(New page: <math>e^{j\omega_0 t} ---> 2\pi \delta (\omega - \omega_0) </math>)
 
 
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<math>e^{j\omega_0 t} ---> 2\pi \delta (\omega - \omega_0) </math>
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<math>x(t)=e^{j\omega_0 t} \longrightarrow {\mathcal X}(\omega)= 2\pi \delta (\omega - \omega_0) </math>

Latest revision as of 12:07, 14 November 2008

$ x(t)=e^{j\omega_0 t} \longrightarrow {\mathcal X}(\omega)= 2\pi \delta (\omega - \omega_0) $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood