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?

  • 2
    $\begingroup$ 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

$\underline{Prologue :}$

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 ($U^*$).


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 $V$.


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 }

  • 2
    $\begingroup$ Thanks Abhilash for such a detailed explanation and intuition.. please keep contributing and sharing.. :).. $\endgroup$ – Vinayak Killedar Jan 27 '19 at 20:28


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.