## 1D MUSIC DOA.

Started by 6 years ago7 replieslatest reply 6 years ago301 views

Hello,

I am currently working on a MUSIC algorithm for DOA of RF emitters in 3D.

At the first stage, interested to apply the 1D MUSIC using a linear array of 4 antennas, and estimate only the azimuth of the target projected on X-Y surface(and not the elevation). The code was written and the relevant simulations performed.

However, there are two issues which require further clarifications.

1. In the described setup, two solutions will be received, as plotted below:

Wondered what may be the methods to eliminate the second solution, given a static array and a moving target.

2. Taking into account that the elevation is not being estimated (as opposed to the case of 2D MUSIC), wondered whether the estimation accuracy in azimuth is being effected, or only the 2D projection of the signal has an impact on the estimated azimuth.

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This ambiguity a well-known problem of all regular structures. Circular arrays have similar phase ambiguity, but there power measurement can help.

Have a look at "Resolving Manifold Ambiguities in Direction-of-Arrival Estimation for Nonuniform Linear Antenna Arrays" by Abramovich, Spencer and Gorokhov for ideas as to how to resolve with linear arrays.

You could consider changing your antenna placement, e.g., adding an antenna on one side of the linear array. Or using directional antennas to create a power difference (which is how we break the symmetry with our two ears).

Y(J)S

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Thank you.

Will review the "Resolving Manifold Ambiguities in Direction-of-Arrival Estimation for Nonuniform Linear Antenna Arrays".

Wondered what is your opinion on the second question above.

Further Elaboration would be:

Given the following 3D scenario, and using 1D MUSIC algorithm(with 4 equally spaced antenna array for instance), can it be assumed that the estimated azimuth of arrival will be the projection on XY as can be seen below:

projection.png

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No, you will not obtain the projection onto the 2d plane, since the vertical distance adds an addition time component that contributes to the phase differences.

The x coordinate will be unchanged, but the y coordinate will be larger than the projected y by the square root of the true y plus the height minus the true y.

Y(J)S

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That came out more complex than it needed to be.

Easier to say it this way - you will obtain a y coordinate that is the square root of the sum of the squares of the projected y coordinate and the height.

Y(J)S

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Thank you Y(J)S.

I understand the geometry.

projection2.png

Seems that the error in the estimated angle can be expressed as:

alpha – alpha’ = atan(y0/x0)- atan(sqrt(y0^2+h^2)/x0)

The DOA is estimated based on difference in phase of the receiving channels/antennas, and limited to half wave. As a result, would expect that the expression of the error would be periodic as the height changes.

Would be glad to understand the physics behind it further.

Can you direct me to a reference elaborating the issue?

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It is not physics, it is simply geometry.

The phase differences are completely determined by the distances from the emitter to the antennas.

The squared distance is R = (delta x)^2 + (delta y)^2 + height^2 = (delta x)^2 + (delta y1)^2 where y1 = sqrt((delta y)^2 + height^2).

To understand this better you can write a simple simulation. Draw circles with radius R around the antennas and find their intersections (you will see two - including the reflected one that you mentioned).

Y(J)S

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