By Poularikas A. D.

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**Extra resources for The Handbook of Formulas and Tables for Signal Processing. “Laplace Transforms”**

**Example text**

To establish the region of convergence, we write the bilateral transform in the form F2 (s) = ∫ ∞ 0 e − st f (t ) dt + ∫ 0 e − st f (t ) dt −∞ σt If the function f(t) is of exponential order (e 1 ), the region of convergence for t > 0 is Re{s} > σ1 If the function f(t) for t < 0 is of exponential order exp(σ2t), then the region of convergence is Re{s} < σ2. Hence, the function F2(s) exists and is analytic in the vertical strip defined by σ1 < Re {s} < σ2 Provided, of course, that σ1 < σ2. If σ1 > σ2, no region of convergence would exist and the inversion process could not be performed.

7 For t > 0, we close the contour to the left, we obtain f (t ) = 3e st 1 = e −2 t (s − 4)(s + 1) s=−1 2 t>0 For t < 0, the contour closes to the right, and now f (t ) = e 4t 3e st 3e st 3 + = − e −t + (s − 4)(s + 2) s=−1 (s + 1)(s + 2) s= 4 5 10 t<0 These examples confirm that we must know the region of convergence to find the inverse transform.

1999 CRC Press LLC Integral I3. Consider a typical hook at s = jnπ. Since (s − jnπ) e st 0 lim = , r→0 s (1 + e − s ) 0 s→ jnπ this expression is evaluated and yields e jnπt/jnπ. Thus, for all poles, I3 = 1 2 πj ∫ π/2 −π/2 r→0 s→ jnπ e st ds s (1 + e − s ) ∞ ∞ sin nπt jπ e jnπt 1 1 1 2 = + = + 2 πj n=−∞ jnπ 2 2 2 π n=1 n n odd n odd ∑ ∑ Finally, the residues enclosed within the contour are ∞ Res e st e jnπt 1 2 1 = + = + −s s (1 + e ) 2 n=−∞ jnπ 2 π ∑ n odd ∞ ∑ n =1 n odd sin nπt n which is seen to be twice the value around the hooks.

### The Handbook of Formulas and Tables for Signal Processing. “Laplace Transforms” by Poularikas A. D.

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