Line 7: Line 7:
 
----
 
----
 
==Answer==
 
==Answer==
Can somebody finish the following computation?
+
The first step is to figure out the period N of x[n].
  
 +
.....please fill in....
 +
 +
 +
Then, one needs to find the coefficients using the summation formula
 
<math>
 
<math>
 
\begin{align}
 
\begin{align}
a_n &= \frac{1}{T} \sum_{2\pi} x[n] e^{-j \frac{2\pi}{T}nk} \\
+
a_n &= \frac{1}{T} \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N}nk}, \text{  by the definition of Fourier series coefficients,} \\
 
& = ...\\
 
& = ...\\
 +
& = ... \text{ please finish }
 
\end{align}
 
\end{align}
 
</math>
 
</math>
  
not finished, I'll come back to this later.  @_@
 
 
-Ty
 
  
  

Revision as of 14:20, 14 September 2010

Exercise: Compute the DT Fourier series coefficients of the following discrete-time signal:

$ x[n]=\sum_{k=-\infty}^\infty \left( u[n-5k]-u[n-4-5k] \right) $

After you have obtained the coefficients, write the Fourier series of x[n].

Help: If the function above does not look periodic to you, please read this page.


Answer

The first step is to figure out the period N of x[n].

.....please fill in....


Then, one needs to find the coefficients using the summation formula $ \begin{align} a_n &= \frac{1}{T} \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N}nk}, \text{ by the definition of Fourier series coefficients,} \\ & = ...\\ & = ... \text{ please finish } \end{align} $




More exercises on computing discrete-time Fourier series

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