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Integral Transform Method CHAPTER 15
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Ch15_2 Contents 15.1 Error Function 15.1 Error Function 15.2Applications of the Laplace Transform 15.2Applications of the Laplace Transform 15.3 Fourier Integral 15.3 Fourier Integral 15.4 Fourier Transforms 15.4 Fourier Transforms 15.5 Fast Fourier Transform 15.5 Fast Fourier Transform
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Ch15_3 15.1 Error Function Properties and Graphs Recalling from Sec 2.3, the error function erf(x) and complementary error function erfc(x) are, respectively (1) With the aid of polar coordinates, we have
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Ch15_4 Moreover This shows erf(x) + erfc(x) = 1(2) The graphs are shown in Fig15.1.
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Ch15_5 Fig 15.1
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Ch15_6 Table 15.1 Table 15.1 shows some useful Laplace transforms. f(t), a > 0 L {f(t)} = F(s) 1. 2. 3.
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Ch15_7 Table 15.1 f(t), a > 0 L {f(t)} = F(s) 4. 5. 6.
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Ch15_8 15.2 Applications of the Laplace Transform Transform of Partial Derivatives The Laplace transform of u(x, t) with respect to t is
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Ch15_9 Also we have that is (3)
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Ch15_10 Example 1 Find the Laplace transform of Solution From (2) and (3),
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Ch15_11 Example 2 Solution We have (5) The Laplace transforms of the boundary condition: (6)
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Ch15_12 Example 2 (2) Since (5) is defined over a finite interval, its complementary function is Let the particular solution be
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Ch15_13 Example 2 (3) But the conditions U(0, s) = 0 and U(1, s) = 0 imply c 1 = 0 and c 2 = 0. We conclude that
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Ch15_14 Example 3 A very long string is initially at rest on the nonnegative x-axis. The string is secured at x = 0 and its distant right end slides down a frictionless vertical support. The string is set in motion by letting it fall under its own weight. Find the displacement u(x, t). Solution The describing equations are
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Ch15_15 Example 3 (2) The second boundary condition indicates that the string is horizontal at a great distance from the left end. From (2) and (3),
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Ch15_16 Example 3 (3) Now by the second translation theorem we have where U s denotes the unit step function. Finally,
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Ch15_17 Fig 15.2
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Ch15_18 Example 4 Solution From (1) and (3)
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Ch15_19 Example 4 (2) The general solution of (7) is
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Ch15_20 Example 4 (3) and using
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Ch15_21 Example 4 (4) From Table 15.1 we have
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Ch15_22 Example 4 (5) The solution (9) can be written in terms of the error function using erfc(x) = 1 – erf(x):
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Ch15_23 Fig 15.3(a)
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Ch15_24 Fig 15.3(b)~(e)
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Ch15_25 15.3 Fourier Integral Fourier Series to Fourier Integral Recall that (1)
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Ch15_26 If we let n = n /p, = n+1 − n = /p, then (1) becomes (2)
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Ch15_27 Letting p , 0, then the first term in (2) is zero and the limit of the sum becomes (3) The result in (3) is called the Fourier Integral of f on (− , ).
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Ch15_28 The Fourier integral of a function f defined in the interval ( - , ) is given by (4) where (5) (6) DEFINITION 15.1 Fourier Integral
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Ch15_29 Let f and f ’ be piecewise continuous on every finite interval, and let f be absolutely integrable on ( - , ). Then the Fourier integral of f on the interval converges to f(x) at a point of continuity. At a point of discontinuity, the Fourier integral will converge to the average Where f(x+) and f(x - ) denote the limit of f at x from the right and from the left, respectively. THEOREM15.1 Conditions for Convergence
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Ch15_30 Example 1 Find the Fourier Integral of the functions Solution
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Ch15_31 Example 1 (2)
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Ch15_32 Cosine and Sine Integrals When f is even on (− , ), B( ) = 0,
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Ch15_33 (i)The Fourier integral of an even function on the interval ( - , ) is the cosine integral (8) where (9) DEFINITION 15.2 Fourier Cosine and Sine Integrals
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Ch15_34 (continued) (ii)The Fourier integral of an odd function on the interval ( - , ) is the sine integral (8) where (9) DEFINITION 15.2 Fourier Cosine and Sine Integrals
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Ch15_35 Example 2 Find the Fourier Integral of the functions Solution See Fig 15.7. We fund that the function f is even.
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Ch15_36 Fig 15.7
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Ch15_37 Example 3 Represent f(x) = e -x, x > 0 (a) by a cosine integral, (b) by a sine integral. Solution The graph is shown in Fig 15.8.
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Ch15_38 Example 3 (2)
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Ch15_39 Fig 15.9
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Ch15_40 Complex Form The Fourier also possesses an equivalent complex form or exponential form.
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Ch15_41
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Ch15_42 Note: (15) is from the fact that the integrand is even of . In (16) we have simply added zero to the integrand, because the integrand is odd of . Finally we have
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Ch15_43 15.4 Fourier Transforms (i)Fourier transform: (5) Inverse Fourier transform: (6) DEFINITION 15.3 Fourier Transform Pairs
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Ch15_44 (continued) (ii)Fourier sine transform: (5) Inverse Fourier sine transform: (6) DEFINITION 15.3 Fourier Transform Pairs
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Ch15_45 (continued) (iii)Fourier cosine transform: (5) Inverse Fourier cosine transform: (6) DEFINITION 15.3 Fourier Transform Pairs
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Ch15_46 Operational Properties Suppose that f is continuous and absolutely integrable on (− , ) and f is piecewise continuous on every finite interval. If f(x) 0 as x , then
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Ch15_47 that is, (13) Similarly (14)
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Ch15_48 Example 1 Solve the heat equations Solution
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Ch15_49 Example 1 (2) From the inverse integral (6) that
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Ch15_50 Example 2 Solve for u(x, y). Solution The domain of y and the prescribed condition at y = 0 indicates that the Fourier cosine transform is suitable.
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Ch15_51 Example 2 (2) The general solution is
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Ch15_52 Example 2 (3)
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Ch15_53 15.5 Fast Fourier Transform Discrete Fourier Transform Consider a function f defined on [0, 2p]. The complex Fourier series is (1) where = 2 /p. When we sample this function with a sampling period T, the discrete signal is (2)
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Ch15_54 Thus the Fourier transform of (2) can then be (3) and is the same as (4) which is called the discrete Fourier transform (DFT).
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Ch15_55 Now consider the values of f(x) at N equally spaced points, x = nT, n = 1, 2, …, N – 1 in the interval [0, 2 ], that is f 0, f 1, f 2, …, f N-1. The finite discrete Fourier series f(x) using the N terms gives us
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Ch15_56 If we let
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Ch15_57 The matrix form is and the N N matrix is denoted by F N.
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Ch15_58 Please verify that
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Ch15_59 Example 1 Let N = 4 so that 4 = e i /2 = i. The matrix F 4 is
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Ch15_60 Fig 15.15
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