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FinalExamPartA

# FinalExamPartA - EGM 4313 Summer 2007 Final Exam Part A...

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EGM 4313 – Summer 2007 NAME:____________________________________ Final Exam – Part A 60 minute time limit. Closed notes – Open book – No calculators 1) Using contour integration find the inverse Fourier transform of ( 29 1 ˆ , 0. f a a i ϖ ϖ = + (Hint: you will need to handle x>0 and x<0 differently.) ( 29 ( 29 1 1 1 1 ˆ 2 2 2 i x i x i x i f x f e dx e dx e dx a i ia ϖ ϖ ϖ ϖ ϖ ϖ π π π -∞ -∞ -∞ - = = = + - If x>o, then consider the following contour: On the curved part ( 29 ( 29 ( 29 sin cos R x iR i x e e e θ θ ϖ - = so the integral on the curved part goes to 0 as R →∞ . Therefore for x>0, ( 29 ( 29 ( 29 2 Re 2 2 2 2 i ia x ax i i f x i sidue at ia i e e π ϖ π π π π - - - = = = × = If x<0, then we must close the contour in the lower half plane so that the integral is zero on the curved part as R →∞ , that is we need sin( θ ) negative in the expression ( 29 ( 29 ( 29 sin cos R x iR i x e e e θ θ ϖ - = . Since the function is analytic everywhere inside the contour the integral is 0, hence ( 29 0 f x = for x<0.

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FinalExamPartA - EGM 4313 Summer 2007 Final Exam Part A...

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The provided text is a part of an exam question from a course in Engineering Mathematics. It contains a mathematical problem related to finding the inverse Fourier transform of a given function using contour integration, and it provides a solution for the problem.
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    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

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TYPE: <one of the following values>
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- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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FinalExamPartA

# FinalExamPartA - EGM 4313 Summer 2007 Final Exam Part A...

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EGM 4313 – Summer 2007 NAME:____________________________________ Final Exam – Part A 60 minute time limit. Closed notes – Open book – No calculators 1) Using contour integration find the inverse Fourier transform of ( 29 1 ˆ , 0. f a a i ϖ ϖ = + (Hint: you will need to handle x>0 and x<0 differently.) ( 29 ( 29 1 1 1 1 ˆ 2 2 2 i x i x i x i f x f e dx e dx e dx a i ia ϖ ϖ ϖ ϖ ϖ ϖ π π π -∞ -∞ -∞ - = = = + - If x>o, then consider the following contour: On the curved part ( 29 ( 29 ( 29 sin cos R x iR i x e e e θ θ ϖ - = so the integral on the curved part goes to 0 as R →∞ . Therefore for x>0, ( 29 ( 29 ( 29 2 Re 2 2 2 2 i ia x ax i i f x i sidue at ia i e e π ϖ π π π π - - - = = = × = If x<0, then we must close the contour in the lower half plane so that the integral is zero on the curved part as R →∞ , that is we need sin( θ ) negative in the expression ( 29 ( 29 ( 29 sin cos R x iR i x e e e θ θ ϖ - = . Since the function is analytic everywhere inside the contour the integral is 0, hence ( 29 0 f x = for x<0.

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FinalExamPartA - EGM 4313 Summer 2007 Final Exam Part A...

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The provided text is a part of an exam question from a course in Engineering Mathematics. It contains a mathematical problem related to finding the inverse Fourier transform of a given function using contour integration, and it provides a solution for the problem.

TYPE: SINGLE_QUESTION_WITH_ANSWER