0
$\begingroup$

I have this question in my homework where I have the following phasor of the electric field (we assume that all waves have $\omega$ frequency): $$\overline{\mathbf{E}}_{1}\left(x,y,z\right)=E_{0}\left[\hat{x}+\left(2+1.5j\right)\hat{y}+\left(2+3j\right)\hat{z}\right]e^{j\left(4x-4y+2z\right)}$$ and we are asked to find another plane wave $\overline{\mathbf{E}}_{2}\left(x,y,z\right)$ s.t their sum will carry power only in the $\hat{y}$ direction. I suggested the following plane wave: $$\overline{\mathbf{E}}_{2}\left(x,y,z\right)=E_{0}\left[-\hat{x}+\left(2+1.5j\right)\hat{y}-\left(2+3j\right)\hat{z}\right]e^{-j\left(4x+4y+2z\right)}$$ and I tried to calculate the average power by calculating: $$\vec{\boldsymbol{S}}_{tot} =\frac{1}{2}\Re\{{\overline{\mathbf{E}}_{tot}\times\overline{\mathbf{H}}_{tot}^{*}}\}$$and show that it has only a $\hat{y}$ non-zero component. I know this must be true but I probably have an arithmetic mistake (or something sillier). I don't know if it's legit to post my calculation here in hope that someone can catch my mistake but after $5$ hours of trying to figure out where's my mistake I am simply desperate. I'm sorry if it's too messy or too elaborate but I wrote it in LyX and converted it to Latex and paste it here. so here it is: \begin{align} \overline{\mathbf{E}}_{tot}\left(x,y,z\right) & =\overline{\mathbf{E}}_{1}\left(x,y,z\right)+\overline{\mathbf{E}}_{2}\left(x,y,z\right)= \\ & =E_{0}\left[\hat{x}+\left(2+1.5j\right)\hat{y}+\left(2+3j\right)\hat{z}\right]e^{j\left(4x-4y+2z\right)} \\ & \quad +E_{0}\left[-\hat{x}+\left(2+1.5j\right)\hat{y}-\left(2+3j\right)\hat{z}\right]e^{-j\left(4x+4y+2z\right)}=\\ & =E_{0}\left[\hat{x}\left(e^{j\left(4x+2z\right)}-e^{-j\left(4x+2z\right)}\right)+\left(2+1.5j\right)\hat{y}\left[e^{j\left(4x+2z\right)}+e^{-j\left(4x+2z\right)}\right] \right.\\ & \qquad \left. +\left(2+3j\right)\hat{z}\left(e^{j\left(4x+2z\right)}-e^{-j\left(4x+2z\right)}\right)\right]e^{-j4y}=\\ & =E_{0}\left[2j\sin\left(4x+2z\right)\hat{x}+\left(4+3j\right)\cos\left(4x+2z\right)\hat{y}+\left(-6+4j\right)\sin\left(4x+2z\right)\hat{z}\right]e^{-j4y} \end{align} The fields themselves: \begin{align} \overline{E}_{tot,x} & =2jE_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \overline{E}_{tot,y} &=\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y} \\ \overline{E}_{tot,z} &=\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \end{align}

Their derivatives:

\begin{align} \partial_{y}\overline{E}_{tot,z}&=\partial_{y}\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=\left(16+24j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \partial_{z}\overline{E}_{tot,y}&=\partial_{z}\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)=-\left(8+6j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \partial_{z}\overline{E}_{tot,x}&=\partial_{z}\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=4jE_{0}\cos\left(4x+2z\right)e^{-j4y} \\ \partial_{x}\overline{E}_{tot,z}&=\partial_{x}\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=-\left(24-16j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y} \\ \partial_{x}\overline{E}_{tot,y}&=\partial_{x}\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)=-\left(16+12j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \partial_{y}\overline{E}_{tot,x}&=\partial_{y}\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=8E_{0}\sin\left(4x+2z\right)e^{-j4y} \end{align}

The magnetic field: \begin{align*} \Longrightarrow\,\,\,\overline{\mathbf{H}}_{tot}\left(x,y,z\right) & =-\frac{1}{j\omega\mu}\nabla\times\overline{\mathbf{E}}_{tot}=\frac{j}{\omega\mu}\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ \partial_{x} & \partial_{y} & \partial_{z}\\ \overline{E}_{tot,x} & \overline{E}_{tot,y} & \overline{E}_{tot,z} \end{vmatrix}\\ & =\frac{j}{\omega\mu}\left[ \hat{x}\left(\partial_{y}\overline{E}_{tot,z}-\partial_{z}\overline{E}_{tot,y}\right)+\hat{y}\left(\partial_{z}\overline{E}_{tot,x}-\partial_{x}\overline{E}_{tot,z}\right) \right.\\ & \qquad \qquad \left. +\hat{z}\left(\partial_{x}\overline{E}_{tot,y}-\partial_{y}\overline{E}_{tot,x}\right) \right] =\\ & =\frac{jE_{0}}{\omega\mu}\left[\left(16+24j+8+6j\right)\sin\left(4x+2z\right)\hat{x}+\left(4j+24-16j\right)\cos\left(4x+2z\right)\hat{y} \right.\\ & \qquad \qquad \left. -\left(16+12j+8\right)\sin\left(4x+2z\right)\hat{z}\right]e^{-j4y}=\\ & =\frac{jE_{0}}{\omega\mu}\left[\left(24+30j\right)\sin\left(4x+2z\right)\hat{x}+\left(24-12j\right)\cos\left(4x+2z\right)\hat{y} \right.\\ & \qquad \qquad \left. -\left(24+12j\right)\sin\left(4x+2z\right)\hat{z}\right]e^{-j4y} \end{align*}

and its components,

\begin{align} \overline{H}_{tot,x}&=-\frac{E_{0}}{\omega\mu}\left(30-24j\right)\sin\left(4x+2z\right)e^{-j4y} \\ \overline{H}_{tot,y}&=\frac{E_{0}}{\omega\mu}\left(12+24j\right)\cos\left(4x+2z\right)e^{-j4y} \\ \overline{H}_{tot,z}&=\frac{E_{0}}{\omega\mu}\left(12-24j\right)\sin\left(4x+2z\right)e^{-j4y} \\ \overline{H}_{tot,x}^{*}&=-\frac{E_{0}^{*}}{\omega\mu}\left(30+24j\right)\sin\left(4x+2z\right)e^{j4y} \\ \overline{H}_{tot,y}^{*}&=\frac{E_{0}^{*}}{\omega\mu}\left(12-24j\right)\cos\left(4x+2z\right)e^{j4y} \\ \overline{H}_{tot,z}^{*}&=\frac{E_{0}^{*}}{\omega\mu}\left(12+24j\right)\sin\left(4x+2z\right)e^{j4y} \end{align}

giving the Poynting vector

\begin{align*} \vec{\boldsymbol{S}}_{tot} & =\frac{1}{2}\overline{\mathbf{E}}_{tot}\times\overline{\mathbf{H}}_{tot}^{*}=\frac{1}{2}\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ \overline{E}_{tot,x} & \overline{E}_{tot,y} & \overline{E}_{tot,z}\\ \overline{H}_{tot,x}^{*} & \overline{H}_{tot,y}^{*} & \overline{H}_{tot,z}^{*} \end{vmatrix}=\\ & =\frac{1}{2}\left[\hat{x}\left(\overline{E}_{tot,y}\cdot\overline{H}_{tot,z}^{*}-\overline{E}_{tot,z}\cdot\overline{H}_{tot,y}^{*}\right)+\hat{y}\left(\overline{E}_{tot,z}\cdot\overline{H}_{tot,x}^{*}-\overline{E}_{tot,x}\cdot\overline{H}_{tot,z}^{*}\right) \right. \\ & \qquad \left. +\hat{z}\left(\overline{E}_{tot,x}\cdot\overline{H}_{tot,y}^{*}-\overline{E}_{tot,y}\cdot\overline{H}_{tot,x}^{*}\right)\right] \end{align*}

with components

\begin{align*} S_{tot,x} & =\frac{1}{2}\left(\overline{E}_{tot,y}\cdot\overline{H}_{tot,z}^{*}-\overline{E}_{tot,z}\cdot\overline{H}_{tot,y}^{*}\right)=\\ & =\frac{1}{2}\left(\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)\cdot\left(\frac{E_{0}^{*}}{\omega\mu}\left(12+24j\right)\sin\left(4x+2z\right)e^{j4y}\right) \right. \\ & \qquad \left. -\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\frac{E_{0}^{*}}{\omega\mu}\left(12-24j\right)\cos\left(4x+2z\right)e^{j4y}\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(\left(4+3j\right)\cdot\left(12+24j\right)-\left(-6+4j\right)\cdot\left(12-24j\right)\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(-24+132j-24-192j\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =-\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(24+60j\right)\sin\left(8x+4z\right) \end{align*}

and

\begin{align*} S_{tot,y} & =\frac{1}{2}\left(\overline{E}_{tot,z}\cdot\overline{H}_{tot,x}^{*}-\overline{E}_{tot,x}\cdot\overline{H}_{tot,z}^{*}\right)=\\ & =\frac{1}{2}\left(\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\left(-\frac{E_{0}^{*}}{\omega\mu}\left(30+24j\right)\sin\left(4x+2z\right)e^{j4y}\right) \right. \\ & \qquad \left. -\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\left(\frac{E_{0}^{*}}{\omega\mu}\left(12+24j\right)\sin\left(4x+2z\right)e^{j4y}\right)\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(-\left(-6+4j\right)\cdot\left(30+24j\right)-2j\cdot\left(12+24j\right)\right)\sin^{2}\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(276+{24j}+48-{24j}\right)\sin^{2}\left(4x+2z\right)=\\ & =\frac{162\left|E_{0}\right|^{2}}{\omega\mu}\sin^{2}\left(4x+2z\right) \end{align*}

and

\begin{align*} S_{tot,z} & =\frac{1}{2}\left(\overline{E}_{tot,x}\cdot\overline{H}_{tot,y}^{*}-\overline{E}_{tot,y}\cdot\overline{H}_{tot,x}^{*}\right)=\\ & =\frac{1}{2}\left(\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\left(\frac{E_{0}^{*}}{\omega\mu}\left(12-24j\right)\cos\left(4x+2z\right)e^{j4y}\right) \right. \\ & \qquad \left. -\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)\cdot\left(-\frac{E_{0}^{*}}{\omega\mu}\left(30+24j\right)\sin\left(4x+2z\right)e^{j4y}\right)\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(2j\cdot\left(12-24j\right)+\left(4+3j\right)\cdot\left(30+24j\right)\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(48+24j+48+186j\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(48+105j\right)\sin\left(8x+4z\right) \end{align*}

So, as you can see the real parts of the $\hat{x}$ and $\hat{z}$ components are not $0$.

Any help would be much appriciated.

$\endgroup$
1
$\begingroup$

I wrote the matlab code below. The values of $E_1$ and $E_2$ are assumed to be at $t=0$ and at $\vec{r}=0$.

The result is that $P=0$ for your choice of $E_2$. If you change the sign of $E_2$ the result is a momentum / energy transport density in the -y direction. A result of P along the positive y-direction seems hard to achieve since the initial wave travels in the -y direction.

r = [0,0,0];

t = 0;

Amp1 = [1,2+1.5j,2+3j]; 

k1 = [4,-4,2];

ph1 = exp(j*(sqrt(sum(k1.*k1))*t+sum(k1.*r)));

E1 = Amp1*ph1;

Amp2 = [-1,2+1.5j,-(2+3j)]; 

k2 = [-4,-4,-2];

ph2 = exp(j*(sqrt(sum(k2.*k2))*t+sum(k2.*r)));

E2 = Amp2*ph2;

Poy = real(cross(E1+E2,cross(k1,conj(E1))+cross(k2,conj(E2))));

display(Poy);
$\endgroup$
0
$\begingroup$

This is almost certainly caused by the minus sign in the exponent in $$ e^{-j\left(4x+4y+2z\right)} = e^{j\left(-4x+4y-2z\right)} $$ in your second wave; the wave that results is precisely the clear answer if you were asked to take the wave $e^{j\left(4x-4y+2z\right)}$ and cancel out the power flow in the $x$ and $z$ directions, but not the $y$ direction, by turning the first two into standing waves.

You then also need to check that the polarizations work out correctly (though they look OK to me at a first glance), and goodness help whoever tries to wade into the thicket of formulas you've posted (which, frankly, should not be done in terms of the total fields ─ calculate everything in terms of $\mathbf E_1$ and $\mathbf E_2$ all the way through, and only put in their explicit expressions when they're absolutely crucial). But the clear place to start is with the correct exponent.

$\endgroup$
  • $\begingroup$ Hi Emilio. firstly, thanks for your comment (I know this question doesn't look so appealing to read). second, I'm not quite sure why the problem would be in the minus sign in the second exponent. the wave vector of the first wave is $\vec{k}_1=(-4,4,-2)$ so the second wave vector should be (for my understanding) $\vec{k}_2=(4,4,2)$ in order to be able to "cancel out" the propagation in the $\hat{x}$ and $\hat{z}$ directions but maintain the propagation in the $\hat{y}$ direction. can you please explain why the minus sign isn't correct? $\endgroup$ – dorsh605 Apr 18 '18 at 18:09
  • 2
    $\begingroup$ Hmmm. My answer is probably wrong, but hopefully this can still be a learning moment: conventions do matter, and futzing about with putting a minus sign in front vs behind the $j$ will do nothing but confuse your audience. If you're working in a convention where spatial phases go as $e^{-j \vec k \cdot \vec r}$, then substituting $e^{-j\left(-4x+4y-2z\right)}$ for $e^{j\left(4x-4y+2z\right)}$ will make your text less clear, not more. If you get a few more dashes on the page it's a price well worth paying for consistency of notation. $\endgroup$ – Emilio Pisanty Apr 18 '18 at 18:19
  • $\begingroup$ So, for clarity (i.e. make this clear in the text of your question, and early on): precisely which wavevectors are you understanding your initial wave to have, and what wavevector did you intend for the second one? What sign convention are you using on $e^{\pm j \vec k \cdot \vec r}$ and on $e^{\pm j \omega t}$? $\endgroup$ – Emilio Pisanty Apr 18 '18 at 18:21
  • $\begingroup$ You are absolutely write about my inconsistent sign convention in the above 2 waves. The sign convention I'm using is $e^{-j\vec{k}\cdot\vec{r}}$ and $e^{j\omega t}$. As I wrote in my previous comment, for my understanding, the initial wave vector is: $\vec{k}_1=(-4,4,-2)$ and the second wave vector is $\vec{k}_2=(4,4,2)$. hope this clarifies the conventions I'm using. $\endgroup$ – dorsh605 Apr 18 '18 at 18:35
0
$\begingroup$

After a lot of effort trying to find a condition for the wave I'm adding ($\overline{\mathbf{E}}_{2}$) it turned out that my math was fine but I chose the wrong relative phases for the different components of $\overline{\mathbf{E}}_{2}$ because I was under the impression that the relative phases of the components of $\overline{\mathbf{E}}_{2}$ doesn't matter (which means that I have more than one d.o.f) but that is wrong in the general case (it is true for example for a wave with linear polarization in the form: $\overline{\mathbf{E}}_{1}\left(x,y,z\right)=E_{0}e^{-j\left(k_xx+k_zz\right)}\hat{y}$)

Anyway, I finally managed to find necessary and sufficient conditions for $\overline{\mathbf{E}}_{2}$ s.t. $\overline{\mathbf{E}}_{1}+\overline{\mathbf{E}}_{2}$ will carry power only in the $\hat{y}$ direction. let's make it general. the most general form of a phasor of planar wave is: $$\overline{\mathbf{E}}_{1}(x,y,z)=[A\hat{x}+B\hat{y}+C\hat{z}]e^{-j(k_xx+k_yy+k_zz)}$$ where: $$A=|A|e^{j\varphi_A}\,\,,B=|B|e^{j\varphi_B}\,\,,C=|C|e^{j\varphi_C}$$ if we want to add another wave $\overline{\mathbf{E}}_{2}$ s.t. their sum will carry power only in the $\hat{y}$ direction we have to choose the following $\mathbf{k}_2$ vector of $\overline{\mathbf{E}}_{2}$: $$\mathbf{k}_2=(-k_x,k_y,-k_z)$$ and we also need to choose the amplitudes of the components of $\overline{\mathbf{E}}_{2}$ such that they equal in absolute value to the corresponding amplitudes of $\overline{\mathbf{E}}_{1}$ because we need to create sinusoidal dependence in the $x$ and $z$ coordinates. thus, the most general form of $\overline{\mathbf{E}}_{2}$ we can offer is: $$\overline{\mathbf{E}}_{2}(x,y,z)=[Ae^{j\varphi_1}\hat{x}+Be^{j\varphi_2}\hat{y}+Ce^{j\varphi_3}\hat{z}]e^{-j(-k_xx+k_yy-k_zz)}$$ now, all that left is to do the (tedious) math and calculate the complex poynting vector of the sum of the 2 waves and finally require that the real parts of it's $x$ and $z$ components will vanish. after doing that I found that $\varphi_1, \varphi_2, \varphi_3$ must hold the following equations: $$\varphi_1 = \varphi_2 + 2(\varphi_B-\varphi_A)+\pi$$ $$\varphi_3 = \varphi_2 + 2(\varphi_B-\varphi_C)+\pi$$ and $\varphi_2$ stays free to choose (which means that we only left with one d.o.f). The above condition doesn't need to hold in the case that $B=0$ or $k_y=0$ or {$A=0$ and $C=0$} (which for example, corresponds to the simple case of $\overline{\mathbf{E}}_{1}\left(x,y,z\right)=E_{0}e^{-j\left(k_xx+k_zz\right)}\hat{y}$ because in that case we have $k_y=0$ and also $A=0$ and $C=0$).

In the specific example of this question we have: $$\overline{\mathbf{E}}_{1}(x,y,z)=E_{0}[\hat{x}+(2+1.5j)\hat{y}+(2+3j)\hat{z}]e^{-j(-4x+4y-2z)}$$ So, if we follow the above condition we get the following wave: $$\overline{\mathbf{E}}_{2}(x,y,z)=E_{0}e^{j\varphi}[-\hat{x}+(2-1.5j)\hat{y}+(-2+3j)\hat{z}]e^{-j(4x+4y+2z)}$$ where $\varphi$ can be any phase (this is our d.o.f) and under that choice we get that $\overline{\mathbf{E}}_{1}+\overline{\mathbf{E}}_{2}$ carries power only in the $\hat{y}$ direction.

$\endgroup$

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.