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(DT signal & its Fourier Coefficients)
(DT signal & its Fourier Coefficients)
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<math>\ x[n] = \frac{5}{2j} (e^{j (3\pi n + \frac{\pi}{4})}-e^{-j (3\pi n + \frac{\pi}{4})}) </math>
 
<math>\ x[n] = \frac{5}{2j} (e^{j (3\pi n + \frac{\pi}{4})}-e^{-j (3\pi n + \frac{\pi}{4})}) </math>
  
<math>\ x[n] = \frac{5}{2j} (e^{j (3\pi n)} e^{\frac{\pi}{4})}-e^{-j (3\pi n)}e^{\frac{\pi}{4}}) </math>
+
<math>\ x[n] = \frac{5}{2j} (e^{j (3\pi n)} e^{\frac{\pi}{4}}-e^{-j (3\pi n)}e^{\frac{\pi}{4}}) </math>
  
 
:<math>e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} </math><br>
 
:<math>e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} </math><br>
  
 
:<math>e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} </math>
 
:<math>e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} </math>

Revision as of 15:38, 26 September 2008

DT signal & its Fourier Coefficients

$ \ x[n] = 5sin(3 \pi n + \frac{\pi}{4}) $

Knowing its Fourier series is

$ \ x[n] = \frac{5}{2j} (e^{j (3\pi n + \frac{\pi}{4})}-e^{-j (3\pi n + \frac{\pi}{4})}) $

$ \ x[n] = \frac{5}{2j} (e^{j (3\pi n)} e^{\frac{\pi}{4}}-e^{-j (3\pi n)}e^{\frac{\pi}{4}}) $

$ e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} $
$ e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} $

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009