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|<math>\mathfrak{F}^{-1}[X(j\omega + omega_0)] = \frac{1}{2\pi} \int_{-\infty}^{\infty}X(j(\omega +\omega_0))e^{j\omega t} d\omega </math><br/>
 
|<math>\mathfrak{F}^{-1}[X(j\omega + omega_0)] = \frac{1}{2\pi} \int_{-\infty}^{\infty}X(j(\omega +\omega_0))e^{j\omega t} d\omega </math><br/>
 
<math>=\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega ')e^{j\omega (\omega ' + \omega_0)} d\omega </math><br/>
 
<math>=\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega ')e^{j\omega (\omega ' + \omega_0)} d\omega </math><br/>
<math>=\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega  ')e^{j\omega(\omega ' +\omega_0) d\omega </math><br/>
 
  
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<math>=e^{j\omega_0 t}\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega ')e^{j\omega '} d\omega' </math><br/>
  
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<math>=x(t)e^{j\omega_0 t}
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Revision as of 22:07, 22 April 2018


Table of CT Fourier Series Coefficients and Properties

Fourier series Coefficients

Function Fourier Series Coefficients


Properties of CT Fourier systems

Property Name Property Proof
Linearity $ \mathfrak{F}(c_1g(t) + c_2h(t) = c_1G(f) + c_2H(f) $ $ \mathfrak{F}(c_1g(t) + c_2h(t) = \int_{-\infty}^\infty c_1g(t) dt + \int_{-\infty}^\infty c_2h(t) dt $

$ =c_1\int_{-\infty}^\infty g(t)e^{i2\pi ft} dt + c_2 \int_{-\infty}^\infty g(t)e^{i2\pi ft} dt $
$ =c_1G(f) + c_2H(f) $

Time Shifting $ \mathfrak{F}(g(t - a)) = e^{-i2\pi fa}*G(f) $ $ \mathfrak{F}(g(t - a)) = \int_{-\infty}^\infty g(t-a)e^{-2\pi ft}dt $

$ =\int_{-\infty}^\infty g(u)e^{-i2\pi f(u+a)}du $
$ =e^{-i2\pi fa}\int_{-\infty}^\infty g(u)e^{-i2\pi fu}du $
$ =e^{-i2\pi fa} G(f) $

Time Scaling $ \mathfrak{F}(g(ct)) = \frac{G(\frac{f}{c})}{|c|} $
$ \mathfrak{F}(g(ct)) = \int_{-\infty}^\infty g(ct)e^{-i2\pi ft}dt $

subtitute : u = ct, du = cdt
$ \mathfrak{F}(g(ct)) = \int_{-c\infty}^{c\infty} \frac{g(u)}{c}e^{-i2\pi f\frac{u}{c}}du $
if c is greater than 0: then no signs change. if c is less than 0: the integration must be flipped as well as the negative from the c so you still get the same equation. therefore the absolute value of c is obtained.

Frequency Shifting $ \mathfrak{F}^{-1}[X(j\omega + omega_0)] = x(t)e^{-j\omega_0t} $ $ \mathfrak{F}^{-1}[X(j\omega + omega_0)] = \frac{1}{2\pi} \int_{-\infty}^{\infty}X(j(\omega +\omega_0))e^{j\omega t} d\omega $

$ =\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega ')e^{j\omega (\omega ' + \omega_0)} d\omega $

$ =e^{j\omega_0 t}\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega ')e^{j\omega '} d\omega' $

$ =x(t)e^{j\omega_0 t} |- | | | |- } $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood