34 J−1 2 DST: a = U sin(πmj=J); m = 1; 2; : : : ; J − 1 (5) m J j Xj=1 J−1 Inverse DST: Uj = am sin(πmj=J); j = 1; 2; : : : ; J − 1 (6) m=1 X Fourier Transforms One may think of (5) as arising from (1) by replacing the integral by a discrete (trapezoid) approximation. Note that if one included the cases m = 0 and m = J in (5) one would find that a0 = aJ = 0. Hence there is not reason to include these. Similarly, U0 and UJ would necessarily be zero by (6). Hence it is not possible to recover non-zero values of U0 and UJ via the inverse DST and these are excluded. In this chapter I provide a summary of various transform pairs. Notation varies in the literature and I present here a self-consistent treatment. The proof that the above pairs are inverses is easy given the orthogonality relation: J−1 πmj πm0j J sin( ) sin( ) = δ 0 (7) J J 2 mm Xj=1 Sine Transform πmj Note that m and j play exactly the same role in sin( J ) so that the discrete analog of the completeness relation (4) holds as a direct consequence of (7) by relabelling j and m: Given a function u(x) on the interval [0; `], the sine transform and its inverse are given by: J−1 πmj πmj0 J sin( ) sin( ) = δjj0 (8) ` J J 2 2 m=1 Sine Transform: am = u(x) sin(πmx=`)dx (1) X ` 0 Equations (7) and (8) are the same thing. 1Z Inverse Sine Transform: u(x) = am sin(πmx=`) (2) m=1 X Cosine Transform To prove that these are indeed inverses of one another, one uses the orthogonality relation for sine functions. Given a function u(x) on the interval [0; `], the cosine transform and its inverse are given by: ` ` 0 ` 2 sin(πmx=`) sin(πm x=`)dx = δmm0 (3) Cosine Transform: am = u(x) cos(πmx=`)dx (9) 0 2 ` 0 Z Z 1 a Inverse Cosine Transform: u(x) = 0 + a cos(πmx=`) (10) which is easy to prove by direct integration. One can also use the completeness relation: 2 m m=1 1 X 1 0 ` 0 = wmam cos(πmx=`) (11) sin(πmx=`) sin(πmx =`) = δ(x − x ) (4) 2 m=0 m=1 X X where the weight function wm is given by: 1 which is not easy to prove. 2 if j = 0 wj = (12) (1 if j > 0 The proof that these are indeed inverses uses the orthogonality relation for cosine functions. Discrete Sine Transform ` 0 ` cos(πmx=`) cos(πm x=`)dx = δ 0 (13) 2 mm Z0 There is a discrete form of the sine transformation, the discrete sine transform (DST). In this case we are which again is easy to prove by direct integration. These is also a difficult to prove completeness relation: 1 given a set of value Uj for j = 1; : : : ; J − 1 generally thought of as a discrete sampling of a function 0 ` 0 u(x) on [0; `], i.e. U = u(jh), which h = `=J. For the DST the function is sampled only on the interior w cos(πmx=`) cos(πmx =`) = δ(x − x ) (14) j m 2 m=0 j = 1; 2; : : : ; J − 1. Then the discrete sine transform (DST) and its inverse are given by: X 33 Numerical methods for PDES Copyright (C) 2002-2005 Dwight Barkley 35 36 Discrete Cosine Transform One can also use the completeness relation: 1 − 0 0 ei2πmx=`e i2πmx =` = `δ(x − x ) (23) For the discrete cosine transform (DST) we consider a set of value Uj for j = 0; : : : ; J again generally m=−∞ thought of as a discrete sampling of a function u(x) on [0; `]. This time however we sample the function at X the end points j = 0 and j = J. Then the discrete cosine transform (DST) and its inverse are given by: which is not easy to prove. J 2 DCT: a = w U cos(πmj=J) (15) In the case which u(x) is a real-valued function, the am will still be complex, but the following symmetry m J j j j=0 holds: X ∗ J a−m = am (24) Inverse DCT: Uj = wmam cos(πmj=J) (16) m=0 where ∗ denotes complex conjugation. Proof: X (17) ∗ ` ` ` ∗ 1 −i2πmx=` 1 ∗ i2πmx=` 1 −i2π(−m)x=` am = u(x)e dx = u (x)e dx = u(x)e dx = a−m where now the weight function is: ` ` ` Z0 Z0 Z0 1 2 if j = 0; J; wj = (18) (1 if 0 < j < J Using this fact, it is possible to re-write the FT pair in such a way that all quantities are real: 1 A One may think of (15) as arising from (9) by replacing the integral by a trapezoid approximation. The actual Inverse Fourier Transform: u(x) = 0 + (A cos(2πmx=`) + B sin(2πmx=`)) 2 m m proof that the above pairs are inverses comes from the following identity. m=1 X J πm0j πmj J with: w w cos( ) cos( ) = δ 0 (19) j m J J 2 mm j=0 2 ` X Fourier Transform: A = 2ar = u(x) cos(2πmx=`)dx m m ` Note that as the the DST, m and j play exactly the same role in this equation so that the discrete analog of Z0 2 ` the completeness relation (14) holds as a direct consequence of (19) by relabelling j and m. B = −2ai = u(x) sin(2πmx=`)dx m m ` Z0 r i r i where am and am are the real and imaginary parts of am: am = am + iam. Fourier Transform Hence the FT of a real-valued function is equivalent to taking both sine and cosine transforms on the interval [0; `] but with argument 2πmx=` rather than πmx=`. Given a function u(x) on the interval [0; `], the Fourier transform (FT) and its inverse are given by: The details are left for the reader, but the first few lines of the derivation are: ` 1 −i2πmx=` Fourier Transform: am = u(x)e dx (20) 1 −∞ ` 0 i2πmx=` i2πmx=` Z1 u(x) = a0 + ame + ame Inverse Fourier Transform: i2πmx=` m=1 m=−1 u(x) = ame (21) X1 X1 −∞ m= i2πmx=` −i2πmx=` X = a0 + ame + a−me m=1 m=1 In this case the function u(x) is often allowed to be complex, but we shall consider only the case of real X1 X u(x). i2πmx=` ∗ −i2πmx=` = a0 + ame + ame m=1 One has the following orthogonality relation for complex exponentials: X ` i2πmx=` −i2πm0x=` e e dx = `δmm0 (22) Discrete Fourier Transform Z0 which as usual is easy to prove by direct integration. For the discrete Fourier transform (DFT) we again consider a set of value Uj for j = 0; : : : ; J thought of as ` ∗ 0 i2πmx=` (Notice that here we have 0 φm(x)φm0 (x)dx = `δmm , where φm(x) = e .) a discrete sampling of a function u(x) on [0; `] including end points j = 0 and j = J. In this case we are Numerical methods for PDESR Copyright (C) 2002-2005 Dwight Barkley Numerical methods for PDES Copyright (C) 2002-2005 Dwight Barkley 37 38 going to consider several forms for the DFT. The first is given by: This is a non-standard form for the DFT. I give it here because it is the form one might guess from the continuous FT (20)-(21). J−1 1 −i2πmj=J DFT: am = Uje ; m = −J=2 + 1; −J=2 + 2; : : : ; J=2; (25) J Probably the most common form for the DFT is given by using the periodicity in am to shift the sum in (26) j=0 X and obtain: J=2 i2πmj=J J−1 Inverse DFT: Uj = ame j = 0; 1; : : : ; J − 1; (26) 1 −i2πmj=J − DFT: am = Uje ; m = 0; 1; : : : ; J − 1; (32) m=XJ=2+1 J Xj=0 where we have assumed J is even. This is the form we shall consider most frequently in this course. J−1 i2πmj=J Inverse DFT: Uj = ame j = 0; 1; : : : ; J − 1; (33) m=0 The proof that the above pairs are inverses comes from the orthogonality of the complex exponentials: X This form has the advantage that both sums are over the same range. However, it has the disadvantages that J−1 −i2πmj=J i2πm0j=J (1) it is not a natural choice given the continuous FT, (2) more importantly, we will consider real Uj and this e e = Jδmm0 (27) form does not make it as obvious that Uj obtained by the inverse transformation will be real. Xj=0 As with the other discrete transforms, this is also equivalent to the completeness relation in the continuous Most libraries that perform discrete Fourier transforms say they compute transforms of type (32)-(33). They case. are all equivalent of course. ∗ Other standard and non-standard forms of the DFT can be derived by noting that if (25) is used to define am Finally, we consider the sine/cosine forms of the DFT. As in the continuous case: Uj real implies a−m = am. for all integer m, then the am are periodic with period J. Similarly, (26) gives periodic Uj with period J. Then (26) gives: J=2 am+kJ = am and Uj+kJ = Uj i2πmj=J Uj = ame (34) − for all integer k. Proof: m=XJ=2+1 J=2−1 −(J=2−1) J−1 i2πmj=J i2πmj=J i2π(J=2)j=J 1 − = a0 + ame + ame + a e (35) a = U e i2π(m+kJ)j=J J=2 m+kJ J j m=1 m=−1 j=0 X X X J=2−1 J−1 i2πmj=J ∗ −i2πmj=J iπj 1 − − = a + a e + a e + a e (36) = U e i2πmj=J e i2πkj 0 m m J=2 J j m=1 j=0 X X J=2−1 J−1 r i 1 − = a + 2a cos(2πmj=J) + (−2a ) sin(2πmj=J) + a cos(πmj=(J=2)) (37) = U e i2πmj=J = a 0 m m J=2 J j m m=1 j=0 X X J=2−1 A AJ=2 = 0 + (A cos(2πmj=J) + B sin(2πmj=J)) + cos(πmj=(J=2)) (38) 2 m m 2 m=1 Using this, the following is an equivalent form for the DFT and its inverse: X So J 1 −i2πmj=J J=2 DFT: am = wjUje ; m = −J=2; −J=2 + 1; : : : ; J=2; (28) J Inverse Fourier Transform: j=0 Uj = (wmAm cos(2πmj=J) + Bm sin(2πmj=J)) X m=0 J=2 X Inverse DFT: U = w^ a ei2πmj=J j = 0; 1; : : : ; J − 1; where j m m (29) 1 m=−J=2 2 if m = 0; J=2; X wm = (39) (1 if 0 < m < J=2 where: 1 and Am and Bm are given by: 2 if j = 0; J; wj = (30) J (1 if 0 < j < J 2 Fourier Transform: A = 2ar = w U cos(2πmj=J) m m J j j and j=0 1 if m = −J=2; J=2; X w^ = 2 J−1 m (31) i 2 (1 if −J=2 < m < J=2 Bm = −2a = Uj sin(2πmj=J) m J j=1 In effect, since U0 = UJ , U0 in (25) can been replaced by (U0 + UJ )=2.