Rowan-Classes/5th-Semester-Fall-2023/Signals-and-Systems/Notes/Chapter5.md

Chapter 5

The Fourier Transform:

X(\omega) = F[x(t)] = \int\limits_{-\infty}^\infty x(t) e^{-j\omega t} dt

The Inverse Fourier Transform:

x(t) = F^{-1}[X(\omega)] = {1 \over 2\pi} \int\limits_{-\infty}^{\infty} X(\omega)e^{j\omega t} d\omega

The Fourier transform exists if:

1. x(t) is abolutely integrable
2. x(t) has only a finite number of discontinuities and a finite number of minima and maxima in any finite interval.

If x(t) is even, X(\omega) is real.

If x(t) is odd, X(\omega) is imaginary.

Otherwise, X(t) has a real and imaginary part.

Example

Consider x(t) = \delta(t):

X(\omega) = \int\limits_{-\infty}^\infty \delta(t)e^{-j\omega t} dt = 1

Example

Consider x(t) = \delta(t - \alpha):

X(\omega) = \int\limits_{-\infty}^\infty \delta(t - \alpha)e^{-j\omega t}dt = e^{-j\omega \alpha}

Example

Consider x(t) = u(t+T) - u(t-T):

This is a pulse whose value is 1 on the interval [-T,T].

X(t) = \int\limits_{-T}^T e^{-j\omega t}dt = -{1 \over j \omega} \left[e^{-j\omega T} - e^{-j\omega (-T)}\right] = {2 \over \omega}\left[{e^{j\omega T} - e^{-j\omega T} \over 2j}\right] = 2T\operatorname{sinc}(\omega T)

Example

Consider x(t) = e^{-|t|}

X(\omega) = \int\limits_

## Method 2 - Laplace Transfor
$$x(t) \to X(s) \to X(\omega)$$
$$s = j\omega$$

Using a Laplace transform requires that $x(t)$ is a causal signal and the region of convergence of $X(s)$ includes the imaginary axis.

### Example - Finite Support Signals
The region of convergence is the entire s-plane (must check $s=0$). It definitely includes $s=j\omeaga$.

$$X(s) = {0.5 \over s^2} - {0.5e^{-2s} \over s^2} - {e^{-2s} \over s}$$
$$X(j\omega) = {0.5 \over j\omega^2} - {0.5e^{-2j\omega} \over j\omega^2} - {e^{-2j\omega} \over j\omega}$$