信號(hào)與系統(tǒng)課件第9章 拉普拉斯變換

1、CHAPTER 9THE LAPLACE TRANSFORM6.0 INTRODUCTIONn With Laplace transform, we expand the application in which Fourier analysis can be used.sten The Laplace transform provides us with a representation for signals as linear combinations of complex exponentials of the form with s= + jn The Laplace transfo

2、rm (拉普拉斯變換) is a generalization of the continuous-time Fourier transform.以傅里葉變換為基礎(chǔ)的頻域分析方法的優(yōu)點(diǎn)在于：它以傅里葉變換為基礎(chǔ)的頻域分析方法的優(yōu)點(diǎn)在于：它給出的結(jié)果有著清楚的物理意義給出的結(jié)果有著清楚的物理意義 ，但也有不足之處。，但也有不足之處。傅里葉變換只能處理符合傅里葉變換只能處理符合狄利赫勒條件狄利赫勒條件的信號(hào)，而實(shí)的信號(hào)，而實(shí)際中會(huì)遇到許多信號(hào)，例如際中會(huì)遇到許多信號(hào)，例如 (t)、t (t)、sint (t)等，它等，它們并不滿足絕對(duì)可積條件，雖然通過求極限的方法可們并不滿足絕對(duì)可積條件，雖然通過

3、求極限的方法可以求得它們的傅里葉變換，但其變換式中常常含有沖以求得它們的傅里葉變換，但其變換式中常常含有沖激函數(shù)，使分析計(jì)算較為麻煩。激函數(shù)，使分析計(jì)算較為麻煩。而有些信號(hào)（如單邊指數(shù)信號(hào)而有些信號(hào)（如單邊指數(shù)信號(hào) ）是不滿足絕對(duì)可積）是不滿足絕對(duì)可積條件的，因而其對(duì)信號(hào)的分析受到限制；條件的，因而其對(duì)信號(hào)的分析受到限制；另外傅氏變換分析法只能求取零狀態(tài)響應(yīng)。另外傅氏變換分析法只能求取零狀態(tài)響應(yīng)。 因此，有必要尋求更有效而簡(jiǎn)便的方法，人們將傅因此，有必要尋求更有效而簡(jiǎn)便的方法，人們將傅里葉變換推廣為拉普拉斯變換（里葉變換推廣為拉普拉斯變換（LT: Laplace Transform）。）。優(yōu)點(diǎn)

4、優(yōu)點(diǎn)：求解比較簡(jiǎn)單，特別是對(duì)系統(tǒng)的微分方程進(jìn)行變換求解比較簡(jiǎn)單，特別是對(duì)系統(tǒng)的微分方程進(jìn)行變換時(shí)，初始條件被自動(dòng)計(jì)入，因此應(yīng)用更為普遍。時(shí)，初始條件被自動(dòng)計(jì)入，因此應(yīng)用更為普遍。缺點(diǎn)缺點(diǎn)：物理概念不如傅氏變換那樣清楚。物理概念不如傅氏變換那樣清楚。拉普拉斯變換（簡(jiǎn)稱拉氏變換）可看作一種廣義的傅拉普拉斯變換（簡(jiǎn)稱拉氏變換）可看作一種廣義的傅氏變換，將頻域擴(kuò)展為復(fù)頻域，簡(jiǎn)化了信號(hào)的變換式氏變換，將頻域擴(kuò)展為復(fù)頻域，簡(jiǎn)化了信號(hào)的變換式，擴(kuò)大了信號(hào)的變換范圍，為分析系統(tǒng)響應(yīng)提供了統(tǒng)，擴(kuò)大了信號(hào)的變換范圍，為分析系統(tǒng)響應(yīng)提供了統(tǒng)一和規(guī)范化的方法。一和規(guī)范化的方法。本章內(nèi)容及學(xué)習(xí)方法本章內(nèi)容及學(xué)習(xí)方法 本

5、章首先由本章首先由傅氏傅氏變換引出變換引出拉氏拉氏變換，然后對(duì)拉氏變換，然后對(duì)拉氏正正變換、拉氏反變換及拉氏變換的變換、拉氏反變換及拉氏變換的性質(zhì)性質(zhì)進(jìn)行討論。進(jìn)行討論。 本章本章重點(diǎn)重點(diǎn)在于，以拉氏變換為工具對(duì)系統(tǒng)進(jìn)行在于，以拉氏變換為工具對(duì)系統(tǒng)進(jìn)行復(fù)頻復(fù)頻域分析域分析。 最后介紹最后介紹系統(tǒng)函數(shù)系統(tǒng)函數(shù)以及以及H(s)零極點(diǎn)零極點(diǎn)概念，并根據(jù)他概念，并根據(jù)他們的分布研究們的分布研究系統(tǒng)特性系統(tǒng)特性，分析，分析頻率響應(yīng)頻率響應(yīng)，還要簡(jiǎn)略介紹，還要簡(jiǎn)略介紹系統(tǒng)系統(tǒng)穩(wěn)定性穩(wěn)定性問題。問題。 注意與傅氏變換的注意與傅氏變換的對(duì)比對(duì)比，便于理解與記憶。，便于理解與記憶。 9.1 THE LAPLAC

6、E TRANSFORMLet s = + j, and using X(s) to denote this integral, we obtaindteetxjXtjt)()(dtetxsXst)()(For some signals which have not Fourier transforms, if we preprocess them by multiplying with a real exponential signal , then they may have Fourier transforms.teThe Laplace transform of x(t) The Lap

7、lace transform is an extension of the Fourier transform; the Fourier transform is a special case of the Laplace transform when = 0.從傅氏變換到拉氏變換從傅氏變換到拉氏變換:傅氏變換信號(hào)不滿足絕對(duì)可積條件的傅氏變換信號(hào)不滿足絕對(duì)可積條件的原因：原因：不趨于零。不趨于零。時(shí)，時(shí)，或或當(dāng)當(dāng))(tftt ( ),ttef te用實(shí)指數(shù)函數(shù)去乘只要 的數(shù)值選取適當(dāng)，可滿足條件，稱為收斂因子。 ( ) e()tf t信號(hào)，乘以衰減因子為任意實(shí)數(shù) 后容易滿足絕對(duì)可積條件，依傅氏

8、變換定義：:j , , s令具有頻率的量綱稱為復(fù)頻率。 es tF sf ttd則則1拉普拉斯正變換 1( ) etFF f tj( )eedttf tt(j )F(j )( ) edtf tt2 2拉氏逆變換拉氏逆變換 j1eed2ttf tFj j1jed2tf tFjj:s 積分限：對(duì)對(duì) ej tf tF是的傅里葉逆變換 et兩邊同乘 以:j ; djdss其中若 取常數(shù)， 則 jj1 e d2 js tf tF ss jj ee ts tFf ttF sf ttdd所以3 3拉氏變換對(duì)拉氏變換對(duì) j1jed 1e d 2 js ts tF sL f tf ttf tLf tF ss 正

9、變換逆變換 : f tF s記作 f tF s稱為原函數(shù)，稱為象函數(shù)。由于實(shí)際信號(hào)都是有始信號(hào)，即由于實(shí)際信號(hào)都是有始信號(hào)，即0)(0 tft時(shí)時(shí)，或者只需考慮或者只需考慮 的部分，此時(shí)的部分，此時(shí)0t 0)()(dtetfsFst積分下限用積分下限用 目的是把目的是把 時(shí)出現(xiàn)的沖激包時(shí)出現(xiàn)的沖激包含進(jìn)去，這樣，利用拉氏變換求解微分方程時(shí)含進(jìn)去，這樣，利用拉氏變換求解微分方程時(shí)，可以直接引用已知的初始狀態(tài)，可以直接引用已知的初始狀態(tài) ，但反，但反變換的積分限并不改變。變換的積分限并不改變。00t)0(f稱為稱為單邊拉氏變換單邊拉氏變換 0, 采用系統(tǒng) 相應(yīng)的單邊拉氏變換為 0j1jed1e d

10、2js ts tF sL f tf ttf tLf tF ss 4. 拉氏變換的收斂域拉氏變換的收斂域乘乘以以收收斂斂因因子子后后，信信號(hào)號(hào))(tf有可能滿足絕對(duì)可積的條件有可能滿足絕對(duì)可積的條件,是否一定滿足，還要看是否一定滿足，還要看 的性質(zhì)與的性質(zhì)與 的相對(duì)關(guān)系的相對(duì)關(guān)系)(tftetf)(通常把使通常把使 滿足絕對(duì)可積條件的滿足絕對(duì)可積條件的 值的范值的范圍稱為拉氏變換的圍稱為拉氏變換的收斂域。收斂域。 存在下列關(guān)系存在下列關(guān)系后后乘以收斂因子乘以收斂因子若若,)(tetf )(0)(lim0 ttetf0 則收斂條件為則收斂條件為的的性性質(zhì)質(zhì)有有關(guān)關(guān)值值，稱稱為為收收斂斂坐坐標(biāo)標(biāo)，與

11、與為為最最低低限限度度的的)(0tf 0 0e)(limtftt 收斂域：收斂域：使使F(s)存在的存在的s的區(qū)域稱為收斂域。的區(qū)域稱為收斂域。記為：記為：ROC(region of convergence)實(shí)際上就是拉氏變換存在的條件；實(shí)際上就是拉氏變換存在的條件；Oj0收斂坐標(biāo)收斂坐標(biāo)收收斂斂軸軸收收斂斂區(qū)區(qū)Example 9.1Consider the signal).()(tuetxtsesesdtedtetuesXtststssttt111)()(lim)(0)(0)(For convergence, we require that Res + 0, or Res , Thus,1(

12、 ),Re X sssaa= -+region of convergence (ROC ) (收斂域)Example 9.2Consider the signal).()(tuetxttststsstttessesdtedttueesXlim)(0)(0)(111)()(For convergence, we require that Res + 0, or Res 0 will also be in the ROC; and the ROC of a right-sided signal is a right-half plane. Property 5: If x(t) is left s

13、ided, and if the line Res = 0 is in the ROC, then all values of s for which Res 0 will also be in the ROC; and the ROC of a left-sided signal is a left-half plane.Property 6: If x(t) is two sided, and if the line Res = 0 is in the ROC, then the ROC will consist of a strip in the s-plane that include

14、s the line Res = 0.RReImLReImLRReImProperty 7: If the Laplace transform X(s) of x(t) is rational, then its ROC is bounded by poles or extends to infinity. In addition, no poles of X(s) are contained in the ROC.Property 8: If the Laplace transform X(s) of x(t) is rational, then if x(t) is right sided

15、, the ROC is the region in the s-plane to the right of the rightmost pole. If x(t) is left sided, the ROC is the region in the s-plane to the left of the leftmost pole.Example 9.6Let)2)(1(1)(sssXReIms-planeROC corresponding to a right-sided signal ROC corresponding to a left-sided signal ROC corresp

16、onding to a two-sided signal 9.3 THE INVERSE LAPLACE TRANSFORM11( )( )2j tFX sX s edwwp+ - =Multiplying both sides by , we obtain tedesXtxst)(21)(Changing the variable of this integration from to s and using the fact that is constant, so that ds = jd. 1( )( )2jstjx tX s e dsjssp+ - =Thus, the basic

17、inverse Laplace transform equation is:The inverse Laplace transform equation states that x(t) can be represented as a weighted integral of complex exponentials. The formal evaluation of the integral for a general X(s) requires the use of contour integration (圍線積分) in the complex plane. For the class

18、 of rational transforms, the inverse Laplace transform can be determined by using the technique of partial-fraction expansion. tetx)(Example 9.7Let . 1Re2,)2)(1(1)(ssssXPerforming the partial-fraction expansion, we obtain )2(1) 1(1)(sssX21( )()( ),2Re 1.(1)(2)LTttx te uteu tsss-= -( )( )* ( )X su tx

19、 ts若若t -桫+In fact, by repeated application of this property, we obtain11( ),Re (1)!()nLTtnteu tsnsaaa-+Example 9.11Use the initial-value theorem to determine the initial-value of)()3(cos)()(2tutetuetxtt220144252)()0(2323limlimsssssssXxss. 1,)2)(102(1252)()3(cos)(222sssssstutetueLTttRe9.6 ANALYSIS AN

20、D CHARACTERIZATION OF LTI SYSTEMS USING THE LAPLACE TRANSFORM The Laplace transforms of the input and the output of an LTI system are related through multiplication by the Laplace transform of the impulse response of the system. Y(s) = H(s) X(s) The ROC associated with the system function for a caus

21、al system is a right-half plane.An ROC to the right of the rightmost pole does not guarantee that a system is causal. For a system with a rational system function, causality of the system is equivalent to the ROC being the right-half plane to the right of the rightmost pole.system function(transfer

22、function)Example 9.12Consider a system with impulse response ).()()21(tuethtjSince h(t) = 0 for t -+It is rational and the ROC is to the right of the rightmost pole, consistent with our statement. Example 9.13Consider the system function ( ),Re1.1seH sss= -+For this system, the ROC is to the right o

23、f the rightmost pole. The impulse response associated with the system),1()()1(tuethtis nonzero for 1 t 1. )()(2tueetxtt Thus,unilateral Laplace transform provide us with information about signals only for .0tExample 9.22Consider the unilateral Laplace transform .23)(2sssX212)(sssXTaking inverse tran

24、sforms of each term results in)()()(2)(2tuetttxt9.9.1 Properties of the Unilateral Laplace Transform Time scaling:0),(1)(aasaatxULXConvolution: assuming that x1(t) and x2(t) are identically zero for t 0. )()()()(2121sstxtxULXXDifferentiation in the time domain :)0()()(xsstxdtdULX).0()()()(101knkknnU

25、LnnxssstxdtdX Proof of this property for first-derivative of x(t):0)()(dtedttdxdttdxstUL0)(tdxest0)()(0dtetxsetxstst)0()(xssX Similarly, the unilateral Laplace transform of second-derivative of x(t) can be obtained by repeating using the property:).0()0()()0()0()()(222xsxssxxsssdttxdXXUL9.9.2 Solvin

26、g Differential Equations Using the Unilateral Laplace TransformExample 9.23Consider the system characterized by the differential equation ),()(2)(3)(22txtydttdydttydwith initial conditions.)0(,)0(yyLet x(t) = u(t). Determine the output y(t).Applying the unilateral transform to both sides of the diff

27、erential equation, we obtain )()(2)0(3)(3)0()0()(2sXsYyssYysysYsor equivalently,ssYssYssYs)(23)(3)(2Thus, we obtain)2)(1()3()(2ssssssYzero-state response zero-input response The unilateral Laplace transform is of considerable value in analyzing causal systems which are specified by linear constant-c

28、oefficient differential equations with nonzero initial conditions (i.e., systems that are not initially at rest).)2)(1()2)(1()2)(1() 3(ssssssss9.9.3 Representation of Circuits in s-domain)()(tRitvRRdttdiLtvLL)()(dttdvCticc)()(The relations between I and V in the time domain for R,L,C are:Respectivel

29、y.Apply unilateral Laplace transform to each equation to obtain )()(sRIsVRR)0()()(LLLissLIsV)0(1)(1)(CCCvssIsCsVFor a circuit, if we obtain the representation for the basic elements in the circuit in the s-domain, then we also obtain the circuit in the s-domain. An inductor with inductance L and ini

30、tial current may be taken as an inductor with inductance L and zero initial current cascaded with a impulse source voltage with area . )0(Li)0(LLiA capacitor with capacitance C and initial voltage may be taken as a capacitor with capacitance C and zero initial voltage cascaded with a step source vol

31、tage with step . )0(Cv)0(CvIR(s)VR (s)R+-IL(s)VL (s)sL+-+ 0LLiIC(s)VC (s)+-+ svC0sC1Representation of the three basic elements in the s-domain with initial state being equalized as a source voltageIR(s)VR (s)R+-VL (s)sL+-)(sIL siL0VC (s)+-)(sIC)0(cCvsC1Representation of the three basic elements in the s-domain with initial state being equalized as a source currentAnother expression of the relation between current and voltage of three basic elements in the s-domain can induce another model:)()(sRIsVRR)0(1)(1)(LLissVsLsI)0()()(CCCCvssCVsIFirst draw the s-domain model of the giv