Why doesn't repeating a signal give rise to a finer resolution of DFT/FFT? If $$x = [1,2,3,4,3,2]$$
and
$$x_1= [x,x,x,x,x,x,x,x,x]$$--that a new vector made of duplicating copies of $x$, then why is it that the FFT of $x$ and $x_1$ are essentially the same.
When I plot the FFTs of each using MATLAB, I get a bunch of zeros for the $x_1$.
Does anyone know why? I mean zero-padding it will increase the resolution of the DFT but why not replicating it?
 A: Let $x^{+} : \{0, \dotsc, dN-1 \} \to \mathbb{C}$ be a concatenation of $d$ copies of a signal $x: \{0, \dotsc, N-1\} \to \mathbb{C}$. Then $x^{+}$ is periodic with period $N$ ($\bmod$ $dN$).  Denote $e^{2\pi i/n}$ by $\zeta_n$ for any $n$. Direct computation of the (unnormalized) Fourier transform gives
$$
\widehat{x^+}(m) = \sum_{k=0}^{dN-1}x_k^+ \zeta_{dN}^{-mk} = \sum_{k = 0}^{N-1}\left( x_k \sum_{j = 0}^{d-1} \zeta_{dN}^{-m(k + jN)}\right) = \sum_{k = 0}^{N-1}\left( x_k \zeta_{dN}^{-mk}\sum_{j = 0}^{d-1} \zeta_{d}^{-mj}\right).
$$
The inner sum simplifies to
$$
\sum_{j = 0}^{d-1}\zeta_d^{-mj} = \begin{cases}
d & \textrm{if } d \textrm{ divides } m\\[1em]
0 & \textrm{otherwise}
\end{cases}.
$$
Plug this in to get
$$
\widehat{x^+}(m) = \begin{cases}
d \cdot \widehat{x}(\tfrac{m}{d}) & \textrm{if } d \textrm{ divides } m\\[1em]
0 & \textrm{otherwise}
\end{cases}.
$$
This shows that $\widehat{x^+}$ is a multiple of a spread out version of $\widehat{x}$ with a bunch of zeroes added in between.
