
#1
Mar912, 03:10 PM

P: 38

Hello,
I was wondering if someone knew if it was physically possible to have a 2x1 optical wave guide combiner which doesn't throw away half of the light. The light must be "single wavelength" and is unpolarised, at least initially. I've drawn a schematic of what I mean. If anyone can give me their experiences on this sort of thing I would be much appreciative. Thank you. 



#2
Mar912, 05:34 PM

P: 19

There is a theorem on this, but I can't remember what its called. The answer, to the best of my knowledge, is no. Picture it in bulk optics: if you have two laser beams, you can use a beam splitter to combine them, but you will only ever get half of each laser beam in each port. It is frustrating, because it seems like it shouldn't be difficult to combine two beams into a single, collinear beam, but it's not possible. I will admit, however, that I have not researched this problem in depth, so there could be something I missed.




#3
Mar1012, 10:25 AM

Sci Advisor
PF Gold
P: 2,020

It's because the two beams are not coherent with each other. If they were ideal plane waves of same frequency and phase, that is, spatially and temporally coherent with each other, then they'd add perfectly and the loss would be 0 dB.
You can see this is so by running the device backwards. A single pure beam is then split into two equal coherent beamsso to run it forwards you must have beams coming in that are mutually coherent. Since two laser beams are not generally mutually coherent, you get the 3 dB loss. 



#4
Mar1112, 07:44 AM

P: 38

Is it possible to have a 2x1 wave guide combiner which doesn't through away 3dB???
Thank you Mr_Physicist, it may just be one of these annoying twists in nature, but if you could remember the name of that theorem then I would be interested to know.
Marcusl, you may have a point but consider this, you split a coherent source into two mutually coherent sources. These then travel the same path and into a splitter in reverse orientation, yet 50% at least must be lost from the combined output. Without relying on wavelength or polarisation can this process be done without such a large loss of light? For example, can two wave guides be tapered into one to achieve this? If not, why not? Thanks to to both of you for your feedback, it is appreciated. narra 



#5
Mar1212, 11:00 AM

Sci Advisor
PF Gold
P: 1,721

Where ports (2) and (3) are the split output ports and port (1) is the single input port. If you work it out, you will find that the impedance mismatch between (1) and ports (2) and (3) results in your 3 dB loss. This can be seen from your scattering matrix. It's been years since I worked the problem, but my recollection is that you will find that, [tex] \overline{\mathbf{S}} = \left[ \begin{matrix} 0 & 0.5 & 0.5 \\ 0.5 & 0.25 & 0.25 \\ 0.5 & 0.25 & 0.25 \end{matrix} \right] [/tex] That is, we see that S11 is 0 as we assumed and that S12 and S13 are both 0.5 as we expect for a splitter. But, S22 and S33 are nonzero, both being 0.25, which means if we feed the splitter in reverse then some of the power is reflected back. In fact we see this from the S21 and S31 which are 0.5. So if I feed on port (2), I would observe the 3 dB loss at the output on port (1). Part of the power is reflected back due to the mismatch of the Y branch with its trunk and another part of the power is coupled to the other Y branch and sent along that waveguide (as seen from the fact that S22 is nonzero and S23 is nonzero if we look from the perspective of port (2)). So you can see that a perfect splitter does not work perfectly in reverse because you can't match all the ports at the same time regardless of how we construct the waveguide. In the above analysis we make no assumptions about the structure of the waveguide but simply see how the scattering matrix falls out. 



#6
Mar1212, 06:29 PM

Sci Advisor
PF Gold
P: 2,020





#7
Mar1212, 08:02 PM

P: 38

Thank you both for your replies.
Born2wire, you have brought a whole new level of complexity to my problem, but it's food for thought and I look forward to investigating it deeper through your matrix approach. But from what you say, it is starting to seem quite convincing as I can begin to imagine the failures in each the different approaches for 3 way splitters. Marcusl, would you be able to explain further your suggestion of: "If you could produce two perfectly coherent beams and a perfect splitter/combiner, you would get full recombination from your back to back devices. ". narra 



#8
Mar1312, 11:27 AM

Sci Advisor
PF Gold
P: 2,020

At microwave frequencies, you can perform your thought experiment of back to back couplers for real:
A <> B Sorry for the low fidelity drawing... All of the power going into port A comes back out of B, except for a few tenths of a dB of ohmic losses. There is no 3dB power loss because signal A is split into coherent inphase waves (assuming a 0° splitter) that recombine inphase into full power at B. This works at microwave frequencies, so I assume the reason it doesn't work optically is that you can't get and maintain full coherence of the waves in the intermediate branches. 



#9
Mar1412, 08:49 PM

P: 38

Hi Marcusl,
Thank you for your reply I still need to look into Born2wire's matrix method to understand things further. I'm not sure if it is limited to the lack of coherence, a laser source can have a coherence length of a >100m while the component is only 60mm. Although I don't fully understand why it works in the microwave regime even with high coherence. I will have a think and a read and maybe post back if I find the solution. Regards, Narra 


Register to reply 
Related Discussions  
Fun Wave Eqn, Seperable Solution, Wave Guide  Calculus & Beyond Homework  0  
Why is the amplitude of wave guide indepentent of z?  Advanced Physics Homework  1  
TE mode in rectangular wave guide  Advanced Physics Homework  0  
Rectangular wave guide, TE01 mode  Classical Physics  23  
Wave Guide  Classical Physics  5 