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<math> \displaystyle e^{j\pi}=-1 \;\;\;\;\;\;\; \cos(t\theta) = (e^{j\theta}+e^{-j\theta})/2 \;\;\;\;\;\;\;\;\;\;\;\; sin(t\theta) = (e^{j\theta}-e^{-j\theta})/(2j) </math>
 
<math> \displaystyle e^{j\pi}=-1 \;\;\;\;\;\;\; \cos(t\theta) = (e^{j\theta}+e^{-j\theta})/2 \;\;\;\;\;\;\;\;\;\;\;\; sin(t\theta) = (e^{j\theta}-e^{-j\theta})/(2j) </math>
  
<math> \Im </math>
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<math> \mathcal{F} </math>
 
[[ 2010 Fall ECE 438 Boutin/ECE438Mid1FormulaSheet Work|Back to 2010 Fall ECE 438 Boutin/ECE438Mid1FormulaSheet Work]]
 
[[ 2010 Fall ECE 438 Boutin/ECE438Mid1FormulaSheet Work|Back to 2010 Fall ECE 438 Boutin/ECE438Mid1FormulaSheet Work]]

Revision as of 06:35, 30 September 2010

2010_Fall_ECE_438_Boutin/ECE438Mid1FormulaSheet_Work_wrk

  • Fourier series of a continuous-time signal x(t) periodic with period T
  • Fourier series coefficients of a continuous-time signal x(t) periodic with period T
$ CTFS $ $ x(t)=\sum_{n=-\infty}^\infty a_n e^{j \frac{2\pi}{T}nt}\;\;\;\;\;\;\;\;\;\;\;\;\;\;a_n=\frac{1}{T} \int_{0}^T x(t) e^{-j \frac{2\pi}{T}nt}dt $
$ CTFT $$ \ x(t) = \int_{-\infty}^{\infty} \chi(f)\ e^{j 2 \pi f t}\,df \;\;\;\;\;\;\;\;\;\;\;\;\;\ \chi(f) = \int_{-\infty}^{\infty} x(t)\ e^{- j 2 \pi f t}\,dt $



$ rep_T [x(t)] = x(t)* \sum_{k=-\infty}^{\infty}\delta(t-kT) \;\;\;\;\;\;\;\;\;comb_T[x(t)] = x(t) . \sum_{k=-\infty}^{\infty}\delta(t-kT) $
$ rep_T [x(t)] \iff \frac{1}{T}comb_\frac{1}{T} [ \mathrm{X}(f)] \;\;\;\;\;\;\;\;\;\; comb_T [x(t)] \iff \frac{1}{T}rep_\frac{1}{T} [ \mathrm{X}(f)] $

$ \displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)\;\;\;\;\;\;for\;\;\alpha>0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;sinc(\theta)=sin(\pi\theta)/(\pi\theta) $

$ \displaystyle e^{j\pi}=-1 \;\;\;\;\;\;\; \cos(t\theta) = (e^{j\theta}+e^{-j\theta})/2 \;\;\;\;\;\;\;\;\;\;\;\; sin(t\theta) = (e^{j\theta}-e^{-j\theta})/(2j) $

$ \mathcal{F} $ Back to 2010 Fall ECE 438 Boutin/ECE438Mid1FormulaSheet Work

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin