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I have been reading through my courses DSP slides and came across something which was not really taught in detail. You can look up here for reference, it is stated almost identical.

Given the following properties for $X[k]$ which the dft of $x[n]$ (sub e for even and sub o for odd):

$$ X_{e}\left(e^{j \omega}\right)=X_{e}^{*}\left(e^{-j \omega}\right) \ (\text{conjugate symmetric FT}) $$

and

$$ X_{o}\left(e^{j \omega}\right)=-X_{o}^{*}\left(e^{-j \omega}\right) (\text{conjugate antisymmetric FT}) $$

Then it states that

$$ X\left(e^{j \omega}\right)=X_{e}\left(e^{j \omega}\right)+X_{o}\left(e^{j \omega}\right) $$

with $X_{e}\left(e^{j \omega}\right)=\frac{1}{2}\left[X\left(e^{j \omega}\right)+X^{*}\left(e^{-j \omega}\right)\right]$ and $X_{o}\left(e^{j \omega}\right)=\frac{1}{2}\left[X\left(e^{j \omega}\right)-X^{*}\left(e^{-j \omega}\right)\right]$.


  • I find the term $X_{e}\left(e^{j \omega}\right)$ a little bit confusing as well. What is it's meaning? I expected it to be a vector of the frequency values.

    • What is the point of using negative exponent for $X(e^{-j \omega})$?
  • I expect the properties to be somewhat related to $X^*[N-k] = X[k]$ though I am not able to see the connection right now


What is the intuition for the formulas and how could it be proofed?


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  • $\begingroup$ Can you provide a reference for these statements? $X(e^{j\omega})$ is continuous but $X[k]$ is discrete, so you seem to be talking about two different flavors of Fourier Transforms (there is a total of 4 flavors). The DFT is discrete in both domains, and typically you would not use a notation like $X(e^{j\omega})$ in the context of a DFT $\endgroup$
    – Hilmar
    Feb 16, 2022 at 19:34
  • $\begingroup$ Oh it's part of my german lecture slides and he is quite close to oppenheim, it was barely introduced for LTI systems after the definition of the frequency response $Y\left(e^{\mathrm{j} \omega}\right)=X\left(e^{\mathrm{j} \omega}\right) H\left(e^{\mathrm{j} \omega}\right)$, the first one is called conjugate symmetric ft and the second conujgate antisymmetrc ft, I will look if I can find something more related. $\endgroup$ Feb 16, 2022 at 19:41
  • $\begingroup$ @Hilmar I found some slides which are almost identical, I added the reference above. $\endgroup$ Feb 16, 2022 at 19:46
  • $\begingroup$ From the source you linked, it looks like the terms $x_e$, $x_o$, $X_e$, and $X_o$ are just convenience terms for deriving the properties of the DTFT. These properties can also be applied to the DFT, as the DFT samples the DTFT. $\endgroup$
    – Ash
    Feb 16, 2022 at 20:56

1 Answer 1

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So it looks you have the DFT and DTFT mixed up. There are four flavors of Fourier Transforms depending on whether the signal in each domain is continuous or discrete. See for example https://ptolemy.berkeley.edu/eecs20/week12/fourier.html

Discrete/continuous also maps to period/aperiodic in the other domain.

So the properties you are listing are DTFT (discrete time, continuous frequency) properties and the properties you are expecting are DFT (discrete/discrete) properties so there's the mismatch. E.g. discrete signals use sums, continuous signals use integrals.

I find the term $X_{e}\left(e^{j \omega}\right)$ a little bit confusing as well.

That is a bit of notation preference. For the DTFT you both see $X\left(e^{j \omega}\right)$ or just $X(\omega)$. The DTFT is just a special case of the (bilateral) Z-transform for $z = e^{-j\omega}$ so the first notation implies "evaluation on the unit circle of the z-plane".

What is the point of using negative exponent for $X(e^{j \omega})$?

This is just flipping the sign of the frequency. Depending on notation preference you either use $X(-\omega)$ or $X(e^{-j \omega})$ to denote negative frequencies.

I expect the properties to be somewhat related to $X^*[N-k] = X[k]$ though I am not able to see the connection right now

That's just the difference between the DFT and the DTFT as described above.

What is the intuition for the formulas

Intuition is in the eyes of the beholder. For some people the "cosine" vs "sine" analogy works, but everyone is different.

and how could it be proofed?

The proof is fairly trivial and can be found in pretty much any text book on the topic. It does vary with the type of Fourier Transform but it's pretty straight forward.

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  • $\begingroup$ Thanks a lot! So would it make sense to interpret $X(e^{jω})$ as a vector? "This is just flipping the sign of the frequency.", so when we talking in related terms for the DFT this would mean the values from $>N/2$ to $N$ (for N even)? I was wondering if it might be a flipping of the values as well like when I compare to $\sin(-x)$. "For some people the "cosine" vs "sine" analogy works, but everyone is different." so might it be just another way for stating what has been shown in this answer for the DTFT or why negative frequencies are required? $\endgroup$ Feb 17, 2022 at 8:15
  • $\begingroup$ $X(e^{j\omega}$ isn't really a vector, it's a continuous function. Then DFT is periodic with $N$ so $X[-n] = X[N-n] = X[17N-n]$ etc. $\endgroup$
    – Hilmar
    Feb 18, 2022 at 12:17

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