4 Further properties of the Fourier transform

We state these properties without proof. As usual $F\left(\omega \right)$ denotes the Fourier transform of $f\left(t\right)$ .

1. Time differentiation property:

   $\mathcal{F}\left\{f^{\prime }\left(t\right)\right\}=\text{i}\omega F\left(\omega \right)$

   (Differentiating a function is said to amplify the higher frequency components because of the additional multiplying factor $\omega$ .)

2. Frequency differentiation property:

   $\mathcal{F}\left\{tf\left(t\right)\right\}=\text{i}\frac{dF}{d\omega }\text{or}\mathcal{F}\left\{\left(-\text{i}t\right)f\left(t\right)\right\}=\frac{dF}{d\omega }$

   Note the symmetry between properties 1. and 2.

3. Duality property:

   If $\mathcal{F}\left\{f\left(t\right)\right\}=F\left(\omega \right)$ then $\mathcal{F}\left\{F\left(t\right)\right\}=2\pi f\left(-\omega \right).$

Informally, the duality property states that we can, apart from the $2\pi$ factor, interchange the time and frequency domains provided we put $-\omega$ rather than $\omega$ in the second term, this corresponding to a reflection in the vertical axis. If $f\left(t\right)$ is even this latter is irrelevant.

For example, we know that if $f\left(t\right)=p_1\left(t\right)=\left\{\begin{array}{ccc}\hfill 1\hfill & \hfill \hfill & \hfill -1<t<1\hfill \\ \hfill 0\hfill & \hfill \hfill & \hfill \text{otherwise}\hfill \end{array}\right.,\text{then}F\left(\omega \right)=2\frac{sin\omega }{\omega }.$

Then, by the duality property, since $p_1\left(\omega \right)$ is even, $\mathcal{F}\left\{2\frac{sint}{t}\right\}=2\pi p_1\left(-\omega \right)=2\pi p_1\left(\omega \right).$

Graphically:

Figure 6

Task!

Recalling the Fourier transform pair

  $f\left(t\right)=\left\{\begin{array}{ccc}{e}^{-2t}\hfill & \hfill \hfill & \hfill t>0\hfill \\ e^{2t}\hfill & \hfill \hfill & \hfill t<0\hfill \end{array}\right.F\left(\omega \right)=\frac{4}{4+{\omega }^2},$

obtain the Fourier transforms of

1. $g\left(t\right)=\frac{1}{4+t^2}$
2. $h\left(t\right)=\frac{1}{4+t^2}cos2t.$

1. Use the linearity and duality properties:

   Answer

   We have $\mathcal{F}\left\{f\left(t\right)\right\}\equiv \mathcal{F}\left\{e^{-2\left|t\right|}\right\}=\frac{4}{4+{\omega }^2}.\therefore \mathcal{F}\left\{\frac{1}{4}{e}^{-2\left|t\right|}\right\}=\frac{1}{4+{\omega }^2}\text{(by linearity)}$

   $\therefore \mathcal{F}\left\{\frac{1}{4+t^2}\right\}=2\pi \frac{1}{4}{e}^{-2\left|-\omega \right|}=\frac{\pi }{2}{e}^{-2\left|\omega \right|}=G\left(\omega \right)\text{(by duality).}$

   

2. Use the modulation property based on the frequency shift property:

   Answer

   We have $h\left(t\right)=g\left(t\right)cos2t.$ $\therefore \mathcal{F}\left\{g\left(t\right)cos{\omega }_{0}t\right\}=\frac{1}{2}\left(G\left(\omega -{\omega }_0\right)+G\left(\omega +{\omega }_0\right)\right),$

   so with ${\omega }_0=2$ $\mathcal{F}\left\{h\left(t\right)\right\}=\frac{\pi }{4}\left\{e^{-2\left|\omega -2\right|}+e^{-2\left|\omega +2\right|}\right\}=H\left(\omega \right)$

   

Exercises

1. Using the superposition and time delay theorems and the known result for the transform of the rectangular pulse $p\left(t\right)$ , obtain the Fourier transforms of each of the signals shown.

   

2. Obtain the Fourier transform of the signal

   $f\left(t\right)={\text{e}}^{-t}u\left(t\right)+{\text{e}}^{-2t}u\left(t\right)$

   where $u\left(t\right)$ denotes the unit step function.

3. Use the time-shift property to obtain the Fourier transform of

   $f\left(t\right)=\left\{\begin{array}{cc}\hfill 1\hfill & \hfill 1\le t\le 3\hfill \\ \hfill \hfill \\ \hfill 0\hfill & \hfill \text{otherwise}\hfill \end{array}\right.$

   Verify your result using the definition of the Fourier transform.

4. Find the inverse Fourier transforms of
   1. $F\left(\omega \right)=20\frac{sin\left(5\omega \right)}{5\omega }{\text{e}}^{-3i\omega }$
   2. $F\left(\omega \right)=\frac{8}{\omega }sin3\omega {\text{e}}^{i\omega }$
   3. $F\left(\omega \right)=\frac{{\text{e}}^{i\omega }}{1-i\omega }$
5. If $f\left(t\right)$ is a signal with transform $F\left(\omega \right)$ obtain the Fourier transform of $f\left(t\right)cos\left({\omega }_{0}t\right)cos\left({\omega }_{0}t\right).$

Answer