5th semester files

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

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+# 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}$$