ece_350_lecture12_fall08
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ece_350_lecture12_fall08

Course Number: TH 350, Fall 2008

College/University: CSU Northridge

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Lecture 12 Lecture Overview End-of-Year Calendar Review Solving State Variable Equations Homogeneous Solution Eigenvalues & Eigenvectors Particular Solution Complete Solution ECE 350 D. van Alphen 1 Repeating: End-of-Year Calendar Tuesday Dec. 2 State Variables Dec. 9 Quiz 3 Thursday Dec. 4 Homework Review ECE 350 D. van Alphen 2 Review Input-output relationship for LTI systems: h(t) x(t) H()...

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12 Lecture Lecture Overview End-of-Year Calendar Review Solving State Variable Equations Homogeneous Solution Eigenvalues & Eigenvectors Particular Solution Complete Solution ECE 350 D. van Alphen 1 Repeating: End-of-Year Calendar Tuesday Dec. 2 State Variables Dec. 9 Quiz 3 Thursday Dec. 4 Homework Review ECE 350 D. van Alphen 2 Review Input-output relationship for LTI systems: h(t) x(t) H() y(t) = x(t) * h(t) Y() = X() H() Y(s) = X(s) H(s) State Variables 2 dt = 1 2 Parsevals Theorem: E x = | x( t) | | X() | d 2 For circuits: assign a state to the voltage across a capacitor, and a state to the current through an inductor. For an Nth order differential equation: assign the states y(t) = q1(t), dy/dt = q2(t), d(N-1y/dtN-1 = qN(t) ECE 350 D. van Alphen 3 Summarizing State Variable Equations for Circuit Problems State variables, continued State equation, vector/matrix form: dq/dt = A q(t) + B x(t) Output equation: y(t) = C q(t) + D x(t) (1) Remaining Problem: How do we solve equation (1) for q(t)? Approach: first find the natural (or homogeneous) solution; i.e., the solution when the input x(t) = 0. Solve dq/dt = A q(t) (2) ECE 350 D. van Alphen 4 Solving the Homogeneous Equation Aside: differential equation review - For the scalar differential equation: & q( t) = a q( t), the solution is: q(t) = q(0) eat I.C. : q(0) Claim: the solution of homogeneous equation (2) is of the same form: q(t) = eAt q(0) where Problem: not a closed-form. k 2 3 At t Ak = 1 + t A + t A 2 + t A3 + K e 2 3! k =0 k! (compare to Maclaurin Series for ex) ECE 350 D. van Alphen 5 Linear Algebra Aside: Eigenvalues & Eigenvectors Multiplying an arbitrary vector x by a conformable square matrix A yields product Ax that is typically a scaled (longer or shorter) and rotated version of the original vector x: Ax 2-dimensional sketch of arbitrary matrix multiplication: x However, for any square matrix A, there will be some vectors, zi, (called the eigenvectors of A) that do not get rotated under the matrix multiplication. They do get scaled, and we call the scalars the eigenvalues, i. ECE 350 D. van Alphen 6 Eigenvalues & Eigenvectors, continued In other words, the eigenvectors of square matrix A are the vectors that do not change direction* when multiplied by A. The eigenvalues are the scalars by which the vectors are shrunk or stretched: 2-dimensional sketch of multiplying an eigenvector of A by A Az = z z Defining equation for eigenvalues () and eigenvectors (z) of square matrix A: Az = z * Here, we do not consider a rotation of 180 a change in direction. ECE 350 D. van Alphen 7 Eigenvalues & Eigenvectors, continued: From the previous equation, we obtain: (A I) z = 0 and det(A - I) = 0 (use to find eigenvalues, s) (called the characteristic eq. of A) 0 I = 0 (use to find eigenvectors, z, once s are known) Example: A = 4 5 , 2 3 1 0 I= , 0 1 4 5 0 A I = 0 = 2 3 ECE 350 _____________ To find A-I: subtract from the diagonal elements of A. 8 D. van Alphen Eigenvalues & Eigenvectors, continued Hence det(A I) = 0 4 2 5 3 = (4 )(3 ) + 10 = 0 2 12 + 10 = 0 2 2 = 0 ( 2) ( + 1) = 0 Now use = 2, -1 Finding the eigenvalues by solving the characteristic equation of A : (A - ) z = 0 to find the eigenvector for each eigenvalue: 5 z1 0 4 2 ( A I)z = z = 0 3 2 2 2 For = 2: 5 2 5 z1 0 = 2z1 5z 2 = 0 z1 = z 2 2 5 z 0 2 2 ECE 350 D. van Alphen 9 Eigenvalues & Eigenvectors, continued 2.5 z= So for = 2: z1 = (5/2) z2 eigenvector is 1 Any multiple of the eigenvector above would also work (eigenvectors are not unique); To standardize our answers, we typically give the unit-length version of the eigenvector ( divide z above by its length): length( z) = (5 / 2)2 + 12 = 7.25 2.69 Unit-length eigenvector: 2.5 / 2.69 .929 z= = .371 1/ 2.69 ECE 350 D. van Alphen 10 Eigenvalues & Eigenvectors, continued Now use : (A - ) z = 0 to find the eigenvector z for = -1: 4 + 1 5 z1 0 ( A I)z = z = 0 3 + 1 2 2 5 5 z1 0 1 = 5z1 5z 2 = 0 z1 = z 2 z = 2 2 z 0 2 1 For standardization, we normalize z above, dividing it by its length: length( z) = (1)2 + 12 = 2 1.414 1/ 1.414 .707 z= = .707 1/ 1.414 ECE 350 (normalized) 11 D. van Alphen MATLAB: Eigenvalues & Eigenvectors To obtain just the eigenvalues of A, type: >> eig(A) or >> e_values = eig(A) To obtain the both eigenvectors and the corresponding eigenvalues, type: >> [V D] = eig(A) Note: V is a matrix whose columns are the eigenvectors of A; D is a diagonal matrix, with eigenvalues on the diagonal. The order of the eigenvalues in D matches the order of the eigenvectors in V. ECE 350 D. van Alphen 12 MATLAB: Eigenvalues & Eigenvectors MATLAB Code for eigenvalues & eigenvectors of A = 2 3 >> A = [4 -5; 2 -3]; >> eig(A) ans = 2 -1 >> [V D] = eig(A) V= 0.9285 0.7071 0.3714 0.7071 D= 20 0 -1 ECE 350 D. van Alphen 13 4 5 Finding Integer Powers of Matrix A Cayley-Hamilton Theorem: Any N-by-N matrix A satisfies its own characteristic equation: det(A I) = 0. Example: 5 4 A= 1 2 5 4 = (5 )(2 ) 4 = 0 2 Characteristic equation: det 1 2 7 + 6 = 0 So, per Cayley-Hamilton: ________________ A2 = _____________ ECE 350 D. van Alphen (3) 14 Finding Integer Powers of Matrix A Repeating equation (3): A2 5 4 A= = 7A 6I, for 1 2 Multiplying (3) by A: A3 = 7A2 6A = 7(7A 6I) 6A = 43A 42I Continuing, we could find any integer power of A as a linear combination of A and I. Claim 1: for any N-by-N matrix A, we can find integer powers of A as a linear combination of Am, m = 0, 1, , N-1. Claim 2: for any N-by-N matrix A, we can evaluate eAt as a timevarying linear combination of Am, m = 0, 1, , N-1. At = N1 ( t) Am e m m=0 ECE 350 D. van Alphen (4) 15 Finding eAt e At = m ( t ) Am m=0 N1 Approach: 1. Find the eigenvalues of A MATLAB function: eig(A) 2. Find the time-varying coefficients, i(t), by writing N equations, replacing A (in eq. (4)) by the (distinct) eigenvalues. t = N1 ( t) m e m m =0 (4) 3. Return to equation (4), and plug in the coefficients, m(t), found in step 2. ECE 350 D. van Alphen 16 eAt Example: 3 / 4 0 A= 1/ 2 1/ 2 Step 1 - Using MATLAB to find the eigenvalues: > A = [3/4 0; 1/2 1/2]; >> eig(A) ans = 0.5000 0.7500 Step 2 - System of equations (from (4)): ( = 1/ 2) : t / 2 = 1 ( t ) m = ( t ) + (1/ 2) ( t ) e m 0 1 m=0 ( = 3 / 4) : e ECE 350 3t / 4 = m ( t) m = 0 ( t) + (3 / 4)1( t) m=0 17 1 D. van Alphen eAt Example: 3 / 4 0 A= 1/ 2 1/ 2 Repeating the system of equations, and subtracting the top equation from the bottom equation: e t / 2 = 0 ( t) + (1/ 2)1( t) 1/ 4 1( t) = e3t / 4 e t / 2 1( t) = 4e3t / 4 4e t / 2 e3t / 4 = 0 ( t) + (3 / 4)1( t) Plugging the answer for 1(t) into the top equation: e t / 2 = 0 ( t) + 2e3t / 4 2e t / 2 ( ) 0 ( t) = 3e t / 2 + 2e3t / 4 Step 3 - Now plug in the values for 0(t), 1(t) into the original equation (4): At = 1 ( t ) Am = ( t ) I + ( t ) A1 e m 0 1 m =0 ECE 350 D. van Alphen 18 eAt Repeating Example: 3 / 4 0 A= 1/ 2 1/ 2 0 t / 2 + 2e3 t / 4 3e t / 2 + 3t / 4 2e At = ( t ) I + ( t ) A1 = 3e e 0 1 0 + ( 4e At = 3e e t / 2 + 3t / 4 2e 3t / 4 4e t / 2 3 / 4 0 ) 1/ 2 1/ 2 0 3e3t / 4 3e t / 2 0 + t / 2 + 2e3 t / 4 2e3 t / 4 2e t / 2 3e 0 t / 2 + 4e 3 t / 4 e D. van Alphen 0 3 t / 4 2e t / 2 2e At = e 5e 3 t / 4 t / 2 + 2e3 t / 4 2e ECE 350 19 Homogeneous Solution for State Variable Problem Recall from p. 5: q(t) = eAt q(0) 3 / 4 0 1 , and q(0) = Suppose, for example, that: A = 1/ 2 1/ 2 2 Then q(t) = eAt q(0) 5e 3 t / 4 = t / 2 + 2e3 t / 4 2e 5e3t / 4 = =5 3t / 4 10e 1 0 t / 2 + 4e3 t / 4 2 e e3 t / 4 3t / 4 2 e ECE 350 D. van Alphen 20 The Particular Solution for State Variable Equations Recall the defining state variable equations: State equation: Output equation: dq/dt = A q(t) + B x(t) y(t) = C q(t) + D x(t) (1) Notation: The complete solution for the states q(t) in equation (1) is the sum of the homogeneous solution (found by letting x(t) = 0) and the particular solution. So far, we can find the homogeneous solution: qh(t) = eAt q(0) We could show that the particular solution is: qp ( t) = e At B x() d 0 ECE 350 D. van Alphen 21 t The Complete Solution for State Variable Equations The complete solution for the state vector is the sum of the homogeneous solution and the particular solution: q( t) = qh ( t) + qp ( t) = e q(0) + e A( t ) B x( ) d At 0 t Another use (in the state variable model) for eAt: Impulse response: h(t) = C eAt B + D (t), t 0 (0 else0 Stability: An LTI system is stable if and only if all of the eigenvalues of matrix A are in the left half of the complex plane; i.e., iff all the eigenvalues have negative real parts. ECE 350 D. van Alphen 22
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