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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 $
$ =a\mathcal{X}(\omega) + b\mathcal{Y}(\omega) $

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} $
$ \int_{-\infty}^{\infty}x(t')e^{-j\omega (t'+t_{o})} dt' $
$ = e^{-j\omega t_{o}}\int_{-\infty}^{\infty}x(t')e^{-j\omega t'} dt' $
$ = e^{-j\omega t_{o}}\mathcal{X}(\omega) $

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 $
$ =\mathfrak{F}(x(t)^{*}) $

Scaling $ x(at) \ $ $ \frac{1}{|a|} \mathcal{X} (\frac{\omega}{a}) $ $ \int_{-\infty}^{\infty} x(at)e^{-j\omega t}dt $

let $ t' = at $
$ = \int_{-\infty}^{\infty} x(t')e^{-j\omega \frac{t'}{a}}\frac{dt'}{a}, a > 0 $
$ = -\int_{-\infty}^{\infty} x(t')e^{-j\omega \frac{t'}{a}}\frac{dt'}{a}, a < 0 $
$ =\frac{1}{|a|} \mathcal{X} (\frac{\omega}{a}) $

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} $
We can replace $ e^{-j\omega n} $ by $ 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 } $
So..
$ = \sum_{n=-\infty}^{\infty}[\sum_{k=-\infty}^{\infty}x[k]*y[n-k]]e^{-j\omega( n - k )} e^{-j\omega k} $
$ = \sum_{k=-\infty}^{\infty}e^{-j\omega k }[\sum_{n=-\infty}^{\infty}x[k]*y[n-k]]e^{-j\omega( n - k )}] $
$ = \chi(\omega)\gamma(\omega) $

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 $

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EISL lab graduate

Mu Qiao