# How to compute the 2N-Point DFT in terms of N-Point DFT with other properties involved

Let's say the N-Point DFT of $$x[n]$$ is $$X(k)$$.

Now from this question, I know how to compute the 2N-Point DFT of the same $$x[n]$$ in terms of $$X(k)$$

However, this idea doesn't translate well if I have to compute the 2N-Point DFT of $$x_1[n]$$, for example, where:

$$x_1[n] = \begin{cases} x[n], & \text{if } n \in [0, N-1]\\ x[n-N], & \text{if } n\in [N, 2N-1]\\ 0 & \text{else} \end{cases}$$

Or for example, $$x_2[n]$$, where:

$$x_2[n] = \begin{cases} x[n], & \text{if } n \in [0, N-1]\\ 0 & \text{else} \end{cases}$$

Although for $$x_2[n]$$, I can intuitively tell that $$X(k) = X_2(k)$$ since there is just zero padding past N. For the case $$x_1[n]$$ it is rather difficult to compute the 2N-Point DFT in terms of Even indexed and Odd indexed samples. So is there a more general way to compute the 2N-Point DFT in terms of N-Point DFT?

The sequence $$x_1[n]$$ is just a concatenation of two periods of $$x[n]$$, which corresponds to upsampling in the frequency domain. I.e., half of the values of the length $$2N$$ DFT of $$x_1[n]$$ will be the (scaled) original values of the length $$N$$ DFT of $$x[n]$$, and the other half will be zero. Of course, this can be shown by writing out the DFT sum:
\begin{align}X_1[k]&=\sum_{n=0}^{2N-1}x_1[n]e^{-j\frac{2\pi}{2N}nk}\\&=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{2N}nk}+\sum_{n=N}^{2N-1}x[n-N]e^{-j\frac{2\pi}{2N}nk}\\&=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{2N}nk}+\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{2N}(n+N)k}\\&=\sum_{n=0}^{N-1}\big(x[n]+x[n]e^{-jk\pi}\big)e^{-j\frac{2\pi}{2N}nk}\\&=\sum_{n=0}^{N-1}\big(x[n]+(-1)^kx[n]\big)e^{-j\frac{2\pi}{N}n\frac{k}{2}}\\&=\begin{cases}2X\left[\frac{k}{2}\right],&k\textrm{ even}\\0,& k\textrm{ odd}\end{cases}\end{align}