(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...) |
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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} | + | \end{align}</math> |
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<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}{ | + | {\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}{ | + | &= \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[ | + | &= \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 | + | {\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[ | + | &= \frac{1}{2}X_N[k] + \frac{1}{2}X_N\left[k-\frac{N}{2}\right], \;\;\; k=0,...,\frac{N}{2}-1 \\ |
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\end{align}\,\!</math> | \end{align}\,\!</math> | ||
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Credit: Prof. Charles Bouman | Credit: Prof. Charles Bouman | ||
| − | Back to [[ | + | 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 18:22, 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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