(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)
Line 11: Line 11:
  
 
<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}
Line 21: Line 24:
  
 
<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} \\
Line 36: Line 39:
  
 
<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>
 +
 +
  
  
Line 47: Line 54:
  
 
<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>
  
Line 57: Line 65:
 
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

Back to Lab Week 7 Quiz Pool

Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett