Several of the most important properties of the Fourier transform are derived.
The Fourier Transform (FT) has several important properties which will be useful:

Linearity:
$$\alpha {x}_{1}\left(t\right)+\beta {x}_{2}\left(t\right)\leftrightarrow \alpha {X}_{1}\left(j\Omega \right)+\beta {X}_{2}\left(j\Omega \right)$$
where
$\alpha $ and
$\beta $ are constants. This property is easy to verify by plugging the left side of
[link] into the definition of the FT.

Time shift:
$$x(t\tau )\leftrightarrow {e}^{j\Omega \tau}X\left(j\Omega \right)$$
To derive this property we simply take the FT of
$x(t\tau )$
$${\int}_{\infty}^{\infty}x(t\tau ){e}^{j\Omega t}dt$$
using the variable substitution
$\gamma =t\tau $ leads to
$$t=\gamma +\tau $$
and
$$d\gamma =dt$$
We also note that if
$t=\pm \infty $ then
$\tau =\pm \infty $ . Substituting
[link] ,
[link] , and the limits of integration into
[link] gives
$$\begin{array}{ccc}\hfill {\int}_{\infty}^{\infty}x\left(\gamma \right){e}^{j\Omega (\gamma +\tau )}d\gamma & =& {e}^{j\Omega \tau}{\int}_{\infty}^{\infty}x\left(\gamma \right){e}^{j\Omega \gamma}d\gamma \hfill \\ & =& {e}^{j\Omega \tau}X\left(j\Omega \right)\hfill \end{array}$$
which is the desired result.

Frequency shift:
$$x\left(t\right){e}^{j{\Omega}_{0}t}\leftrightarrow X\left(j(\Omega {\Omega}_{0})\right)$$
Deriving the frequency shift property is a bit easier than the time shift property. Again, using the definition of FT we get:
$$\begin{array}{ccc}\hfill {\int}_{\infty}^{\infty}x\left(t\right){e}^{j{\Omega}_{0}t}{e}^{j\Omega t}dt& =& {\int}_{\infty}^{\infty}x\left(t\right){e}^{j(\Omega {\Omega}_{0})t}dt\hfill \\ & =& X\left(j(\Omega {\Omega}_{0})\right)\hfill \end{array}$$

Time reversal :
$$x(t)\leftrightarrow X(j\Omega )$$
To derive this property, we again begin with the definition of FT:
$${\int}_{\infty}^{\infty}x(t){e}^{j\Omega t}dt$$
and make the substitution
$\gamma =t$ . We observe that
$dt=d\gamma $ and that if the limits of integration for
$t$ are
$\pm \infty $ , then the limits of integration for
$\gamma $ are
$\mp \gamma $ . Making these substitutions into
[link] gives
$$\begin{array}{ccc}\hfill {\int}_{\infty}^{\infty}x\left(\gamma \right){e}^{j\Omega \gamma}d\gamma & =& {\int}_{\infty}^{\infty}x\left(\gamma \right){e}^{j\Omega \gamma}d\gamma \hfill \\ & =& X(j\Omega )\hfill \end{array}$$
Note that if
$x\left(t\right)$ is real, then
$X(j\Omega )=X{\left(j\Omega \right)}^{*}$ .

Time scaling: Suppose we have
$y\left(t\right)=x\left(at\right),a>0$ . We have
$$Y\left(j\Omega \right)={\int}_{\infty}^{\infty}x\left(at\right){e}^{j\Omega t}dt$$
Using the substitution
$\gamma =at$ leads to
$$\begin{array}{cc}\hfill Y\left(j\Omega \right)& =\frac{1}{a}{\int}_{\infty}^{\infty}x\left(\gamma \right){e}^{j\Omega \gamma /a}d\gamma \hfill \\ & =\frac{1}{a}X\left(\frac{\Omega}{a}\right)\hfill \end{array}$$

Convolution: The convolution integral is given by
$$y\left(t\right)={\int}_{\infty}^{\infty}x\left(\tau \right)h(t\tau )d\tau $$
The convolution property is given by
$$Y\left(j\Omega \right)\leftrightarrow X\left(j\Omega \right)H\left(j\Omega \right)$$
To derive this important property, we again use the FT definition:
$$\begin{array}{ccc}\hfill Y\left(j\Omega \right)& =& {\int}_{\infty}^{\infty}y\left(t\right){e}^{j\Omega t}dt\hfill \\ & =& {\int}_{\infty}^{\infty}{\int}_{\infty}^{\infty}x\left(\tau \right)h(t\tau ){e}^{j\Omega t}d\tau dt\hfill \\ & =& {\int}_{\infty}^{\infty}x\left(\tau \right)\left[{\int}_{\infty}^{\infty},h,(t\tau ),{e}^{j\Omega t},d,t\right]d\tau \hfill \end{array}$$
Using the time shift property, the quantity in the brackets is
${e}^{j\Omega \tau}H\left(j\Omega \right)$ , giving
$$\begin{array}{ccc}\hfill Y\left(j\Omega \right)& =& {\int}_{\infty}^{\infty}x\left(\tau \right){e}^{j\Omega \tau}H\left(j\Omega \right)d\tau \hfill \\ & =& H\left(j\Omega \right){\int}_{\infty}^{\infty}x\left(\tau \right){e}^{j\Omega \tau}d\tau \hfill \\ & =& H\left(j\Omega \right)X\left(j\Omega \right)\hfill \end{array}$$
Therefore, convolution in the time domain corresponds to multiplication in the frequency domain.

Multiplication (Modulation):
$$w\left(t\right)=x\left(t\right)y\left(t\right)\leftrightarrow \frac{1}{2\pi}{\int}_{\infty}^{\infty}X\left(j(\Omega \Theta )\right)Y\left(j\Theta \right)d\Theta $$
Notice that multiplication in the time domain corresponds to convolution in the frequency domain. This property can be understood by applying the inverse Fourier Transform
[link] to the right side of
[link]
$$\begin{array}{ccc}\hfill w\left(t\right)& =& \frac{1}{2\pi}{\int}_{\infty}^{\infty}\frac{1}{2\pi}{\int}_{\infty}^{\infty}X\left(j(\Omega \Theta )\right)Y\left(j\Theta \right){e}^{j\Omega t}d\Theta d\Omega \hfill \\ & =& \frac{1}{2\pi}{\int}_{\infty}^{\infty}Y\left(j\Theta \right)\left[\frac{1}{2\pi},{\int}_{\infty}^{\infty},X,\left(j(\Omega \Theta )\right),{e}^{j\Omega t},d,\Omega \right]d\Theta \hfill \end{array}$$
The quantity inside the brackets is the inverse Fourier Transform of a frequency shifted Fourier Transform,
$$\begin{array}{ccc}\hfill w\left(t\right)& =& \frac{1}{2\pi}{\int}_{\infty}^{\infty}Y\left(j\Theta \right)\left[x,\left(t\right),{e}^{j\Theta t}\right]d\Theta \hfill \\ & =& x\left(t\right)\frac{1}{2\pi}{\int}_{\infty}^{\infty}Y\left(j\Theta \right){e}^{j\Theta t}d\Theta \hfill \\ & =& x\left(t\right)y\left(t\right)\hfill \end{array}$$

Duality: The duality property allows us to find the Fourier transform of timedomain signals whose functional forms correspond to known Fourier transforms,
$X\left(jt\right)$ . To derive the property, we start with the inverse Fourier transform:
$$x\left(t\right)=\frac{1}{2\pi}{\int}_{\infty}^{\infty}X\left(j\Omega \right){e}^{j\Omega t}d\Omega $$
Changing the sign of
$t$ and rearranging,
$$2\pi x(t)={\int}_{\infty}^{\infty}X\left(j\Omega \right){e}^{j\Omega t}d\Omega $$
Now if we swap the
$t$ and the
$\Omega $ in
[link] , we arrive at the desired result
$$2\pi x(\Omega )={\int}_{\infty}^{\infty}X\left(jt\right){e}^{j\Omega t}dt$$
The righthand side of
[link] is recognized as the FT of
$X\left(jt\right)$ , so we have
$$X\left(jt\right)\leftrightarrow 2\pi x(\Omega )$$