I have a problem with expressing odd samples of X2 in terms of X1. I understand that the resulting DFT will be more precise in terms of expressing the exact spectrum of signal x[n], due to more samples. Moreover I know that even samples of X2(k) are copy of the spectrum of X1(k), however I do not know how to mathematically compute the odd ones.
The task is as follows:
Let $x[n]$ be a periodic sequence with period $N_1$ . Thus $x[n]$ is also periodic for period $N_2 = 2 N_1$ . We may compute $X_1[k]$: $N$-point DFT of $x[n]$ and $X_2[k]$: $2N$-point DFT of $x[n]$.
- Express $X_2$in terms of $X_1$ Hint: it is easy with even samples $X_2[2m]$ , harder with odd ones $X2_[2m+ 1]$ , for $m$ integer.