Continuous-Time Fourier Transform

Definition

Forward Continuous-Time Fourier Transform

$$X(\omega )=\mathcal{F}\left\{x(t)\right\}={\int }_{-\infty }^{\infty }x(t)\exp (-j\omega t)\mathrm{d}t$$

Inverse Continuous-Time Fourier Transform as

$$x(t)={\mathcal{F}}^{-1}\left\{X(\omega )\right\}=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X(\omega )\exp (j\omega t)\mathrm{d}\omega$$

Periodic Data

For a period function with Fourier Series $x(t)=\sum _{k=-\infty }^{\infty }{a}_k\exp (jk{\omega }_{0}t)$, we have

$$X(\omega )=2\pi \sum _{k=-\infty }^{\infty }{a}_k\delta (\omega -k{\omega }_0)$$

Properties

1. If $x(t)\stackrel{F}{\leftrightarrow }X(\omega )$ and $y(t)\stackrel{F}{\leftrightarrow }Y(\omega )$ , then

$$x(t)\ast y(t)\stackrel{F}{\leftrightarrow }X(\omega )Y(\omega )$$

and

$$x(t)y(t)\stackrel{F}{\leftrightarrow }\frac{1}{2\pi }X(\omega )\ast Y(\omega )$$

2. If $x(t)\stackrel{F}{\leftrightarrow }X(\omega )$, then

$$x(t-t_0)=x(t)\ast \delta (t-t_0)\stackrel{F}{\leftrightarrow }\exp (-j\omega t_0)X(\omega )$$

3. If $x(t)\stackrel{F}{\leftrightarrow }X(\omega )$, then

$$\exp (j{\omega }_{0}t)x(t)\stackrel{F}{\leftrightarrow }X(\omega -{\omega }_0)$$

4. If $x(t)\stackrel{F}{\leftrightarrow }X(\omega )$, then

$$x^{\prime }(t)=x(t)\ast {\delta }^{\prime }(t)\stackrel{F}{\leftrightarrow }j\omega X(\omega )$$

5. If $x(t)\stackrel{F}{\leftrightarrow }X(\omega )$, then

$$x(\alpha t)\stackrel{F}{\leftrightarrow }\frac{1}{\left|\alpha \right|}X\left(\frac{\omega }{\alpha }\right)$$

Particularly

$$x(-t)\stackrel{F}{\leftrightarrow }X(-\omega )$$

Examples

Delta function

The Fourier transform of the Dirac delta function $\delta (t-t_0)$ is given by

$$\mathcal{F}\left\{\delta (t-t_0)\right\}={\int }_{-\infty }^{\infty }\delta (t-t_0)\exp (-j\omega t)\mathrm{d}t=\exp \left(-j\omega t_0\right)$$

That means, the Fourier transform of $\delta (t-t_0)$ has a fixed norm $1$ and $t_0$ only affects its phase.

As for the derivative of the Dirac delta function,

$$\mathcal{F}\left\{{\delta }^{\prime }(t-t_0)\right\}=j\omega \exp (-j\omega t_0)$$

using the property

$$\left[\frac{\mathrm{d}}{\mathrm{d}x}\delta (x-a)\right]f(x)=-\delta (x-a)f^{\prime }(x)$$

Exponential function

The Fourier transform of $\exp (j{\omega }_{0}t)$ is

$$\mathcal{F}\left\{\exp (j{\omega }_{0}t)\right\}={\int }_{-\infty }^{\infty }\exp (j{\omega }_{0}t)\exp (-j\omega t)\mathrm{d}t=2\pi \delta (\omega -{\omega }_0)$$

Square waves

Recalling the Fourier Series of square waves, for a square wave with valid length of $\lambda$, we have the Fourier Series $\lambda \mathrm{sinc}(\lambda k\pi )$, therefore the Fourier Transform is

$$X(\omega )=\sum _{k=-\infty }^{\infty }2\pi \lambda \mathrm{sinc}(\lambda k\pi )\delta \left(\omega -k{\omega }_0\right),{\omega }_0=\frac{2\pi }{T}$$

where $T$ is the period of the original square wave

Dirac Comb

Recalling the Fourier Series of Dirac Comb $x(t)=\sum _{k=-\infty }^{\infty }\delta (t-kT)$ is $a_k\equiv \frac{1}{T}$, therefore

$$X(\omega )=\sum _{k=-\infty }^{\infty }\frac{2\pi }{T}\delta (\omega -k{\omega }_0),{\omega }_0=\frac{2\pi }{T}$$

Trigonometric functions

For $x(t)=\sin ({\omega }_{0}t)$, we have

$$X(\omega )=2\pi \left[-\frac{1}{2j}\delta (\omega +{\omega }_0)+\frac{1}{2j}\delta (\omega -{\omega }_0)\right]=-\frac{\pi }{j}\delta (\omega +{\omega }_0)+\frac{\pi }{j}\delta (\omega -{\omega }_0)$$

And for $x(t)=\cos ({\omega }_{0}t)$,

$$X(\omega )=2\pi \left[\frac{1}{2}\delta (\omega +{\omega }_0)+\frac{1}{2}\delta (\omega -{\omega }_0)\right]=\pi \delta (\omega +{\omega }_0)+\pi \delta (\omega -{\omega }_0)$$

Rectangular Pulse

$$x(t)=\{\begin{array}{ll}1,& \left|t\right|<T_1\\ 0,& \left|t\right|>T_1\end{array}⟹X(\omega )=\frac{2\sin \omega T_1}{\omega }$$

Unit step

For $x(t)$ being unit step $u(t)$

$$X(\omega )=\frac{1}{j\omega }+\pi \delta (\omega )$$

Furthermore, If $\mathcal{R}e\left\{a\right\}>0$, then

1. $\exp (-at)u(t)\stackrel{F}{\leftrightarrow }\frac{1}{a+j\omega }$
2. $t\exp (-at)u(t)\stackrel{F}{\leftrightarrow }\frac{1}{(a+j\omega {)}^2}$
3. $\frac{t^{n-1}}{(n-1)!}\exp (-at)u(t)\stackrel{F}{\leftrightarrow }\frac{1}{(a+j\omega {)}^n}$

Sampling Function

If $x(t)=\frac{\sin Wt}{\pi t}$, then

$$X(j\omega )=\{\begin{array}{ll}1,& \left|\omega \right|<W\\ 0,& \left|\omega \right|>W\end{array}$$

Proportional Function

$$x(t)=t⟹X(\omega )=j\cdot 2\pi \cdot {\delta }^{\prime }(\omega )$$

Other Functions

• $\frac{1}{t^2}\to -\pi \left|\omega \right|$