If $X(k)$ is the $N$-point DFT of $x(n)$, and $y(n)= x\left(\frac {n+1}{2}\right)$ for odd $n$, and $0$ for even $n$.

What is the $2N$ point DFT of $y(n)$ in terms of $X(k)$?

So far, I've noticed that $y(2n+1)=x(n+1)$, but I'm not sure what to do with that information in terms of the DFT summation. Can someone help me out here?

  • $\begingroup$ Hint: Write down explicitly, using numbers and not symbols such as $n$ or $N$, the six elements of the sequence $y$ for the case $N=3$. You can have $y$ and $x$ in your answer but no symbol $n$ or $N$. Then write down the six-point DFT of $y$, (no summation signs $\displaystyle\Sigma$ allowed) and replace each $y$ by the corresponding value from your first answer. You should be having six equations here. Then write the 3-point DFT of $x$ using the same rules. Finally stare very hard at the results. $\endgroup$ Oct 19, 2019 at 3:32
  • $\begingroup$ @DilipSarwate, I got the answer, but I want to know how to derive it directly. $\endgroup$
    – S'Danc
    Oct 19, 2019 at 5:15
  • $\begingroup$ @S'Danc: Dilip's suggestion was supposed to help you derive it yourself. Have you tried it? $\endgroup$
    – Matt L.
    Oct 19, 2019 at 6:51

1 Answer 1


I would follow a signal block diagram based solution for this problem.

First as suggested in the comments, it's very helpful to investigate a few values of $y[n]$ and $x[n]$ for some $n$ :

$$ \begin{align} y[0] &= 0 ~~~,~~~ y[1] = x[1] \\ y[2] &= 0 ~~~,~~~ y[3] = x[2] \\ y[4] &= 0 ~~~,~~~ y[5] = x[3] \\ y[6] &= 0 ~~~,~~~ y[7] = x[4] \\ \end{align} $$

with some experience, or by a direct investigation of this above sequence it can be seen that the system that produce $y[n]$ from $x[n]$ is the folowing :

$$ x[n] \longrightarrow \boxed{ \uparrow 2} \longrightarrow w[n] \longrightarrow \boxed{ z^1} \longrightarrow y[n]$$

where the up arrow indicates an expansion by $2$ and the $z^1$ indicates a left shift (advance) by one sample.

Write down the DTFT relations between those signals $x[n]$,$w[n]$ and $y[n]$ :

$$ \begin{align} W(\omega) &= X(2\omega) \\ Y(\omega) &= e^{j\omega} W(\omega) \\ \\ Y(\omega) &= e^{j\omega} X(2\omega) \\ \end{align} $$

And relate $2N$ point DFT $Y[k]$ of $y[n]$ to $N$ point DFT $X[k]$ of $x[n]$.

$$ \begin{align} Y[k] &= Y(\omega_k) = Y(\frac{2\pi}{2N} k) &, ~~ k=0,1,...,2N-1\\ Y[k] &= e^{j \frac{2\pi}{2N} k} X(\frac{2\pi}{N} k) &, ~~ k = 0,1,...,2N-1\\ Y[k] &= e^{j \frac{\pi}{N} k} X[k] &, ~~ k = 0,1,...,2N-1 \end{align} $$

It can be seen that the length $2N$ DFT sequence $Y[k]$ consists of two fold replica of length $N$ DFT sequence $X[k]$ weighted by $e^{j \frac{\pi}{N} k}$. Note that $X[k]$ is periodic in $k$ by $N$, hence repeats twice during $k = 0,1,...,2N-1$.


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