1 The Laplace transform The unilateral or one-sided Laplace transform function of real single valued signal ()ftis defined by the analysis formula as; () () ()0stFs ft fte dt+∞−==∫L [1.1] where()Fsdenotes Laplace transform of ()ft, the time function ste− is complex exponential whose power st is unitless with time t in second and frequency in 1sec Hertz=. s is also a complex variable, whose real part is σand imaginary part jω. 2sj jfσωσ π=+ =+ [1.2] []Re s σ= [1.3] []Im s ω= [1.4] The s-plane or Complex frequency plane is shown in figure 1-1.2 Signal Processing []Re sσ=[]Imjsω= Figure 1-1: s-plane The Laplace transform is called one-sided because ()0ft= for 0t <. The Laplace transform has meaning if the integral converges. () ()00st t j tfte dt fte e dtσω∞∞−−−=<∞∫∫ [1.5] The jteω− has magnitude of unity ()00tfte dtσ∞−<∫ [1.6] If the functions ()ft are of exponential order ()otft Aeσ< [1.7] Then: ()()00000ttttfte dt Ae e dt Ae dtσσσσσ∞∞∞−−−−<=∫∫∫ [1.8] The above integral converges for 000σσ σσ−> ⇒ > [1.9] The range of the values of the complex variables s for which the integral converges is called the existence region or the region of convergence (ROC).The Laplace transform 3 The unilateral Laplace transform is restricted to causal time functions. Consider the initial conditions in the solution of differential equations as well as in the analysis of systems. The bilateral or two-sided Laplace transform of real, and single valued signal()ftis defined by the analysis formula as; () () ()stFs ft ft e dt+∞−−∞==∫L [1.10] where ()Fsis defined for regions in s called the region of convergence ROC, for which the integral exists and s is a complex variable, 2sj jfσωσ π=+ =+ [1.11] and the Laplace transform is called two-sided because the signal()ft exist for x−∞ < < ∞. The bilateral and unilateral transforms are equivalent when ()0 for 0ft t=<. The relation between the signal()ftand its Laplace transform ()Fsis denoted as; () ()ft Fs⇔ [1.12] When sis purely imaginary, thus sjω=, () ()jtFj ft e dtωω+∞−−∞=∫ [1.13] corresponds to the Fourier transform of ()ft. () ()sjFs ftω==F [1.14] the Laplace transform of the signal ()ftis equivalent to the Fourier transform of ()ftmultiplied by the exponential function teσ−. () ( ) ()()jtFs F j fte dtσωσω+∞−+−∞=+=∫ [1.15] () () ()tjt tFs fte e dt fteσω σ+∞−− −−∞ == ∫F [1.16] The Laplace transform of ()ft is the Fourier transform of ()tfteσ−4 Signal Processing Example 1-1 Find the Laplace transform of the unit step function() ()ft ut=. Solution: () () ()0110st st stFs ut ute dt e dt e dtss+∞ ∞−− −−∞∞== ==− =∫∫L [1.17] For all 0s > Example 1-2 Find the Laplace transform of the time function() ()atft e ut−=. Solution: () () ()0at at st at stFs eut eutedt eedt∞∞−−−−−−∞== =∫∫L [1.18] ()()110satFs esa sa−+∞−==++ [1.19] The integral will converge for 0sa+>. Hence the region of convergence issa>−, or in another word, ()lim 0satte−+→∞= [1.20] only if 0sa+>orsa>−. If ais real ()() ( ) () ()00 0 0jts at j at at atFs e dt e dt e e dt e dtωσω σ σ−∞∞ ∞ ∞−+ − + + − + − +== = =∫∫ ∫ ∫ [1.21] The term ()ateσ−+ grows exponentially if aσ<−, so the integral diverges. The necessary condition for convergence of the integral is []Re saσ=>−as shown in Figure 1.2.The Laplace transform 5 σjωa−σjωa−0a>0a< Figure 1.2: Region of Convergence for () ()atft e ut−= Example 1-3 Find the Laplace transform of the time function() ( )atft e u t−=− −. Solution () ( )()0011satat at stFs e u t e e dt esa sa−+−−−−∞−=− −=− =− =−∞++∫L [1.22] For []Re 0sa+<the exponential term has a magnitude of zero and the integral converges. The region of convergence,[] []Re Resa<− is shown in Figure 1.3 σjωa−σjωa−0a>0a< Figure 1.3 : Region of convergence for () ( )atft e u t−=− −6 Signal Processing The algebraic expressions for ()Fsfor two different signals in example 1-2 and 1-3 are identical except for the ROC s. therefore, the ROC must be specified as a part of the Laplace transform. The uniqueness theorem states that if()1ftand()2fthave the same Laplace transform ()Fsand the same region of convergence, then () ()12ft ft= Example 1-4 Find the Laplace transform of the Real exponential time causal function () ()atft eut=. Solution: The Laplace transform of ()ft is ()at stFs ee dt∞−−∞=∫ [1.23] since ()ut is zero for 0t< so the non zero values of signal ()ft is set to zero for 0t< and the lower limit of integral becomes 0. () ()()0110satat at stFs eut ee dt esa sa∞−−−∞−=== =−−∫L [1.24] For []Re 0sa−> the exponential term has a magnitude of zero and the integral converges. The region of Convergence for [] []Re Resa>, is shown in Figure 1.4 σjωaσjωa0a>0a< Figure 1-4: Region of convergence for () ()atft eut=The Laplace transform 7 Example 1-5 Find the Laplace transform of the unit impulse function,() ()ft tδ=. Solution: () () ()stFs t te dtδδ∞−−∞==∫L [1.25] Using the sifting property of the unit impulse function which states: () ( ) ( )00ft t t dt ftδ∞−∞−=∫ [1.26] If the ()stft e−=and 00t =, then: () () ()()01sstFs t te dt eδδ∞−−−∞== ==∫L [1.27] ROCis all s-plane and it is independent of s. Example 1-6 Find the Laplace transform of the real exponential anti casual time function () ( )atft eu t=− Solution: () () ()at at stFs eu t eu te dt∞−−∞=−=−∫L [1.28] Since anti causal unit step function ()ut− is zero for 0t>, so nonzero values of signal ()ft is set to zero for 0t> and the upper limit of integral becomes zero. () ( )0at at stFs eu t ee dt−−∞=−=∫L [1.29] ()()01sateFssa sa−−−=− =−−∞−− [1.30] []() is Re ReROC s a<8 Signal Processing The inverse Laplace transform The original time function can be recovered from ()Fsby the contour integral () () () ()112cjstcjft Fs Fseds ftjπ+∞−−∞==∫=L [1.31] The real part of the upper and lower limit, c of the integral should be selected such that if the ROC of F(s) is 12Re[ ]sσσσ<=<, then 12cσσ<<. The inverse