EE 422G Notes: Chapter 5 Instructor: Cheung Page 5-1Chapter 5 The Laplace Transform 5-1 Introduction (1) System analysis Linear Dynamic system: )(...)()(...)()()(01)1(1)(txbdttdxbtyadttdyadttdymmmnnnnn++=+++−− Question: Can we determine y(t) for a given x(t)? Answer: Use Fourier transform to convert the ODE into algebraic equation! ))(()()()()(1fYFtyfXfHfY−== (2) Two Problems 1. Some common signals do not have a Fourier Transform ! Example: What is the Fourier transform of )()( tuetxt= ? ∫∫∞−∞∞−−==022)()( dteedtetuefXftjtftjtππ does not exist for any f as te blows up when ∞→t. Even though )(tx grows unbounded as ∞→t, it may still exist as an intermediate step in a larger system. Consider the output when )(tx is fed into a system with impulse response ()).(23)(2tueethtt −−−= Dynamic System x(t) input y(t) output A processor which processes the input signal to produce the output Dynamic System X(f) Y(f ) H(f ) PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 5 Instructor: Cheung Page 5-2()( )( ))(23)(23)()(2022tueedeeedueetuetyttttt−−−−−∞∞−−−−−=−=−−=∫∫ττττττττττ which certainly decays to 0 as ∞→t. 2. Initial Condition Problem Say we know the output y(t) of the above dynamic system is 5 at t=0. Nowhere in the Fourier system equations below we could insert this information: ))(()()()()(1fYFtyfXfHfY−== (3) Solution: Laplace Transform Even though x(t) does not go to zero (when ∞→t), but tetxσ−)( may for large enough α.  We will assume all signals are “causal” : 0)(=tx for 0<t  Fourier transform of tetxσ−)( : )( of Transform Laplace where)()())((00)(0txjsdtetxdtetxdteetxsttjtjt=+===∫∫∫∞−∞+−∞−−ωσωσωσ • We will see how Laplace transform takes care of initial conditions later. • Even though inverse Laplace Transform exists, it involves more sophisticated concepts from complex number theory. Just like Inverse Fourier Transform, we will just use table (and Matlab). • Laplace transform is just as nice as Fourier Transform: ))(()()()()(1sYLtyconditionsinitialsXsHsY−=+= • Fourier Transform of )()( tutx can obtained by substituting ωjs= (i.e. setting )0=σ in ).(sX1. 1 If the Region of Converge of X(s) does not include the imaginary axis (ωjs=), then its Fourier Transform does not exist. (More later) PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 5 Instructor: Cheung Page 5-35-2 Examples of Evaluating Laplace Transforms using the definition (1) Step function x(t)=u(t) ( )( )==∞ →<∞ →>=+−=+−=−=−−==−∞→∞→−∞→−−−∞→−∞→∞==−∞−∞−∫∫?lim as 0)Re( ifsurenot as 0)Re( if0 as 0)Re( if/11lim11lim1)(1)]([)Im()Re()Re()Im()Re(000tsjtttsttstsjtststtttstststesesessseessessestdesdtetuL When Re(s) = 0, define ω=Im(s). The integral becomes ∫∫∞∞−−∞−= dtetudtetjtjωω)(0 which is the Fourier transform of u(t). From chapter 4, we know that )(21)]([ωπδω+=jtuF Note that for ω≠0, the Fourier Transform can be evaluated by substituting s=jω in the expression 1/s. (2) Exponential )()( tuetxtα−= ( )=−=++∞ →−<∞ →−>+=+++−=+++−=+−+−===+−∞→∞→+−∞→+−+−+−∞→+−∞→∞+−∞+−∞−−−∫∫∫?lim as )Re()Re( if))(Im(2)/(1 as )Re()Re( if0 as )Re()Re( if)/(11lim11lim1)(1)]([)Im()Re()Re()Im()Re()(0)(0)(0tsjtttsttstsjtsttsttstsstttesssesessseessestsdesdtedteetueLαααααααααααπδαααααααααα PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 5 Instructor: Cheung Page 5-4(3) )()(ttxδ= Recall that the impulse function is represented as a limit of convention functions straddling the origin. To incorporate the “full” delta function, we define the lower limit of our Laplace integral to be 0-. 1)sin(cos)()]([0000=−====−−−=−=−−=−∞−∫ttttjttststtjteeeedtettLωωδδσωσ No constraint on s. 5-2B Region of Convergence (ROC) ROC: Pictorial description of the value s where the Laplace Transform of a function exists. Anatomy of ROC Properties of ROC 1. It is bordered by the RIGHTMOST POLES of X(s). A pole is defined by the complex value s’ such that the algebraic X(s’) is infinite. 2. If s’ is in the interior of the ROC, then all s with Re(s)=Re(s’) are inside the ROC. 3. It extends to positive infinity Re(s) Im(s) Im(s) Re(s) Re(-α) ROC of 1/s ROC of 1/(s+α) α Region where X(s) is finite. In fact, it converges absolutely (see below). Region where X(s) is infinite. Region where X(s) is finite, but not absolute convergent. PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.comEE 422G Notes: Chapter 5 Instructor: Cheung Page 5-5In the interior of ROC (not boundary), not only does the Laplace integral converge, it converges ABSOLUTELY. Normal convergent (or X(s) is finite): ∞<=∫∞−0)()( dtetxsXst Absolute convergent: ∞<∫∞−0|)(| dtetxst We cannot prove that without venturing into a branch of mathematics called the complex variable theory. However, knowing that X(s) converges absolutely inside the ROC, we can explain some of the properties: For example: 2. If s’ is in the interior of the ROC, then all s with Re(s)=Re(s’) are inside the ROC. Why? dtstjsttxdtstjsttxdtetxst))Im(exp())Re(exp()())Im()Re(exp()()(000−⋅−=−−=∫∫∫∞∞∞− Note that 1))(Im(sin))(Im(cos))Im(exp(22=+=− stststj, we have dtsttxdtetxst∫∫∞∞−−=00))Re(exp()()(which does not depend on Im(s). 3. ROC extends to positive infinity. Assume p is in the ROC. Let q be a complex number with )Re()Re( pq>. ∞<−≤−=∫∫∫∞∞∞−dtpttxdtqttxdtetxqt000))Re(exp()())Re(exp()()(. The second last inequality is