Line 23: Line 23:
 
<math>\begin{align}
 
<math>\begin{align}
 
x[n] &= \frac{1}{N}\sum_{n=0}^{N-1}e^{\frac{j2\pi nk}{N}} \\
 
x[n] &= \frac{1}{N}\sum_{n=0}^{N-1}e^{\frac{j2\pi nk}{N}} \\
\frac{1}{2}e^{\frac{j\pi n}{2}} + \frac{1}{2}e^{\frac{j\pi n}{12}} &= \frac{1}{12}\sum_{n=0}^{11}e^{\frac{j2\pi nk}{12}}
+
\frac{1}{2}e^{\frac{j\pi n}{2}} + \frac{1}{2}e^{\frac{j\pi n}{12}} &= \frac{1}{12}\sum_{n=0}^{11}e^{\frac{j2\pi nk}{12}}  
 
\end{align}</math>
 
\end{align}</math>
  
For <math>k=0,1,...,N-1</math>
+
By comparison, we know for <math>k=0,1,...,11</math>
 +
 
 +
<math class="inline">
 +
X_{12}[k] = \left\{
 +
\begin{array}{ll}
 +
6, & k=1,6 \\
 +
0, & otherwise.
 +
\end{array}
 +
\right.
 +
</math>  
 +
 
 +
c)
  
  

Revision as of 14:31, 13 October 2011

Homework 4, ECE438, Fall 2011, Prof. Boutin


Question 1

a) For $ k=0,1,...,N-1 $

$ \begin{align} X_N(k) &= \sum_{k=0}^{N-1}x[n]e^{-\frac{j2\pi nk}{N}} \\ &= x[0]e^{-\frac{j2\pi 0\cdot k}{N}} \\ &= 1 \end{align} $

b) Using Euler Formula, we have

$ \begin{align} x[n] &= e^{\frac{j\pi n}{3}}(\frac{ e^{\frac{j\pi n}{6}} + e^{-\frac{j\pi n}{6}} }{2}) \\ &= \frac{1}{2}e^{\frac{j\pi n}{2}} + \frac{1}{2}e^{\frac{j\pi n}{12}} \end{align} $

Observing that $ x[n] $ has fundamental period $ N=12 $. Using IDFT, we have

$ \begin{align} x[n] &= \frac{1}{N}\sum_{n=0}^{N-1}e^{\frac{j2\pi nk}{N}} \\ \frac{1}{2}e^{\frac{j\pi n}{2}} + \frac{1}{2}e^{\frac{j\pi n}{12}} &= \frac{1}{12}\sum_{n=0}^{11}e^{\frac{j2\pi nk}{12}} \end{align} $

By comparison, we know for $ k=0,1,...,11 $

$ X_{12}[k] = \left\{ \begin{array}{ll} 6, & k=1,6 \\ 0, & otherwise. \end{array} \right. $

c)



Question 2


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