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|<math>\frac{1}{\alpha + j\omega} </math>
|<math> \frac{1}{\alpah T + j2\pi k} </math>
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|<math> \frac{1}{\alpha T + j2\pi k} </math>
 
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Revision as of 22:47, 22 April 2018


Table of CT Fourier Series Coefficients and Properties

Fourier series Coefficients

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Properties of CT Fourier systems

Function Fourier Series Coefficients
$ sin(\omega_0t) $ $ \frac{1}{2j}e^{j\omega_0 t} - \frac{1}{2j}e^{-j\omega_0 t} $ $ a_1 = \frac{1}{2j} a_{-1} = -\frac{1}{2j} and a_k = 0 for k not 1,-1 $
$ cos(\omega_0t) $ $ \frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omega_0 t} $ $ a_1 = \frac{1}{2} a_{-1} = \frac{1}{2} and a_k = 0 for k not 1,-1 $
1 2\pi \delta[\omega] a_0 = 1 else a_k = 0
$ e^{\alpha t}u(t) $ $ \frac{1}{\alpha + j\omega} $ $ \frac{1}{\alpha T + j2\pi k} $
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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} $

Time Reversal $ \mathfrak{F}[g(-t)] = G(-\omega) $ $ \mathfrak{F}[g(-t)] = \int_{-\infty}^{\infty}g(-t)e^{-j\omgea t} dt $

replace t with -t
$ \mathfrak{F}[g(-t)] = -\int_{\infty}^{-\infty} g(t')e^{-j\omega t' } dt' $
$ =\int_{-\infty}^{\infty}g(t') e^{-j\omega t'} dt' $
$ =G(-\omega) $

Complex Conjugate $ \mathfrak{F}(g*(t) = G*(-j\omega) $ $ g*(t) = [\frac{1}{2\pi} \int_{-\infty}^{\infty} G(-\omega)e^{j\omega t} d\omega]^* $

$ =[\frac{1}{2\pi} \int_{-\infty}^{\infty} G*(-\omega)e^{j\omega' t} d\omega]^* $
$ x*(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} G*(-\omega')e^{j\omega' t} d\omega $
$ = \mathfrak{F}[G*(-\omega)] $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn