信号与系统课件第9章 拉普拉斯变换

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 (拉普拉斯变换) is a generalization of the continuous-time Fourier transform.以傅里叶变换为基础的频域分析方法的优点在于：它以傅里叶变换为基础的频域分析方法的优点在于：它给出的结果有着清楚的物理意义给出的结果有着清楚的物理意义 ，但也有不足之处。，但也有不足之处。傅里叶变换只能处理符合傅里叶变换只能处理符合狄利赫勒条件狄利赫勒条件的信号，而实的信号，而实际中会遇到许多信号，例如际中会遇到许多信号，例如 (t)、t (t)、sint (t)等，它等，它们并不满足绝对可积条件，虽然通过求极限的方法可们并不满足绝对可积条件，虽然通过

3、求极限的方法可以求得它们的傅里叶变换，但其变换式中常常含有冲以求得它们的傅里叶变换，但其变换式中常常含有冲激函数，使分析计算较为麻烦。激函数，使分析计算较为麻烦。而有些信号（如单边指数信号而有些信号（如单边指数信号 ）是不满足绝对可积）是不满足绝对可积条件的，因而其对信号的分析受到限制；条件的，因而其对信号的分析受到限制；另外傅氏变换分析法只能求取零状态响应。另外傅氏变换分析法只能求取零状态响应。 因此，有必要寻求更有效而简便的方法，人们将傅因此，有必要寻求更有效而简便的方法，人们将傅里叶变换推广为拉普拉斯变换（里叶变换推广为拉普拉斯变换（LT: Laplace Transform）。）。优点

4、优点：求解比较简单，特别是对系统的微分方程进行变换求解比较简单，特别是对系统的微分方程进行变换时，初始条件被自动计入，因此应用更为普遍。时，初始条件被自动计入，因此应用更为普遍。缺点缺点：物理概念不如傅氏变换那样清楚。物理概念不如傅氏变换那样清楚。拉普拉斯变换（简称拉氏变换）可看作一种广义的傅拉普拉斯变换（简称拉氏变换）可看作一种广义的傅氏变换，将频域扩展为复频域，简化了信号的变换式氏变换，将频域扩展为复频域，简化了信号的变换式，扩大了信号的变换范围，为分析系统响应提供了统，扩大了信号的变换范围，为分析系统响应提供了统一和规范化的方法。一和规范化的方法。本章内容及学习方法本章内容及学习方法 本

5、章首先由本章首先由傅氏傅氏变换引出变换引出拉氏拉氏变换，然后对拉氏变换，然后对拉氏正正变换、拉氏反变换及拉氏变换的变换、拉氏反变换及拉氏变换的性质性质进行讨论。进行讨论。 本章本章重点重点在于，以拉氏变换为工具对系统进行在于，以拉氏变换为工具对系统进行复频复频域分析域分析。 最后介绍最后介绍系统函数系统函数以及以及H(s)零极点零极点概念，并根据他概念，并根据他们的分布研究们的分布研究系统特性系统特性，分析，分析频率响应频率响应，还要简略介绍，还要简略介绍系统系统稳定性稳定性问题。问题。 注意与傅氏变换的注意与傅氏变换的对比对比，便于理解与记忆。，便于理解与记忆。 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.从傅氏变换到拉氏变换从傅氏变换到拉氏变换:傅氏变换信号不满足绝对可积条件的傅氏变换信号不满足绝对可积条件的原因：原因：不趋于零。不趋于零。时，时，或或当当)(tftt ( ),ttef te用实指数函数去乘只要 的数值选取适当，可满足条件，称为收敛因子。 ( ) e()tf t信号，乘以衰减因子为任意实数 后容易满足绝对可积条件，依傅氏

8、变换定义：:j , , s令具有频率的量纲称为复频率。 es tF sf ttd则则1拉普拉斯正变换 1( ) etFF f tj( )eedttf tt(j )F(j )( ) edtf tt2 2拉氏逆变换拉氏逆变换 j1eed2ttf tFj j1jed2tf tFjj:s 积分限：对对 ej tf tF是的傅里叶逆变换 et两边同乘 以:j ; djdss其中若 取常数， 则 jj1 e d2 js tf tF ss jj ee ts tFf ttF sf ttdd所以3 3拉氏变换对拉氏变换对 j1jed 1e d 2 js ts tF sL f tf ttf tLf tF ss 正

9、变换逆变换 : f tF s记作 f tF s称为原函数，称为象函数。由于实际信号都是有始信号，即由于实际信号都是有始信号，即0)(0 tft时时，或者只需考虑或者只需考虑 的部分，此时的部分，此时0t 0)()(dtetfsFst积分下限用积分下限用 目的是把目的是把 时出现的冲激包时出现的冲激包含进去，这样，利用拉氏变换求解微分方程时含进去，这样，利用拉氏变换求解微分方程时，可以直接引用已知的初始状态，可以直接引用已知的初始状态 ，但反，但反变换的积分限并不改变。变换的积分限并不改变。00t)0(f称为称为单边拉氏变换单边拉氏变换 0, 采用系统 相应的单边拉氏变换为 0j1jed1e d

10、2js ts tF sL f tf ttf tLf tF ss 4. 拉氏变换的收敛域拉氏变换的收敛域乘乘以以收收敛敛因因子子后后，信信号号)(tf有可能满足绝对可积的条件有可能满足绝对可积的条件,是否一定满足，还要看是否一定满足，还要看 的性质与的性质与 的相对关系的相对关系)(tftetf)(通常把使通常把使 满足绝对可积条件的满足绝对可积条件的 值的范值的范围称为拉氏变换的围称为拉氏变换的收敛域。收敛域。 存在下列关系存在下列关系后后乘以收敛因子乘以收敛因子若若,)(tetf )(0)(lim0 ttetf0 则收敛条件为则收敛条件为的的性性质质有有关关值值，称称为为收收敛敛坐坐标标，与

11、与为为最最低低限限度度的的)(0tf 0 0e)(limtftt 收敛域：收敛域：使使F(s)存在的存在的s的区域称为收敛域。的区域称为收敛域。记为：记为：ROC(region of convergence)实际上就是拉氏变换存在的条件；实际上就是拉氏变换存在的条件；Oj0收敛坐标收敛坐标收收敛敛轴轴收收敛敛区区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 ) (收敛域)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 (围线积分) 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