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| <math>x(t)*y(t) \ </math> | | <math>x(t)*y(t) \ </math> | ||
| <math> \mathcal{X}(\omega)\mathcal{Y}(\omega) \!</math> | | <math> \mathcal{X}(\omega)\mathcal{Y}(\omega) \!</math> | ||
| + | | Remember that <math> x(t)*y(t) =\int_{-\infty}^{\infty} x(t)y(t-t')dt' </math><br /> | ||
| + | <math>\mathfrak{F}(x[n]*y[n]) = \sum_{n=-\infty}^{\infty}[\sum_{k=-\infty}^{\infty}x[k]*y[n-k]]e^{-j\omega n}</math><br /> | ||
| + | We can replace <math>e^{-j\omega n}</math> by <math>e^{-j\omega n} = e^{-j\omega n + j\omega k -j\omega k }= e^{-j\omega( n - k ) - j\omega k }= e^{-j\omega( n - k )} e^{-j\omega k }</math><br /> | ||
| + | So..<br /> | ||
| + | <math>= \sum_{n=-\infty}^{\infty}[\sum_{k=-\infty}^{\infty}x[k]*y[n-k]]e^{-j\omega( n - k )} e^{-j\omega k}</math><br /> | ||
| + | <math>= \sum_{k=-\infty}^{\infty}e^{-j\omega k }[\sum_{n=-\infty}^{\infty}x[k]*y[n-k]]e^{-j\omega( n - k )}] </math><br /> | ||
| + | <math>= \chi(\omega)\gamma(\omega)</math><br /> | ||
|- | |- | ||
Revision as of 23:10, 14 November 2018
CTFT of periodic signals with properties
| Function | CTFT | ||
|---|---|---|---|
| $ sin(\omega_0t) $ | $ \frac{\pi}{j}(\delta(\omega - \omega_0) - \delta(\omega+\omega_0)) $ | ||
| $ cos(\omega_0t) $ | $ \pi(\delta(\omega - \omega_0) + \delta(\omega+\omega_0)) $ | ||
| $ e^{j\omega_0t} $ | $ 2\pi\delta(\omega - \omega_0) $ | ||
| $ \sum^{\infty}_{k=-\infty} a_{k}e^{ikw_{0}t} $ | $ 2\pi\sum^{\infty}_{k=-\infty}a_{k}\delta(w-kw_{0}) \ $ | ||
| $ \sum^{\infty}_{n=-\infty} \delta(t-nT) \ $ | $ \frac{2\pi}{T}\sum^{\infty}_{k=-\infty}\delta(w-\frac{2\pi k}{T}) $ |
| Name | $ x(t) \longrightarrow \ $ | $ \mathcal{X}(\omega) $ | Proof |
|---|---|---|---|
| Linearity | $ ax(t) + by(t) \ $ | $ a \mathcal{X}(\omega) + b \mathcal{Y} (\omega) $ | $ \mathfrak{F}(ax(t) + by(t)) = \int_{-\infty}^{\infty}[ax(t) + by(t)]e^{-j\omega t} dt $ $ \int_{-\infty}^{\infty}ax(t)e^{-j\omega t} dt + \int_{-\infty}^{\infty}by(t)e^{-j\omega t} dt $ |
| Time Shifting | $ x(t-t_0) \ $ | $ e^{-j\omega t_0}X(\omega) $ | $ \mathfrak{F}(x(t - t_{o})) = \int_{-\infty}^{\infty}x(t-t_{0})e^{-j\omega t} dt $ let $ t' = t - t_{o} $ |
| Frequency Shifting | $ e^{j\omega_0 t}x(t) $ | $ \mathcal{X} (\omega - \omega_0) $ | Refer to Time Shifting section |
| Conjugation | $ x^{*}(t) \ $ | $ \mathcal{X}^{*} (-\omega) $ | $ \mathcal{X}^{*}(-\omega) = \int_{-\infty}^{\infty} (x(t)e^{j\omega t}dt)^{*} $ $ = \int_{-\infty}^{\infty} x^{*}(t)e^{-j\omega t}dt $ |
| Scaling | $ x(at) \ $ | $ \frac{1}{|a|} \mathcal{X} (\frac{\omega}{a}) $ | $ \int_{-\infty}^{\infty} x(at)e^{-j\omega t}dt $ let $ t' = at $ |
| Multiplication | $ x(t)y(t) \ $ | $ \frac{1}{2\pi} \mathcal{X}(\omega)*\mathcal{Y}(\omega) $ | |
| Convolution | $ x(t)*y(t) \ $ | $ \mathcal{X}(\omega)\mathcal{Y}(\omega) \! $ | Remember that $ x(t)*y(t) =\int_{-\infty}^{\infty} x(t)y(t-t')dt' $ $ \mathfrak{F}(x[n]*y[n]) = \sum_{n=-\infty}^{\infty}[\sum_{k=-\infty}^{\infty}x[k]*y[n-k]]e^{-j\omega n} $ |
| Differentiation | $ tx(t) \ $ | $ j\frac{d}{d\omega} \mathcal{X} (\omega) $ | |
| Duality | $ \mathcal{X} (-t) $ | $ 2 \pi x (\omega) \ $ | |
| Parseval's Relation | $ \int_{-\infty}^{\infty} |x(t)|^2 dt = $ | $ \frac{1}{2\pi} \int_{-\infty}^{\infty} |\mathcal{X}(w)|^2 dw $ |
