October 29, 200112- Introduction to Laplace Transform12.2 The Step Functionece309-07-1 October 29, 2001 12- Introduction to Laplace Transform 12.1 Introduction • Frequency-domain analysis is limited to circuit with sinusoidal analysis • Laplace transform is very powerful tools for analyzing circuit with sinusoidal or non-sinusoidal input. • Laplace transformation transforms the circuit from time domain to the frequency domain, obtain solution and apply to inverse Laplace transform to the result to transform it back to the time domain. Reasons: 1. It can be applied to a wider variety of inputs then phasor analysis 2. Provide an easy way to solve circuit problem. (working with algebraic equation instead of differential equation) 3. Can provide us, in one single operation the total response of the circuit. 12.2 Definition of Laplace Transform The Laplace transform of a function is given by {}0()() ()stftFs fte dtL−∞−==∫ Lower limit in the integral is zero. We are interesting positive values of t. This is known as the one-sided (unilateral) Laplace transform Both sided (bilateral) Laplace transform has lower limit -∝. Two types of Laplace transform • A functional transform is the Laplace transform of a specific function, such as sinωt, t, e-at • An Operational transform defines a general mathematical property of the Laplace transform, such as finding the transform of the derivatives of f(t). 12.2 The Step Function Mathematical definition of the Step function is () 0Ku t= 0t < ()Ku t K= 0t > where K is a constant. t f(t) Kece309-07-2 A step function is called an unit step function if K=1. In mathematical terms 0, 0()1, 0tutt<=> A step function is not defined at t=0, We assume that (0) 0.5Ku K= If step accurs at t=a is expressed as ()Ku t a− Mathematical definition of the Step function is ()0Ku t a−= ta< ()Ku t a K−= ta> A step function equal to K for t<a is expressed as ()Ku a t− Mathematical definition of the Step function is ()Ku a t K−= ta< ()0Ku a t−= ta> t f(t) K θ+ θ - t f(t) K t f(t) K aece309-07-3 A Finite-width pulse can be expressed as []() ( ) ( )ft Kut a ut b K=−−−= atb<< []() ( ) ( ) 0ft Kut a ut b=−−−= andta tb<> Example: 502() 10 2 640 5 6 8ttft ttt≤≤=≤≤−≤≤ [][ ][ ]( ) 5 ( ) ( 2) 10 ( 2) ( 6) (40 5 ) ( 6) ( 8)ft tut ut ut ut t ut ut=−−+−−+−−−− Lets find the Laplace transform of unit stpe function {}00()1() ()0st st stutF s u t e dt e dt esL−+∞∞−− −+=∞===−∫∫ {}()1()utFssL == The Laplace transform of ()Ku t {}()()Ku tKFssL == The Laplace transform of ()ut a− {}()1()st staut aFs e dt easL∞−−−=∞==−∫ {}()()asut aeFssL−−== t f(t) K a b t f(t) 10 2 6