Complex Valued Electric Field

1. Aug 4, 2011

john1989

Hi

I am new to EM and i have a question.

When using Maxwell equations to compute time harmonic electric fields we use phasors for an easier solution.

The result is a complex valued electic field E(z) where E(z) is a vector. To get the time harmonic field we use E=Re{E(z)*exp(-i*2*pi*f*t)}

My question is: in the complex valued field E(z) what is the physical meaning of the real and the imaginary part the may come as a result of the equations.

For example if we get E(z)=(4+5i)x where x is the vector of the x-axis what is the physical meaning of 4 and 5 of its amplitude

2. Aug 19, 2011

jsgruszynski

The complex coefficient, exp(j omega t), is simply a time/phase delay added to the real solution.

3. Aug 19, 2011

john1989

i'm asking if the amplitude of exp(i*w*t) is a complex number

thanks though

4. Aug 19, 2011

yungman

We always use $\cos(\omega t -kz)$ to represent harmonic wave.

$$e^{j(\omega t-kz)}= \cos (\omega t -kz)+ j\sin (\omega t -kz)\;\Rightarrow \; \Re e [e^{j\omega t-kz}]=\cos (\omega t -kz)$$

Nothing more than that.

5. Aug 31, 2011

john1989

ok
but what if the $$e^{j(\omega t-kz)}$$ has a complex amplitude
what is the physical meaning of this amplitude
for example does it mean that it has to components of different phase?

6. Aug 31, 2011

yungman

That is a very good question that I wonder about. I hope someone can answer. All I know is the imaginary part plays a role, for example if the function is

$$je^{j\omega t-kz}\;\Rightarrow\; \Re e [je^{j\omega t-kz}]=-\sin (\omega t -kz)$$

The sine term comes out in this case. I hope someone can explain this.

7. Aug 31, 2011

Floid

Guessing here without really thinking through:

cos(x+90) = -sin(x)

So your complex amplitude is still just a phase shift...

8. Aug 31, 2011

yungman

I understand the phase shift part. The question is even more fundamental, why not just use phase shift and eliminate all the complex stuff? This is where I get loss. I just use the complex notation because everyone use it.

As as practising engineer, I like to deal with "real" things. Seem like this is more driven by math than real life.

9. Aug 31, 2011

dlgoff

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/impcom.html" [Broken]

Last edited by a moderator: May 5, 2017
10. Aug 31, 2011

yungman

I have no issue using all the formulas in the link, in fact as I said, we are forced to learn using it. Books pretty much give out the formulas as is and live with it. I understand the complex number are just a way to provide of phase shift calculation and all that.

The only thing I can think of is if we use phase angle in trig function to do calculation, it can get more complicate and long. by using:

$$e^{jx}=\cos x +j\sin x$$

I think you make calculation easier in complex form. If so, I sure wish the books will explain this. The best explanation I ever saw was what I posted in #4. You go read most of the physics, engineering books, they pretty much say "use this".........in a nice way!!!

To be honest, I yet to find an advanced level EE book explain things in English yet. You get into electromagnetics, fourier transform, laplace transform, all books start out in Calculus equation from the get go. In fact, after studying PDE, I really wonder how undergrad student really appreciate electromagnetics or electrodynamics without a semester of PDE that most of the colleges don't require for undergrad. All the explanation are in calculus, ODE and PDE. Why can't they write in English so people can understand it. I question most of the EE student gone through EM really know what it is.

On this note, I have been into heavy duty circuit and system design for almost 30 years, RF, data acq, HV etc, all sort of analog, transistors and IC design. I can tell you I never even use any calculus beyond the first semester or maybe a little of the integration. Usually never exceed basic stuff like super position, thevenin?? etc. All the math that I learn is nothing more than to be able to understand the text books. Once I understand the book, I almost never have to look back at the math again. there is something really wrong with this picture. In fact I never have calculus knowledge beyond the second semester until I stop working!!! I only study all the calculus in the last few years.

This is the reason you see people graduated with good grades never able to transition into real life career. I have seen so many lousy engineer with advanced degree.....even from MIT and UC Berkley!!!!

Last edited by a moderator: May 5, 2017
11. Sep 1, 2011

Floid

Because sometimes it is easier to work with the complex notation versus the phase shift. But in the end it is two ways of saying the same thing with phase shift being much easier for most people to conceptualize.

It is sort of like having cartesian, spherical, cylinderical coordinate systems. You can solve any problem using just one, but you can transition between them to make solving some problems and conceptualizing the situation easier.

12. Sep 1, 2011

yungman

Problem is it takes me a while to realize that. Books don't usually explain that, they just start out using complex notation.

That is the reason of my long ranting post, books don't explain the reason, they start out with all the math instead of explaining in English. Look at EM books, starting out with Divergence and Curl right about on the first page!!! I had to study 3 times over on three different books to try to understand in English. You can really go through the whole class, getting an A in the class without understanding it. I first started out with Ulaby and did ALL the problems, then I went on to Cheng and also did a lot of the problems. It was not until I study the third time with Griffiths that is really for Physics major that I feel I finally get the feel of it. I don't even dare to say I understand EM, just a feel of it. Griffiths is about the best book NOT used in EE major, I have 8 books on EM, if EE student don't use Griffiths, I really really question EE student really know much about EM. Yes you can get A's, I am sure I can get an A in college after the first time studying Ulaby ( which is an easy book). You can learn how to plug in the formulas and get the right answer!!!! That's why I feel there is such a big disconnect and that is the reason why so many good student never transition into a good engineer.

13. Sep 14, 2013

fishguy

thanks for explaining such a wonderful notation, especially introducing Griffiths's book here

14. Sep 15, 2013

meBigGuy

$$e^{j(\omega t-kz)}$$ has an amplitude of 1 and a phase. The amplitude is real. The quantity being expressed can also be represented by two numbers, a "real" part and an "imaginary" part which are actually just the sin and cos of the magnitude/phase that was originally expressed. If you want to give it a complex amplitude, you have to multiply it by a complex number.

Complex numbers are simply a "rectangular" way to represent a magnitude and phase. Sometimes you can relate easily to what the parts represent, sometime it's pretty abstract. Take Impedance, for example, which is resistance and reactance. But other times the "magnitude in the X direction" is just that.

Last edited: Sep 15, 2013
15. Oct 28, 2014

Jeffrey Yang

Hi John:

I hope you can see this "so late" reply. :-)

Yes, I'm also thinking this "fundamental" question. When we have used the time-harmonic approximation, why we still get a complex value amplitude. Some books I've read also "emphasise" that spatial part E(r) is complex. In my opinion, such a "complex valued" amplitude means "it can be complex valued" if the fields contains an extra phase. Exp(i*w*t) has an initial phase equals zero, therefore if the fields has an extra phase it must be wrote into the spatial part. Of course, if there has no extra phase, the spatial amplitude would be real. The so-called extra phase here can be determined by the source of the field, or ourselves, for physical and mathematical reasons.

This is my understanding. Overall, the complex valued amplitude is just a complete statement.

16. Oct 29, 2014

sophiecentaur

This is also a late entry but what do you actually mean by a Complex Amplitude? The complex notation for periodic functions always implies "the real part of" when you get to the end of your calculation. So what situation will generate the need for a 'complex' amplitude? What would you be trying to represent in that mathematical model? If your interest is 'just' mathematical then just follow the rules and you will obtain an answer but it could be a case of GIGO, scientifically.

17. Oct 29, 2014

Jeffrey Yang

To my own experience, the "complex amplitude" will be a little bit "physically" meaningful when we discuss the resonance system. In a well-confined system the resonance mode can only be real valued (even if we use complex notation) with real resonance frequency. However, for a real open resonance system, mode is partially leaking away. In such a case, both the amplitude and resonant frequency will be complex. In such a case, the complex amplitude meaning the system is non-Hermitian, or free propagating in some degree.

But anyway, the complex amplitude just means a spatial phase information is contained here.

18. Oct 29, 2014

sophiecentaur

All that can go 'inside' the exponential bit, surely?? ( as in ωt - kx ) When there are multiple travelling waves (a standing wave pattern), that will just be a sum of such exponential terms. Are you saying that you can represent that in terms of a complex amplitude function? I guess I shouldn't be surprised.

19. Oct 29, 2014

Jeffrey Yang

I mean even you using the exponential form exp[i*(wt-kx)], which is complex due to the phase information, to construct a standing wave pattern, the imaginary part "of the spatial component", or "of amplitude" we discussed here, will finally disappear. Such as

sin( πx/nL) *exp(i*wt)= 1/2i * {exp[i*(πx/nL)] + exp[-i*(πx/nL)]} * exp(i*wt)

This is just because the standing wave will only has in phase part or anti-phase part in the space, and therefore the spatial mode pattern will always be real. But once the boundary condition becomes weaker, which means the mode is leaky, the spatial pattern will be complex again. It also corresponds to whether the system is Hermitain or not, although they are not directly shown.

20. Oct 29, 2014

sophiecentaur

That makes sense now.
Cheers.