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! Coefficients
 
! Coefficients
 
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|<math>sin(\omega_0t) </math>
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|<math>\frac{1}{2j}e^{j\omega_0 t} - \frac{1}{2j}e^{-j\omega_0 t} </math>
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|<math>a_1 = \frac{1}{2j} a_{-1} = -\frac{1}{2j} and a_k = 0 for k not 1,-1 </math>
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|-
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|<math>cos(\omega_0t) </math>
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|<math>\frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omega_0 t} </math>
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|<math>a_1 = \frac{1}{2} a_{-1} = \frac{1}{2} and a_k = 0 for k not 1,-1 </math>
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|-
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|1
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|2\pi \delta[\omega]
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| a_0 = 1 else a_k = 0
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|-
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|<math>e^{\alpha t}u(t) </math>
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|<math>\frac{1}{\alpha + j\omega} </math>
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|<math> \frac{1}{\alpha T + j2\pi k} </math>
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|-
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}
  
  
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<math>=e^{-i2\pi fa}\int_{-\infty}^\infty g(u)e^{-i2\pi fu}du </math><br/>
 
<math>=e^{-i2\pi fa}\int_{-\infty}^\infty g(u)e^{-i2\pi fu}du </math><br/>
 
<math>=e^{-i2\pi fa} G(f)</math><br/>
 
<math>=e^{-i2\pi fa} G(f)</math><br/>
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|-
  
 
|Time Scaling  
 
|Time Scaling  
|<math>\mathfrak{F}(g(ct)) = |fraq{G(\fraq{f}{c}}{|c|} </math><br/>
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|<math>\mathfrak{F}(g(ct)) = \frac{G(\frac{f}{c})}{|c|} </math><br/>
 
|<math>\mathfrak{F}(g(ct)) = \int_{-\infty}^\infty g(ct)e^{-i2\pi ft}dt </math><br/>
 
|<math>\mathfrak{F}(g(ct)) = \int_{-\infty}^\infty g(ct)e^{-i2\pi ft}dt </math><br/>
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subtitute : u = ct, du = cdt <br/>
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<math> \mathfrak{F}(g(ct)) = \int_{-c\infty}^{c\infty} \frac{g(u)}{c}e^{-i2\pi f\frac{u}{c}}du </math><br/>
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if c is greater than 0: then no signs change.
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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.
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|-
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|Frequency Shifting
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|<math>\mathfrak{F}^{-1}[X(j\omega + omega_0)] = x(t)e^{-j\omega_0t} </math>
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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/>
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<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}</math><br/>
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|-
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|Time Reversal
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|<math>\mathfrak{F}[g(-t)] = G(-\omega) </math>
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|<math>\mathfrak{F}[g(-t)] = \int_{-\infty}^{\infty}g(-t)e^{-j\omega t} dt</math><br/>
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replace t with -t <br/>
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<math>\mathfrak{F}[g(-t)] = -\int_{\infty}^{-\infty} g(t')e^{-j\omega t'
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} dt' </math><br/>
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<math> =\int_{-\infty}^{\infty}g(t') e^{-j\omega t'} dt' </math><br/>
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<math> =G(-\omega) </math>
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|-
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|Complex Conjugate
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|<math>\mathfrak{F}(g*(t) = G*(-j\omega) </math>
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|<math> g*(t) = [\frac{1}{2\pi} \int_{-\infty}^{\infty} G(-\omega)e^{j\omega t} d\omega]^* </math><br/>
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<math>=[\frac{1}{2\pi} \int_{-\infty}^{\infty} G*(-\omega)e^{j\omega' t} d\omega]^* </math><br/>
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<math> x*(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} G*(-\omega')e^{j\omega' t} d\omega </math><br/>
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<math> = \mathfrak{F}[G*(-\omega)]</math><br/>
 
|-
 
|-
 
}
 
}
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----
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[[2018_Spring_ECE_301_Boutin|Back to ECE301 Spring 2018 Prof. Boutin]]
 +
 +
--[[User:Plomada|Plomada]] 10:51 22 April 2018 (UTC)

Latest revision as of 21:52, 22 April 2018


Table of CT Fourier Series Coefficients and Properties

Fourier series Coefficients

}

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} $
}

Back to ECE301 Spring 2018 Prof. Boutin

--Plomada 10:51 22 April 2018 (UTC)

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\omega 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