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I've seen many examples of ultrasonic phased arrays being used with non-linear phase relationship between elements to concentrate all the power at a specific point, for example here:

             Ultrasound Phased Array.

As I understand, this is different to how a typical RF phased array works, which takes advantage of linear phase differences between elements in order to steer the beam such that gain is highest in a desired direction, though without a focal point. However as far as I can tell the physics should be the same, so can an RF phased array be used to focus RF energy to a particular point in space? If so, does this mean that the gain for such a configuration then varies as a function of distance from the source (assuming constant direction from the source)?

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To the extent that "focus" and "point" are meaningful, and for "phased array" of sufficient size:

Sure.

Waves are waves, as long as the medium is linear, they work the same, superposition applies. An optical focus is the sum of myriad waves incident from many directions, phased in such a way that they cancel out outside of the focus, and superimpose within it.

Therein lies the answer: focus below a wavelength (or fraction thereof) isn't meaningful. Focus below a fraction of the antenna aperture isn't meaningful either (that fraction being determined, in part, by the number of elements in the array), that is, the spacial/imaging resolution of the system, as defined in the usual way for optical systems.

You can't make an arbitrary E/M field in some given volume of space at any distance from an array, it can't be any sharper than the wavelength used (or, again, a modest fraction thereof).

Ultrasonic frequencies have quite short wavelength in liquid media, allowing "point" dimensions of say fractional mm; that, and converging sources from all around a patient, allows enough peak power say to ablate kidney stones or whatnot.

So on the other hand: if you wanted energy, of any given frequency, focused down to a point (in the mathematical sense, a singular location of zero width/area/volume/etc.): that's a very strong "no".

The exact phasing depends on all the reflection/refraction in the system (which is obviously quite complex in something like a patient's body, but also for applications like around-the-corner imaging), and phase and amplitude must be set so as to invert that relationship.

I don't know what nonlinearity you're referring to specifically, but the gain terms across all elements in the array need not be a straight line (or simple curve i.e. sinusoid), and in general can be just whatever.

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  • \$\begingroup\$ Thank you for your answer, I think it pretty much answers my question. Regarding the nonlinear term I used, I simply mean that the phase difference between elements of the array is not constant, which as I understand it produces a flat wave front in theory. This is opposed to the focused configuration, which as I understand would initially produce a converging wave front up to the focal point beyond which it diverges - hence my thinking that the gain would vary with distance? \$\endgroup\$
    – Christian
    Commented Aug 1 at 15:35
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    \$\begingroup\$ Yes, and in addition the array itself need not be planar, and compensation can be added to account for refraction/reflection on the way. Plane waves from a plane array are a special case (which might also be expressed as, focus at infinity). \$\endgroup\$ Commented Aug 1 at 15:40
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    \$\begingroup\$ Can you clarify what you mean by " Focus below a fraction of the antenna aperture isn't meaningful either"? Generally the focal point size is not proportional to the antenna aperture, it's actually inversely related: A larger antenna can focus to a smaller spot size. \$\endgroup\$
    – The Photon
    Commented Aug 1 at 16:00
  • \$\begingroup\$ @ThePhoton Yes, more like an inverse fraction would be more correct to say there -- suggestions welcome. I don't feel it necessary to go into depth of an optical imaging system, but just making a nod to a en.wikipedia.org/wiki/Diffraction-limited_system is probably enough. \$\endgroup\$ Commented Aug 1 at 16:09

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