I have an assignment where I'm given the DFT of a sequence $x[n]$ as $X[k]=\{4,3,2,1,0,1,2,3\}$ and also $$y[n] = \left\{ \begin{array}[cc] xx[n/2] & \text{if n is even} \\ 0 & \text{otherwise} \end{array} \right. $$

and I'm supposed to find and sketch the DFT of $y[n]$.

So $y[n] = \{x[0], 0, x[1], 0 ... x[7], 0\}$ and it's not complicated to find $Y[k]$ if we know $x[n]$

I know how to use the definition of the DFT and IDFT to calculate $x[n]$ but it's a tedious task to do by hand, especially when the sequence is longer than a few items. Is there a quicker way to calculate the DFT and IDFT by hand without using a program like Matlab?

  • $\begingroup$ Hint: start by figuring out the $z$-transform of this system. $\endgroup$
    – Phonon
    Oct 2, 2014 at 19:33
  • 1
    $\begingroup$ I'm sure your class covered that special case of inserting zeros into a time domain representation as it is used for upsampling. The effect this procedure has on the spectrum of the signal is rather simple. So I'd suggest you look that up and use it to answer the question. If you know how it works this question takes 10 seconds to do. $\endgroup$
    – Jazzmaniac
    Oct 3, 2014 at 10:24

1 Answer 1


If you use the DFT formula, you get:
$$ Y[k] = \sum_{n=0}^{2N-1}y[n]e^{\frac{-2\pi k n}{2N}} $$ Now, substituting the definition of $y[n]$ you get:
$$ Y[k] = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi k (2n)}{2N}} = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi k n}{N}} $$
So, for $0\leq k < N$ you get that $$ Y[k] = X[k]$$ and for $k\geq N$ you get $$ Y[k] = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi k n}{N}} = \sum_{n=0}^{N-1}x[n]e^{\frac{-2\pi (k-N) n}{N}} = Y[k-N] $$
Therfore, $$ Y[k] = \begin{cases}X[k] & 0\leq k < N \\ X[k-N] & k\geq N\end{cases} $$ or $X[k]=\{4,3,2,1,0,1,2,3,4,3,2,1,0,1,2,3\}$

  • $\begingroup$ Hi, I just came across this answer and had a few questions. How come you replaced N+1 with 2N +1 (is this because the length of Y(k) is twice as long?) Why is there no "i" term in the DFT formula? How come the 2N+1 disappeared the the second step. Is the final result X[k] supposed to be Y[k]? $\endgroup$
    – Tim M
    May 24, 2018 at 9:03

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