I've been told that complex signals are a "notational convenience for easily making two signals orthogonal so that they can go on the same wire." Is this accurate/what does this mean?

Is there a physical meaning to complex signals? Is multiplying by j actually shorthand for multiplying the real part and imaginary part by orthogonal carriers? (is that the way this would be observed in real life?)

  • \$\begingroup\$ Where did you read the source of this quote? It doesn't sound very fluent which is really watering down what could be a decent question about orthogonality of signals. \$\endgroup\$
    – Andy aka
    Commented Oct 23, 2013 at 11:03
  • \$\begingroup\$ Is it not a phase shift of the resultant signal with regards to the source signal? \$\endgroup\$ Commented Oct 23, 2013 at 11:34
  • \$\begingroup\$ It's an excellent question! The wording simply reflects that fact that it is a confusing subject and the OP can't get his head around it. If he understood it he probably wouldn't need to ask an excellently formed question. \$\endgroup\$ Commented Oct 23, 2013 at 12:14
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    \$\begingroup\$ @Andyaka The quote itself appears grammatically correct and lexically valid as I see it. Which part would you be referring to as insufficiently fluent, please? \$\endgroup\$ Commented Oct 23, 2013 at 12:17
  • \$\begingroup\$ @AnindoGhosh "a notational convenience" doesn't appear to be an adequate description of something that can "easily make two signals orthogonal so that they can go on the same wire". If I included the OP's words before this "quote" I could construe this meaning: "complex signals easily make two signals orthogonal etc.." and this has the ring of a half twisted, 2nd hand quote that the OP may possibly have re-jigged unintentionally. \$\endgroup\$
    – Andy aka
    Commented Oct 23, 2013 at 12:28

2 Answers 2


Using complex numbers to express sinusoidal signals is hardly "just a notational convenience".

On what it means for a sinusoid to have two orthogonal components:

First, realise that "orthogonal" is just a fancy word for "separate" or "fully independent".

Assume that you're dealing with a sinusoidal signal of fixed frequency \$ \omega \$. Such signals have two degrees of freedom - amplitude \$A\$ and phase \$\phi\$. That is:

$$ x(t) = \operatorname{Re}(A e^{j \phi}\times e^{j\omega t}) = A\cos ( \omega t + \phi ) $$

Information can be conveyed by either varying the amplitude or varying the phase, so there are two separate "channels" for information.

Equivalently, you can express the same fixed-frequency sinusoidal signal as the sum of two signals, phase-displaced by 90 degrees:

$$ x(t) = A_1 \sin (\omega t) + A_2 \cos (\omega t) $$

Think of the sin term as a "vertical" wiggle, and the cos term as a "horizontal" wiggle. Again, these form two separate "channels" for communicating information.

It is fairly easy to build equipment that separates the sine component from the cosine component, so this is used as the basis of practical communications schemes. See quadrature amplitude modulation (QAM).

On the physical meaning of "multiplying by \$j\$":

In phasor form, the phase of the signal is given by a complex number \$e^{j\phi}\$ like so:

$$ e^{j\phi} = \cos{\phi} + j \sin \phi $$

If you multiply by \$j\$ you get:

$$ j\times e^{j\phi}=j \cos \phi - sin \phi $$ $$ j \times e^{j\phi} = j\sin (\phi+90^\circ)+cos(\phi + 90^\circ) $$ $$ j \times e^{j\phi} = e^{j\phi+90^\circ} $$

Which is to say that multiplying a phasor by \$j\$ changes its phase by \$+90^\circ\$. I like to think that two phasors \$A\$ and \$jA\$ are at right angles to each other, i.e. they are orthogonal.


Complex numbers are used to represent complex signals. From the complex numbers you can tell both amplitude and phase of the signal.

In regard to the quote. Using a technique like Phase shifting, you can have more than more signal flowing simultaneously. Ever wondered how more than one phone call could be transmitted from the same phone line?

The quote doesn't make much sense really - if I understood the underlying meaning of it correctly that is.

But from phase modulation you can make two signals orthogonal.


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