I am trying to solve the above question. I am not sure how to proceed. I know the formula for 64 point DFT of $x[n]$.
$X[k]=\sum_{n=0}^{63} x[n] e^{-j2\pi nk/64}$
But how can I find $R[k]$ and $Y[k]$? Can anyone please tell the relation?
I am trying to solve the above question. I am not sure how to proceed. I know the formula for 64 point DFT of $x[n]$.
$X[k]=\sum_{n=0}^{63} x[n] e^{-j2\pi nk/64}$
But how can I find $R[k]$ and $Y[k]$? Can anyone please tell the relation?
Given a finite length input sequence $x[n]$ of $N=64$ points, its 64-point DFT be $X[k]$. Then according to the block diagram you've provided, the signal $r[n]$ will be of $N/2$ points long and its $N/2$ - point DFT $R[k]$ will be related to $X[k]$ as :
$R[k] = X[2k]$, which will be manipulated as:
$$ \begin{align} X[k] &= \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} kn} ~~~,~~~ k=0,1,...,N-1 \\ R[k] & = X[2k] = \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} 2kn} ~~~,~~~ k=0,1,...,N/2-1\\ &=\sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N/2} kn} \\ &=\sum_{n=0}^{\frac{N}{2}-1} x[n] e^{-j \frac{2\pi}{N/2} kn} + \sum_{n=\frac{N}{2}}^{N-1} x[n] e^{-j \frac{2\pi}{N/2} kn} \\ &=\sum_{n=0}^{\frac{N}{2}-1} x[n] e^{-j \frac{2\pi}{N/2} kn} + \sum_{n=0}^{\frac{N}{2}-1} x[n+\frac{N}{2}] e^{-j \frac{2\pi}{N/2} kn} \\ &=\sum_{n=0}^{\frac{N}{2}-1} \left( x[n]+x[n+\frac{N}{2}] \right) e^{-j \frac{2\pi}{N/2} kn} \\ &= \text{N/2-point DFT} \{ r[n] \}\\ &= \sum_{n=0}^{\frac{N}{2}-1} r[n] e^{-j \frac{2\pi}{N/2} kn} \\ \end{align} $$
hence we see that $r[n]$ is the sequence whose length (or period) is reduced to $N/2$ and is generated by adding the second half of $x[n]$ onto the first half.