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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?

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    $\begingroup$ Can I please ask what do you think about this? Where exactly is the difficulty? Have you handled any complex $f(t)$ so far? Did you notice anything "strange" about them, compared to real $f(t)$s? $\endgroup$ – A_A Dec 8 '17 at 9:06
  • $\begingroup$ maybe look up the meaning of inner product in what we normally call a Hilbert space. an example: $$ \langle x, y \rangle = \sum_{n} x_n \overline{y_n} $$ then they define the norm of the vector $x$ as the square root of the inner product of $x$ with itself. $\endgroup$ – robert bristow-johnson Dec 8 '17 at 18:28
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$\underline{Prologue :}$

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 |$z$| and other for argument $\theta$ ) any comparison involves 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

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

If we seek maximum value out of the above multiplication then

$|U$x$V|_{max} = |MN|_{max} * |e^{j ( \theta + \phi )}|_{max} $

Suppose our comparison is only about the relative location of $V$ with respect to $U$ in the complex argument plane, for the time being, forget about the magnitudes $M$ and $N$.

maximum value of the second term $\;e ^{j ( \theta + \phi )}$ occurs at 1.

for that to happen
$\theta+\phi = 2\pi k \;\;\;\;$ where $\:k\:\epsilon\;( 0,1,2...$

if $k = 0$,

$\theta + \phi = 0$

$\phi = - \theta $

ie, when two complex numbers with bounded magnitudes get multiplied, their maximum value occurs, when the second number ($V$) is in the negative argument direction of the first number ($U$), ie in the direction of its conjugate ($U^*$).

[picture]

ie maximum of $U \times V$ occurs when $V$ points in the direction of $U^*$ { the orange dashed line in the picture }

$\textit{-or conversely-}$

If $V$ is getting multiplied by $U^*$ then their maximum occurs when $V \textit{points in the direction of } U$

In another words $U^* \times V$ gives nearness of $V$ w.r.t $U$ in terms of angle ( complex argument ) and it will decrease as a function of $cos \:\alpha$, where $ \alpha$ is the angle between $U$ and $V$.

$\underline{Epilogue:}$

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

$\hspace{1cm} (f \otimes g)(\tau) = \int_{-\infty}^\infty f^*(t) g(t + \tau) dt$

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 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 on the whole '$t$', the value of the integration shows the how much similarity is between signal $f$ and signal $g$. that is 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 }

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    $\begingroup$ Thanks Abhilash for such a detailed explanation and intuition.. please keep contributing and sharing.. :).. $\endgroup$ – Vinayak Killedar Jan 27 '19 at 20:28
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HINT:

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

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