MATHEMATICAL METHODS FOR ENGINEERS II (EE203) Chapter 3 LAPLACE 1.1 Introduction 1.2 Laplace Transforms 1.3 Inverse Laplace Transform 1.4 Solution of Differential Equations 1.5 Application to Step Functions and Delta Functions Lecture #3 1.50 Application to Step Functions and Delta Functions INTRODUCTION This section is important since we shall reach the point where Laplace Transform method shows its superiority in engineering applications. We will introduce two functions: A. Unit step function: u(t) B. Unit impulse function (dirac’s delta):δ(t) These functions are suitable for solving ODEs with complicated right sides of considerably engineering interest such as single waves, input (driving forces) that are discontinuous or act for some time only. 1.51 Application to Step Functions A. UNIT STEP FUNCTION: Definition 1. Unit step function, u(t-a) is a right shifted or delayed version of u(t) by amount of a (unit time). 0, u (t ) 1, t0 t 0 0, u (t a ) 1, ta ta A. UNIT STEP FUNCTION: Definition 2. A step function is a piecewise continuous function of the form g1 0 t a1 g 2 a1 t a 2 g 3 a 2 t a3 g (t ) g a t an2 n 2 n 3 g n 1 a n 2 t a n 1 g n t a n 1 g1 [ g 2 g1 ] H (t a1 ) [ g 3 g 2 ] H (t a 2 ) .... [ g n 1 g n 2 ] H (t a n 2 ) [ g n g n 1 ] H (t a n 1 ) For n=3 the step function of the form is g1 0 t a g (t ) g 2 a t b g t b 3 g1 [ g 2 g1 ]H (t a) [ g3 g 2 ]H (t b) 1.51 Application to Step Functions Laplace Transform of Unit Step Function Laplace transform of u(t) is given by: 1 Lu (t ) u (t )e dt 1.e dt s 0 0 st st For unit step function u(t-a): Lu (t a ) u (t a )e st dt 0 a 0 a u (t a )e st dt u (t a )e st dt as 1 e Lu (t a ) 0 1.e st dt e e as s s a 1.51 Application to Step Functions Second Shifting Theorem The first shifting Theorem (s-shifting) is: L e f (t ) F ( s a ) e f (t ) L F ( s a ) Laplace Inverse Laplace at at 1 Now, we have second shifting Theorem (t-shifting): L f (t a ) H (t a ) e as F ( s ) Laplace f (t a ) H (t a ) L1 e as F ( s ) Inverse Laplace 1.51 Application to Step Functions EXAMPLE: 1. Find the Laplace Transform of the function: 1 t 4 f (t ) t4 1 2. Describe this function in term of unit step function a) 1 (2 t ) 0 t 2 f (t ) 2 0 t2 b) 1 f (t ) 2 1 0t 4 4t 7 t7 g1 0 t a g (t ) g 2 t b g1 [ g 2 g1 ]H (t a ) g1 0 t a g (t ) g 2 a t b g t b 3 g1 [ g 2 g1 ]H (t a) [ g3 g 2 ]H (t b) 3. Find the Laplace transform of the expression: a) 2(t 1) H (t 1) (t 2) H (t 2) (t 2) e H (t 2) b) c) sin 2(t ) H (t ) Answer 2e s e 2 s a) 2 2 s s b) e 2 s s 1 2e s c) 2 s 4 L f (t a ) H (t a ) e as F ( s ) 4. Find the Laplace transform of the expression: 1) f (t ) (2t 2 3t 1) H (t 3) 2) f (t ) e t H (t 2) Answer 1) e 3 s 4 15 26 3 2 s s s e 2 ( s 1) 2) s 1 1.52 Application to Delta Functions B. UNIT IMPULSE (Dirac’s Delta) FUNCTION: Definition The Dirac’s delta can be loosely thought of as a function on the real line which is zero everywhere except at the δ(t) origin, where it is infinite. , (t ) 0, t 0 t0 and (t )dt 1 t Similarly, for the shifted case: , (t a) 0, δ(t-a) ta ta a t 1.52 Application to Delta Functions Laplace Transform of Delta Function If f(t) is a continuous function at t=a, then f (t ) (t a ) dt f (a) 0 This result is known as the shift of an impulse function, because all values of f(t) are shifted except at t=a st 1)If f ( t ) e , then the Laplace transform of the unit impulse function is L{ (t a)} e st (t a)dt e as When a=0 0 L { ( t )} 1 1.52 Application to Delta Functions Laplace Transform of Delta Function 2) By replacing f(t) with est f (t) the Laplace transform of f (t ) (t a ) is 0 0 L{ f (t ) (t a)} e st [ f (t ) (t a)dt [e st f (t )] (t a)dt e as f (a) Therefore L{ f (t ) (t a)} f (a)e When a=0 L{ f (t ) (t )} f (0) as Laplace transform of Step and Impulse function Example 1) (t ) (t 3 ) 1 2) (t ) 2 3) 7 (t 3) 1 3 4) t (t ) 3 5)t 2 (t 2) Answer 1) e s e 3s 2)e 1 s 2 3) 7e 3 s 3 4) e 3 5)4e 2 s 1 s 3