| Line 14: | Line 14: | ||
&x(n)=\frac{1}{N}\sum_{k=0}^{N-1}X(k)e^{j\frac{j2\pi kn}{N}} \\ | &x(n)=\frac{1}{N}\sum_{k=0}^{N-1}X(k)e^{j\frac{j2\pi kn}{N}} \\ | ||
&\text{Switch n and k} \\ | &\text{Switch n and k} \\ | ||
| − | \Leftrightarrow &x(k)=\frac{1}{N}\sum_{ | + | \Leftrightarrow &x(k)=\frac{1}{N}\sum_{n=0}^{N-1}X(n)e^{j\frac{j2\pi kn}{N}} \\ |
| − | \Leftrightarrow &Nx(-k)=\sum_{ | + | \Leftrightarrow &Nx(-k)=\sum_{n=0}^{N-1}X(n)e^{-j\frac{j2\pi kn}{N}}=\text{ DFT of }X(n) |
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 06:40, 21 October 2010
ECE438 Week9 Quiz Question6 Solution
The Duality Property of DFT is descriped below:
$ \text{If the DFT of }x[n]\text{ is denoted as }X(k)\text{ with a length of N} $
$ \text{Then the DFT of }X(n)\text{ is }Nx(-k) $
Derivation:
$ \begin{align} &x(n)=\frac{1}{N}\sum_{k=0}^{N-1}X(k)e^{j\frac{j2\pi kn}{N}} \\ &\text{Switch n and k} \\ \Leftrightarrow &x(k)=\frac{1}{N}\sum_{n=0}^{N-1}X(n)e^{j\frac{j2\pi kn}{N}} \\ \Leftrightarrow &Nx(-k)=\sum_{n=0}^{N-1}X(n)e^{-j\frac{j2\pi kn}{N}}=\text{ DFT of }X(n) \end{align} $
Conclusion:
if $ x(n)\xrightarrow{DFT} X(k) $
then $ X(n)\xrightarrow{DFT} Nx(-k) $
