9. ft - The Fourier Transform As we have seen, any...

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The Fourier Transform As we have seen, any (sufciently smooth) Function f ( t ) that is periodic can be built out oF sin’s and cos’s. We have also seen that complex exponentials may be used in place oF sin’s and cos’s. We shall now use complex exponentials because they lead to less writing and simpler computations, but yet can easily be converted into sin’s and cos’s. IF f ( t ) has period 2 , its (complex) ±ourier series expansion is f ( t ) = s k = −∞ c k e ik π t with c k = 1 2 i f ( t ) e ik π t dt (1) Not surprisingly, each term c k e ik π t in this expansion also has period 2 , because c k e ik π ( t +2 ) = c k e ik π t e i 2 = c k e ik π t . We now develop an expansion For non-periodic Functions, by allowing complex exponentials (or equivalently sin’s and cos’s) oF all possible periods, not just 2 , For some ²xed . So, From now on, do not assume that f ( t ) is periodic. ±or simplicity we’ll only develop the expansions For Functions that are zero For all sufciently large | t | . With a little more work, one can show that our conclusions apply to a much broader class oF Functions. Let L > 0 be sufciently large that f ( t ) = 0 For all | t | ≥ L . We can get a ±ourier series expansion For the part oF f ( t ) with L < t < L by using the periodic extension trick. De²ne F L ( t ) to be the unique Function determined by the requirements that i ) F L ( t ) = f ( t ) For L < t L ii ) F L ( t ) is periodic oF period 2 L Then, For L < t < L , f ( t ) = F L ( t ) = s k = −∞ c k ( L ) e ik π L t where c k ( L ) = 1 2 L i L L f ( t ) e ik π L t dt (2) IF we can somehow take the limit L → ∞ , we will get a representation oF f that is is valid For all t ’s , not just those in some ²nite interval L < t < L . This is exactly what we shall do, by the simple expedient oF interpreting the sum in (2) as a Riemann sum approximation to a certain integral. ±or each integer k , de²ne the k th Frequency to be ω k = k π L and denote by Δ ω = π L the spacing, ω k +1 ω k , between any two successive Frequencies. Also de²ne ˆ f ( ω ) = I −∞ f ( t ) e iωt dt . Since f ( t ) = 0 For all | t | ≥ L c k ( L ) = 1 2 L i L L f ( t ) e ik π L t dt = 1 2 L i −∞ f ( t ) e i ( k π L ) t dt = 1 2 L i −∞ f ( t ) e k t dt = 1 2 L ˆ f ( ω k ) = 1 2 π ˆ f ( ω k ) Δ ω In this notation, f ( t ) = F L ( t ) = 1 2 π s k = −∞ ˆ f ( ω k ) e k t Δ ω (3) For any L < t < L . As we let L → ∞ , the restriction L < t < L disappears, and the right hand side converges exactly to the integral 1 2 π I −∞ ˆ f ( ω ) e iωt . To see this, cut the domain oF integration into ω y y = 1 2 π ˆ f ( ω ) e iωt 0 ω 1 ω 2 ω 3 Δ ω February 25, 2007 The Fourier Transform 1
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9. ft - The Fourier Transform As we have seen, any...

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