
#1
Jan2213, 11:52 AM

P: 3,842

For RHC EM wave travel in +z direction, the unit vector is [itex]\hat {E}=\frac{\hat {x}+\hat{y}j}{\sqrt{2}}[/itex] ignoring the ωtkz.
What if the RHC EM wave travels in z direction? The unit vector should be [itex]\hat {E}=\frac{\hat {x}\hat{y}j}{\sqrt{2}}[/itex] ignoring the ωt+kz. Am I correct? 



#2
Jan2313, 09:06 AM

P: 861

This is an apology for my earlier post which was here for a few hours. I'd failed to spot the j in your vectors, so what I wrote was misleading.




#3
Jan2313, 12:12 PM

P: 3,842

To make it easy, I put the receiving antenna at the origin facing the +z direction. Then I put the transmitting antenna at great distant on the +z axis transmitting towards the origin. With this setup, I can find the degree of matching ( Loss factor) in this system of two antennas. Polarization characteristics of an antenna is by the EM wave it produces when transmitting, which in this case is RHC ( right hand circular) polarization and is circular polarization. So I simplify the issue by just looking at the two plane wave facing each other. Back to your question, Polarization of EM wave is direction dependent, ie: travel in +z or z. So we first have to set up ONE reference coordinates using the receiving antenna at the origin and put the two antennas in z axis. With this the transmitting wave is z direction, and the ( pretended) transmitting wave from the receiving antenna is traveling in +z direction. That's where I come up with the two waves traveling in opposite direction. The Loss factor is defined as: [tex]LF=\left(\hat {E}_w \cdot \hat{E}_a\right)^2[/tex] Where [itex]\hat{E}_w[/itex] is unit vector of the plane wave by the transmitting antenna towards the origin along z axis. [itex]\hat{E}_a[/itex] is unit vector of plane wave by the receiving antenna ( pretend) transmitting from origin towards +z direction. 



#4
Jan2313, 01:49 PM

Sci Advisor
PF Gold
P: 1,721

Characterize EM wave propagate in +z and z
The direction of the circular polarization is taken in context to how the polarization rotates when viewed from the rear of the wave. So if you switch the direction of propagation, you need to correspondingly adjust the direction of rotation. To view the rotation for a +z travelling wave, you view the rotation of the electric vector in time in the xy plane where the upper vertical axis is +x and the right hand horizontal axis is +y. To view the rotation for a z travelling way, you look at the xy plane where the upper vertical axis is the +y and the right hand horizontal axis is the +x. Doing this we can see that neither of the two waves proposed by the OP is a RHCP.
The correct waves, using the [itex]j\omega t[/itex] time convention are: +z: [itex] \left( \hat{x}  j\hat{y} \right) e^{j\omega t  \beta z} [/itex] z: [itex] \left( \hat{x} + j\hat{y} \right) e^{j\omega t + \beta z} [/itex] Obviously of course to find the alignment of the electric field we simply take [itex] Re \left\{ \left( \hat{x}  j\hat{y} \right) e^{j\omega t} \right\} [/itex] when observing at z=0 to see how the z propagating wave rotates and we see that this is a RHCP as we advance t. 



#5
Jan2313, 07:08 PM


#6
Jan2413, 10:34 AM

Sci Advisor
PF Gold
P: 1,721

Balanis's "Advanced Engineering Electromagnetics" does a good job of explicitly explaining the concept in his fourth chapter with the figures as you have desired. The simple way of looking at this is by constructing a 2D plot where the crossproduct of the axes gives you the direction of the wave propagation (hence the specifications I gave for the axes in my previous post). Then you can simply plot out the progression of the polarization as a function of increasing t using the real part of the time harmonic expression. For example, [itex] Re \left\{ \left( \hat{x}  j\hat{y} \right) e^{j\omega t} \right\} [/itex] at t= 0 is [itex]\hat{x}[/itex] and at say t = 0.01 and a unity wavelength is [itex]0.998\hat{x} + 0.0628\hat{y}[/itex]. From those two points we can easily discern the handedness of the propagation.
A more rigorous and general way of finding the polarization is by use of the Poincaré Sphere. Using the Poincaré Sphere, we find that [itex]\gamma = \tan^{1} \frac{1}{1} = 45^{\circ}[/itex] and [itex]\delta = 90^{\circ}  0^{\circ} = 90^{\circ}[/itex]. This gives the point on the sphere to be [itex] \epsilon = 45^{\circ} [/itex] and [itex]\tau = 22.5^{\circ}[/itex]. From the [itex]\epsilon[/itex], we deduce that we have an RHCP since [itex]2\epsilon[/itex] means that the point lies at the southern pole of the sphere. I'll say that your figures seem to be correct though I do not think that mapping out the explicit orthogonal components of the electric field has much utility when compared with the superposition of the two since the latter clearly demonstrates the circular polarization. 



#7
Jan2413, 03:39 PM

P: 3,842

Thanks for your detail reply. I just ordered the Advanced EEM by Balanis from Amazon. I am using the Antenna theory by Balanis also, so we are talking about the same convention. I have both 2nd and the 3rd edition.
There is more confusion!!! I am using the 2nd edition only because I have downloaded the solution manual on line. But the convention in the example inside the book and the solution of the problems are using a different convention. Attached is a scan of Example 2.10 in p71 of the Antenna Theory 2nd edition. Just ignore all my writing on the book, just read the text only. The standard convention of a RHC antenna transmitting in +z direction should be [itex]\hat{\rho}_w=\frac{\hat{\theta}\hat{\phi}j}{\sqrt{2}}[/itex] as we both agreed. And the receiving antenna has to be along the z axis with transmitting wave in ve z direction. Therefore [itex]\hat{\rho}_w=\frac{\hat{\theta}+\hat{\phi}j}{\sqrt{2}}[/itex]. But as you see, the answer in the book is exactly opposite. The answers in the solution manual are exactly opposite from what our understanding. I am confused. The solution manual is written by someone else, but I take it that the book is not wrong!!! I draw the wave just for my own understanding in the physical sense. Also the book did say the sense of rotation is from the wave with leading phase to the lagging one. By drawing the two orthogonal components out, I can see the rotation. I have a hard time finding the relation between the Poincare Sphere to ellipse translation and the derivation. If you have any suggestion on materials relate to this, please let me know. I have joined two antenna forums thinking they should know these kind of things, but apparently engineers don't really care about this, all they care is RHC match to RHC and everything is fine!!! Really appreciate your help. 



#8
Jan2413, 03:48 PM

P: 3,842

BTW
The example used in Balanis 3rd edition agrees with the convention we both agreed: Don't tell me there is a change of heart!!! But I read the text in both books, they are exactly the same. Hey Born2Wire, I really appreciate your help on this. It is not easy to find materials on this. The only other book is by Kraus and he uses a different convention where the propagation is towards you ( out of paper) rather into the paper.....Or as you put it, looking at the head of the wave instead of the tail of the wave. Kraus uses Poincare sphere. I looked at other EM books, they mostly use Kraus' convention. 



#9
Jan2513, 06:55 PM

P: 3,842

I have read through P146 to 166 in Advanced EEM by Balanis, I am more sure I am right in the last two posts.




#10
Jan2613, 02:12 AM

Sci Advisor
Thanks
P: 2,132

This naming drives me always nuts. It sometimes even depends on the book! It's easier to use modern physics to name the circularly polarized waves. As massless vector fields despite energy and momentum (density) the standard modes are characterized by the helicity, which is the projection of the total angular momentum to the direction of momentum and takes the values [itex]\pm 1[/itex].




#11
Jan2613, 03:46 AM

P: 3,842

Thanks for your reply. The issue is all the inconsistent are from the same author....Constantine Balanis. He made it very clear how to classify RHC and LHC and their unit vector according to the direction of propagation. It is very straight as both Born2wire and me agree on the same convention. But then in one of the example, it is opposite. I just want to make sure I interpreted the question correctly. I have to assume I am wrong and the example of the book is right.




#12
Jan3113, 11:01 AM

P: 3,842

I bring this one up again to show all the inconsistency from Balanis.




#13
Mar213, 03:56 AM

P: 3,842

First look at the rotation of the vector ##\hat {\theta}+\hat{\phi}j## [tex]\vec E(0,t)\;=\;Re[\hat {\theta}E_{\theta 0}e^{j\omega t}+\hat{\phi}E_{\phi 0}e^{j\omega t}e^{j\frac{\pi}{2}}]\;=\;\;Re[\hat {\theta}E_{\theta 0}e^{j\omega t}+\hat{\phi}jE_{\phi 0}e^{j\omega t}]\;=\;\hat {\theta}E_{\theta 0}\cos\omega t+\hat {\phi}E_{\phi 0}\cos(\omega t +\frac{\pi}{2})[/tex] Let##\Psi## be the angle from ##\theta## axis to ##\vec E(0,t)## [tex]\Rightarrow \;\tan\Psi\;=\;\frac {\cos(\omega t+\frac{\pi}{2})}{\cos \omega t}\;=\;\frac {\cos \omega t \cos(\frac{\pi}{2})\sin \omega t\sin(\frac{\pi}{2})}{\cos \omega t}\;=\;\tan\omega t[/tex] [tex]\Rightarrow\;\Psi\;=\;\omega t[/tex] Therefore ##\Psi## decreases as time t increases. With this, if the plane wave is traveling in +z direction ( in spherical coordinates, in +R direction) AND draw the diagram where the propagation is INTO the paper, this is LHC ( left hand rotation). Therefore, when the book said the RHC transmitting antenna at origin radiating wave at +z direction ( +R), the unit vector HAS to be [tex]\hat{\rho}_w\;=\;\frac{\hat{\theta}\hat{\phi}j}{\sqrt{2}}[/tex] But the book claimed: [tex]\hat{\rho}_w\;=\;\frac{\hat{\theta}+\hat{\phi}j}{\sqrt{2}}[/tex] The book is wrong!!! There is no if and buts about this, there is no convention confusion. The example 2.12 in post #8 copied from the latest edition is correct. Anyone can comment on this? 



#14
Mar313, 06:52 PM

P: 3,842

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