State-space analysis of control systems: Part I

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Caitlin George
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1 Why a different approach? State-pace analyi of control ytem: Part I Uing a tate-variable approach give u a traightforward way to analyze MIM multiple-input, multiple output ytem. A tate variable model help u undertand ome complex general concept about control ytem, uch a controllability and obervability. Signal-flow graph In order to introduce ome key idea in tate-variable ytem modeling, we need to ue ignal-flow graph. Thee graph allow for only three type of operation: Addition of incoming ignal at a node: Here the node i a mall circle. Any ignal flowing out of a node i the um of all the ignal flowing in. Amplification by a fixed factor, poitive or negative: the gain i jut written above the ignal path. Integration: Thi i denoted by a box containing the integral ign, or /. The tate variable model for any linear ytem i a et of firt-order differential equation. Therefore, the output of each integrator in a ignal-flow graph of a ytem are the tate of that ytem. For any ytem, an infinite number of ignal graph are poible, but only a few are of interet. Let look at ome procee for obtaining a ignal-flow graph for a given ytem. Thi i bet done by mean of a pecific example. Conider the tranfer function, and it equivalent differential equation: ince thi i a econd-order ytem, it tate model will have two tate, which will appear at the output of the two integrator in any ignal flow graph. Next, we will conider three form of the tate model for thi ytem, each of which reult from a lightly different approach: Control canonical form: Thi form get it name from the fact that all of the tate are fed back to the input in the ignal flow graph. For thi tate-variable model, olve the differential equation for the highet-order derivative of the output a

2 Thi olution i for a particular econd order ytem, but you can ee how to extend thi idea to a higher-order ytem To begin to draw the ignal graph, we connect two integrator in cacade, and identify the output and it two derivative uing prime to denote differentiation, which give Note that the ignal are drawn immediately to the right of on the output ide of the node. The differential equation for the highet derivative of ot identifie thi derivative a the um of everal term. Two of thee term depend on lower-order derivative of the output, and one depend on the input. You draw the ignal path correponding to the lower output derivative a feedback loop a hown here Thi diagram obviouly repreent, in a graphic way, the firt two term on the right ide of the equation for o t. Jut like any diagram of a ignal-proceing ytem, the input hould be on the left and the output hould be on the right. Putting the input into the diagram give All term of the equation for o t but the lat one are now repreented in the diagram. In order to ue only integration, addition and multiplication in out ignal graph, we have

3 to repreent term which are proportional to firt and higher derivative in the following way: uppoe we rewrite the tranfer function a Thi clearly how that the output arie from two term, and the firt of thee term could be obtained from the ignal graph we have o far. The econd term i proportional to the derivative of the firt one. The ignal graph ha a node from which we can get the derivative of the output, namely o t. To finih our ignal graph, we jut move the input gain to the output ide, and take an additional ignal proportional to o t to the new output via a feed-forward loop with the required proportionality contant. The reult i You can now identify each tate with an integrator output, to yield the tate x and x, a hown next: The firt derivative of each tate i the ignal jut back on the uptream ide of each integrator. Thu, we can write two differential tate equation and an additional equation called the output equation, which relate the tate to the ytem output, a

4 Thee equation can be organized into a compact et of matrix equation which look like thi: And have the general form, dx Ax Bi dt o Cx Di In thi general form for the tate equation model, if there are n tate, r input, and p output, then the matrice will have the following name and form row x column: Sytem matrix, A: n n, Input matrix, B: n r, utput matrix, C: p n, Feed-forward matrix, D: p r. Note on tranfer function normalization: Notice how the highet-order for thi tranfer function keep appearing in denominator everywhere. Tranfer function coefficient are not unique, and you can alway divide numerator and denominator of any tranfer function by the highet-order to obtain a normalized tranfer function of the form m... m G, n... where the highet- order i unity. Thi obviouly make for cleaner matrice. If you normalize the tranfer function firt, the control canonical form tate equation look like thi for a normalized 4 th -order ytem. Extenion to higher order i traightforward: and D contain only zero. A [ ] Modal or modal canonical form B, C, Suppoe you converted the econd order tranfer function we are uing a an example here into pole reidue form,

5 You could convert any one of the term into the following imple ignal flow graph and tate- and output equation: For a tranfer function with n ditinct pole two pole with the ame value can be artificially eparated by ome very mall difference, you would get the following graph and tate- and output equation: The intereting thing about thi form i the appearance of the ytem pole a the element in a diagonalized ytem matrix. Thi ay that the eigenvalue of the ytem matrix regardle of what form it i in are the pole of the tranfer function. Alo, note that in thi form, the coefficient in the equation will generally be complex. Thi wa not the cae for the control canonical form earlier, ince the coefficient in the equation there were ratio of real tranfer function coefficient. Here i the general matrix modal form for a fourth-order ytem: p p A, B, C [ A A A A4 ] p p4 with D containing all zero. berver canonical form There i one more pecial form of the tate equation that i of interet. In thi cae the feedback i from the output to the tate variable. For thi form, we tart with a normalized, rd -order tranfer function,

6 G, and draw the following ignal-flow graph: which lead to the matrix form, A, B, [ ] C, and D contain all zero. For the control canonical form, we jutified the form of the ignal-flow graph by olving the differential equation for the highet-order derivative of the output. For the modal form, we did thi by firt looking at a ingle term of the reidue-pole form of the tranfer function, then adding imilar term. For the oberver canonical form, uppoe we multiply the normalized tranfer function by raied to that power, thereby creating a rational polynomial in / a follow: I G. Thi lead to the following LaPlace tranform equation relating the input, I to the output : I I I. Thi equation correpond exactly to the ignal-flow graph: The firt term on the right ide get integrated three time, the econd twice, and the third once. Tranformation to other tate-pace repreentation How are the different tate-pace repreentation related, other than in repreenting the ame phyical ytem? If a linear ytem can be repreented by two tate vector, u and v, the two vector mut be related through a tranformation T by utv, and vt - u The invere of T mut exit, that i T mut be non-ingular. We can ue thi relation to tranform the tate and output equation a well, for example,

7 if with one tate vector, u & Gu Hi, o Pu Qi, then uing the tranformation, T, u & GTv Hi and o PTv Qi. Pre-multiplying by the invere of T give a new et of tate equation, v& T u& T GTv T Hi and o PTv { Qi. A where A, B, and C are repectively the new ytem, input, and output matrice for the ytem uing the tate vector v. The availability of the tranformation, T, mean that an infinite number of tate repreentation for a ytem are poible. nly a few of thee are intereting. B C Solving State Equation: LaPlace domain Since we ve gone to all thi work to develop a wide variety of available tate equation, it might be intereting at thi point to actually olve one! If we aume zero initial condition uually the cae in control ytem deign and take the LaPlace tranform of both ide of the tate equation, we get X AX BI, which i olvable a X [ Ι A] BI note here that I i the LaPlace tranform of the input, while I i the identity matrix. Thi almot look like a tranfer function, but one more thing i needed: the output in term of the tate, which we get from the output lower equation, dx Ax Bi dt o Cx Di LaPlace tranforming the output equation and ubtituting in the reult for X above give a matrix verion of an input-output relationhip, { C Ι A B D} I. The quantity in {} i really a tranfer function in matrix form. For example, in a ytem with three output and two input, it ay H H I H H. I H H The element of the matrix, H, how the effect that each input ha on each output. Since the ytem i linear, uperpoition applie and the effect of each input add. Thi way of organizing the treatment of multiple-input, multiple-output MIM ytem i one characteritic that make tate variable o ueful.

MECH 2110 - Statics & Dynamics

MECH 2110 - Statics & Dynamics Chapter D Problem 3 Solution 1/7/8 1:8 PM MECH 11 - Static & Dynamic Chapter D Problem 3 Solution Page 7, Engineering Mechanic - Dynamic, 4th Edition, Meriam and Kraige Given: Particle moving along a traight