Inserting zeros: effect on DFT

Question

Let N be an even integer, and $x[n]$ be a finite length signal over the interval $n\in [0,N-1]$; let $X[k]$ be the N-pt DFT of $x[n]$.

Find the 2N-Pt DFT of $x_1[n]$ in terms of $x[n]$, where:

$$x_1[n] = \begin{cases} x[\frac{n}{2}], & \text{if } n \text{ is even}\\ 0 & \text{else} \end{cases} $$

Now, the idea I had in mind was to write out the DFT sum and then change the summing variable $n\rightarrow 2n$ but that doesn't work:

$$X_1(k) = \sum_{n=0}^{2N-1} x[n/2]e^{-j\frac{2\pi}{2N}nk} $$

$$\downarrow n \text{ to } 2n$$

$$X_1(k) = \sum_{n=0}^{N-1/2} x[n]e^{-j\frac{2\pi}{N}nk} \neq X(k)$$

This doesn't work for two reasons, 1) The form of the DFT is not correct and 2) the fact that $x_1[n] = x[n/2]$ for only even $n$'s is not reflected in the sum!

Any help would be appreciated!

Answer 1

You're very close. We have

$$\begin{align}X_1[k]&=\sum_{{n=0,\;n\textrm{ even}}}^{2N-1}x[n/2]e^{-j\frac{2\pi} {2N}nk}\\&=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}nk}\\&=X[k],\qquad k=0,1,\ldots, 2N-1\end{align}$$

Naturally, the DFTs of $x[n]$ and $x_1[n]$ have the same values. The only difference is that $X_1[k]$ has length $2N$, i.e., the values of $X[k]$ are simply repeated. This is what upsampling (insertion of zeros) does: the period in the other domain is multiplied by the number of inserted zeros plus $1$, i.e., one inserted zero results in twice the period, two zeros result in three times the period, etc.

The dual problem (concatenating two periods in the sample domain) was discussed in this question.