Is it possible to have a 2x1 wave guide combiner which doesn't through away 3dB?

[narra]

Is it possible to have a 2x1 wave guide combiner which doesn't through away 3dB?

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.

 

[Mr_Physicist]

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.

 

[marcusl]

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 beams--so 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.

 

[narra]

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

 

[Born2bwire]

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

This is going to happen regardless of how you design the waveguide. Try working this exercise which I believe is relevant here. I'm taking this from Hermann Haus' "Electromagnetic Noise and Quantum Optical Measurements."

2.9 A lossless "Y", [...], is a three-port. The three-port can be matched from port (1) by slow tapering. Show that if it is matched as seen from port (1), it cannot appear matched as seen from ports (2) and (3). Find the scattering matrix.

You can solve this merely on the basis of how S parameters are defined and assuming that the each of the three branches of the waveguide have some unknown impedance.

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,

\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]

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 non-zero, 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 non-zero and S23 is non-zero 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.

 

[marcusl]

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

Even laser light is not single mode, single frequency, and an ideal plane wave over its beamwidth--there are annoying effects like speckle indicating imperfection. If you could produce two perfectly coherent beams and a perfect splitter/combiner, you would get full recombination from your back to back devices. What you propose works at RF where it possible to produce signals in two paths that are truly coherent.

 

[narra]

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

 

[marcusl]

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 in-phase waves (assuming a 0° splitter) that recombine in-phase 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.

 

[narra]

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