Fourier Transform

Fact Instead of discrete amplitude $C_n$ for each frequency $k_n$ at intervals of $k_0$, we have a continuous function $F(K)$ giving the amplitude of frequency $k$ over infinitesimal frequency intervals $\text{d}k$

Def Fourier Transform

$$\mathcal{F}[f(x)]=F(k)={\int }_{-\infty }^{\infty }f(x)e^{-ikx}dx$$

And the inverse Fourier transform is given by:

$${\mathcal{F}}^{-1}[F(k)]=f(x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }F(k)e^{ikx}dk$$

Prop $\mathcal{F}[cf(x)]=c\mathcal{F}[f(x)]$

Tophat, DiracDelta and Shah Function

Def Tophat Function Define the tophat function as:

$$\text{п}(\frac{x}{2b})=\{\begin{array}{l}1\text{if }x\in (-b,b)\\ 0\text{otherwise}\end{array}$$

Def DiracDelta Function Delta function $\delta (x)$ is an even function that is zero everywhere except at the origin and has area one. That is:

$${\int }_{-\infty }^{\infty }\delta (x)\text{d}x=1$$

Prop By the property of delta function, we can get that:

$$\begin{array}{rlr}{\int }_{-\infty }^{\infty }\delta (x)f(x)dx& ={\int }_{-\infty }^{\infty }\delta (x)f(0)dx& \\ & =f(0){\int }_{-\infty }^{\infty }\delta (x)dx& \\ & =f(0)& \end{array}$$

Generally, we can offset the delta function and get the following:

$${\int }_{-\infty }^{\infty }\delta \left(x-x_0\right)f(x)dx=f\left(x_0\right)$$

Prop

• $\delta (ax)=\frac{\delta (x)}{|a|}$
• ${\int }_{-\infty }^{\infty }f(x){\delta }^{\prime }(x)dx=-f^{\prime }(0)$
• The Fourier Transform of the delta function is 1: $\mathcal{F}[\delta (t)]=1$
• The Inverse Fourier Transform of the delta function is $\frac{1}{2\pi }$: ${\mathcal{F}}^{-1}[\delta (t)]=\frac{1}{2\pi }$
• $\mathcal{F}[1]=2\pi \cdot \delta (t)$

Def Shah Function The Shah function is defined by an infinite series of delta functions:

$${⨿}_T(t)=\sum _{n=-\infty }^{\infty }\delta (t-nT)$$

Prop The Fourier Transform of a Shah is a Shah.

$$\mathcal{F}\left[{⨿}_T(t)\right]=\frac{2\pi }{T}{⨿}_{\frac{2\pi }{T}}(\omega )$$

Linearity, Symmetry and Duality

Prop The Fourier transform is linear:

$$\mathcal{F}[af(t)+bg(t)]=a\mathcal{F}[f(t)]+b\mathcal{F}[g(t)]$$

Prop The scaling of Fourier Transform:

$$\mathcal{F}[f(ax)]=\frac{1}{|a|}F\left(\frac{k}{a}\right)$$

Corollary $\mathcal{F}[f(-x)]=F(-k)$

Prop Symmetries

f(t) is real	Real part of F(\omega) is even; Imaginary part is odd
f(t) is real and even	F(\omega) is real and even
f(t) is real and odd	F(\omega) is imaginary and odd

It is equivalent to say that:

$$F(\omega )=\tilde{F}(-\omega )$$

where $\tilde{F}$ is the complex conjugate of $F$. For a real-valued function $f(t)$, the negative frequency components of the Fourier transform must be complex conjugate of the positive frequency components.

Prop Duality If $\mathcal{F}[f(t)]=F(\omega )$ then $\mathcal{F}[F(t)]=2\pi f(-\omega )$

Prop Coordinate Shift

$$\mathcal{F}[f(x-x_0)]=e^{-ik{x}_0}F(k),\mathcal{F}[e^{i{k}_{0}x}f(x)]=F(k-k_0)$$

Convolution

Recall that convolution is the integral of the product of two functions after one is reversed and shifted:

The convolution of two functions $f,g\in L^1({\mathbb{R}}^d)$ is defined by $(f\ast g)(x):={\int }_{{\mathbb{R}}^d}f(y)g(x-y)\text{dd}y.$

Link to original

It has the following properties:

Circular transclusion detected: Mathematics/Measure-Theory/The-Space-of-Integrable-Functions

Thrm Convolution Theorem The convolution is function multiplication in Fourier space: $\mathcal{F}[f(x)\ast g(x)]=F(k)G(k),\mathcal{F}[f(x)g(x)]=\frac{1}{2\pi }F(k)\ast G(k)$where $\ast$ is the convolution operator.

Derivatives of Fourier

Prop If $\mathcal{F}[f(t)]=F(\omega )$ and the first derivative of $f$ exists, then $\mathcal{F}[f^{\prime }(t)]=i\omega F(\omega )$. More generally, $\mathcal{F}[f^{(n)}(t)]=(i\omega {)}^{n}F(\omega )$ Proof

$$\begin{array}{rl}& f(t)={\mathcal{F}}^{-1}[F(\omega )]=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }F(\omega )e^{i\omega t}d\omega \\ ⟹& f^{\prime }\left(t\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }i\omega F(\omega )e^{i\omega t}d\omega ={\mathcal{F}}^{-1}[i\omega F(\omega )]\\ ⟹& \mathcal{F}[f^{\prime }(t)]=i\omega F(\omega )\end{array}$$

Energy and Parseval’s Theorem

Thrm Parseval’s Theorem If $v(t)$ is some finite signal, then the energy will be finite. Parseval’s theorem says:

$${\int }_{-\infty }^{\infty }|f(t){|}^2\mathrm{d}t=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }|F(\omega ){|}^2\mathrm{d}\omega$$