MEC702_Wk_11_Lect_2

# MEC702_Wk_11_Lect_2 - MEC702 Wk 11 Lect 2 Fourier Integrals...

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Unformatted text preview: MEC702 Wk 11 Lect 2 Fourier Integrals Fourier series are powerful tools for problems involving functions that are periodic or are of interest on a nite interval only. Since, of course, many problems involve functions that are nonperiodic and are of interest on the whole x-axis, we ask what can be done to extend the method of Fourier series to such functions. This idea will lead to  Fourier integrals. Consider any periodic function f (x) of period p = 2L that can be represented by a Fourier series ∞ P f (x) = a0 + (an cos wn x + bn + sin wn x) wn = n=1 Taking L → ∞, we obtain the following integrals: nπ L . (The details of this derivation can be found in the text book) A(w) = B(w) = and ˆ f (x) = 1 π ˆ 1 π ∞ f (v) cos wvdv −∞ ˆ ∞ f (v) sin wvdv −∞ ∞ [A(w) cos wx + B(w) sin wx]dw 0 This representation of Theorem. f (x) If a function is called the f (x) Fourier integral of f (x). is a piece-wise continuous function and can be represented by a Fourier integral with A and B, then the value of the Fourier integral at discontinuities equals the average of the left-hand and right-hand limits of f (x). Example. Single Pulse, Sine Integral. Dirichlet's Discontinuous Factor. Gibbs Phenomenon Find the Fourier integral representation of the function ( f (x) = 1, |x| < 1 . 0, |x| > 1 1 Solution. We solve for A(w) A(w) = = = = B(w) B(w): ˆ 1 ∞ f (v) cos wvdv π −∞ ˆ 1 +1 cos wvdv π −1  +1 1 sin wv π w −1 2 sin w . πw and = = = 1 π 1 π 0. ˆ ∞ f (v) sin wvdv −∞ ˆ +1 sin wvdv −1 hence the Fourier integral becomes ˆ f (x) = ∞ [A(w) cos wx + B(w) sin wx]dw  ˆ ∞ 2 sin w cos wx + 0 dw πw 0 ˆ 2 ∞ cos wx sin w dw. π 0 w 0 = = x and variable We do not need to evaluate the innite integral. This is since Note that there are two variables to consider here, the variable w which is being integrated, but there is a way around it. Remark. we already know that the innite integral is equal to f (x), which we already know, hence we just equate them. So looking at the function f (x) (which is dened for all values x between −∞ and +∞) and the innite integral (which is dened for all values from 0 to +∞), we see that we have to break the integral down into intervals. From above, we already know that 2 π ˆ 0 ∞ cos wx sin w dw = f (x), w 2 but f (x) is simply 2 π ˆ ∞ 0 f (x) Adding the value of 2 π Then cos wx sin w dw = f (x) = w ˆ ∞ 0 ˆ ∞ 0 x=0 x = 1, we end up having 1, 0 < x < 1 cos wx sin w dw = 21 , x = 1 w 0, x > 1. π 2, 0 ≤ x < 1 cos wx sin w dw = π4 , x = 1 w 0, x > 1. This integral is known as The case at Dirichlet's discontinuous factor. is of particular interest. If ˆ ∞ 0 which is known as the ( 1, 0 < x < 1 0, x > 1 x = 0, then sin w π dw = , w 2 sine integral Si(u) as u → ∞. Just as in the case of Fourier series, approximations of Fourier integral are obtain by replacing numbers a. ∞ by Hence the integral 2 π ˆ 0 a cos wx sin w dw. w f (x), that at x = a approaches innity. Note the oscillations near the points of discontinuity of −1, +1. We might expect these oscillations to disappear as But this is not true. x = ±1. known as With increasing a, they are shifted closer to the points This unexpected behaviour also happens with Fourier series and is Gibbs phenomenon. 3 Fourier Cosine Integral and Fourier Sine Integral Just as Fourier series simplify if a function is even or odd, so do Fourier integrals. If f (x) is even, then B(w) = 0. cosine integral Then the representation reduces to ˆ Fourier ∞ A(w) cos wxdx f (x) = 0 where A(w) = ˆ 2 π If f (x) is odd, then A(w) = 0. sine integral ∞ f (v) cos wvdv 0 Then the representation reduces to ˆ ∞ f (x) = B(w) sin wxdx 0 where B(w) = 2 π ˆ ∞ f (v) sin wvdv 0 4 Fourier Fourier Cosine and Sine Transforms An integral transform is a transformation in the form of an integral that produces from given functions new functions depending on a dierent variable. One is mainly interested in these transforms because they can be used as tools in solving ODEs, PDEs, and integral equations and can often be of help in handling and applying special functions. Fourier Cosine Transform The Fourier Cosine transform concerns even functions r fˆc (w) = as the ˆ We have ∞ f (x) cos wxdx 0 Fourier Cosine transform and r f (x) = as the 2 π f (x). 2 π ˆ ∞ fˆc (w) cos wxdx 0 inverse Fourier Cosine transform. Fourier Sine Transform The Fourier Cosine transform concerns odd functions r fˆs (w) = as the ˆ We have ∞ f (x) sin wxdx 0 Fourier Sine transform and r f (x) = as the 2 π f (x). 2 π ˆ ∞ fˆs (w) sin wxdx 0 inverse Fourier Sine transform. Remark. • The derivations of these formala can be found in the text book. • Other notations for these transforms are Fc (f ) = fˆc , 5 Fs (f ) = fˆs Example. Find the Fourier cosine and Fourier sine transforms of the function ( f (x) = k, 0, 0<x<a . x>a Solution. r fˆc (w) = r fˆs (w) = Example. Verify that 2 π 2 π ˆ ˆ r a k cos wxdx = 0 r a k sin wxdx = 0 Fc (e−x ) = √2 π 1+w2 . 6 2 k π 2 k π   sin aw w  1 − cos aw w  . ...
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