What is least number of beam combiners required to combine 'n' number of beams?

[raghavsol]

Hi to All,

Here is a puzzle:

If I have to combine 2 optical beams (of same spectrum) using a beam splitter (division of ampiltude), like; dichroic dual beam combiner etc...then I need 1 such dual beam combiner.

In my experiment I have 4 beams (of same spectrum) from 4 different sources and I am using 3 beam combiners to combine them finally. In general I can say that in my geometry If there are 'n' number of beams then I will need 'n-1' number of beam combiners. Can there be a different geometry, so that I can reduce number of beam combiners, so that my transmission can be increased, because at each beam combiner 50% loss is there. Or 'n-1' is the least one? I can send send my experiment's geometry if required.

Hope all will enjoy this simple problem.

Raghav
email: raghavsol@gmail.com

[wfwhitaker]

Simple for you, oh wise one of the East.

[Donald Miller]

In your example of four beams, it initially seems that N-1 is optimal...I assume here that you are combining two beams into separate bundles and recombining these initial bundles a second time. If, however, one were to take your example and increase the number of beams to six or eight or some higher number than four...so long as they are an equal number than the N-1 parameter no longer is valid. In the case of six beams a total of four recombiners would suffice (n-2) and in the case of 8 beams the total necessary recombiners would be five (n-3).

I hope that I am following your reasoning and have interperted your question properly.

[percepts]

In your example of four beams, it initially seems that N-1 is optimal...I assume here that you are combining two beams into separate bundles and recombining these initial bundles a second time. If, however, one were to take your example and increase the number of beams to six or eight or some higher number than four...so long as they are an equal number than the N-1 parameter no longer is valid. In the case of six beams a total of four recombiners would suffice (n-2) and in the case of 8 beams the total necessary recombiners would be five (n-3).

I hope that I am following your reasoning and have interperted your question properly.

Donald,

If your 6 beams are numbered 1 to 6 what are the pairings of beams you use to arrive at 4 recombiners.

[Ole Tjugen]

It is actually n-1. Think about it - each beam combiner removes one beam. To remove all but one beam takes n-1 combiners.

[Jim Michael]

Can a combiner take as input a previously combined beam?

[Dan Fromm]

I'm with Ole. You have what looks like a strictly bifurcating tree, reversed. These beasts have n-1 nodes where n is the number of leaves.

[Athiril]

It has to be n-1

but the effiency is different the way you link it up.

If they are all of the same intensity/signal/value, then there will be no difference, twice the value and half again is the value.

Say we have 6 nodes.

Lets call them a, b, c, d, e, f, and lets assign these 'stengths'/values to them
a=1
b=2
c=3
d=4
e=5
f=6

Joining a+b gives 1.5 (50% loss).

So lets try it like this, join a+b to start with then the joined (a+b) to c and so on.

So ((((a+b))+c)+d)+e)+f = 5.125

Instead of a+b c+d e+f

(a+b)+(c+d) and finally ((a+b)+(c+d))+(e+f) which equals 4 given a 50% loss at each joining.

You must do this in order of weakest to strongest signals for it to work best. Otherwise if its in total reverse (strong to weak).. you'll end up with a signal nearly twice as weak as the method which gives 4.

edit: though if you do that the strongest signals will be contributing the most, wont be much left of the weak ones.

[Dr Klaus Schmitt]

Well, there are beam combiner cubes which accept 3 input beams to give 1 output beam, so if you use one of those plus a regular beam combiner cube, you could get away with just 2 cubes.