# Why doesn't repeating a signal give rise to a finer resolution of DFT/FFT?

If x = [1 2 3 4 3 2]; and x1= [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 x1 are essentially the same.

When I plot the FFTs of each using MATLAB, I get a bunch of zeros for the x1.

Does anyone know why? I mean zero-padding it will increase the resolution of the DFT but why not replicating it?

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The functions $x$ and $x_1$ are essentially the same, the input is (implicitly) assumed to be periodic anyway. Adding zeros will change the input signal, but can not add more data... (The transform of $x_1$ will be the sum of time-shifted copies of $x$ padded with an appropriate number of zeros.) –  copper.hat Nov 4 '12 at 2:56
FFT already assumes the input signal is periodic. –  chaohuang Nov 4 '12 at 5:26
@copper.hat: thanks for responding. It was my understanding that since $x$ is finite length, it cannot be periodic. Can you elaborate on the statement "The transform of $x_1$ will be the sum of time-shifted copies of $x$ padded with an appropriate number of zeros"? I think this where I am failing to see this. If you could kindly illustrate with a simpler example signal, it will be very helpful to the discussion. Also, how do I promote your response as the answer? –  user43901 Nov 4 '12 at 18:49

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.

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+1 Nicely expressed. –  copper.hat Nov 5 '12 at 5:18