ME45: Control Systems Lecture 2 Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Transfer function Models for systems electrical mechanical electromechanical Block diagrams Linearization Course roadmap Analysis Time response Transient Steady state Frequency response Bode plot Stability Routh-Hurwitz (Nyquist) Design Design specs Root locus Frequency domain PID & Lead-lag Design examples (Matlab simulations &) laboratories Spring 2 Spring 2 2 One of most important math tools in the course! Definition: For a function f(t) (f(t)= for t<), f(t) We denote of f(t) by F(s). Spring 2 3 t (s: complex variable) F(s) Example of Step function f (t) = 5u(t) = 5 t t < Remember L(u(t)) = /s f(t) F(s) = f (t)e st dt = 5e st dt = 5 e st dt = 5 s st e Spring 2 4 5 = 5 s = 5 s t
Integration is Hard table Tables are Easier Spring 2 5 Spring 2 6 Inverse Laplace Transform Properties of Linearity Properties of Differentiation Proof. Proof. Spring 2 7 Spring 2 8
Properties of Integration Properties of Final value theorem if all the poles of sf(s) ) are in the left half plane (LHP) Proof. Poles of sf(s) ) are in LHP,, so final value thm applies. Spring 2 9 Some poles of sf(s) ) are not in LHP,, so final value thm does NOT apply. Spring 2 Properties of Initial value theorem if the limits exist. Properties of Convolution Convolution Remark: In this theorem, it does not matter if pole location is in LHP or not. IMPORTANT REMARK L ( F (s)f 2 (s)) f (t) f 2 (t) Spring 2 Spring 2 2
An advantage of We can transform an ordinary differential equation (ODE) into an algebraic equation (AE). ODE Solution to ODE AE Partial fraction expansion Spring 2 3 3 2 Example st Order ODE with input and Initial Condition 5 y '() t + y() t = 3 u() t y() = Take Laplace Transform 5[ sy (s) y() ]+ [ Y (s)]= 3 s Solve for Y(s) ( 5s + ) 5y() + 3 s 5 5s + s 5s + s + 2 ss+ 2 + 3 = +.6 (Initial Condition) + (Input) Spring 2 4 Example (cont) Use table to Invert Y(s) term by term to find y(t) From the Table: So that ( s + 2) +.6 ss+ ( 2) L s + a = e at L s + 2 = e 2t a (.3)2 L ss+ a = ( e at) L ss+ 2 y(t) = e 2t +.3( e 2t ) (Initial Condition) + (Input) =.3( e 2t) Spring 2 5 Properties of Differentiation (review) Spring 2 6
Example 2 s 2 Y (s) s ( s 2 ) (s + ) + s 2 s 2 (s + ) s 2 3/2 + s 2 ( s 2 ) A ( s ) + B s + + C (Find A, B and C) s 2 ( s ) + ( /2) ( s + ) + ( ) s 2 y(t) = ( 3/2)e t + ( /2)e t + ( )t How do we do that??? Spring 2 7 Partial fraction expansion Multiply both sides by (s-) & let s Similarly, Example 2 (cont d) (s + ) ( s 2 ) + s 2 s 2 ( s )Y (s) s = ( s ) (s + ) so A = ( s ) s 2 = A A ( s ) + B ( s + ) + C s 2 + s 2 ( s 2 ) B = ( s + )Y (s) s = 2 C = ()Y s 2 (s) = s ( s ) + B s + + C s 2 = A = + s s 2 s + unknowns = + 2 = 3 2 Spring 2 8 Example 3 ODE with initial conditions (ICs) Example 3 (cont d) Inverse If we are interested in only the final value of y(t), apply Final Value Theorem: (This also isn t in the table ) Spring 2 9 Spring 2 2
Example: Newton s s law M EX. Air bag and accelerometer Tiny MEMS accelerometer Microelectromechanical systems (MEMS) We want to know the trajectory of x(t). By, (Total response) = (Forced response) + (Initial condition response) Spring 2 2 Spring 2 (Pictures from various websites) 22 Summary & Exercises In this way, we can find a rather complicated solution to ODEs easily by using table! (Important math tool!) Definition table Properties of Solution to ODEs via Exercises Read Chapters and 2. Solve Quiz problems Spring 2 23 Spring 2 24