# Can anyone help with the properties of the DFT?

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?

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:

\begin{align}X_1[k]&=\sum_{n=0}^{N-1}x[N-1-n]e^{-j\frac{2\pi}{N}nk}\\&=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}(N-1-n)k}\\&=e^{j\frac{2\pi}{N}k}\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}n(-k)}\\&=e^{j\frac{2\pi}{N}k}X[-k]\end{align}

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 $$N/2$$.

• 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 $\text{mod }N$?
– user67157
Mar 29 at 14:11
• @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. Mar 29 at 14:15