Class 12 (The Continuous-Time Fourier Transform)

Continuous-time Fourier Transform

For the Fourier Series, we showed we could express any continuous-time, power signal by a discrete summation of harmonic cosines and sines (or complex exponentials). For the Fourier Transform, we now express any continuous-time, energy signal as a continuous summation (i.e., an integral) of complex exponentials.

Continuous-time Fourier transform analysis, also simply known as the Fourier transform, is defined by $$X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j \omega t } dt$$

Continuous-time Fourier transform synthesis, also known as the inverse Fourier transform is defined by $$x(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} X(\omega) e^{j \omega t } d \omega$$

Note that the constant (i.e., the normalization) in front of the Fourier transform and inverse Fourier transform may sometimes differ depending on your field of study.

An impulse

Question: Given $x(t) = \delta(t-\tau)$, find $X(\omega)$.

Solution: The Fourier Transform of $x(t)$ is defined by $$ \begin{eqnarray} X(\omega) &=& \int_{-\infty}^{\infty} x(t) e^{-j \omega t } dt \\ &=& \int_{-\infty}^{\infty} \delta(t-\tau) e^{-j \omega t } dt \\ &=& \int_{-\infty}^{\infty} \delta(t-\tau) e^{-j \omega t } dt \\ \end{eqnarray} $$ At this point $\delta(t-\tau)$ is equal to zero unless, $t=\tau$. As a result, all values in the integral (a continuous summation) area $0$ except for $t = \tau$. Therefore, our expression simplifies to $$ \begin{eqnarray} X(\omega) &=& e^{-j k \omega \tau } \end{eqnarray} $$

A Centered Box

Question: Given $x(t) = u(t+\tau/2) - u(t-\tau/2)$, find $X(\omega)$.

Solution: The Fourier Transform of $x(t)$ is defined by $$ \begin{eqnarray} X(\omega) &=& \int_{-\infty}^{\infty} x(t) e^{-j \omega t } dt \\ &=& \int_{-\infty}^{\infty} (u(t+\tau) - u(t-\tau)) e^{-j \omega t } dt \\ &=& \int_{-\tau/2}^{\tau/2} e^{-j \omega t } dt \\ &=& \left. \frac{-1}{j \omega} e^{-j \omega t } \right|_{-\tau/2}^{\tau/2} \\ &=& \frac{-1}{j \omega} \left( e^{j \omega \tau / 2 } - e^{-j \omega \tau / 2 } \right) \\ &=& \frac{-1}{j \omega} \left( e^{j \omega \tau / 2 } - e^{-j \omega \tau / 2 } \right) \\ &=& \frac{2 \sin \left( \frac{\omega \tau}{2} \right)}{\omega} \\ &=& \tau \frac{\sin \left( \frac{\omega \tau}{2} \right)}{\left( \frac{\omega \tau}{2} \right)} \\ &=& \tau \, \textrm{sinc}\left( \frac{\omega \tau}{2} \right) \end{eqnarray} $$