Designing Information Devices and Systems II Spring 2018 J. Roychowdhury and M. Maharbiz Discussion 8A

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1 EECS 6B Designing Information Devices an Systems II Spring 28 J Roychowhury an M Maharbiz Discussion 8A Change of Basis Review Figure : Left: basis vectors x an y Right: basis vectors u an v T We can think of an n-imensional vector as a point in an n-imensional space (here, p = 3 is in a 2-imensional space) The vector s coorinates represent how we can scale an sum basis vectors in this space to reach the point Usually, we will use orthogonal, unit-length basis vectors like x = T T y = Then, p s coorinates are given by 3 because p can be written as 3 p = 3 x + y = T an However, we can also think about writing the coorinates of p in terms of a ifferent set of basis vectors, like u an v shown on the right, where u = x + y an v = 2 x + y If we solve for x an y in terms of u an v, we have x = 2 v an y = u + 2 v an we can re-write p as p = u v Since the relationship between the { x, y} basis to the { u, v} basis is linear, we can escribe the change of basis with a transformation matrix T : T = The columns of T are the representations of x an y in terms of u an v Then, 3 = 2 2 }{{}}{{}}{{} new coorinates T 2 2 ol coorinates EECS 6B, Spring 28, Discussion 8A

2 2 Controller Canonical Form When working with systems in state-space, you may have notice that a single system can be represente in many ifferent forms, epening on factors, such as how you orere your state vector Writing out systems in certain canonical forms often allows engineers to quickly etermine system behavior The controller canonical form, which guarantees controllability an simplifies eigenvalue placement, takes on the following form: Ã =, B = a a 2 a n a n 2 Change of Basis to Controller Canonical Form Given a controllable system of the form t x(t) = A x(t)+bu(t), we can transform it into controller canonical form by choosing some T, such that: z = T x,ã = TAT, an B = T B for matrices Ã an B of the form shown above We can calculate this T using R n = B AB A n B, the controllability matrix of the original form using A an B Note that R n is full rank, an therefore invertible, because the original system is controllable Calculate R n an take the last row of R n to be q T Then, using q T, we can construct this matrix T : q T q T A T = q T A 2 q T A n Also notice that when we place our system in feeback using u(t) = K z = K x with K = KT, we get the close loop matrix T (A BK)T = Ã B K The eigenvalues of both systems are the same, an we can arbitrarily assign the eigenvalues of Ã B K with the choice of K Therefore, we just prove that controllability enables arbitrary eigenvalue assignment in any state space system Note that it is not necessary to bring the system to the controller canonical form to assign its eigenvalues You can still use what we i in the last section to choose K in orer to obtain esirable eigenvalues 22 Coefficient Matching to Controller Canonical Form For smaller systems, we can use a technique calle coefficient matching to bring the system to controller canonical form Different forms of a system are similarity transforms of each other This means that the () EECS 6B, Spring 28, Discussion 8A 2

3 eigenvalues of a system are preserve Thus, we can fin the characteristic polynomial of our system in its original form, an then map the coefficients of this polynomial into the state matrix of our system in controller canonical form Let s walk through an example to convert to controller canonical form x(t) = A x(t) + Bu(t) = t t z(t) = Ã z(t) + Bu(t) = x(t) + z(t) + a a 2 Our first step shoul always be to check for controllability to ensure that a transformation into controllable canonical form actually exists R n = B AB = Now that we have verifie controllability, we can calculate the characteristic polynomial of A u(t) u(t) The characteristic polynomial of Ã is et(a λi) = ( λ) 2 = λ 2 2λ + ) et (Ã λi = λ(a 2 λ) a = λ 2 λa 2 a From here we can see that a 2 = 2 an a =, giving us the controller canonical form t z(t) = Ã z(t) + Bu(t) = z(t) + 2 u(t) Controller Canonical Form Introuction (a) Show that the system in controller canonical form given in Equation () is always controllable The controllability matrix is R n = B Ã B Ã n B = Since it s a triangular matrix, its eigenvalues are the entries on its main iagonal Therefore, it always has rank n EECS 6B, Spring 28, Discussion 8A 3

4 (b) OPTIONAL: Show that the characteristic polynomial of A = a a 2 a n a n is et(a λi) = λ n a n λ n a n λ n 2 a Let T n = λi A Observe that T n can be written as follows λ T n = T n a, where λ λ T n = λ a 2 a 3 a n λ a n The reason we o this is because T n an T n have the exact same structure an T n is neste within T n We can exploit this because when we calculate the eterminant, we fin the eterminant of neste matrices as well Specifically, if we expan the eterminant on the first column of T n, we get et(t n ) = λ et(t n ) + ( ) n λ ( a )et λ or et(t n ) = λ et(t n ) + ( ) n ( a )( ) n = λ et(t n ) a We can follow the same process again for T n by observing T n 2 neste within it λ λ T n 2 = λ a 3 a 4 a n λ a n It follows that et(t n ) = λ et(t n 2 ) a 2 EECS 6B, Spring 28, Discussion 8A 4

5 an or et(t n ) = λ(λ et(t n 2 ) a 2 ) a et(t n ) = λ 2 et(t n 2 ) a 2 λ a T n 2 gives T n 3, which gives T n 4, an so on We keep oing this until we reach T T = λ a n with et(t ) = λ a n Observing the structure of the T n,t n,,t an how the eterminants are relate to each other, we get that et(t n ) = λ n a n λ n a n λ n 2 a This is the characteristic polynomial of A 2 Controller Canonical Form (a) Show that for any square matrix A an any invertible matrix T, the matrix TAT has i the same characteristic polynomial as A ii the same eigenvalues ( ) ( et TAT λi = et TAT λtt ) ( = et T (A λi)t ) ( = et(t )et(a λi)et T ) = et(a λi) ( ) Since et TAT λi = et(a λi), A an TAT have the same characteristic polynomials an therefore the same eigenvalues (b) Put the system 2 t x(t) = 2 x(t) + u(t) 2 in controller canonical form without computing the transformation matrix T The characteristic polynomial for the A matrix is λ 3 6λ 2 + 2λ 7 controller canonical form is t z(t) = z(t) + u(t) Therefore, this system in EECS 6B, Spring 28, Discussion 8A 5

6 (c) Design a state feeback controller (ie, fin K) to put the eigenvalues of the above system at λ =, ± j The characteristic polynomial for A BK, or Ã B K, shoul be Matching it with we get that is (λ + )(λ + + j)(λ + j) = λ 3 + 3λ 2 + 4λ + 2 λ 3 (a 3 k 3 )λ 2 (a 2 k 2 )λ (a k ), a k = 2,a 2 k 2 = 4,a 3 k 3 = 3 k = 9,k 2 = 8,k 3 = 9, K = () Now fin the transformation matrix T that transforms the original state space representation into controller canonical form R n = B AB A B = = q T = q T T = q T A = 2 q T A R n = You can check that Ã = TAT an that B = T B (e) Calculate K = KT Where are the eigenvalues of A BK? K = KT = = The eigenvalues shoul be exactly the same as the eigenvalues for the system in controller canonical form, ie, at λ =, ± j Contributors: Justin Yim Jane Liang EECS 6B, Spring 28, Discussion 8A 6

7 Deborah Soung John Maiens Titan Yuan EECS 6B, Spring 28, Discussion 8A 7

Designing Information Devices and Systems II Spring 2018 J. Roychowdhury and M. Maharbiz Discussion 2A

EECS 6B Designing Information Devices an Systems II Spring 208 J. Roychowhury an M. Maharbiz Discussion 2A Secon-Orer Differential Equations Secon-orer ifferential equations are ifferential equations of