# Expressing 2N point DFT in terms of N point DFT

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.

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.

Consider a sequence $$x[n]$$ of length $$N$$ whose $$N$$-point DFT is $$X[k]$$. Then let $$X_2[k]$$ be the $$2N$$-point DFT of $$x[n]$$.
As you have stated, the even indexed samples of $$X_2[k]$$ will be easily shown to be: $$X_2[k] = X[k/2] ~~~,~~~k = 2m, m=0,1,...,N-1$$
Then the odd indexed samples of $$X_2[k]$$ will be given by the $$N$$-point DFT of the signal $$x[n] e^{-j \frac{\pi}{N} n }$$, $$n=0,1,...,N-1$$.
$$X_{2}[k] = \sum_{n=0}^{N-1} x[n] e^{-j \frac{\pi}{N} n} e^{-j \frac{2\pi}{N} k n} = X[k + 0.5] ~~~~ ,~~~ k =2m+1, m=0,1,...,N-1$$
• Can you explain why we use here $$e^{-j\frac{\pi}{N}n}$$ and not $$e^{-j\frac{2\pi}{N}n}$$? – Celeborth Nov 25 '18 at 12:18