What does conjugation in the time-domain of a signal mean?

Question

I've never been explicitly told what the conjugation of a signal in the time-domain means. I'm mainly asking because in my signals class, my professor stated that for a signal x(t) to be real: x(t) = x(t)*, and I don't understand why.

Answer 1

If $$ z\equiv a+ib $$ is a complex number ($a$ and $b$ are real numbers), its conjugate is defined to be $$ \overline{z}\equiv a-ib. $$

Note that we can write a real number $x$ as $$ x=x+i0. $$ The conjugate of $x$ is equal to itself: $$ \overline{x}=x-i0=x. $$ Therefore, if a number is real, it is equal to its conjugate.

For the converse, if $z\equiv a+ib$ is equal to its conjugate, $$ a+ib=a-ib $$ which implies $$ ib=-ib. $$ If $b$ is anything other than zero, we arrive at a contradiction, since $i\neq-i$. Therefore, if a complex number is equal to its conjugate, it is real.

All that remains to be done is to apply your result pointwise (i.e. for every possible time $t$), and you arrive at the result $x(t)=x(t)^*$ for all $t$ if and only if $x(t)$ is real for all $t$.

Answer 2

Let $z$ be some fixed but arbitrary complex number with Cartesian representation $z=\Re{(z)}+i\,\Im{(z)}$ for some real pair $(\Re{(z)},\Im{(z)})\in\mathbb{R}^2$. The complex conjugate of $z$ is then,

$$z^*=\Re{(z)}-i\,\Im{(z)}.$$

Subtracting the conjugate of $z$ from $z$ itself then yields:

$$\begin{align} z-z^* &=(\Re{(z)}+i\,\Im{(z)})-(\Re{(z)}-i\,\Im{(z)})\\ &=2i\,\Im{(z)}. \end{align}$$

Solving for the imaginary component of $z$, we have:

$$\Im{(z)}=\frac{z-z^*}{2i}.$$

Thus, if $z\neq z^*$, or equivalently if $z-z^*\neq 0$, then $\Im{(z)}\neq 0$. Contrapositively, if $\Im{(z)}=0$ then $z=z^*$. $\blacksquare$