Generalized verison of Fourier series for non-periodic functions. Transforms a time-domain function into frequency-domain representation.
Fourier transforms are widely used in applied mathematics, physics, and engineering. Specifically:
Signal processing
Quantum mechanics
Heat transfer and diffusion
Wave propgation
If :
f f f ( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ ) f f f f ′ f' f ′
Then :
f ( t ) = 1 2 π ∫ − ∞ ∞ F ( ω ) e i ω t d ω f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\text{d}\omega
f ( t ) = 2 π 1 ∫ − ∞ ∞ F ( ω ) e iω t d ω Where:
F ( ω ) = ∫ − ∞ ∞ f ( t ) e − i ω t d t F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\text{d}t
F ( ω ) = ∫ − ∞ ∞ f ( t ) e − iω t d t Provided that the integral converges.
At a point of discontinuity of f f f
Suppose f f f ( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ )
F { f ( t ) } = F ( ω ) = ∫ − ∞ ∞ f ( t ) e − i ω t d t \mathcal{F}\{f(t)\} = F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\text{d}t
F { f ( t )} = F ( ω ) = ∫ − ∞ ∞ f ( t ) e − iω t d t If the above integral diverges, Fourier transform of f f f
f ( t ) f(t) f ( t ) F ( ω ) F(\omega) F ( ω ) 1 1 1 2 π δ ( ω ) 2\pi \delta(\omega) 2 π δ ( ω ) u ( t ) exp ( − α t ) u(t) \exp(-\alpha t) u ( t ) exp ( − α t ) 1 α + i ω \dfrac{1}{\alpha + i\omega} α + iω 1 { 1 , − α ≤ t ≤ α 0 , otherwise \begin{cases} 1,& -\alpha \le t \le\alpha\\0,& \text{otherwise}\end{cases} { 1 , 0 , − α ≤ t ≤ α otherwise 2 sin ( ω α ) ω \dfrac{2\sin(\omega \alpha)}{\omega} ω 2 sin ( ω α ) u ( t ) u(t) u ( t ) π δ ( ω ) − i ω \pi\delta(\omega) - \dfrac{i}{\omega} π δ ( ω ) − ω i δ ( t ) \delta(t) δ ( t ) 1 1 1 δ ( t − a ) \delta(t - a) δ ( t − a ) exp ( − i ω a ) \exp (-i\omega a) exp ( − iωa ) cos ( a t ) \cos(at) cos ( a t ) π ( δ ( ω + a ) + δ ( ω − a ) ) \pi\big(\delta(\omega + a) + \delta(\omega - a)\big) π ( δ ( ω + a ) + δ ( ω − a ) ) sin ( a t ) \sin(at) sin ( a t ) π i ( δ ( ω − a ) − δ ( ω + a ) ) \dfrac{\pi}{i}\big(\delta(\omega - a) - \delta(\omega + a)\big) i π ( δ ( ω − a ) − δ ( ω + a ) ) exp ( − α ∣ t ∣ ) \exp(-\alpha \lvert t \rvert) exp ( − α ∣ t ∣) 2 α α 2 + ω 2 \dfrac{2\alpha}{\alpha^{2} + \omega^{2}} α 2 + ω 2 2 α sgn ( t ) \operatorname{sgn}(t) sgn ( t ) 2 i ω \dfrac{2}{i\omega} iω 2 1 t \dfrac{1}{t} t 1 − i π sgn ( ω ) -i\pi\,\operatorname{sgn}(\omega) − iπ sgn ( ω ) exp ( − α t 2 ) \exp(-\alpha t^{2}) exp ( − α t 2 ) π α e − ω 2 / ( 4 α ) \sqrt{\dfrac{\pi}{\alpha}}\,e^{-\omega^{2}/(4\alpha)} α π e − ω 2 / ( 4 α )
Here A ∈ R , α > 0 A \in \mathbb{R}, \alpha \gt 0 A ∈ R , α > 0
sgn ( t ) = { 1 , t > 0 0 , t = 0 − 1 , t < 0 \operatorname{sgn}(t)=\begin{cases}
1, & t>0\\
0, & t=0\\
-1, & t<0
\end{cases}
sgn ( t ) = ⎩ ⎨ ⎧ 1 , 0 , − 1 , t > 0 t = 0 t < 0 c f ( t ) + g ( t ) = c F ( ω ) + G ( ω ) \mathcal{cf(t) + g(t)} = cF(\omega) + G(\omega)
c f ( t ) + g ( t ) = c F ( ω ) + G ( ω ) Aka. 1st shift.
F { e i a t f ( t ) } = F ( ω − a ) \mathcal{F}\{e^{iat}f(t)\}=F(\omega-a)
F { e ia t f ( t )} = F ( ω − a ) Aka. 2nd shift.
F { f ( t − t 0 ) } = e − i ω t 0 F ( ω ) \mathcal{F}\{f(t-t_0)\}=e^{-i\omega t_0}F(\omega)
F { f ( t − t 0 )} = e − iω t 0 F ( ω ) Similar to in Laplace Transform . When α > 0 \alpha \gt 0 α > 0
F { f ( α t ) } = 1 α F ( ω α ) \mathcal{F}\{f(\alpha t)\}=\frac{1}{\alpha}F\left(\frac{\omega}{\alpha}\right)
F { f ( α t )} = α 1 F ( α ω ) F { f ( − t ) } = F ( − ω ) \mathcal{F}\{f(-t)\}=F(-\omega)
F { f ( − t )} = F ( − ω ) F { f ( t ) ‾ } = F ( − ω ) ‾ \mathcal{F}\{\overline{f(t)}\} = \overline{F(-\omega)}
F { f ( t ) } = F ( − ω ) Suppose f f f n n n ∀ i ∈ Z ∪ [ 0 , n ] , lim t → ± ∞ f ( i ) ( t ) = 0 \forall i \in \mathbb{Z}\cup [0,n], \lim\limits_{t \to \pm \infty} f^{(i)}(t) = 0 ∀ i ∈ Z ∪ [ 0 , n ] , t → ± ∞ lim f ( i ) ( t ) = 0
F { d n f ( t ) d t n } = ( i ω ) n F ( ω ) \mathcal{F}\left\{\frac{\text{d}^n f(t)}{\text{d}t^n} \right\}=(i\omega)^n F(\omega)
F { d t n d n f ( t ) } = ( iω ) n F ( ω ) F { t f ( t ) } = i d F ( ω ) d ω \mathcal{F}\{t f(t)\}= i\frac{dF(\omega)}{d\omega}
F { t f ( t )} = i d ω d F ( ω ) Similar to in Laplace Transform . When α > 0 \alpha \gt 0 α > 0
F { f ∗ g } = F ( ω ) G ( ω ) \mathcal{F}\{f*g\}=F(\omega)G(\omega)
F { f ∗ g } = F ( ω ) G ( ω ) F { f ⋆ g } = F ( ω ) G ( − ω ) \mathcal{F}\{f\star g\}=F(\omega)G(-\omega)
F { f ⋆ g } = F ( ω ) G ( − ω ) If F { f ( t ) } = F ( ω ) \mathcal{F}\{f(t)\}=F(\omega) F { f ( t )} = F ( ω ) then F { F ( t ) } = 2 π f ( − ω ) \mathcal{F}\{F(t)\}=2\pi f(-\omega) F { F ( t )} = 2 π f ( − ω )
Using Euler expansion:
F ( ω ) = ∫ − ∞ ∞ f ( t ) cos ( ω t ) d t − i ∫ − ∞ ∞ f ( t ) sin ( ω t ) d t F(\omega)=\int_{-\infty}^{\infty} f(t)\cos(\omega t)\,\text{d}t - i\int_{-\infty}^{\infty} f(t)\sin(\omega t)\,\text{d}t
F ( ω ) = ∫ − ∞ ∞ f ( t ) cos ( ω t ) d t − i ∫ − ∞ ∞ f ( t ) sin ( ω t ) d t
If f ( t ) f(t) f ( t )
If f ( t ) f(t) f ( t )
Suppose f f f [ 0 , ∞ ) [0, \infty) [ 0 , ∞ )
Denoted by F c ( ω ) F_c(\omega) F c ( ω )
F c ( ω ) = ∫ 0 ∞ f ( t ) cos ( ω t ) d t F_c(\omega)=\int_0^\infty f(t)\cos(\omega t)\,\text{d}t
F c ( ω ) = ∫ 0 ∞ f ( t ) cos ( ω t ) d t Inverse:
f ( t ) = 2 π ∫ 0 ∞ F c ( ω ) cos ( ω t ) d ω f(t)=\frac{2}{\pi}\int_0^\infty F_c(\omega)\cos(\omega t)\,\text{d}\omega
f ( t ) = π 2 ∫ 0 ∞ F c ( ω ) cos ( ω t ) d ω Denoted by F s ( ω ) F_s(\omega) F s ( ω )
F s ( ω ) = ∫ 0 ∞ f ( t ) sin ( ω t ) d t F_s(\omega)=\int_0^\infty f(t)\sin(\omega t)\,\text{d}t
F s ( ω ) = ∫ 0 ∞ f ( t ) sin ( ω t ) d t Inverse:
f ( t ) = 2 π ∫ 0 ∞ F s ( ω ) sin ( ω t ) d ω f(t)=\frac{2}{\pi}\int_0^\infty F_s(\omega)\sin(\omega t)\,\text{d}\omega
f ( t ) = π 2 ∫ 0 ∞ F s ( ω ) sin ( ω t ) d ω Fourier transforms are used to solve PDEs, especially when the domain is infinite or semi-infinite (start at a point, but goes to infinity in one direction).
For one-dimensional heat flow:
∂ 2 θ ∂ x 2 = 1 k ∂ θ ∂ t \frac{\partial^2 \theta}{\partial x^2}=\frac 1k\frac{\partial \theta}{\partial t}
∂ x 2 ∂ 2 θ = k 1 ∂ t ∂ θ Applying Fourier transform in x x x t t t
∂ Θ ∂ t = − k ω 2 Θ \frac{\partial \Theta}{\partial t} = -k\omega^2 \Theta
∂ t ∂ Θ = − k ω 2 Θ Here Θ ( ω , t ) = F { θ ( x , t ) } \Theta(\omega,t) = \mathcal{F}\{\theta(x,t)\} Θ ( ω , t ) = F { θ ( x , t )}
The above is a separable 1st order ODE. The solution is:
Θ ( ω , t ) = Θ ( ω , 0 ) e − k ω 2 t \Theta(\omega,t)=\Theta(\omega,0)e^{-k\omega^2 t}
Θ ( ω , t ) = Θ ( ω , 0 ) e − k ω 2 t θ ( x , t ) = 1 2 π ∫ − ∞ ∞ Θ ( ω , 0 ) e − k ω 2 t e i ω x d ω \theta (x, t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \Theta(\omega,0) e^{-k\omega^2 t} e^{i\omega x} \,\text{d}\omega
θ ( x , t ) = 2 π 1 ∫ − ∞ ∞ Θ ( ω , 0 ) e − k ω 2 t e iω x d ω ∂ 2 y ∂ t 2 = c 2 ∂ 2 y ∂ x 2 \frac{\partial^2 y}{\partial t^2}=c^2\frac{\partial^2 y}{\partial x^2}
∂ t 2 ∂ 2 y = c 2 ∂ x 2 ∂ 2 y By applying Fourier transform w.r.t. x x x
∂ 2 Y ∂ t 2 + c 2 ω 2 Y = 0 \frac{\partial^2 Y}{\partial t^2} + c^2 \omega^2 Y = 0
∂ t 2 ∂ 2 Y + c 2 ω 2 Y = 0 Here Y ( ω , t ) = F y ( x , t ) ( ω ) Y(\omega,t) = \mathcal{F}{y(x,t)}(\omega) Y ( ω , t ) = F y ( x , t ) ( ω )
Y = F f cos ( c ω t ) + F g sin ( c ω t ) c ω Y = F{f} \cos(c\omega t) + F{g}\frac{\sin(c \omega t)}{c\omega}
Y = F f cos ( c ω t ) + F g c ω sin ( c ω t ) ∂ 2 y ∂ t 2 = c 2 ∂ 2 y ∂ x 2 \frac{\partial^2 y}{\partial t^2}=c^2\frac{\partial^2 y}{\partial x^2}
∂ t 2 ∂ 2 y = c 2 ∂ x 2 ∂ 2 y Invert transform to obtain y(x,t).