| Line 5: | Line 5: | ||
| − | + | ==Question 1== | |
| + | |||
| + | a) For <math>k=0,1,...,N-1</math> | ||
| + | |||
| + | <math>\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}</math> | ||
| + | |||
| + | b) Using Euler Formula, we have | ||
| + | |||
| + | <math>\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}{6}} | ||
| + | \end{align}</math> | ||
| + | |||
| + | Observing that <math>x[n]</math> has fundamental period <math>N=12</math>. Using IDFT, we have | ||
| + | |||
| + | <math>\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}{6}} &= \frac{1}{12}\sum_{n=0}^{11}e^{\frac{j2\pi nk}{12}} | ||
| + | \end{align}</math> | ||
| + | |||
| + | By comparison, we know for <math>k=0,1,...,11</math> | ||
| + | |||
| + | <math class="inline"> | ||
| + | X_{12}[k] = \left\{ | ||
| + | \begin{array}{ll} | ||
| + | 6, & k=1,3 \\ | ||
| + | 0, & otherwise. | ||
| + | \end{array} | ||
| + | \right. | ||
| + | </math> | ||
| + | |||
| + | c) | ||
| + | |||
| + | <math>x[n]=(\frac{1}{\sqrt 2} + j\frac{1}{\sqrt 2})^n = (e^{\frac{j\pi}{4}})^n</math> | ||
| + | |||
| + | Then <math>x[n]</math> has fundamental period <math>N=8</math>. Using IDFT, we have | ||
| + | |||
| + | <math>\begin{align} | ||
| + | x[n] &= \frac{1}{N}\sum_{n=0}^{N-1}e^{\frac{j2\pi nk}{N}} \\ | ||
| + | e^{\frac{j\pi n}{4}} &= \frac{1}{8}\sum_{n=0}^{7}e^{\frac{j2\pi nk}{8}} | ||
| + | \end{align}</math> | ||
| + | |||
| + | By comparison, we know for <math>k=0,1,...,7</math> | ||
| + | |||
| + | <math class="inline"> | ||
| + | X_{8}[k] = \left\{ | ||
| + | \begin{array}{ll} | ||
| + | 8, & k=1 \\ | ||
| + | 0, & otherwise. | ||
| + | \end{array} | ||
| + | \right. | ||
| + | </math> | ||
Revision as of 08:45, 30 September 2013
HW6_Solution_ECE438F13
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}{6}} \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}{6}} &= \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=1,3 \\ 0, & otherwise. \end{array} \right. $
c)
$ x[n]=(\frac{1}{\sqrt 2} + j\frac{1}{\sqrt 2})^n = (e^{\frac{j\pi}{4}})^n $
Then $ x[n] $ has fundamental period $ N=8 $. Using IDFT, we have
$ \begin{align} x[n] &= \frac{1}{N}\sum_{n=0}^{N-1}e^{\frac{j2\pi nk}{N}} \\ e^{\frac{j\pi n}{4}} &= \frac{1}{8}\sum_{n=0}^{7}e^{\frac{j2\pi nk}{8}} \end{align} $
By comparison, we know for $ k=0,1,...,7 $
$ X_{8}[k] = \left\{ \begin{array}{ll} 8, & k=1 \\ 0, & otherwise. \end{array} \right. $
