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

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?

• 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?
– A_A
Dec 8, 2017 at 9:06
• 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. Dec 8, 2017 at 18:28

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

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).

• Thanks Abhilash for such a detailed explanation and intuition.. please keep contributing and sharing.. :).. Jan 27, 2019 at 20:28
• "maximum value of the second term $|e^{j ( \theta + \phi )}|$ occurs at 1. for that to happen, $\theta + \phi = 2 \pi k$". This does not make sense. $|e^{j ( \theta + \phi )}| = 1$ regardless of $\theta$ or $\phi$. Sep 6, 2021 at 23:03
• The whole point of discussion is about the real value. Sep 8, 2021 at 1:35

HINT:

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