I ml. g l. sin. l 0,,2,...,n A. ( t vs. ), and ( vs. ). d dt. sin l

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1 Advanced Control Theory Homework #3 Student No.597 Name : Jinseong Kim. The simple pendulum dynamics is given by g sin L A. Derive above dynamic equation from the free body diagram. B. Find the equilibrium point of the simple pendulum. C. Linearize above system with respect to each equilibrium point. Express in the form of x( t. Ax D. Simulate the original and linearized system with Runge-Kutta 4 th order. Draw the graph of ( t vs., ( t vs., and ( vs.. - Solution A. Free body diagram은아래그림과같이그렸다. T I lf lmgsin where, I ml B. g l sin 문제와같은식을얻으려면 의기준을 3 시방향으로잡으면된다. For the equilibrium points, we must solve d f ( g dt sin l g and sin l,,,...,n C. D. f f x x A g g f f cos l l x x Case study - (with l.5, g 9.8, =, =.3

2 Advanced Control Theory Homework #3 Student No.597 Name : Jinseong Kim Case study - (with l.5, g 9.8, =, =.. For the following nonlinear input-output system y( t y( y ( y uu A. Obtain a nonlinear state-space representation. B. Linearize this system around its equilibrium output trajectory when u(, and write it in state space form. C. Simulate the original and linearized system with Runge-Kutta 4 th order. Draw the graph of ( t vs. y, ( t vs. y, and ( y vs. y. - Solution A. d y y dt y y y y y yuu B. For the equilibrium points, we must solve y f( y y y y y yuu y yy y y y u u y y ( y(y y u(, y Linearize using the Jacobian f f y y A f f yy y y y, y y y f f u u B f f y, y u u State space form

3 ECE 55, Fall 5 Problem Set # Solution Solutions: (Remember: generally, there are more than one state-space realizations of the same system.. Nonlinear system ( points (a A straightforward attempt is to denote x = y and x = ẏ, then one can write { x = x, x = ÿ = x (x + (x + + u + u. The main problem in defining the state variables in this case lies in the derivative terms of the input signal u. The state variables must be such that they will eliminate the derivatives of u in the state equation. One way to obtain a correct equation is to define (by direct examination of the given equation: x = y, x = x u. Hence the corresponding state-space equations are x = ẏ = x + u, x = ẍ u = ÿ u = x (x + (x + u + + u, y = x. A more systematic way to obtain the state-space equation is using the integration technique (compare to picking state variables by direct examination. From ÿ = y (y + (ẏ + + u + u, integrate to obtain ẏ = u + (y + u (y + (ẏ +. Let x = (y +u (y +(ẏ +. Then ẏ = u+x and ẋ = (y +u (y +(ẏ +. Let x = y, then we have a state-space realization { x + u which is the same as (. ẋ = x (x + (x + u + + u (b The first step is to obtain its equilibrium. Set u and x, x in equation (, and we have { { x = x (x + (x + = x (x + = x = x = Linearize the system ( at (,, and we get ( ( f f δ δx + x u x=(, u= δy = [ δx. x=(, u= δu = [ [ δx + δu, 3 (

4 . Satellite Problem ( points Note, in the system: r = r θ k r + u ( θ = θṙ r + r u (3 that θ does not explicitly appear in the equations. Hence we may pick the states to be x = [x, x, x 3 T = [ṙ, r, θ T R 3 and denote the inputs to be u = [u, u T R. The state space representation of the system is: r ṙ θ = f (ṙ, r, θ, u, u f (ṙ, r, θ, u, u f 3 (ṙ, r, θ, u, u = r θ k/r + u ṙ θṙ/r + u /r (4 Then f x = f ṙ f ṙ f 3 ṙ f r f r f 3 r f f θ f θ 3 θ = θ + k/r 3 r θ θ/r θṙ/r u /r ṙr = 3ω pω ω/p evaluated at ṙ =, r = p, θ = ω, u = u = and k = p 3 ω. Similarly we have: f u =. (5 /p The linearized equation is therefore: 3ω pω δ ω/p δx + /p δu. (6 3. A Multiple Input and Multiple Output (MIMO System ( points Assume that the system is relaxed at t =. Then integrate the differential equations to obtain: ẏ = u + ( y + u + ( 3y + u (7 ẏ = ( y + 3y + u 3 + ( y y + u u 3 (8 Denote the inputs to be u = [u, u, u 3 T R 3 and the outputs y = [y, y T R. Define the state variables to be x = [x, x, x 3, x 4, x 5 T R 5 where: x = y, x = ( y + u + x 3, x 3 = ( 3y + u, x 4 = y, x 5 = ( y y + u u 3.

5 Then the state equations are: Or in matrix form: y = ẋ = x + u ẋ = x + x 3 + u ẋ 3 = 3x 4 + u ẋ 4 = x + 3x 4 + x 5 + u 3 ẋ 5 = x x 4 + u u 3 y = x y = x [ x + u (9 x ( 4. A Linear Time Varying System ( points (a Let us first assume that the system is relaxed at time t =, integrate the system and we get: ÿ ẏ y = t u + (t + u ẏ = (y + tu + t (y + τu dτ. ( Now let x = [x, x T R where we define x = y and x = t (y + τu dτ = ẏ y tu. Then we have: [ [ t x + u t y = [ ( x (b Define w = tu. Then ÿ ẏ y = w + ẇ, which is a linear system with input w and output y. So, the transfer function is Y (s W (s = s + s s, which gives the state space in the controllable canonical form: [ [ [ [ x + w = x + t y = [ x. u (3 3

6 In fact, the state-space in part (a is the observable canonical form with respect to input w and output y. Yet another solution: Let x = y, x = ẏ tu. Then ( ( ẏ x = + tu = ÿ t u u x + x + 3tu y = [ x 5. State-space vs. Input-output Representation ( points For the given transfer function, we have: G(s = [ [ t x + u 3t (4 s + s 3 + 4s + 5s + = s + (s + (s + = B(s A(s. (5 (a Controllable canonical form. Let A(sW (s = U(s, Y (s = B(sW (s, i.e. w (3 + 4ẅ + 5ẇ + w = u and y = ẇ + w. Choose the states to be x = [w, ẇ, ẅ T R 3. Then we have x + u 5 4 y = [ x (b Observable canonical form. From y (3 + 4ÿ + 5ẏ + y = u + u we define the states to be x = [x, x, x 3 T : x = y x = ẏ + 4y x 3 = ÿ + 4ẏ + 5y u Then we obtain: 4 5 y = [ x x + u (c Using partial fractions, we have: s + (s + (s + = = (s + (s + = A (s + + A (s + (s + + (s + Let X (s = (s+ U(s, X (s = (s+ U(s, then Y (s = X (s X (s, or in the time domain: [ y = [ x x + [ u 4