Note #07_Fourier Transforms

# Note #07_Fourier Transforms - Note #07. Fourier Transform...

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Note #07. Fourier Transform Pairs ECSE-2410 Signals & Systems (Wozny) February, 2008 p. 1 } The Fourier transform of a signal is defined as ; also written as ) ( t x = dt e t x X t j ) ( ) ( ω { ) ( ) ( t x F X = . The inverse Fourier transform is defined as = ωω π d e X t x t j 2 1 ) ( ) ( ; also written as . The signal and its transform { ) ( ) ( 1 X F t x = } ) ( t x ) ( X represent a Fourier transform pair, ) ( ) ( X t x (text, p. 30). All transform pairs can be found by direct integration of the defining integrals above. Sometimes this integration is messy and cannot be avoided. However, there are many problems where integration is unnecessary. We solve such problems by expressing unknown complex signals in terms of simpler signals and operations whose transforms are known. When you have finished studying this note, you will have derived all the transform pairs in Table 4.1 (text) and properties in Table 4.2, as well as learning an approach for finding other transform pairs not in tables. 1. Find the Fourier transform of { } 0 ), ( ) ( > = a e t u e t x at . [Why do we need the condition, ?] {} 0 > a e This result is found by integration as shown in Example 4.1 in the text, resulting in j a X a e t u e t x at + = > = 1 ) ( 0 ), ( ) ( 2. Find the Fourier transform of ) ( ) ( t u t x = . Again we find this result by direct integration. = = = 0 0 1 ) ( ) ( t j t j t j e j dt e dt e t u X But the upper limit, , is indeterminate, since So we can’t evaluate the integral. j e ? ) sin( ) cos( = + = j e j We get around this problem by using a convergence factor, which ultimately results in delta functions, since that is the only way we can force the integral to converge. Use convergence factor, { } 0 ), ( ) ( ~ > = a e t u e t x at . Then ) ( ~ lim ) ( 0 t x t u a = and ) ( ~ lim ) ( 0 X X a = , where

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2 2 2 2 ) ( ~ ) ( ~ 1 ) ( ~ ω ωω + + = + = + = a j a a X j X j a X I R Now as , 0 a = = + = 0 , 0 , 0 ) ( ~ 0 0 2 2 2 a a a X R = = + = 0 , undefined 0 , 1 ) ( ~ 1 2 2 j I j a j X j . Using the above results, we will define the transform of a step as {} = = = = = 0 , ) ( lim 0 , ) ( lim ) ( ~ lim ) ( 0 0 0 a 1 0 0 R j I a a X X j X t u F , but need to show that the indeterminate term 0 0 0 ) ( lim = R a X for 0 = makes sense. Consider 2 2 ) ( ~ + = a a X R as and 0 a 0 . ) ( ~ R X 0 a 1 a 2 1 a a We can see that as , tends to a pulse of zero width and infinite height . These are two of the defining conditions of a delta function. The third is that the area under the pulse remains constant as . The area in this case is 0 a ) ( ~ R X 0 a π = = + = a d a a d X R 1 2 2 tan ) ( ~ . Since this area is independent of a , then the three conditions define a delta function. Pictorially, ) ( ~ R X 0 a 1 a 2 1 a a ) ( δ 0 = ) ( ) ( ~ lim 0 ωπ δω R a X p. 2
Thus the Fourier transform of a step function is {} = = 0 ), ( 0 , ) ( 1 ωω πδ ω j t u F .

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## This note was uploaded on 05/03/2009 for the course ECSE 2410 taught by Professor Wozny during the Spring '07 term at Rensselaer Polytechnic Institute.

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Note #07_Fourier Transforms - Note #07. Fourier Transform...

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