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Suppose $x[n]=[x_0,x_1, \ldots,x_{N-1}]$ and $y[n]=[x_0,0,x_1,0,\ldots,0,x_{N-1}]$. What is the relationship between the Fourier Transforms of both the signals?

When I try for some specific examples in MATLAB, I see that $Y[k]$ has an image of $X[k]$ but how do we show this mathematically?

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Let's be more general and consider the insertion of $L-1$ zeroes between each sample. Mathematically, this can be written as $$ y[m]= \begin{cases} x[n] & \text{if } m = nL\\ 0 & \text{otherwise} \end{cases}. $$ The DTFT of $y[n]$ is then given by $$Y(e^{j\Omega}) = \sum_{m=-\infty}^{\infty} y[m]e^{-jm\Omega}.$$ Because $y[m]$ is non-zero only when $m$ is a multiple of $L$, that is $m = nL$, this becomes $$Y(e^{j\Omega}) = \sum_{n=-\infty}^{\infty} x[n]e^{-jnL\Omega} = X(e^{jL\Omega}).$$ From this, you can see that the DTFT of $y$ is the DTFT of $x$ with a period $L$ times smaller (i.e. $2\pi/L$ instead of $2\pi$). Intuitively, this makes sense: you know that "expanding" a signal in the time-domain is equivalent to "squeezing" its spectrum in the frequency domain (see the time scaling property of the Fourier transform).

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  • $\begingroup$ Thank you Anapar. Between, instead of DFT in the answer, shouldn't it be DTFT? $\endgroup$ – user7080 Jun 1 '17 at 6:17
  • $\begingroup$ Indeed, that's fixed now. $\endgroup$ – anpar Jun 1 '17 at 6:22

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