CTFT conjugation and conjugate symmetry

Conjugation property

Statement: if $x(t) \xleftrightarrow{F} X(\omega)$, then

$$x^*(t) \xleftrightarrow{F} X^*(-\omega)$$

Proof: start from the definition and take the conjugate:

$$X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \;\Rightarrow\; X^*(\omega) = \int_{-\infty}^{\infty}x^*(t)e^{j\omega t}dt$$

Replace $\omega$ by $-\omega$:

$$X^*(-\omega) = \int_{-\infty}^{\infty}x^*(t)e^{-j\omega t}dt = F\{x^*(t)\}$$

So the Fourier transform of $x^*(t)$ is $X^*(-\omega)$.

Conjugate symmetry for real signals

Statement: if $x(t)$ is real, then

$$X(-\omega) = X^*(\omega)$$

Proof: a real signal satisfies $x(t) = x^*(t)$. Taking the Fourier transform of both sides and using the conjugation property:

$$X(\omega) = X^*(-\omega)$$

Replacing $\omega$ by $-\omega$ gives $X(-\omega) = X^*(\omega)$.

The same result follows directly:

$$X(-\omega) = \int x(t)e^{j\omega t}dt = \left[\int x(t)e^{-j\omega t}dt\right]^* = X^*(\omega) \quad (x \text{ real})$$

Consequences

Write $X(\omega) = \mathrm{Re}\{X(\omega)\} + j\,\mathrm{Im}\{X(\omega)\} = |X(\omega)|e^{j\angle X(\omega)}$. Then for real $x(t)$:

Further, if $x(t)$ is real and even, $X(\omega)$ is real and even. If $x(t)$ is real and odd, $X(\omega)$ is purely imaginary and odd. Because of this symmetry, only the positive frequencies need to be plotted for real signals.

Example

$x(t) = e^{-at}u(t)$ ($a > 0$) is real, and $X(\omega) = \frac{1}{a + j\omega}$:

$$X(-\omega) = \frac{1}{a - j\omega} = \left[\frac{1}{a + j\omega}\right]^* = X^*(\omega)$$

$|X(\omega)| = 1/\sqrt{a^2 + \omega^2}$ is even, and $\angle X(\omega) = -\tan^{-1}(\omega/a)$ is odd.