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2.161 Signal Processing: Continuous and Discrete Fall 2008 MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing – Continuous and Discrete 1 The Laplace Transform 1 Introduction In the class handout Introduction to Frequency Domain Processing we introduced the Fourier trans­ form as an important theoretical and practical tool for the analysis of waveforms, and the design of linear filters. We noted that there are classes of waveforms for which the classical Fourier integral does not converge. An important function that does not have a classical Fourier transform are is the unit step (Heaviside) function ( 0 t · 0; us(t) = 1 t > 0; Clearly Z 1 jus(t)j dt = 1; ¡1 and the forward Fourier integral Z 1 Z 1 ¡j­t ¡j­t Us(jΩ) = us(t)e dt = e dt (1) ¡1 0 does not converge. Aside: We also saw in the handout that many such functions whose Fourier integrals do not converge do in fact have Fourier transforms that can be defined using distributions (with the use of the Dirac delta function ±(t)), for example we found that 1 F fu (t)g = ¼±(Ω) + : s jΩ Clearly any function f(t) in which limj tj!1 f(t) 6= 0 will have a Fourier integral that does not converge. Similarly, the ramp function ( 0 t · 0; r(t) = t t > 0: is not integrable in the absolute sense, and does not allow direct computation of the Fourier trans­ form. The Laplace transform is a generalized form of the Fourier transform that allows convergence of the Fourier integral for a much broader range of functions. The Laplace transform of f(t) may be considered as the Fourier transform of a modified function, formed by multiplying f(t) by a weighting function w(t) that forces the product x(t)w(t) to zero as time jtj becomes large. In particular, the two-sided Laplace transform uses an exponential weighting 1D. Rowell, September 23, 2008 1 1.0 f(t) w(t) 0 t f(t)w(t) Figure 1: Modification of a function by a multiplicative weighting function w(t). function ¡σjtj w(t) = e (2) where σ is real. The Laplace transform L fx(t)g is then n o ¡σjtj L fx(t)g = F x(t)e Figure ?? shows how this function will force the product w(t)x(t) to zero for large values of t. Then for a given value of σ, provided Z 1 ¯ ¯ ¯ ¡σjtj ¯ ¯e x(t)¯ dt < 1; ¡1 ¡σjtj the Fourier transform of x(t)e (Eq. (??)): n o ¡σjtj X(jΩjσ) = F x(t)e (3) will exist. The modified transform is not a function of angular frequency Ω alone, but also of the value of the chosen weighting constant σ. The Laplace transform combines both Ω and σ into a single complex variable s s = σ + jΩ (4) and defines the two-sided transform as a function of the complex variable s Z 1 ¡σjtj ¡j­t X(s) = (x(t)e )e dt (5) ¡1 In engineering analysis it is usual to restrict the application of the Laplace transform to causal functions, for which x(t) = 0 for t < 0. Under this restriction the integrand is zero for all negative time and the limits on the integral may be changed Z 1 Z 1 ¡σt ¡j­t ¡st X(s) = x(t)e e dt = x(t)e dt; (6) 0¡ 0¡ which is commonly known as the one-sided Laplace transform. In this handout we discuss only the properties and use of the one-sided transform, and refer to it generally as the Laplace transform. Throughout the rest of this handout it should be kept clearly in mind that the requirement x(t) = 0 for t < 0¡ must be met in order to satisfy the definition of the Laplace transform. 2 For a given x(t) the integral may converge for some values of σ but not others. The region of convergence (ROC) of the integral in the complex s-plane is an important qualification that should be specified for each transform X(s). Notice that when σ = 0, so that w(t) = 1, the Laplace transform reverts to the Fourier transform. Thus, if x(t) has a Fourier transform X(jΩ) = X(s) js =j­ : (7) Stated another way, a function x(t) has a Fourier transform if the region of convergence of the Laplace transform in the s-plane includes the imaginary axis. The inverse Laplace transform may be defined from the Fourier transform. Since n o ¡σt X(s) = X(σ + jΩ) = F x(t)e the inverse Fourier transform of X(s) is Z 1 ¡σt 1 j­t x(t)e = F fX(s)g = X(σ + jΩ)e dΩ: (8) 2¼ ¡1 σt If each side of the equation is multiplied by e Z 1 1 st x(t) = X(s)e dΩ: (9) 2¼ ¡1 The variable of integration may be changed from Ω to s = σ + jΩ, so that ds = jdΩ, and with the corresponding change in the limits the inverse Laplace transform is Z σ+j1 1 st x(t) = X(s)e ds (10) 2¼j σ¡j1 The evaluation of this integral requires integration along a path parallel to the jΩ axis in the complex s plane. As will be shown below, in practice it is rarely necessary to compute the inverse Laplace transform. The one-sided Laplace transform pair is defined as Z 1 ¡st X(s) = x(t)e dt (11) 0¡ Z σ+j1 1 st x(t) = X(s)e ds: (12) 2¼j σ¡j1 The equations are a transform pair in the sense that it is possible to move uniquely between the two representations. The Laplace transform retains many of the properties of the Fourier transform and is widely used throughout engineering systems analysis. We adopt a nomenclature similar to that used for the Fourier transform to indicate Laplace transform relationships between variables. Time domain functions are designated by a lower-case letter, such as y(t), and the frequency domain function use the same upper-case letter, Y (s). For one-sided waveforms we distinguish between the Laplace and Fourier transforms by the argument X(s) or X(jΩ) on the basis that X(jΩ) = X(s)j s=j­ if the region of convergence includes the imaginary axis. A bidirectional Laplace transform rela­ tionship between a pair of variables is indicated by the nomenclature x(t) (L) X(s); and the operations of the forward and inverse Laplace transforms are written: L fx(t)g = X(s) L¡1 fX(s)g = x(t): 3 1.1 Laplace Transform Examples Example 1 Find the Laplace transform of the unit-step (Heaviside) function ( 0 t · 0 us(t) = 1 t > 0: Solution: From the definition of the Laplace transform Z 1 ¡st Us(s) = us(t)e dt (i) 0¡ Z 1 ¡st = e dt ¡ · 0 ¯ ¯1 1 ¡st¯ = ¡ e ¯ s 0¡ 1 = ; (ii) s provided σ > 0. Notice that the integral does not converge for σ = 0, and therefore Eq. (ii) cannot be used to define the Fourier transform of the unit step. Example 2 Find the Laplace transform of the one-sided real exponential function ( 0 t · 0 x(t) = at e t > 0: Solution: Z 1 ¡st X(s) = x(t)e dt (i) 0¡ Z 1 ¡(s¡a)t = e dt ¡ · 0 ¯ ¯1 1 ¡(s¡a)t¯ = ¡ e ¯ s ¡ a 0¡ 1 = (ii) s ¡ a The integral will converge only if σ > a and therefore the region of convergence is the region of the s-plane to the right of σ = a, as shown in Fig. ??. Clearly the imaginary axis lies within the ROC, and the Fourier transform exists, only if a < 0. 4 j W j W s - p l a n e ROC s - p l a n e ROC x s x s 0 a a 0 ( a ) a > 0 ( b ) a < 0 Figure 2: Definition of the region of convergence for Example ?? for (a) a > 0, and (b) a < 0. Example 3 Find the Laplace transform of the one-sided ramp function ( 0 t < 0 r(t) = t t ¸ 0: Solution: The Fourier integral for the ramp function does not converge, but its Laplace transform is Z 1 ¡st R(s) = te dt; (i) 0¡ and integrating by parts · ¯ Z ¯1 1 1 ¡st¯ 1 ¡st R(s) = ¡ te ¯ + e dt (ii) s 0¡ s 0¡ h ¯ 1 ¯1 = 0 + e¡st¯ s2 0¡ 1 = (iii) s2 The region of convergence is all of the s-plane to the right of σ = 0, that is the right-half s plane. Example 4 Find the Laplace transform of the Dirac delta function ±(t). Solution: Z 1 ¡st Δ(s) = ±(t)e dt (i) 0¡ = 1 (ii) 5 by the sifting property of the impulse function. Thus ±(t) has a similar property in the Fourier and Laplace domains; its transform is unity and it converges everywhere. Example 5 Find the Laplace transform of a one-sided sinusoidal function ( 0 t · 0: x(t) = sin Ω0t t > 0: Solution: The Laplace transform is Z 1 ¡st X(s) = sin(Ω0t)e dt; (i) 0¡ and the sine may be expanded as a pair of complex exponentials using the Euler formula Z 1 h i 1 ¡(σ+j(­¡­ ))t ¡(σ+j(­+­ ))t X(s) = e 0 ¡ e 0 dt (ii) j2 0¡ · ¯1 · ¯1 1 ¯ 1 ¯ ¡(σ+j(­¡­0))t¯ ¡(σ+j(­+­0))t¯ = ¡ e ¯ ¡ ¡ e ¯ σ + j(Ω ¡ Ω0) 0¡ σ + j(Ω + Ω0) 0¡ Ω0 = 2 2 (σ + jΩ) + Ω0 Ω0 = 2 2 (iii) s + Ω0 for all σ > 0.
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