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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?

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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:

$$\begin{align}X_1[k]&=\sum_{n=0}^{2N-1}x_1[n]e^{-j\frac{2\pi}{2N}nk}\\&=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{2N}nk}+\sum_{n=N}^{2N-1}x[n-N]e^{-j\frac{2\pi}{2N}nk}\\&=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{2N}nk}+\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{2N}(n+N)k}\\&=\sum_{n=0}^{N-1}\big(x[n]+x[n]e^{-jk\pi}\big)e^{-j\frac{2\pi}{2N}nk}\\&=\sum_{n=0}^{N-1}\big(x[n]+(-1)^kx[n]\big)e^{-j\frac{2\pi}{N}n\frac{k}{2}}\\&=\begin{cases}2X\left[\frac{k}{2}\right],&k\textrm{ even}\\0,& k\textrm{ odd}\end{cases}\end{align}$$

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    $\begingroup$ Thanks for the step-by-step solution! So I shall continue to always consider such a question in two halves. $\endgroup$ Mar 30, 2023 at 12:28

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