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Spherical coordinates describe a vector or point in space with
a distance and two angles. The distance, *R*, is
the usual Euclidean norm. There are multiple conventions regarding
the specification of the two angles. They include:

Azimuth and elevation angles

Phi and theta angles

*u*and*v*coordinates

Phased Array System Toolbox™ software natively supports the azimuth/elevation representation. The software also provides functions for converting between the azimuth/elevation representation and the other representations. See Phi and Theta Angles and U and V Coordinates.

In Phased Array System Toolbox software, the predominant convention for spherical coordinates is as follows:

Use the azimuth angle,

*az*, and the elevation angle,*el*, to define the location of a point on the unit sphere.Specify all angles in degrees.

List coordinates in the sequence (

*az*,*el*,*R*).

The *azimuth angle* of a vector is the angle between
the *x*-axis and the orthogonal projection of the vector onto the
*xy* plane. The angle is positive in going from the
*x* axis toward the *y* axis. Azimuth angles lie
between –180 and 180 degrees. The *elevation angle* is the angle
between the vector and its orthogonal projection onto the *xy*-plane. The
angle is positive when going toward the positive *z*-axis from the
*xy* plane. By default, the boresight direction of an element or array
is aligned with the positive *x*-axis. The boresight direction is the
direction of the main lobe of an element or array.

**Note**

The elevation angle is sometimes defined in the literature as the angle a vector makes
with the positive *z*-axis. The MATLAB^{®} and Phased Array System Toolbox products do not use this definition.

This figure illustrates the azimuth angle and elevation angle for a vector shown as a green solid line.

As an alternative to azimuth and elevation angles, you can use angles denoted by
*φ* and *θ* to express the location of a point
on the unit sphere. To convert the *φ/θ* representation to and from
the corresponding azimuth/elevation representation, use coordinate conversion
functions, `phitheta2azel`

and `azel2phitheta`

.

The phi angle (*φ*) is the angle from the positive
*y*-axis to the vector’s orthogonal projection onto the
*yz* plane. The angle is positive toward the positive
*z*-axis. The phi angle is between 0 and 360 degrees. The theta angle
(*θ*) is the angle from the *x*-axis to the vector
itself. The angle is positive toward the *yz* plane. The theta angle is
between 0 and 180 degrees.

The figure illustrates phi and theta for a vector that appears as a green solid line.

The coordinate transformations between φ/θ and *az/el* are described by
the following equations

$$\begin{array}{l}\mathrm{sin}el=\mathrm{sin}\varphi \mathrm{sin}\theta \\ \mathrm{tan}az=\mathrm{cos}\varphi \mathrm{tan}\theta \\ \mathrm{cos}\theta =\mathrm{cos}el\mathrm{cos}az\\ \mathrm{tan}\varphi =\mathrm{tan}el/\mathrm{sin}az\end{array}$$

In radar applications, it is often useful to parameterize the
hemisphere x ≥ 0 using coordinates denoted by *u* and *v*.

To convert the φ/θ representation to and from the corresponding

*u*/*v*representation, use coordinate conversion functions`phitheta2uv`

and`uv2phitheta`

.To convert the azimuth/elevation representation to and from the corresponding

*u*/*v*representation, use coordinate conversion functions`azel2uv`

and`uv2azel`

.

You can define *u* and *v* in
terms of φ and θ:

$$\begin{array}{l}u=\mathrm{sin}\theta \mathrm{cos}\varphi \\ v=\mathrm{sin}\theta \mathrm{sin}\varphi \end{array}$$

In these expressions, φ and θ are the phi and theta angles, respectively.

In terms of azimuth and elevation, the *u* and *v* coordinates
are

$$\begin{array}{l}u=\mathrm{cos}el\mathrm{sin}az\\ v=\mathrm{sin}el\end{array}$$

The values of *u* and *v* satisfy
the inequalities

$$\begin{array}{l}-1\le u\le 1\\ -1\le v\le 1\\ {u}^{2}+{v}^{2}\le 1\end{array}$$

Conversely, the phi and theta angles can be written in terms
of *u* and *v* using

$$\begin{array}{l}\mathrm{tan}\varphi =v/u\\ \mathrm{sin}\theta =\sqrt{{u}^{2}+{v}^{2}}\end{array}$$

The azimuth and elevation angles can also be written in terms of *u* and
*v*:

$$\begin{array}{l}\mathrm{sin}el=v\\ \mathrm{tan}az=\frac{u}{\sqrt{1-{u}^{2}-{v}^{2}}}\end{array}$$

The following equations define the relationships between rectangular
coordinates and the (*az*,*el*,*R*)
representation used in Phased Array System Toolbox software.

To convert rectangular coordinates to (*az*,*el*,*R*):

$$\begin{array}{l}R=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}\\ az={\mathrm{tan}}^{-1}(y/x)\\ el={\mathrm{tan}}^{-1}(z/\sqrt{{x}^{2}+{y}^{2}})\end{array}$$

To convert (*az*,*el*,*R*)
to rectangular coordinates:

$$\begin{array}{l}x=R\mathrm{cos}(el)\mathrm{cos}(az)\\ y=R\mathrm{cos}(el)\mathrm{sin}(az)\\ z=R\mathrm{sin}(el)\end{array}$$

When specifying a target’s location with respect to a
phased array, it is common to refer to its distance and direction
from the array. The distance from the array corresponds to *R* in
spherical coordinates. The direction corresponds to the azimuth and
elevation angles.

*Broadside angles* are useful when describing the response of a uniform
linear array (ULA). The array response depends directly on the broadside angle and
not on the azimuth and elevation angles. Start with a ULA and draw a plane
orthogonal to the ULA axis as shown in blue in the figure. The broadside angle, β,
is the angle between the plane and the signal direction. To compute the broadside
angle, construct a line from any point on the signal path to the plane, orthogonal
to the plane. The angle between these two lines is the broadside angle and lies in
the interval [–90°,90°]. The broadside angle is positive when measured toward the
positive direction of the array axis. Zero degrees indicates a signal path
orthogonal to the array axis. ±90° indicates paths along the array axis.
All signal paths having the same broadside angle form a cone around the ULA
axis.

The conversion from azimuth angle, *az*, and elevation angle,
*el*, to broadside angle, *β*, is

$$\beta ={\mathrm{sin}}^{-1}(\mathrm{sin}(az)\mathrm{cos}(el))$$

This equation shows that

For an elevation angle of zero, the broadside angle equals the azimuth angle.

Elevation angles equally above and below the

*xy*plane result in identical broadside angles.

You can convert from broadside angle to azimuth angle but you must specify the elevation angle

$$az={\mathrm{sin}}^{-1}\left(\frac{\mathrm{sin}\beta}{\mathrm{cos}(el)}\right)$$

Because the signals
paths for a given broadside angle, *β*, form
a cone around the array axis, you cannot specify the elevation angle
arbitrarily. The elevation angle and broadside angle must satisfy

$$\left|el\right|+\left|\beta \right|\le 90$$

The following figure depicts a ULA with elements spaced *d* meters apart
along the *y*-axis. The ULA is irradiated by a plane wave emitted
from a point source in the far field. For convenience, the elevation angle is zero
degrees. In this case, the signal direction lies in the *xy*-plane.
Then, the broadside angle reduces to the azimuth angle.

Because of the angle of arrival, the array elements are not
simultaneously illuminated by the plane wave. The additional distance
the incident wave travels between array elements is *d
sinβ* where *d* is
the distance between array elements. The constant time delay, *τ*,
between array elements is

$$\tau =\frac{d\mathrm{sin}\beta}{c},$$

where *c* is the speed of the wave.

For broadside angles of ±90°, the signal is incident
on the array parallel to the array axis and the time delay between
sensors equals *±d/c*. For a broadside angle
of zero, the plane wave illuminates all elements of the ULA simultaneously
and the time delay between elements is zero.

Phased Array System Toolbox software provides functions `az2broadside`

and `broadside2az`

for
converting between azimuth and broadside angles.

The following examples show how to use the `az2broadside`

and `broadside2az`

functions.

A target is located at an azimuth angle of 45° and at an elevation angle of 60° relative to a ULA. Determine the corresponding broadside angle.

bsang = az2broadside(45,60)

bsang = 20.7048

Calculate the azimuth for an incident signal arriving at a broadside angle of 45° and an elevation of 20°.

az = broadside2az(45,20)

az = 48.8063