Chap 13. Fourier Transform

Review : Fourier integral

f(x)를 실수에서 적분한 값이 유한함.

∫−∞∞f(x)dx < ∞

${f(x+)+f(x-)\over 2} = \int_0^\infty A(\omega) \cos{\omega x} + B(\omega) \sin{\omega x} d\omega$

$= \int_0^\infty \bigg(\int_{-\infty}^\infty {1\over \pi} f(v) \cos {\omega v} dv \cdot \cos{\omega x}+ \int_{-\infty}^\infty {1\over \pi} f(v) \sin{\omega v} dv\cdot \sin{\omega x}\bigg) d\omega$

여기서 v, ω 적분은 f(x) piecewise continuous 조건에 의해 Fubini’s theorem 사용가능

${1\over \pi} \int_0^\infty \int_{-\infty}^\infty f(v) \cos{(\omega(x-v))}dv d\omega$

(coscos + sinsin = cos(a-b))

우함수이므로 _ω_도 (−∞, ∞)로 확장

$= {1\over 2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \bigg(f(v) \cos{\big(\omega(x-v)\big)}d\omega\bigg)dv$

한편 같은 꼴의 sin 함수는 기함수이므로 적분값은 0임. 즉 i를 곱하여 더하여도 식의 값은 변하지 않음.

\[= {1\over 2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \bigg(f(v) \cos{\big(\omega(x-v)\big)} + i\cdot f(v) \sin{\big(\omega(x-v)\big)}d\omega\bigg)dv\] \[= {1\over 2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \bigg(f(v) e^{i\omega (x-v)}d\omega\bigg)dv\]

• pf) eiθ = cos θ + i_sin _θ

  f(θ) = cos θ + i_sin _θ(f(0) = 1)

  f′(θ) = −sin θ + i_cos _θ = i(cos θ + i_sin _θ) = i × f(θ)

  \[\int {f'(\theta)\over f(\theta)} d\theta = \int i d\theta \\ = \log \vert f(\theta) \vert = i\theta + C\]

  ∴ f(θ) = eiθ + C, θ → 0, 1 = eC, C = 0

  for Linear ODE, one IVP, identical solution

다시 증명으로 돌아가서, altering ω, v

\[= {1\over 2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \bigg(f(v) e^{i\omega (x-v)} d\omega \bigg)dv \\ = {1\over 2\pi} \int_{-\infty}^\infty \bigg[e^{i\omega x} \int_{-\infty}^\infty f(v) e^{-i\omega v} dv \bigg]d\omega\]

${1\over 2\pi}$를 나눠갖기 위해 $\sqrt{1\over 2\pi}$ 2개로 나누어 분배하자.

\[= \int_{-\infty}^\infty \sqrt{1\over 2\pi} e^{i\omega x} \bigg(\int_{-\infty}^\infty \sqrt{1\over 2\pi} e^{-i\omega v} dv\bigg) d\omega\]

괄호 안의 integral을 f̂(ω), Fourier transform of f(x)라 하자.

\[\hat f(\omega) := \int_{-\infty}^\infty \sqrt{1\over 2\pi} f(v) e^{-i\omega v} dv\]

그러면, 준 식은 간단히 다음과 같이 변환된다.

\[f(x) = \sqrt{1\over 2\pi} \int_{-\infty}^\infty e^{i\omega x} \hat f (\omega ) d\omega\]

Fourier cosine / sine Transform

\[\hat f(\omega) = \sqrt{1\over 2\pi} \int_{-\infty}^\infty f(x) e^{-\omega x} dx \\ = \sqrt{1\over 2\pi} \int_{-\infty}^\infty f(v)\bigg(\cos{\omega x} + i \cdot \sin{\omega x}\bigg)dx\]

f : even funtion,

\[= \sqrt{2\over \pi} \int_0^\infty f(x) \cos{\omega x} dx =: \hat f_c(x)\]

f : odd function,

\[= -i \times \sqrt{2\over \pi} \int_0^\infty f(x) \sin{\omega x} dx =: \hat f_s(x)\]

ex)

f(x) = e−ax(a > 0)

$f’(x) = -ae^{-ax}, \\f’‘(x) = a^2e^{-ax} = a^2 f(x)$

f̂c(f″) = a_2_f̂c(f)

\[\hat f_c(f'') = \sqrt{2\over \pi}\int_0^\infty f''(x)\cos\omega x dx\]

integral by part,

\[= \sqrt{2\over \pi } \bigg[f'(x)\cos\omega x\bigg]_0^\infty - \sqrt{2\over \pi} \int_0^\infty f'(x) \omega \sin\omega xdx\] \[= \sqrt{2\over \pi} \cdot -f'(0) -\sqrt{2\over \pi} \omega \bigg[ f(x) \sin\omega x \bigg]_0^\infty +\sqrt{2\over \pi} \omega \int_0^\infty f(x) -1 \cdot \omega f(x) \cos\omega x dx\] \[= -\sqrt{2\over \pi} f'(0) -\omega ^2 \hat f_c(f)\]

$(a^2+\omega^2) \hat f_c(f) = -f’(0) \times \sqrt{2\over \pi}, \\ \hat f_c(e^{-ax}) = \sqrt{2\over \pi } \bigg({a\over a^2+\omega^2}\bigg)$

Fourier Transform의 성립.

Linearity of F.T

suppose,

∫−∞∞f(x)dx < ∞, ∫−∞∞g(x)dx < ∞

Fourier transform of a ⋅ f(x) + b ⋅ g(x)

\(= {a\over \sqrt{2\pi} }\int_{-\infty}^\infty f(x) e^{-i\omega x} dx + {b\over \sqrt{2\pi} }\int_{-\infty}^\infty g(x) e^{-i\omega x} dx \\ = a\cdot \hat f(\omega) + b \cdot \hat g(\omega)\)Linear

\[\cal{F} (f'(x)) = \sqrt{1\over 2\pi} \int_{-\infty} ^\infty f'(x) e^{-i\omega x} dx = \\ \sqrt{1\over 2\pi} \bigg[f(x) e^{-i\omega x}\bigg]_{-\infty}^\infty + \int_{-\infty}^\infty \sqrt{1\over 2\pi} i\omega f(x) e^{-i\omega x} dx\]

if lim_x_ → ±∞f(x) = 0,

$= 0 + i\omega \cal{F} (f(x))$

$\therefore \cal{F} (f’(x)) = i\omega \cal{F} (f(x))$

Convolution of F.T

• Convolution

(f(x)  g_(x))∫−∞∞f(p)g(x − p)_dp*

결론부터 말하면,

$\cal{F} (f(x)*g(x)) = \sqrt{2\pi} \cal{F} (f(x)) \cal{F} (g(x))$

\[\cal{F^{-1}} \big(\sqrt{2\pi} \hat f(\omega) \hat g(\omega )\big) = \sqrt{1\over 2\pi} \int_{-\infty}^\infty \sqrt{2\pi} \hat f(\omega ) \hat g(\omega) e^{i\omega x} d\omega \\ = \int_{-\infty}^\infty \hat f(\omega) \hat g(\omega) e^{i\omega x}d\omega\]

pf)

\[\cal{F} \big((f*g) (x)\big) = \sqrt{1\over 2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty f(p) g(x-p) dp \cdot e^{-i\omega x} dx\]

let $ q = x-p$ then p = x − q

\[= \sqrt{1\over 2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty f(p) g(q) e^{-i\omega(p+q)} dqdp\]

∵p + q = x

\[= \sqrt{1\over 2\pi} \bigg( \int_{-\infty}^\infty f(p) e^{-i\omega p} dp \bigg)\bigg( \int_{-\infty}^\infty g(q) e^{-i\omega q} dq \bigg) \\ = \cal{F} (f(x)) \sqrt{2\pi} \cal{F} (g(x))\]

• Fourier Transform of xf(x)

\[\cal{F} (xf(x)) = i \times \hat f'(x)\]

pf)

${d\over d\omega} \hat f(\omega) = {d\over d\omega} \bigg(\sqrt{1\over 2\pi} \int_{-\infty}^\infty f(x) e^{-i \omega x} dx \bigg) \\ = \sqrt{1\over 2\pi} \int_{-\infty}^\infty f(x)(-ix) e^{-i\omega x} dx \\= -i \sqrt{1\over 2 \pi } \int_{-\infty} ^\infty xf(x) e^{-i\omega x} dx \\ = -i \times \cal{F} (xf(x))$

eg)

$\cal{F} (xe^{-x}) = i {d\over d\omega} (\hat f(\omega)) \\ = i\omega \cal{F} (e^{-x^2})\times -{1\over 2}$

pf)

lim_x_ → ±∞f(x) = 0

이라 가정하면,

$\cal{F} (e^{-ax^2})$

을 구해보자.

(e−ax_2)′ = −2_axe−ax_2 − 2_axf(x)

$\cal{F} (f’(x)) = -2a\cal{F}(xf(x))$

\[i\omega \cal {F} (f(x)) = -2ai (\hat f(\omega))'\] \[-{2a\over \omega} \hat f'(\omega) = \hat f(\omega)\]

분자 분모로 넘겨서 적분해주면,

\[\int -{\omega \over 2a} = \int {\hat f'(\omega) \over \hat f(\omega)} d\omega\] \[\hat f(\omega) = C e^{-{\omega^2 \over 4a} }, \hat f(0) = C\] \[\hat f(0) = \sqrt{1\over 2\pi} \int_{-\infty}^\infty f(x) 1 dx = \sqrt{1\over 2\pi} \int_{-\infty}^\infty e^{-ax^2} dx = \sqrt{1\over 2\pi} \int_{-\infty}^\infty e^{-t^2} \sqrt{1\over a} dt\]

$let \sqrt{a} x = t$

\[= \sqrt{1\over 2a}\] \[\therefore \hat f(\omega) = \sqrt{1\over 2a} e^{-{\omega^2\over 4a} } = \cal{F} (e^{-ax^2}) \\ \cal{F}(e^{-x^2}) = \sqrt{1\over 2} e^{-{\omega^2\over 4} }\]

f̂(ω) = f(ω)

if f(x) is even,

$\hat f_c (\omega) = \hat f (\omega) \\ = \sqrt{2\over \pi} \int_0^\infty f(x) \cos{\omega x} dx$

eg)

$\int_0^\infty e^{-x^2} \cos \omega x dx = {\sqrt{\pi}\over 2} e^{-{\omega^2 \over 4} } \\ \omega = 2b \\ \int_0^\infty e^{-s^2} \cos {(2bs)} ds = e^{-b^2} \times {\sqrt{\pi}\over 2}$