《信号与系统 Signals and Systems》课程教学资料（英文版）lecture17 Motivation for the Laplace transform

H AA Signals and systems Fall 2003 Lecture #17 4 November 2003 Motivation and Definition of the bilateral) Laplace transform 2. Examples of Laplace Transforms and Their Regions of Convergence(RoCs 3. Properties of rocs

Signals and Systems Fall 2003 Lecture #17 4 November 2003 1. Motivation and Definition of the (Bilateral) Laplace Transform 2. Examples of Laplace Transforms and Their Regions of Convergence (ROCs) 3. Properties of ROCs

Motivation for the laplace transform CT Fourier transform enables us to do a lot of things, e.g Analyze frequency response oflti systems Sampling Modulation Why do we need yet another transform? One view of Laplace Transform is as an extension of the Fourier transform to allow analysis of broader class of signals and systems particular, Fourier transform cannot handle large(and important) classes of signals and unstable systems i e. when

• CT Fourier transform enables us to do a lot of things, e.g. — Analyze frequency response of LTI systems — Sampling — Modulation # Motivation for the Laplace Transform • In particular, Fourier transform cannot handle large (and important) classes of signals and unstable systems, i.e. when • Why do we need yet another transform? • One view of Laplace Transform is as an extension of the Fourier transform to allow analysis of broader class of signals and systems

Motivation for the Laplace Transform(continued In many applications, we do need to deal with unstable systems, e. g Stabilizing an inverted pendulum Stabilizing an airplane or space shuttle Instability is desired in some applications, e.g. oscillators and asers How do we analyze such signals/systems? Recall from Lecture #5, eigenfunction property of lti systems h(s st H( h(te s dt(assuming this converges) est is an eigenfunction of any lti system s=o+jo can be complex in general

Motivation for the Laplace Transform (continued) • How do we analyze such signals/systems? Recall from Lecture #5, eigenfunction property of LTI systems: — est is an eigenfunction of any LTI system — s = σ + j ω can be complex in general • In many applications, we do need to deal with unstable systems, e.g. — Stabilizing an inverted pendulum — Stabilizing an airplane or space shuttle # — Instability is desired in some applications, e.g. oscillators and lasers

The(bilateral) Laplace Transform (t)←→X(s) Aes dt=Ca(th s=0+jw is a complex variable- Now we explore the full range of s Basic ideas absolute integrability needed (1)X(S)=X(a+ju) a(t) e-o e yu dt=F{x(t)e吗} (2)A critical issue in dealing with Laplace transform is convergence X(s generally exists only for some values of s located in what is called the region of convergence(roc) ROC=[s=o+jw so that la(t)e - atl dt oo 3)If s=j@ is in the roc (i.eo=0), then only on a not on w absolute integrability X(sls=jw= Fla(ty condit

(2) A critical issue in dealing with Laplace transform is convergence: — X(s) generally exists only for some values of s, located in what is called the region of convergence (ROC) The (Bilateral) Laplace Transform (3) If s = jω is in the ROC (i.e. σ = 0), then Basic ideas: (1) absolute integrability condition absolute integrability needed

Example #1: a- an arbitrary real or complex number Ix(t)I Ix(t)I Unstable no Fourier Transform Refa>0 Rea]o but Laplace Transform exists X1(s) S eute (8+a) 0 tat (8+a) +a s+a This converges only if Re(sta)>0, 1.e. Re(s)>-Re(a) X1(s) 9e{s}>-9e{a +a ROC

Example #1: This converges only if Re(s+a) > 0, i.e. Re(s) > -Re(a) Unstable: • no Fourier Transform • but Laplace Transform exists

Example #2: Ix(t) Ix(t)I (t)=eu(t) Real Rea>0 (t) dt tat dt 0 (s+a)t S十a)∞ S+ a This converges only if Resta)<0, i.e. Re(s)<-Re(a) X2(5) S+a, tels]<-efar Same as X1(s), but different ROC ROC Key point(and key difference from FT): Need both x(s) and roc to uniquely determine x(t). No such an issue for FT

Example #2: This converges only if Re(s+a) < 0, i.e. Re(s) < -Re(a) Key Point (and key difference from FT): Need both X(s) and ROC to uniquely determine x(t). No such an issue for FT

Graphical visualization of the roc Example #1 Example #2 X1(s) 9e{s}>-9e{a} s+a +a at rig signal a2(t)=-eatu(t)-left-sided signal ga 9 plane plane g

Graphical Visualization of the RO C Example #1 Example #2

Rational transforms Many (but by no means all) Laplace transforms of interest to us are rational functions of s(e.g, Examples #1 and #2 In genera impulse responses of lti systems described by lCCDes, where X(S=D(s), N(s),D(s)polynomials in s Roots of N(s)=zeros of X(s) Roots of D(s)=poles of X(s) Any x(t) consisting of a linear combination of complex exponentials for t>0 and for t<0(e.g, as in Example #1 and #2) has a rational Laplace transform

Rational Transforms • Many (but by no means all) Laplace transforms of interest to us are rational functions of s (e.g., Examples #1 and #2; in general, impulse responses of LTI systems described by LCCDEs), where • Roots of N(s) = zeros of X(s) • Roots of D(s) = poles of X(s) • Any x(t) consisting of a linear combination of complex exponentials for t > 0 and for t < 0 (e.g., as in Example #1 and #2) has a rational Laplace transform

Example #3 a(t)= 3eu(t)-2e u(t) X( 3e2t-2e-te-stdt 0 3/c-(s-2)tat-2 0e-(s+1)tt BOTH required Requires es>2 Requires es>-1 ROC intersection 小 +7 +7 X 8-2)(s+1)s2-8 9e{s}>2 gn Notation S-plane o—zerO g Q: Does x(t have FT?

Example #3 Notation: × — pole ° — zero Q: Does x ( t) have FT? BOTH required → ROC intersection

Laplace transforms and rocs Some signals do not have Laplace Transforms(have no ROC) (a)c(t)=Ce-t for all t since/| c(t)e-at dt=oo for all o (a+1) (b)c(t)=e3wot for all t FT: X(w)=2 (w-wo) t for all X(s)is defined only in ROC; we dont allow impulses in LTS

Laplace Transforms and ROCs X( s) is defined only in ROC; we don’t allow impulses in LTs • Some signals do not have Laplace Transforms (have no ROC ) (a) (b)