTI, so the cases of interest here are really those with N � 2.) Now fo cus on the state vector x(mN) for integer m, i.e., the state of the LPV system sampled regularly once every p erio d. Evidently ih x(mN + N) � A(N ; 1)A(N ; 2) � � � A(0) x(mN) � �(N � 0) x(mN) (10.8) The sampled state thus admits an LTI state-space mo del. The pro cess of con-structing this sampled mo del for an LPV system is referred to as lifting. Driven Resp onse Now let us consider the driven system, i.e., u(k) 6Referring back to � 0 for at least some k. (10.1), we have x(1) � A(0)x(0) + B(0)u(0) x(2) � A(1)x(1) + B(1)u(1) � A(1)A(0)x(0) + A(1)B(0)u(0) + B(1)u(1) (10.9) which leads to k1;Xx(k) � �(k � 0)x(0) + �(k � ` + 1)B(`)u(`) `�0 � �(k � 0)x(0) + ;(k � 0)U(k � 0) � (10.10) where 10 ih ;(k � 0) � �(k � 1)B(0) j �(k � 2)B(1) j � � � j B(k ; 1) � U(k � 0) � BBBB@ u(0) u(1) . . . CCCCA (10.11) u(k ; 1) What (10.10) shows is that the solution of the system over k steps has the same form as the solution over one step, which is given in the �rst equation of (10.1). Also note that the system resp onse is divided into two terms: one dep ends only on the initial state x(0) and the other dep ends only on the input. These terms are resp ectively called the natural or unforced or zero-input resp onse, and the zero-state resp onse. Note also that the zero-state resp onse has a form that is reminiscent of a convolution sum� this form is sometimes referred to as a superposition sum.If (10.10) had b een simply claimed as a solution, without any sort of derivation, then its validity could b e veri�ed by substituting it back into the system equations: kX x(k + 1) � �(k + 1� 0)x(0) + �(k + 1� ` + 1)B(`)u(`) `�0 k;1X � �(k + 1� 0)x(0) + �(k + 1� ` + 1)B(`)u(`) + B(k)u(k) `�0 " #k;1X � A(k) �(k � 0)x(0) + �(k � ` + 1)B(`)u(`) + B(k)u(k) `�0 � A(k)x(k) + B(k)u(k) : (10.12) Clearly, (10.12) satis�es the system equations (10.1). It remains to b e veri�ed that the pro-p osed solution matches the initial state at k � 0. We have x(0) � �(0� 0)x(0) � x(0) � (10.13) which completes the check. If Y(k � 0) is de�ned similarly to U(k � 0), then following the sort of derivation that led to (10.10), we can establish that Y(k � 0) � �(k � 0)x(0) + �(k � 0)U(k � 0) (10.14) for appropriately de�ned matrices �(k � 0) and �(k � 0). We leave you to work out the details. Once again, (10.14) for the output over k steps has the same form as the expression for the output at a single step, which is given in the second equation of (10.1). 10.3 Linear Time-Invariant Mo dels In the case of a time-invariant linear discrete-time system, the solutions can b e simpli�ed considerably. We �rst examine a direct time-domain solution, then compare this with a transform-domain solution, and �nally return to the time domain, but in mo dal co ordinates. Direct Time-Domain Solution For a linear time-invariant system, observe that )A(k) � A for all k � 0� (10.15)B(k) � B where A and B are now constant matrices. Thus �(k � `) � A(k ; 1) : : : A(`) � Ak;` � k � ` (10.16)so that, substituting this back into (10.10), we are left with …