Let N be an even integer, and $x[n]$ be a finite length signal over the interval $n\in [0,N-1]$; let $X[k]$ be the N-pt DFT of $x[n]$.

Analytically find the DFT of each sequence below in terms of $X[k]$. (The DFT size is N unless specified otherwise):

a) $x_1[n] = x[N-1-n]$

b) $x_2[n] = (-1)^n\cdot x[n]$

For part a) I know that the DFT of a reversed sequence is as follows:

$$D\{f_{-n}\}_k = D\{f_{n}\}_{-k} = F_{-k}$$

where $F_{-k}$ is the sequence of DFT coefficients. However, I do no know how to combine it with the shifting by $N-1$.

And for part b) I know that $e^{-j\pi n} = (-1)^n$ but I don't know how I can convert it to a from where I can use the circular shift in frequency property:

$$W_N^{-k_0n}x_n \xrightarrow{DFT_N} X_{(k-k_0){\text{mod } N}}$$

where $W_N^{-k_0n} = e^{j\frac{2\pi}{N}k_0n}$.

Perhaps, by taking $-\frac{2\pi}{N}k_0 = \pi$, I can bring it to a similar form but how can I showcase the $\text{mod } N$ operation?


1 Answer 1


You don't necessarily need to rely on knowing or remembering any properties of the DFT in such cases. Just write down the DFT sum and see what happens:


Now try to derive the expression for $X_2[k]$ yourself, noting that for even $N$ we can write

$$(-1)^n=e^{-j\pi n}=e^{-j\frac{2\pi}{N}\frac{N}{2}n}$$

with integer $N/2$.

  • $\begingroup$ That was a very slick summing variable change for $X_1(k)$. If I follow a similar procedure for $X_2(k)$ I end up with $$X_2(k) = \sum_{n=0}^{N-1} x[n] e^{-j\frac{2\pi}{N}n(k+\frac{N}{2}) }$$ which gives us $X_2(k) = X(k+\frac{N}{2})$ right and there are no issues with having to do $\text{mod }N$? $\endgroup$
    – user67157
    Mar 29, 2023 at 14:11
  • $\begingroup$ @AcerbicNarcissist: That looks good to me. Note that $X[k]$ is $N$-periodic, so it doesn't matter if you use a modulo operation or not. If you insist on having all your indices in the interval $[0,N-1]$ then you need to use modulo, otherwise you don't. $\endgroup$
    – Matt L.
    Mar 29, 2023 at 14:15

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