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) $
