Wednesday, 20 January 2016

linear algebra - Recovering symmetry in coupled oscillators


Consider a pair of LC oscillators, one with capacitance C1 and inductance L1 and the other with capacitance C2 and inductance L2. Suppose they're connected through a capacitor Cg. We want to find the normal modes and frequencies.



If we write out Kirchhoff's laws, we find V1+¨V1(1+ϵ1)/ω21(ϵ1/ω21)¨V2=0V2+¨V2(1+ϵ2)/ω22(ϵ2/ω22)¨V1=0 where ϵiCg/Ci and ω2i1/LiCi. These equations can be written in matrix form as (V1V2)=((1+ϵ1)/ω21ϵ1/ω21ϵ2/ω22(1+ϵ2)/ω22)(¨V1¨V2). Now if L1=L2 and C1=C2 then ϵ1=ϵ2ϵ and ω1=ω2ω0 and the matrix equation becomes (V1V2)=((1+ϵ)/ω20ϵ/ω20ϵ/ω20(1+ϵ)/ω20)(¨V1¨V2). In this particular case, the matrix can be written in the nice form 1+ϵω20Iϵω20σx and it's pretty easy to find the normal modes and normal frequencies.[a]


However, when the oscillators aren't identical, e.g. Eq. (), expressions for the normal modes and frequencies are pretty messy. Is there a transformation we can apply to () to bring it into a simple form like () so that the mode analysis results in simpler equations?


Perhaps another way to ask this would be to ask for a systematic way to rescale the variables so that the matrix in the equations of motion is symmetric or perhaps Hermitian.


[a] The frequencies are ω0 (even mode) and ω0/1+2ϵ (odd mode).




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...