48 lines
1.7 KiB
Markdown
48 lines
1.7 KiB
Markdown
# Chapter 5
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The Fourier Transform:
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$$X(\omega) = F[x(t)] = \int\limits_{-\infty}^\infty x(t) e^{-j\omega t} dt$$
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The Inverse Fourier Transform:
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$$x(t) = F^{-1}[X(\omega)] = {1 \over 2\pi} \int\limits_{-\infty}^{\infty} X(\omega)e^{j\omega t} d\omega$$
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The Fourier transform exists if:
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1. $x(t)$ is abolutely integrable
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2. $x(t)$ has only a finite number of discontinuities and a finite number of minima and maxima in any finite interval.
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If $x(t)$ is even, $X(\omega)$ is real.
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If $x(t)$ is odd, $X(\omega)$ is imaginary.
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Otherwise, $X(t)$ has a real and imaginary part.
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### Example
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Consider $x(t) = \delta(t)$:
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$$X(\omega) = \int\limits_{-\infty}^\infty \delta(t)e^{-j\omega t} dt = 1$$
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### Example
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Consider $x(t) = \delta(t - \alpha)$:
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$$X(\omega) = \int\limits_{-\infty}^\infty \delta(t - \alpha)e^{-j\omega t}dt = e^{-j\omega \alpha}$$
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### Example
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Consider $x(t) = u(t+T) - u(t-T)$:
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This is a pulse whose value is 1 on the interval $[-T,T]$.
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$$X(t) = \int\limits_{-T}^T e^{-j\omega t}dt = -{1 \over j \omega} \left[e^{-j\omega T} - e^{-j\omega (-T)}\right] = {2 \over \omega}\left[{e^{j\omega T} - e^{-j\omega T} \over 2j}\right] = 2T\operatorname{sinc}(\omega T)$$
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### Example
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Consider $x(t) = e^{-|t|}$
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$$X(\omega) = \int\limits_
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## Method 2 - Laplace Transfor
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$$x(t) \to X(s) \to X(\omega)$$
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$$s = j\omega$$
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Using a Laplace transform requires that $x(t)$ is a causal signal and the region of convergence of $X(s)$ includes the imaginary axis.
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### Example - Finite Support Signals
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The region of convergence is the entire s-plane (must check $s=0$). It definitely includes $s=j\omeaga$.
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$$X(s) = {0.5 \over s^2} - {0.5e^{-2s} \over s^2} - {e^{-2s} \over s}$$
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$$X(j\omega) = {0.5 \over j\omega^2} - {0.5e^{-2j\omega} \over j\omega^2} - {e^{-2j\omega} \over j\omega}$$
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