Prove that, in DFT, Upsampling in time domain is equal to replication in frequency domain
We start with
$$X[k] = \sum_{n=0}^{N-1} x[n] \cdot W_{N}^{nk}$$
Let's define a new sequence of $y[r]$ of length $R=M \cdot N$, where $R, M, N \in \mathbb{Z} &gt; 0$

$$y[r] = \begin{cases}
y[n \cdot M] = x[n] \qquad &amp; r = n \cdot M \\ 
\\ 0 \qquad &amp; \text{otherwise} \\
\end{cases}$$

Transform it
$$Y[k] = \sum_{r=0}^{R-1} y[r] \cdot W_{R}^{rk}$$

Now most of the samples of $y[r]$ are zero, so we can eliminate them from the sum

$$Y[k] = \sum_{r=0}^{R-1} y[r] \cdot W_{R}^{rk} = \sum_{n=0}^{N-1} y[n \cdot M] \cdot W_{R}^{nMk} = \sum_{n=0}^{N-1} x[n] \cdot W_{R}^{nMk} $$

Next we take a look at the twiddle factor
$$W_{R}^{nMk} = e^{-j\frac{2 \pi nMk}{R}} = e^{-j\frac{2 \pi nk}{N}} = W_{N}^{nk} $$
 and put this back into  the Fourier Transform

$$Y[k] =  \sum_{n=0}^{N-1} x[n] \cdot W_{N}^{nk} $$

That's pretty much it. Comparing to the first equation we see directly that
$$Y[k]=X[k]$$

We also see that $Y[k]$ is periodic in $N$ and not just in $R$. Since both $x[n]$ and $W_{N}^{nk}$ are periodic in $N$, so must be $Y[k]$.