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Now this is a simple question, but I still ask it for clarification:

I know that an even signal $$h[n] = h[-n]$$ results in a real-valued DTFT (we have proven that in class). Now my question is the following: does a real-valued DTFT also result in an even signal? Would this mean, that the signal of the real DTFT is always acausal?

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The proof is quite straightforward. With

$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{jn\omega}d\omega\tag{1}$$

and with $X(e^{j\omega})=X^*(e^{j\omega})$ (i.e., a real-valued DTFT) we get

$$\begin{align}x[-n]&=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{-jn\omega}d\omega\\&=\frac{1}{2\pi}\left[\int_{-\pi}^{\pi}X^*(e^{j\omega})e^{jn\omega}d\omega\right]^*\\&=\frac{1}{2\pi}\left[\int_{-\pi}^{\pi}X(e^{j\omega})e^{jn\omega}d\omega\right]^*\\&=x^*[n]\end{align}$$

i.e., if $X(e^{j\omega})$ is real-valued we obtain

$$x[n]=x^*[-n]$$

(or $x[n]=x[-n]$ if $x[n]$ is real-valued).

So it's true that a real-valued DTFT cannot correspond to a causal sequence, other than the trivial sequence $x[n]=a\delta[n]$ with $a\in\mathbb{R}$.

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  • $\begingroup$ Thank you a lot! : ) $\endgroup$ – Phobos Mar 9 '20 at 9:24

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