(New page: Category:ECE Category:ECE 301 Category:2007 Category:Fall Category:mboutin Category:MISC Sometimes is is difficult to get answers to fit the solution key. Here is...)
 
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<math> \frac{3}{(k \pi)^2} \left ( cos \left ( \frac{\pi}{3} k \right ) - cos \left ( \frac {2 \pi}{3} k \right ) \right ) </math>
 
<math> \frac{3}{(k \pi)^2} \left ( cos \left ( \frac{\pi}{3} k \right ) - cos \left ( \frac {2 \pi}{3} k \right ) \right ) </math>
  
<math> \begin{align} &= \frac{3}{(k \pi)^2} \left ( \frac{1}{2}e^{jk \frac{\pi}{3}} + \frac{1}{2}e^{-jk \frac{\pi}{3}} - \frac{1}{2}e^{jk \frac{2 \pi}{3}} - \frac{1}{2}e^{-jk \frac{2 \pi}{3}} \right ) \cdot \frac{2}{2} \\
+
<math> \begin{align} &= \frac{3}{(k \pi)^2} \left ( \frac{1}{2} e^{jk \frac{\pi}{3}} + \frac{1}{2} e^{-jk \frac{\pi}{3}} - \frac{1}{2} e^{jk \frac{2 \pi}{3}} - \frac{1}{2} e^{-jk \frac{2 \pi}{3}} \right ) \cdot \frac{2}{2} \\
  
 
&= \frac{6}{(k \pi)^2} \left ( \frac{1}{4} e^{jk \frac{\pi}{3}} + \frac{1}{4} e^{-jk \frac{\pi}{3}} - \frac{1}{4} e^{jk \frac{2 \pi}{3}} - \frac{1}{4} e^{-jk \frac{2 \pi}{3}} \right ) \\
 
&= \frac{6}{(k \pi)^2} \left ( \frac{1}{4} e^{jk \frac{\pi}{3}} + \frac{1}{4} e^{-jk \frac{\pi}{3}} - \frac{1}{4} e^{jk \frac{2 \pi}{3}} - \frac{1}{4} e^{-jk \frac{2 \pi}{3}} \right ) \\
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&= \frac{6}{(k \pi)^2} \left ( -\frac{1}{4} e^{jk \frac {2 \pi}{3}} + \frac{1}{4} e^{jk \frac{\pi}{3}} + \frac{1}{4} e^{-jk \frac{\pi}{3}} - \frac{1}{4} e^{-jk \frac{2 \pi}{3}} \right ) \\
 
&= \frac{6}{(k \pi)^2} \left ( -\frac{1}{4} e^{jk \frac {2 \pi}{3}} + \frac{1}{4} e^{jk \frac{\pi}{3}} + \frac{1}{4} e^{-jk \frac{\pi}{3}} - \frac{1}{4} e^{-jk \frac{2 \pi}{3}} \right ) \\
  
&= \frac{6}{(k \pi)^2} \left ( -\frac{1}{4} e^{jk \frac{\pi}{2}} e^{jk \frac{\pi}{6}} + \frac{1}{4} e^{jk \frac{\pi}{2}} e^{-jk \frac{\pi}{6}} + \frac{1}{4} e^{-jk \frac{\pi}{2}} e^{jk \frac{\pi}{6}} - \frac{1}{4} e^{-jk \frac{\pi}{2}}e^{-jk \frac{\pi}{6]} \right )  
+
&= \frac{6}{(k \pi)^2} \left ( -\frac{1}{4} e^{jk \frac{\pi}{2}} e^{jk \frac{\pi}{6}} + \frac{1}{4} e^{jk \frac{\pi}{2}} e^{-jk \frac{\pi}{6}} + \frac{1}{4} e^{-jk \frac{\pi}{2}} e^{jk \frac{\pi}{6}} - \frac{1}{4} e^{-jk \frac{\pi}{2}}e^{-jk \frac{\pi}{6}} \right ) \\
 +
 
 +
&= \frac{6}{(k \pi)^2} \left ( \frac{1}{2j} e^{jk \frac{\pi}{2}} - \frac{1}{2j} e^{-jk \frac{1\pi}{2}} \right ) \left ( \frac{1}{2j} e^{jk \frac{\pi}{6}} - \frac{1}{2j}e^{-jk \frac{\pi}{6}} \right ) \\
 +
 
 +
&= \frac{6}{(k \pi)^2} sin \left ( \frac{\pi}{2} k \right ) sin \left ( \frac{\pi}{6} k \right )
  
 
\end{align} </math>
 
\end{align} </math>

Revision as of 12:37, 18 December 2008


Sometimes is is difficult to get answers to fit the solution key. Here is a place to post methods for changing seemingly natural answers into the solution key's answers...

Hw 4 Problem 22.a (b)

$ \frac{3}{(k \pi)^2} \left ( cos \left ( \frac{\pi}{3} k \right ) - cos \left ( \frac {2 \pi}{3} k \right ) \right ) $

$ \begin{align} &= \frac{3}{(k \pi)^2} \left ( \frac{1}{2} e^{jk \frac{\pi}{3}} + \frac{1}{2} e^{-jk \frac{\pi}{3}} - \frac{1}{2} e^{jk \frac{2 \pi}{3}} - \frac{1}{2} e^{-jk \frac{2 \pi}{3}} \right ) \cdot \frac{2}{2} \\ &= \frac{6}{(k \pi)^2} \left ( \frac{1}{4} e^{jk \frac{\pi}{3}} + \frac{1}{4} e^{-jk \frac{\pi}{3}} - \frac{1}{4} e^{jk \frac{2 \pi}{3}} - \frac{1}{4} e^{-jk \frac{2 \pi}{3}} \right ) \\ &= \frac{6}{(k \pi)^2} \left ( -\frac{1}{4} e^{jk \frac {2 \pi}{3}} + \frac{1}{4} e^{jk \frac{\pi}{3}} + \frac{1}{4} e^{-jk \frac{\pi}{3}} - \frac{1}{4} e^{-jk \frac{2 \pi}{3}} \right ) \\ &= \frac{6}{(k \pi)^2} \left ( -\frac{1}{4} e^{jk \frac{\pi}{2}} e^{jk \frac{\pi}{6}} + \frac{1}{4} e^{jk \frac{\pi}{2}} e^{-jk \frac{\pi}{6}} + \frac{1}{4} e^{-jk \frac{\pi}{2}} e^{jk \frac{\pi}{6}} - \frac{1}{4} e^{-jk \frac{\pi}{2}}e^{-jk \frac{\pi}{6}} \right ) \\ &= \frac{6}{(k \pi)^2} \left ( \frac{1}{2j} e^{jk \frac{\pi}{2}} - \frac{1}{2j} e^{-jk \frac{1\pi}{2}} \right ) \left ( \frac{1}{2j} e^{jk \frac{\pi}{6}} - \frac{1}{2j}e^{-jk \frac{\pi}{6}} \right ) \\ &= \frac{6}{(k \pi)^2} sin \left ( \frac{\pi}{2} k \right ) sin \left ( \frac{\pi}{6} k \right ) \end{align} $

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