EM plane waves are represented using the equation (ignoring amplitude) $$E(x,t) = e^{j (wt - kx)}$$

They are also represented in this form: $$E(x,t) = \cos(wt - kx)$$

I want to get the electric field squared. In the first form, I set $$E^2(x,t) = e^{2j(wt - kx)} = \cos(2(wt-kx)) + j\sin(2(wt-kx))$$

Taking the real part, I get \$\cos(2(wt-kx))\$. For the latter form, I instead get \$\cos^2(wt-kx)\$, which these two do not equal.

Since, the above is false, why is it valid to represent plane waves with $$e^{j (wt - kx)}\:?$$ Performing a simple operation on this yields a false answer. It's been a while since I've worked with waves, so I've forgotten a lot.


What you are actually trying to find out here, is for what operators \$H\$ you have the following property:

$$\mathcal{R}\left\{H\left\{f(x,t)\right\}\right\} = H\left\{\mathcal{R}\left\{f(x,t)\right\}\right\}$$

Or otherwise, which operators \$H\$ are commutative with taking the real part. Any operator that does not have this property (like squaring) will give rise to the issues you are addressing.

So using the exponential notation is only valid if the problem never requires you to use an operator that doesn't satisfy that property. Most problems are linear though, and all linear operations are commutative with taking the real part.

Can someone clear up my confusion? Since, the above is false, how can we simply work with \$e^{j(\omega t - kx)}\$

So to answer your question: you can't in your example.

  • \$\begingroup\$ I don't understand. Every book I've seen represents an electromagnetic wave as e^(j(wt - kx)), so what do you mean we can't? \$\endgroup\$ Jan 26 '19 at 22:32
  • \$\begingroup\$ It means that the problems they solve do not involve "squaring" a wave in the sense that you're suggesting. \$\endgroup\$
    – Sven B
    Jan 26 '19 at 22:33
  • \$\begingroup\$ You are saying, I am mistaken in squaring an electric field? I've updated my question for clarity \$\endgroup\$ Jan 26 '19 at 22:50
  • \$\begingroup\$ No, but if you do square an electric field you can't use the exponential notation because the real part won't represent the actual solution anymore - as you've demonstrated in your question. If your books use the notation, I doubt you will find an exercise or example where they square the electric field. \$\endgroup\$
    – Sven B
    Jan 27 '19 at 1:10

You went from \$\left ( Re\{e^{j\theta}\} \right )^2\$ to \$Re \left \lbrace \left (e^{j\theta}\right )^2\right \rbrace\$. As you just found out, you can't do that.

Based on your comment to this answer, your confusion lies in not remembering the Euler identity: $$ e^{j\theta} = \cos \theta + j \sin \theta$$ From this you can get $$\cos \theta = \frac{e^{j\theta} + e^{-j\theta}}{2}, \, \sin \theta = \frac{j e^{j\theta} - j e^{-j\theta}}{2} $$

So $$\cos^2 \theta = \left (\frac{e^{j\theta} + e^{-j\theta}}{2} \right )^2$$ But \$\left ( Re\{a + j b\} \right )^2 \ne Re \left \lbrace \left (a + j b\right )^2\right \rbrace\$ for \$a, b\$ real and \$b \ne 0\$ (this is a good one to work out if you haven't done complex numbers in a bit). That means that for \$\sin \theta \ne 0\$ the relation in your question does not hold.

  • \$\begingroup\$ I understand the derivation above. I just don't understand how we can let e^(i theta) represent cos(theta), if squaring e^(i theta) and then taking the real part does not give us cos^2(theta). \$\endgroup\$ Jan 26 '19 at 22:27
  • 1
    \$\begingroup\$ We cannot let \$e^{j \theta}\$ represent \$\cos \theta\$, we can just use the real part of \$e^{j \theta}\$, under certain constrained conditions. Then we need to be mindful of what the constraints are. \$\endgroup\$
    – TimWescott
    Jan 27 '19 at 22:33

Since Tim hasn't responded yet, I'll add the following. Hopefully, some clarity arrives.

You probably already know that \$\operatorname{cos}\left(2\theta\right)=2\operatorname{cos}^2\left(\theta\right)-1\$. There are several ways to derive this result, but here's one:

$$\begin{align*} \operatorname{cos}^2\left(\theta\right)&=\left[\frac{e^{j\theta}+e^{-j\theta}}{2}\right]^2\\\\ &=\frac14\left(e^{j2\theta}+e^{-j2\theta}+2\right)\\\\ &=\frac14\left(2\operatorname{cos}\left(2\theta\right)+2\right)\\\\\therefore\\\\\operatorname{cos}\left(2\theta\right)&=2\operatorname{cos}^2\left(\theta\right)-1 \end{align*}$$

You also can very easily perform the following:

$$\begin{align*} \left[e^{i\theta}\right]^2&=\left(\operatorname{cos} \theta+i\operatorname{sin} \theta\right)^2\\\\ &=\operatorname{cos}^2 \theta-\operatorname{sin}^2 \theta+2 i\cdot\operatorname{cos} \theta\cdot\operatorname{sin} \theta \end{align*}$$

The real portion of that is obviously \$\operatorname{cos}^2 \theta-\operatorname{sin}^2\theta=2\operatorname{cos}^2\left(\theta\right)-1=\operatorname{cos}\left(2\theta\right)\$.

I'm not sure what's left to discuss. Is that clear enough?

Sidebar: Physicists use \$\Phi e^{i\left(kx-\omega t\right)}\$ and engineers use the notation you provided. Mostly, just a sign convention difference. (The 3D version in physics is \$\Phi e^{i\left(\vec{k}\cdot\vec{r}-\omega t\right)}\$ -- note that \$\vec{k}\cdot\vec{r}\$ is a dot product with a scalar result, which combines well with \$\omega t\$.) I like this page: Linear and nonlinear waves.

  • \$\begingroup\$ I understand the derivation. My confusion is that we were using e^(i theta) to represent cos(theta). So, when we square e^(i theta), we expect to get cos^2(theta) with some imaginary term we discard, but this is not the case. Then, how are we allowed to use e^(i theta) to represent a real life sinusoid? \$\endgroup\$ Jan 26 '19 at 22:25
  • \$\begingroup\$ @Goldname No. It's not "we expect," but instead should be you expect to get that (currently, anyway.) And that's what needs fixing. I think you see a conundrum that the rest of us do not see. The above I wrote shows exactly why. Perhaps you are failing to present your case clearly enough? Or, perhaps we just are communicating at cross-purposes and neither of us yet realizes it. That's possible, too. But I've struggled to see how that might be possible and I don't see it, yet. \$\endgroup\$
    – jonk
    Jan 26 '19 at 22:30
  • \$\begingroup\$ In that case, the representation: e^(j (wt -kx)) of any EM wave, is false and inconsistent with the correct representation: cos(wt - kx)?? \$\endgroup\$ Jan 26 '19 at 22:33
  • \$\begingroup\$ @Goldname I added a page for you to examine at the very end of what I wrote. Perhaps you could visit it and see if you can find what you are looking for there? But I'm not seeing an inconsistency yet in my own understanding. But perhaps it's because I don't possess your entire context. Can you provide a web page for me to examine that is causing you trouble? \$\endgroup\$
    – jonk
    Jan 26 '19 at 22:35
  • \$\begingroup\$ Thanks for the response. I might be misunderstanding you, but I'll try to rephrase my confusion in a better way. There are two ways to represent a plane wave: E(x,t) = Ae^(j*(kx - wt)) and also E(x,t) = Acos(kx - wt). If I want to square a plane wave, the former and latter real parts do not equal each other. So, how can both these representations be correct? \$\endgroup\$ Jan 26 '19 at 22:44

When solving the wave equation of \$\mathbf{E}\$, we are basically solving a differential equation, e.g. $$ \frac{\partial^2E}{\partial x^2} -k\frac{\partial^2E}{\partial t^2}=0 $$

Our aim is to find a solution of the form: \$E= E_0\cos(wt-\beta x)\$.

To arrive at \$E\$, we try to solve a different problem: $$ \frac{\partial^2E_c}{\partial x^2} -k\frac{\partial^2E_c}{\partial t^2}=0, $$ whose solution \$E_c\$ is a complex number (containing our actual required solution \$E\$, as its real part), i.e. \$E_c = E_0 e^{j(wt-\beta x)}\$.

It's easy to see, the derivative operator (in the new problem) will act linearly with the real part and the imaginary part, and the {real part of derivative} equals {derivative of real part}. So the real part of LHS should be equal to the real part of RHS; so the real part of the solution \$E_c\$ we get, should be equal to the solution \$E\$ to our original problem.

Solving the 2nd equation is easier than the first one, so we solve that first, find \$E_c\$, then find \$H_c\$, and finally take the real part of each, knowing that all our operations here are linear, so real part would have come out intact. See for yourself that the above holds even when E is vector. This should emphasize the limitations of this trick: non-linear operation, like finding square, will not give faithful result.

Thus, to answer your question, EM plane waves are actually of form: \$E(x,t)=\cos(wt−kx)\$ and, are equivalently (not exactly, i.e. with some constraints) represented using: \$E(x,t)=e^{j(wt−kx)}\$. Now it should be clear- to find \$E^2(x,t)\$, the first form will give correct result, while the second won't.


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