Help solving a System of Linear Equations

In summary: There are many counter examples, but they are all special cases of the solution that assumes the arbitrary constant \phi is zero. For instance, if ##x_n=0## for all ##n##, then the solution is of the form b=[1 0 0 ... 0] and so \phi=0.
  • #1
marcusl
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I have the following equation to solve for the coefficients b:
[itex]\sum\limits_{m,n=0}^N b_m b_n^* (x_{mn}+y_{mn}) e^{ ik (ux_{mn} + vy_{mn}) } = 0[/itex]
which must be satisfied for all u and v in the interval [-1,1]. Here k is a constant, b is a vector of length N with unit norm
[itex]||b||^2=1[/itex],
and x and y are vectors of constants with
[itex]x_{mn}=x_m-x_n \text{ and } y_{mn}=y_m-y_n[/itex].
This equation arises in an analysis of plane waves incident on an array of sensors; the question (not homework) is to find weights on the various sensors such that the weighted element voltages sum to create an omni-directional sensitivity power pattern. I am happy to provide the derivation of how the physical system leads to this equation if it would be helpful.

Arguing from physical intuition, the sole solution appears to me that all [itex]b_m=0[/itex] except for one (call it [itex]b_0[/itex], it doesn't matter which we choose) that has unity modulus, [itex]b_0=e^{i\phi}[/itex]. This works because [itex]x_{00}=y_{00}=0[/itex] and is equivalent physically to turning M-1 sensors off and receiving with just a single active sensor. Choosing the arbitray constant as [itex]\phi=0[/itex] results in
[itex]\mathbf{b}\rm=\array{[1&0&&0&...&0]}^T[/itex]
where, again, the indexing is arbitrary so we can assign the active element to any element in the array.

Can anyone suggest a more rigorous argument?
 
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  • #2
Your problem isn't linear in ##\mathbf{b}##.

Things that come to mind are inverting the FFT?
 
  • #3
I see issues with an FFT. First, x and y don't ncessarily have regular spacing, so this expression is not a DFT. We could take a DFT of both sides, but I don't think that simplifies anything.

As for the title of my post--Oops, obviously the equation is nonlinear. I had started with an earlier version of the problem involving the electric field (rather than power) pattern
[itex]\sum\limits_{n=0}^{N-1} b_n e^{ -ik (ux_n + vy_n) } = \alpha e^{i\phi(u,v)}[/itex]
then failed to notice that the title needed to change. Perhaps this linear form is easier to deal with, but it has a constant [itex]\alpha[/itex] and a function [itex]\phi(u,v)[/itex] that are arbitrary, so I wasn't sure what to do with it.

EDIT: corrected upper limit
 
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  • #4
Well, the linear form is a general problem of the ##A x = b## type where ##x_n = b_n## and ##b = \alpha e^{i\phi(u,v)}## discretely sampled. You could try one of many approaches to linear systems LU or QR or singular value decomposition. Depends of the size of ##N##. How badly do you want a solution?
 
  • #5
Paul, these are numerical techniques that won't work when your A is a function and b is unspecified. I'm looking for a general demonstration or proof that the solution is of the form b=[1 0 0 ... 0].
 
  • #6
marcusl said:
I'm looking for a general demonstration or proof that the solution is of the form b=[1 0 0 ... 0].
I'm confused. Aren't there many counter examples given ##\phi## is arbitrary? Proof, assume a solution not of this form. Compute the corresponding ##\phi##?
 

FAQ: Help solving a System of Linear Equations

How do I solve a system of linear equations?

Solving a system of linear equations involves finding the values of the variables that make all of the equations in the system true. This can be done by using various methods such as substitution, elimination, or graphing. It is important to carefully analyze the equations and choose the most appropriate method for the given system.

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How do I know if a system of linear equations has no solution?

If a system of linear equations has no solution, it means that the lines represented by the equations are parallel and will never intersect. This can be determined by graphing the equations or by using the elimination method, where you will end up with a contradiction (such as 0=2) when trying to solve for the variables.

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