10.8. Fourier Integrals - Application of Fourier series to nonperiodic function ∞ Use Fourier series of a function fL with period L (L ) Ex. 1) Square wave 0 if − L < x < −1 = − < < fL (x) 1 if 1 x 1 (2L > 2) 0 if 1< x < L − < < = = 1 if 1 x 1 f (x) lim fL (x) L→∞ 0 other regions 1 1 π π = 1 = 1 = 1 n x = 2 sin(n / L) Amplitude spectrum a 0 dx , a n cos dx 2L −1 L L −1 L L nπ/ L From Fourier Series to the Fourier Integral ∞ nπ Fourier series of f(x) (period 2L): f (x) = a + ()a cos w x + b sin w x , w = L 0 n n n n n L L ∞, f(x) ? n=1 ∞ = 1 L + 1 L + L fL (x) fL (v)dv cos w n x fL (v)cos w n vdv sin w n x fL (v)sin w n vdv −L −L −L 2L L n=1 π 1 Δw Δw = w + − w = = n 1 n L L π L ∞ ()Δ cos w n x w fL (v)cos w n vdv = 1 L + 1 −L fL (x) fL (v)dv −L π L 2L n=1 + ()Δ sin w n x w fL (v)sin w n vdv −L ∞ = If f ( x ) lim f L ( x ) is absolutely integrable, f ( x ) dx exists L→∞ −∞ 1 ∞∞∞ L ∞, then f (x) = cos wx f (v)cos wvdv + sin wx f (v)sin wvdv dw π 0 −∞ −∞ 1 ∞ 1 ∞ A(w) = f (v)cos wvdv, B(w) = f (v)sin wvdv π −∞ π −∞ ∞ f (x) = []A(w)cos wx + B(w)sin wx dw Fourier Integral 0 Theorem 1: Fourier Integral - f(x): piecewise continuous f(x) can be represented right-hand / left-hand derivatives exist ∞ by Fourier integral. integral f ( x ) dx exists −∞ Applications of the Fourier Integral - Solving differential equations (see 11.6) & integration, … Ex. 2) Single pulse, sine integral f (x) =1 if x <1, 0 if x >1 1 1 2sin w 1 1 A(w) = cos wvdv = , B(w) = sin wvdv = 0 π −1 πw π −1 2 ∞ cos wxsin w f (x) = dw π 0 w Fluctuation caused by the Si func π/ 2 if 0 ≤ x <1 ∞ cos wxsin w dw = π/ 4 if x =1 0 w 0 if x >1 Dirichlet’s discontinuous factor ∞ sin w π u sin w dw = at x = 0 Sine integral: Si(u) = dw 0 w 2 0 w Fourier Cosine and Sine Integrals 2 ∞ Fourier cosine integral: For even function f(x): B(w)=0, A(w) = f (v)cos wvdv ∞ π 0 f (x) = A(w)cos wx dw 0 2 ∞ Fourier sine integral: For odd function f(x): A(w)=0, B (w) = f (v)sin wvdv ∞ π 0 f (x) = B(w)sin wx dw 0 Evaluation of Integrals - Fourier integrals for evaluating integrals Ex. 3) Laplace integrals f (x) = e−kx (x,k > 0) ∞ 2 − 2k / π (a) Fourier cosine integral: A(w) = e kv cos wvdv = π 0 k 2 + w 2 ∞ ∞ − 2k cos wx cos wx π − f (x) = e kx = dw dw = e kx π 0 k 2 + w 2 0 k 2 + w 2 2k ∞ 2 − 2w / π (b) Fourier sine integral: B(w) = e kv sin wvdv = π 0 k 2 + w 2 ∞ ∞ − 2 w sin wx w sin wx π − f (x) = e kx = dw dw = e kx π 0 k 2 + w 2 0 k 2 + w 2 2 10.9. Fourier Cosine and Sine Transforms - Integral transforms: useful tools in solving ODEs, PDEs, integral equations, and special functions … Laplace transforms Fourier transforms from Fourier integral expressions - Fourier cosine transforms, Fourier sine transforms (for real…) Fourier transforms (for complex… ) Fourier Cosine Transforms ∞ Fourier cosine integral for even function f(x): f (x) = A(w)cos wx dw 0 ∞ = 2 = 2 A(w) f (v)cos wvdv, A(w) FC (w) π 0 π ∞ = 2 → FC (w) f (x)cos wxdx : f (x) FC (w) Fourier cosine transform of f(x) π 0 ∞ = 2 → Inverse Fourier cosine transform f (x) FC (w)cos wxdw : FC (w) f (x) π 0 of FC(x) Fourier Sine Transforms ∞ Fourier sine integral for even function f(x): f (x) = B(w)sin wx dw 0 ∞ = 2 = 2 B(w) f (v)sin wvdv, B(w) FS (w) π 0 π ∞ = 2 → FS (w) f (x)sin wxdx : f (x) FS (w) Fourier sine transform of f(x) π 0 ∞ = 2 → Inverse Fourier sine transform f (x) FS (w)sin wxdw : FS (w) f (x) π 0 of FC(x) Ex. 1) Fourier cosine and sine transforms f (x) = k, 0 < x < a; 0, x > a See Table I & II (in 10.11) a a − = 2 = 2 sin aw = 2 = 2 1 cosaw FC (w) k cos wxdx k , FS (w) k sin wxdx k π 0 π w π 0 π w Ex. 2) Fourier cosine transform of the exponential function: f(x) = e-x ∞ 2 π = 2 −x = FC (w) e cos wxdx π 0 1+ w 2 ∞ Linearity, Transforms of Derivatives ℑ + = 2 ()+ C (af bg) af bg cos wxdx ℑ = ℑ = π 0 C (f ) FC , S (f ) FS ℑ + = ℑ + ℑ ℑ + = ℑ + ℑ C (af bg) a C (f ) b C (g); S (af bg) a S (f ) b S (g) Theorem 1: Cosine and sine transforms of derivatives f(x): continuous & absolutely integrable, f(x) 0 as x ∞ f’(x) piecewise continuous 2 ℑ (f '(x)) = wℑ (f (x)) − f (0); ℑ (f '(x)) = −wℑ (f (x)) C S π S C 2 2 ℑ (f ''(x)) = −w 2ℑ (f (x)) − f '(0); ℑ (f ''(x)) = −w 2ℑ (f (x)) + wf (0) C C π S S π ∞ ∞ ℑ = 2 = 2 ∞ + = C (f '(x)) f '(x)cos wxdx f (x)cos wx w f (x)sin wxdx π 0 π 0 0 2 ℑ (f ''(x)) = wℑ (f '(x)) − f '(0) ← ℑ (f '(x)) = −wℑ (f (x)) (See section 11.6) C S π S C Ex. 3) Fourier cosine transforms of exp. Function: f(x)=e-ax (a > 0) ℑ = 2ℑ = − 2ℑ + 2 ℑ = 2 a C (f '') a C (f ) w C (f ) a C (f ) π π a 2 + w 2 10.10. Fourier Transform - from Fourier integral in complex form Complex Form of the Fourier Integral ∞ f (x) = []A(w)cos wx + B(w)sin wx dw 0 1 ∞ 1 ∞ A(w) = f (v)cos wvdv, B(w) = f (v)sin wvdv π −∞ π −∞ 1 ∞∞ f (x) = f (v)()cos wv cos wx + sin wvsin wx dvdw π 0 −∞ 1 ∞∞ even function of w ! = f (v)cos(wx − wv)dvdw π 0 −∞ 1 ∞ ∞ = f (v)cos(wx − wv)dvdw 2π −∞ −∞ 1 ∞ ∞ f (v)sin(wx − wv)dvdw = 0 2π −∞ −∞ Use Euler’s formula: eix = cos x + isin x f (v)[]cos(wx − wv) + isin(wx − wv) = f (v)ei(wx−wv) ∞ ∞ 1 − f (x) = f (v)ei(wx wv)dvdw 2π −∞ −∞ Fourier Transform and Its Inverse ∞ ∞ 1 1 − f (x) = f (v)e iwvdv eiwxdw −∞ −∞ 2π 2π ∞ 1 − Fourier transform of f(x): F(w) = f (x)e iwxdx 2π −∞ 1 ∞ Inverse Fourier transform of F(w): f (x) = F(w)eiwxdw 2π −∞ Ex. 1) Fourier transform of f(x)=k (0<x<a) and f(x)=0 otherwise −aw 1 a − k(1− e ) F(w) = ke iwxdx = 2π 0 iw 2π 2 Ex. 2) Fourier transform of f ( x ) = e − ax (a>0) Linearity. Fourier Transform of Derivatives Theorem 1: Linearity ∞ ℑ + = 1 []+ −iwx C (af bg) af (x) bg(x) e dx ℑ(f (x)) = F(w) 2π −∞ ℑ(af + bg) = aℑ(f ) + bℑ(g) Theorem 2: Fourier transform of the derivative of f(x) f(x): continuous, f(x) 0 as IxI ∞, f’(x) absolutely integrable ℑ(f '(x)) = iw ℑ(f (x)) ℑ(f ''(x)) = −w 2 ℑ(f (x)) ∞ ∞ 1 − 1 − ∞ − ℑ(f ') = f '(x)e iwxdx = fe iwx − (−iw) f (x)e iwxdx 2π −∞ 2π −∞ −∞ ℑ(f '') = iwℑ(f ') = −i2w 2ℑ(f ) = w 2ℑ(f ) − 2 Ex. 3) Fourier transform of f (x) = xe x − 2 1 − 2 1 − 2 iw − 2 ℑ(xe x ) = − ℑ()(e x )' = − iwℑ()e x = − e w / 4 2 2 2 2 from Ex. 2 Convolution ∞ ∞ - Convolution of f * g: h(x) = (f *g)(x) = f (p)g(x − p)dp = f (x − p)g(p)dp −∞ −∞ Theorem 3: Convolution theorem ℑ(f *g) = 2π ℑ(f )ℑ(g) ∞ −∞ 1 − ℑ(f *g) = f (p)g(x − p)e iwxdp dx 2π −∞ −∞ ∞ −∞ 1 − + Interchange of integration order = f (p)g(q)e iw(p q)dq dp 2π −∞ −∞ x-p=q ∞ −∞ 1 − − = f (p) iwp dp g(q)e iwqdp = 2πℑ(f )ℑ(g) 2π −∞ −∞ ∞ Inverse Fourier transform: ( f * g )( x ) = F ( w ) G ( w ) e iwx dw (see 11.6) −∞.