Integral Transform Method CHAPTER 15

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

Ch15_ 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

Ch15_4 Moreover This shows erf(x) + erfc(x) = 1(2) The graphs are shown in Fig15.1.

Ch15_5 Fig 15.1

Ch15_6 Table 15.1  Table 15.1 shows some useful Laplace transforms. f(t), a > 0 L {f(t)} = F(s)

Ch15_7 Table 15.1 f(t), a > 0 L {f(t)} = F(s)

Ch15_ Applications of the Laplace Transform  Transform of Partial Derivatives The Laplace transform of u(x, t) with respect to t is

Ch15_9  Also we have that is (3)

Ch15_10 Example 1 Find the Laplace transform of Solution From (2) and (3),

Ch15_11 Example 2 Solution We have (5) The Laplace transforms of the boundary condition: (6)

Ch15_12 Example 2 (2) Since (5) is defined over a finite interval, its complementary function is Let the particular solution be

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

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

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),

Ch15_16 Example 3 (3) Now by the second translation theorem we have where U s denotes the unit step function. Finally,

Ch15_17 Fig 15.2

Ch15_18 Example 4 Solution From (1) and (3)

Ch15_19 Example 4 (2) The general solution of (7) is

Ch15_20 Example 4 (3) and using

Ch15_21 Example 4 (4) From Table 15.1 we have

Ch15_22 Example 4 (5) The solution (9) can be written in terms of the error function using erfc(x) = 1 – erf(x):

Ch15_23 Fig 15.3(a)

Ch15_24 Fig 15.3(b)~(e)

Ch15_ Fourier Integral  Fourier Series to Fourier Integral Recall that (1)

Ch15_26  If we let  n = n  /p,  =  n+1 −  n =  /p, then (1) becomes (2)

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 (− ,  ).

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

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

Ch15_30 Example 1 Find the Fourier Integral of the functions Solution

Ch15_31 Example 1 (2)

Ch15_32 Cosine and Sine Integrals  When f is even on (− ,  ), B(  ) = 0,

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

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

Ch15_35 Example 2 Find the Fourier Integral of the functions Solution See Fig We fund that the function f is even.

Ch15_36 Fig 15.7

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.

Ch15_38 Example 3 (2)

Ch15_39 Fig 15.9

Ch15_40 Complex Form  The Fourier also possesses an equivalent complex form or exponential form.

Ch15_41

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

Ch15_ Fourier Transforms (i)Fourier transform: (5) Inverse Fourier transform: (6) DEFINITION 15.3 Fourier Transform Pairs

Ch15_44 (continued) (ii)Fourier sine transform: (5) Inverse Fourier sine transform: (6) DEFINITION 15.3 Fourier Transform Pairs

Ch15_45 (continued) (iii)Fourier cosine transform: (5) Inverse Fourier cosine transform: (6) DEFINITION 15.3 Fourier Transform Pairs

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

Ch15_47 that is, (13) Similarly (14)

Ch15_48 Example 1 Solve the heat equations Solution

Ch15_49 Example 1 (2) From the inverse integral (6) that

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.

Ch15_51 Example 2 (2) The general solution is

Ch15_52 Example 2 (3)

Ch15_ 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)

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).

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

Ch15_56  If we let

Ch15_57 The matrix form is and the N  N matrix is denoted by F N.

Ch15_58 Please verify that

Ch15_59 Example 1 Let N = 4 so that  4 = e i  /2 = i. The matrix F 4 is

Ch15_60 Fig 15.15