(A periodic CT signal)
(A periodic CT signal)
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Input CT signal: <math> x(t) = cos2t+sin2t</math>
 
Input CT signal: <math> x(t) = cos2t+sin2t</math>
  
<math>\,x(t)={e^{2j\pi t}+e^{-2j \pi t}}{2} + {e^{j2 \pi t}-e^{-j2 \pi t}}{2j}</math>
+
<math>\,x(t)=\frac {e^{2j\pi t}+e^{-2j \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j}</math>
 +
 
  
 
<math>x(t)=e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{j2\pi t}+\frac{-1-j}{2j}e^{-j2\pi t}</math>
 
<math>x(t)=e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{j2\pi t}+\frac{-1-j}{2j}e^{-j2\pi t}</math>

Revision as of 15:19, 26 September 2008

A periodic CT signal

Fourier series of x(t):
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $

, where $ a_k $ is
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.


Input CT signal: $ x(t) = cos2t+sin2t $

$ \,x(t)=\frac {e^{2j\pi t}+e^{-2j \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j} $


$ x(t)=e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{j2\pi t}+\frac{-1-j}{2j}e^{-j2\pi t} $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman