(New page: Category:2010 Fall ECE 438 Boutin ---- == Solution to Q3 of Week 7 Quiz Pool == ---- <math>\begin{align} \text{(a)} \quad & y[n]=e^{j\frac{2\pi}{N}n}x[n], \;\;\; n=0,...,N-1 \\ & X_N...)
 
 
(8 intermediate revisions by 2 users not shown)
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<math>\begin{align}
 
<math>\begin{align}
−
Y_N[k] &= \sum_{n=0}^{N-1}e^{j\frac{2\pi}{N}n}x[n]e^{-j\frac{2\pi}{N}kn}, \;\;\; k=0,...,N-1 \\
+
{\color{White}abcde} Y_N[k] &= \sum_{n=0}^{N-1}e^{j\frac{2\pi}{N}n}x[n]e^{-j\frac{2\pi}{N}kn}, \;\;\; k=0,...,N-1 \\
 
&= \sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}(k-1)n} \\
 
&= \sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}(k-1)n} \\
 
&= X_N[k-1] \\
 
&= X_N[k-1] \\
−
\end{align}\,\!</math>
+
\end{align}</math>
 +
 
 +
 
 +
 
  
 
<math>\begin{align}
 
<math>\begin{align}
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<math>\begin{align}
 
<math>\begin{align}
−
Y_N[k] &= x[N-1] + \sum_{n=1}^{N-1}x[n-1]e^{-j\frac{2\pi}{N}kn}, \;\;\; \text{Let } m=n-1  \\
+
{\color{White}abcde} Y_N[k] &= x[N-1] + \sum_{n=1}^{N-1}x[n-1]e^{-j\frac{2\pi}{N}kn}, \;\;\; \text{Let } m=n-1  \\
 
&= x[N-1] + \sum_{m=0}^{N-2}x[m]e^{-j\frac{2\pi}{N}k(m+1)} \\
 
&= x[N-1] + \sum_{m=0}^{N-2}x[m]e^{-j\frac{2\pi}{N}k(m+1)} \\
 
&= x[N-1] + e^{-j\frac{2\pi}{N}k}\sum_{m=0}^{N-2}x[m]e^{-j\frac{2\pi}{N}km} \\
 
&= x[N-1] + e^{-j\frac{2\pi}{N}k}\sum_{m=0}^{N-2}x[m]e^{-j\frac{2\pi}{N}km} \\
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<math>\begin{align}
 
<math>\begin{align}
−
Y_{2N}[k] &= \sum_{n=0, n\text{ even}}^{2N-1}x\left[\frac{n}{2}\right]e^{-j\frac{2\pi}{N}kn}, \;\;\; \text{Let } n=2m \\
+
{\color{White}abcde} Y_{2N}[k] &= \sum_{n=0}^{2N-1}y[n]e^{-j\frac{2\pi}{2N}kn} \\
−
&= \sum_{m=0}^{N-1}x[m]e^{-j\frac{2\pi}{N}k2m} \\
+
&= \sum_{n=0, n\text{ even}}^{2N-1}x\left[\frac{n}{2}\right]e^{-j\frac{2\pi}{2N}kn}, \;\;\; \text{Let } n=2m \\
−
&= X_N[2k], \;\;\; k=0,...,2N-1 \\
+
&= \sum_{m=0}^{N-1}x[m]e^{-j\frac{2\pi}{2N}k2m} \\
 +
&= \sum_{m=0}^{N-1}x[m]e^{-j\frac{2\pi}{N}km}, \;\;\; k=0,...,2N-1 \\
 +
&= X_N[k], \;\;\; k=0,...,2N-1 \\
 
\end{align}\,\!</math>
 
\end{align}\,\!</math>
 +
 +
  
  
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<math>\begin{align}
 
<math>\begin{align}
−
Y_{N/2}[k] &= \sum_{n=0}^{\frac{N}{2}-1}x[2n]e^{-j\frac{2\pi}{N/2}kn} = \sum_{n=0, n\text{ even}}^{N-1}x[n]e^{-j\frac{2\pi}{N}kn} \\
+
{\color{White}abcde} Y_{\frac{N}{2}}[k] &= \sum_{n=0}^{\frac{N}{2}-1}x[2n]e^{-j\frac{2\pi}{N/2}kn} = \sum_{n=0, n\text{ even}}^{N-1}x[n]e^{-j\frac{2\pi}{N}kn} \\
 
&= \sum_{n=0}^{N-1} \frac{1}{2}\left(1+(-1)^n\right)x[n]e^{-j\frac{2\pi}{N}kn} \\
 
&= \sum_{n=0}^{N-1} \frac{1}{2}\left(1+(-1)^n\right)x[n]e^{-j\frac{2\pi}{N}kn} \\
 
&= \frac{1}{2}\sum_{n=0}^{N-1} x[n]e^{-j\frac{2\pi}{N}kn} + \frac{1}{2}\sum_{n=0}^{N-1} (-1)^n x[n]e^{-j\frac{2\pi}{N}kn} \\
 
&= \frac{1}{2}\sum_{n=0}^{N-1} x[n]e^{-j\frac{2\pi}{N}kn} + \frac{1}{2}\sum_{n=0}^{N-1} (-1)^n x[n]e^{-j\frac{2\pi}{N}kn} \\
 
&= \frac{1}{2}\sum_{n=0}^{N-1} x[n]e^{-j\frac{2\pi}{N}kn} + \frac{1}{2}\sum_{n=0}^{N-1} e^{-j\frac{2\pi}{N}\left(\frac{N}{2}\right)n} x[n]e^{-j\frac{2\pi}{N}kn}, \;\;\; (e^{j\pi n}=(-1)^n) \\
 
&= \frac{1}{2}\sum_{n=0}^{N-1} x[n]e^{-j\frac{2\pi}{N}kn} + \frac{1}{2}\sum_{n=0}^{N-1} e^{-j\frac{2\pi}{N}\left(\frac{N}{2}\right)n} x[n]e^{-j\frac{2\pi}{N}kn}, \;\;\; (e^{j\pi n}=(-1)^n) \\
−
&= \frac{1}{2}X_N[k] + \frac{1}{2}X_N[K-\frac{N}{2}], \;\;\; k=0,...,\frac{N}{2}-1 \\
+
&= \frac{1}{2}X_N[k] + \frac{1}{2}X_N\left[k-\frac{N}{2}\right], \;\;\; k=0,...,\frac{N}{2}-1 \\
 +
 
 
\end{align}\,\!</math>
 
\end{align}\,\!</math>
  
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Credit: Prof. Charles Bouman
 
Credit: Prof. Charles Bouman
  
−
Back to [[ECE438_Week6_Quiz|Lab Week 6 Quiz Pool]]
+
Back to [[ECE438_Week7_Quiz|Lab Week 7 Quiz Pool]]
  
 
Back to [[ECE438_Lab_Fall_2010|ECE 438 Fall 2010 Lab Wiki Page]]
 
Back to [[ECE438_Lab_Fall_2010|ECE 438 Fall 2010 Lab Wiki Page]]
  
 
Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]]
 
Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]]

Latest revision as of 19:22, 26 February 2012


Solution to Q3 of Week 7 Quiz Pool

$ \begin{align} \text{(a)} \quad & y[n]=e^{j\frac{2\pi}{N}n}x[n], \;\;\; n=0,...,N-1 \\ & X_N[k]=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}kn} \\ \end{align}\,\! $

$ \begin{align} {\color{White}abcde} Y_N[k] &= \sum_{n=0}^{N-1}e^{j\frac{2\pi}{N}n}x[n]e^{-j\frac{2\pi}{N}kn}, \;\;\; k=0,...,N-1 \\ &= \sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}(k-1)n} \\ &= X_N[k-1] \\ \end{align} $



$ \begin{align} \text{(b)} \quad & y[n]=\left\{\begin{array}{ll}x[N-1], & n=0,\\ x[n-1], & n=1,...,N-1\end{array} \right.\\ \end{align}\,\! $

$ \begin{align} {\color{White}abcde} Y_N[k] &= x[N-1] + \sum_{n=1}^{N-1}x[n-1]e^{-j\frac{2\pi}{N}kn}, \;\;\; \text{Let } m=n-1 \\ &= x[N-1] + \sum_{m=0}^{N-2}x[m]e^{-j\frac{2\pi}{N}k(m+1)} \\ &= x[N-1] + e^{-j\frac{2\pi}{N}k}\sum_{m=0}^{N-2}x[m]e^{-j\frac{2\pi}{N}km} \\ &= x[N-1] + e^{-j\frac{2\pi}{N}k}\sum_{m=0}^{N-1}x[m]e^{-j\frac{2\pi}{N}km} - e^{-j\frac{2\pi}{N}k}x[N-1]e^{-j\frac{2\pi}{N}k(N-1)} \\ &= x[N-1](1-e^{-j2\pi k}) + e^{-j\frac{2\pi}{N}k}\sum_{m=0}^{N-1}x[m]e^{-j\frac{2\pi}{N}km}, \;\;\; ( e^{-j2\pi k} = 1, \; \forall \; \text{integer} \; k ) \\ &= e^{-j\frac{2\pi}{N}k}\sum_{m=0}^{N-1}x[m]e^{-j\frac{2\pi}{N}km} \\ &= e^{-j\frac{2\pi}{N}k} X_N[k], \;\;\; k=0,...,N-1 \\ \end{align}\,\! $


$ \text{(c)} \quad y[n]=\left\{\begin{array}{ll}x[n/2], & n \text{ is even},\\ 0, & n \text{ is odd},\end{array} \right. n=0,...,2N-1 $

$ \begin{align} {\color{White}abcde} Y_{2N}[k] &= \sum_{n=0}^{2N-1}y[n]e^{-j\frac{2\pi}{2N}kn} \\ &= \sum_{n=0, n\text{ even}}^{2N-1}x\left[\frac{n}{2}\right]e^{-j\frac{2\pi}{2N}kn}, \;\;\; \text{Let } n=2m \\ &= \sum_{m=0}^{N-1}x[m]e^{-j\frac{2\pi}{2N}k2m} \\ &= \sum_{m=0}^{N-1}x[m]e^{-j\frac{2\pi}{N}km}, \;\;\; k=0,...,2N-1 \\ &= X_N[k], \;\;\; k=0,...,2N-1 \\ \end{align}\,\! $



$ \text{(d)} \quad y[n]=x[2n], \;\;\; n=0,...,\frac{N}{2}-1, \;\; N\text{ even.} $

$ \begin{align} {\color{White}abcde} Y_{\frac{N}{2}}[k] &= \sum_{n=0}^{\frac{N}{2}-1}x[2n]e^{-j\frac{2\pi}{N/2}kn} = \sum_{n=0, n\text{ even}}^{N-1}x[n]e^{-j\frac{2\pi}{N}kn} \\ &= \sum_{n=0}^{N-1} \frac{1}{2}\left(1+(-1)^n\right)x[n]e^{-j\frac{2\pi}{N}kn} \\ &= \frac{1}{2}\sum_{n=0}^{N-1} x[n]e^{-j\frac{2\pi}{N}kn} + \frac{1}{2}\sum_{n=0}^{N-1} (-1)^n x[n]e^{-j\frac{2\pi}{N}kn} \\ &= \frac{1}{2}\sum_{n=0}^{N-1} x[n]e^{-j\frac{2\pi}{N}kn} + \frac{1}{2}\sum_{n=0}^{N-1} e^{-j\frac{2\pi}{N}\left(\frac{N}{2}\right)n} x[n]e^{-j\frac{2\pi}{N}kn}, \;\;\; (e^{j\pi n}=(-1)^n) \\ &= \frac{1}{2}X_N[k] + \frac{1}{2}X_N\left[k-\frac{N}{2}\right], \;\;\; k=0,...,\frac{N}{2}-1 \\ \end{align}\,\! $

Credit: Prof. Charles Bouman

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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva