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A left-sided signal x(t)=-e¯bt u(-t): 0 == X(s) -e-bu(t)e-st dt =- -Le-c 1 -(b+o+jw)t dt = = -00 -∞ (a + b) + jw 1 s+b For this integral to converge, it is necessary that b +σ <0; i.e., ROC: Re[s]=σ < −b. 2 How ?

A left-sided signal x(t)=-e¯bt u(-t): 0 == X(s) -e-bu(t)e-st dt =- -Le-c 1 -(b+o+jw)t dt = = -00 -∞ (a + b) + jw 1 s+b For this integral to converge, it is necessary that b +σ <0; i.e., ROC: Re[s]=σ < −b. 2 How ?

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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A left-sided signal x(t)=-e¯bt u(-t): 0 == X(s) -e-bu(t)e-st dt =- -Le-c 1 -(b+o+jw)t dt = = -00 -∞ (a + b) + jw 1 s+b For this integral to converge, it is necessary that b +σ <0; i.e., ROC: Re[s]=σ < −b. 2 How ?
Transcribed Image Text:A left-sided signal x(t)=-e¯bt u(-t): 0 == X(s) -e-bu(t)e-st dt =- -Le-c 1 -(b+o+jw)t dt = = -00 -∞ (a + b) + jw 1 s+b For this integral to converge, it is necessary that b +σ <0; i.e., ROC: Re[s]=σ < −b. 2 How ?
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