Why do we need to conjugate complex signals in autocorrelation and cross correlation

Question

Why is it necessary to conjugate $f(t)$ while performing auto correlation or cross correlation with respect to $g(t)$, if $f(t)$ and $g(t)$ are complex signals?

Answer 1

$\underline{\text{Prologue :}}$

Let me ask you another question. How will you compare two complex numbers $U$ ($a+jb$) and $V$ ($c+jd$)? By comparing magnitude? Subtract them and take real part? Multiply them and compare?

Since any complex number involves two entities (one for magnitude $\lvert z \rvert$ and other for argument $\theta$), any comparison involves a comparison of those two entities. Suppose we represent $U$ and $V$ in a complex argument plane as $M e^{j \theta}$ and $N e^{j \phi }$. If we multiply both of them together, we get

$$U V = (M e^{j \theta}) (N e^{j \phi }) = MN e^{j (\theta + \phi)}$$

If we seek the maximum value out of the above multiplication, then

$$\lvert U V \rvert_{\max} = \lvert MN \rvert_{\max} \lvert e^{j (\theta + \phi)} \rvert_{\max}$$

For the time being forget about the magnitudes $M$ and $N$ and suppose our comparison is only about the relative location of $V$ with respect to $U$ in the complex argument plane. The maximum value of the second term $e^{j (\theta + \phi )}$ occurs at 1. For that to happen

$$\theta + \phi = 2 \pi k \quad \text{for } k = 0, 1, 2, \dots$$

If $k = 0$, $\theta + \phi = 0 \implies \phi = - \theta$ i.e when two complex numbers with bounded magnitudes get multiplied, their maximum value occurs when the argument of the second number ($V$) is the negative of the first ($U$), which is the same as the complex conjugate of $U$, $U^{*}$.

The maximum of $U \times V$ occurs when $V$ points in the direction of $U^{*}$ (the orange dashed line in the picture), or conversely, if $V$ is getting multiplied by $U^{*}$ then the maximum occurs when $V$ points in the direction of $U$

In another words, $U^{*} \times V$ gives the 'nearness' of $V$ with respect to $U$ in terms of the angle (complex argument) and it will decrease as a function of $cos alpha$, where $\alpha$ is the angle between $U$ and $V$.

$\underline{\text{Epilogue:}}$

Considering two signals $f$ and $g$, correlation is given as

$$(f \otimes g)(\tau) = \int_{-\infty}^\infty f^*(t) g(t + \tau) dt$$

At any instant '$t$', a signal is just a point in the complex plane. So at any moment $f$ and $g$ are just two points in the complex plane. Then the job of comparison of two signal becomes a mere comparison of two complex points.

With this logic, to compare two signals with a given lag $\tau$ between them, just multiply point by point in $t$ by taking the complex conjugate of the other signal. If you get a big number that means both reference and compared signals are just looking in the same direction in the complex plane. When integrated over $t$, the value of the integration shows the how much similarity is between signal $f$ and signal $g$, which is known as the cross-correlation between two signals.

(I didn't talk about magnitudes $M$ and $N$, right? The method in which everything is put in a single benchmark by avoiding magnitudes $M$ and $N$ is known as normalized cross-correlation).

Answer 2

HINT:

What is the physical meaning of the autocorrelation evaluated at lag $\tau=0$?