Conjugate Symmetry

Claim

if a signal is real-valued, then its fourier transform is conjugate symmetric. In other words,
$$ x(n) \in \R \mbox{ } \forall n \Rightarrow X(\omega) = X^*(\omega) $$

Proof

We can use the definition of the fourier transform of a signal $x(n)$: \begin{align*} X(\omega) &= \sum \limits_{n=-\infty}^{\infty} x(n) e^{-i\omega n} \\ X(-\omega) &= \sum \limits_{n=-\infty}^{\infty} x(n) e^{i\omega n} \\ X^*(-\omega) &= \sum \limits_{n=-\infty}^{\infty} x^*(n) e^{-i\omega n} \\ X^*(-\omega) &= \sum \limits_{n=-\infty}^{\infty} x(n) e^{-i\omega n} \quad \mbox{since $x(n)$ is real} \\ X^*(-\omega) &= X(\omega) \\ \end{align*}
Here is a more simplistic proof that uses the fact that $\ftrans{x^*(n)} = X^*(-\omega)$:
\begin{align*} x(n) &\ftp X(\omega) \\ x^*(n) &\ftp X^*(\omega) \\ x(n) &\ftp X^*(\omega) \\ \end{align*}
$\qed$
This implies that the frequency response is equal to its conjugate (for real-valued signals $x(n)$):
$$x \in \left[ \Z \to \R \right]$$ $$\Rightarrow $$ $$X(\omega) = X^*(\omega)$$

Dan Lynch © Mathapedia 2025