# Real-valued DTFT

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?

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 $$a\in\mathbb{R}$$.

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