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| | <math>\int_{-\infty}^{\infty} |x(t)|^2 dt = </math>
 
| | <math>\int_{-\infty}^{\infty} |x(t)|^2 dt = </math>
 
| <math> \frac{1}{2\pi} \int_{-\infty}^{\infty} |\mathcal{X}(w)|^2 dw</math>
 
| <math> \frac{1}{2\pi} \int_{-\infty}^{\infty} |\mathcal{X}(w)|^2 dw</math>
| <math> \int_{-\infty}^{\infty} x(t)x(t) =  \int_{-\infty}^{\infty}x(t)dt(\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}d\omega)</math><br />
+
| <math> \int_{-\infty}^{\infty} x(t)x(t) dt =  \int_{-\infty}^{\infty}x(t)dt(\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}d\omega)</math><br />
 
<math>= \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)d\omega [\int_{-\infty}^{\infty}x(t)\frac{1}{2\pi}e^{j\omega t}]dt</math><br />
 
<math>= \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)d\omega [\int_{-\infty}^{\infty}x(t)\frac{1}{2\pi}e^{j\omega t}]dt</math><br />
 
<math>= \frac{1}{2\pi}\int_{-\infty}^{\infty}\chi(\omega)[\chi(-\omega)]d\omega</math><br />
 
<math>= \frac{1}{2\pi}\int_{-\infty}^{\infty}\chi(\omega)[\chi(-\omega)]d\omega</math><br />
 
<math>= \frac{1}{2\pi }\int_{-\infty}^{\infty} | \mathcal{X} (\omega) |^{2}d\omega</math><br />
 
<math>= \frac{1}{2\pi }\int_{-\infty}^{\infty} | \mathcal{X} (\omega) |^{2}d\omega</math><br />
 
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Revision as of 00:41, 15 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 $
$ =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}) $

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(t)*y(t)) = \int_{-\infty}^{\infty}[\int_{-\infty}^{\infty} x(t')y(t-t')dt' ]e^{-j\omega t}dt $
Replace $ e^{-j\omega t} $ by $ e^{-j\omega( t - t' )} e^{-j\omega t' } $
$ = \int_{-\infty}^{\infty}[\int_{-\infty}^{\infty} x(t')y(t-t')dt' ]e^{-j\omega( t - t' )} e^{-j\omega t' }dt $
$ = \int_{-\infty}^{\infty}e^{-j\omega t' }dt'[\int_{-\infty}^{\infty} x(t')y(t-t')e^{-j\omega( t - t' )}dt'] $
$ = \chi(\omega)\gamma(\omega) $

Multiplication $ x(t)y(t) \ $ $ \frac{1}{2\pi} \mathcal{X}(\omega)*\mathcal{Y}(\omega) $ Refer to Convolution section
Differentiation $ tx(t) \ $ $ j\frac{d}{d\omega} \mathcal{X} (\omega) $
Duality $ \mathcal{X} (-t) $ $ 2 \pi x (\omega) \ $ $ 2\pi \mathcal{X} (-\omega) = \int_{-\infty}^{\infty} \mathcal{X}(t')e^{-j\omega t'}dt' $

let t = t'
$ = \int_{-\infty}^{\infty} \mathcal{X}(t)e^{-j\omega t}dt $
= CTFT[x(t)]

Parseval's Relation $ \int_{-\infty}^{\infty} |x(t)|^2 dt = $ $ \frac{1}{2\pi} \int_{-\infty}^{\infty} |\mathcal{X}(w)|^2 dw $ $ \int_{-\infty}^{\infty} x(t)x(t) dt = \int_{-\infty}^{\infty}x(t)dt(\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}d\omega) $

$ = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)d\omega [\int_{-\infty}^{\infty}x(t)\frac{1}{2\pi}e^{j\omega t}]dt $
$ = \frac{1}{2\pi}\int_{-\infty}^{\infty}\chi(\omega)[\chi(-\omega)]d\omega $
$ = \frac{1}{2\pi }\int_{-\infty}^{\infty} | \mathcal{X} (\omega) |^{2}d\omega $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn