Fourier Transform Applications: Radiation from Electric Currents

Far Field Radiation from Electric Current

Fourier Transform Applications

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The world of electromagnetics and radiation is governed by Maxwell's Equations, a set of complicated vector differential equations for determining Electric and Magnetic fields. On this page, we are interested in determining the fields radiated by electric currents flowing on a surface. Specifically, we want to know the Electric and Magnetic fields far from the source currents.
This problem has application in antennas, particularly with determining the radiation patterns of horn antennas. In addition, determining the fields radiated from electric currents is useful in scattering applications, with the most prevalent example being RCS (radar cross section).
Defining the Problem: Electric Current Flowing on a Surface

[This page will require a good understanding of spherical coordinates].
Consider a conductive surface (i.e. metal), enclosed within the purple boundary as shown in Figure 1. There is an electric current, written by the vector Js, flowing on the surface as shown. We are intersted in the radiated fields far from the surface.

application of fourier transforms to surface current radiation
Figure 1. An Electric Current (written with the vector Js) flows on a Surface.

We will write the radiated electric fields as the vector E in spherical coordinates:

electric field in spherical coordinates [Equation 1]

Note that the radial component of the E-field (the field pointing in the direction of propagation) will always be zero for radiated fields: radiated fields will be zero .
Before we determine the radiated fields from the full surface, let's note what the radiated fields are from very small surface section. The radiated Electric Fields (E) from a very small section of the surface is given by:

electric field from a small surface point [Equation 2]

In Equation [2], R is the distance from the source point to the observation point (i.e., where we are measuring the Electric Fields at), is the angular frequency in radians/second, is the permeability of free space (a constant), and k is the wavenumber. We also use the convention of j being the square root of negative 1, the imaginary number (which gives us phase information). Note that in Equation [2], the radially-directed E-fields (in the direction) are zero.
Now, to find the radiated fields from an area of surface current, we must simply sum up all the individual small sections - that is, use integration:
total electric field [Equation 3]

To do this exactly, we would need to choose an observation point, and then calculate R for each section of the surface to the observation point. This would make the integration very difficult. Just as in the diffraction section, we will make some approximations.
Recall that we are interested in the far-field radiation. This means the radiation far from the source current. Since the function 1/R is slowly varying for large values of R, we can approximate this as a constant for the entire surface, and pull it out of the integral in Equation [3]. This approximation should be understood - if you disagree, take the time to make sure it makes sense.
The term R also occurs in the exponential in Equation [3]. However, it is not ok to treat this as constant, because the exponential is multiplied by the wavenumber, and is not necessarily a slowly varying function across the surface. To make a proper approximation, let's walk through some geometry. First, assume we are interested in the fields at some arbitrary point far from the surface. This point is defined in terms of R, theta and phi:
calculating the electric fields far from the source
Figure 2. An arbitrary observation point, far from the surface. The point is defined in terms of R, theta and phi.
Now consider an arbitrary point on the surface, P, as shown in Figure 3:

two vectors from different locations on the surface

Figure 3. Showing rays from two distinct points (O and P) towards the observation point in the far field.

In Figure 3, I have drawn the vectors R and r with the same magnitude (size) and direction. Because the observation point (where we want to calculate the E-fields) is very far away (and much larger than the surface where the electric currents flow), these two vectors will be approximately parallel. However, because they travel different distances, one ray will arrive quicker than the other (a shorter distance).

The green line in Figure 3 is perpendicular to the direction at the observation point. This line helps us to see geometrically how much extra phase the ray from point P (the vector r) is than the ray from the origin O (the vector R). We need to factor in this extra distance in calculating the phase in Equation [3]. This can be done by viewing Figure 4:

determining the relative distance or phase

Figure 4. Graphical illustration of determining the relative phase between the two vectors.

In Figure 4, we define fourier transform applications as the angle between the vectors r and R. By using some geometry, we see that the gold line has a distance equal to extra distance (where |P| is the distance from the origin [O] to the point [P]).

By observing the geometrical relationships in Figure 4, you see that the vector r travels a longer distance than the vector r, where the extra distance is equal to .

Therefore, we need to subtract out this phase from the complex exponential in Equation [3]. That is, for an arbitrary point on the surface, the phase will be modified as:

complex phase term for far field [Equation 4]

In Equation [4], R is now a constant, which is equal to the distance between the origin of our coordinate system and the observation point. We can now use Equation [4] in Equation [3] to determine the far fields from the electric surface currents:

result begins to look like a fourier transform [Equation 5]

In Equation [5], P represents the distance of an arbitrary point on the surface (that we are integrating over) and the origin. The result in Equation [5] may not look helpful or simplified. But as we will see, the radiated far fields from a surface current are equal to the Fourier Transform of that current!

Example

Let's look at an example where the surface is a region in the x-y plane, where electric current is flowing. We can use Equation [5] to determine the radiated fields far from the source. Take the surface in Figure 1 (in the x-y plane), and let's assume the surface current Js is given by:

example of a uniform surface current [Equation 6]

Equation [6] is that of a surface current that travels in the +x-direction, with a constant magnitude (Jo) everywhere over a square region of width a in the x-direction and length b in the y-direction. We now need to evaluate Equation [5] to with this surface current function to determine the radiated fields.

We first note that dS = dx dy, since we are integrating over a flat region in the x-y plane:

evaluating the integral with differential surface element [Equation 7]

Now we want simplify the exponential. Note that extra distance is mathematically equivalent to the dot product of the vector P (any source location) and the unit vector (see Figure 4). Remember, we are interested in the fields at an arbitrary far away observation point, which are defined by the spherical coordinate variables theta and phi. We can write this out then:

evaluating the phase term [Equation 8]

If we first define the two dimensional Fourier Transform as:

two dimensional fourier transform [Equation 9]

We can now insert Equation [8] into Equation [7], and we see the Fourier Transform emerge:

fourier transform of surface current [10]

In addition, if we rewrite our surface current as the product of two functions of one variable:

separable function [11]

Then we are left with two one dimensional Fourier Transforms:

two one-dimensional fourier transforms [12]

In Equation [12], the standard one-dimensional fourier transform definition is used:

standard fourier transform [13]

From Equations [12] and [13] and the solution for the Fourier Transform of a box function, we can compute the integrals in [12]:

fourier transforms of surface current [14]

Note that in Equation [14], sinc(-x)=sinc(x), so we can drop the minus signs in the sinc() functions.

There is one final step before we can evaluate the radiated fields. We were able to pull out the constant unit vector in the x-direction, unit vector in x-direction . We must convert the vector into it's equivalent representation in terms of the , and vectors (unit vector in the theta and phi directions). This is a simple transformation, but we need to do it because the component of the radiated fields will be zero:

rewriting the xhat vector in terms of spherical unit vectors [15]

Finally, by incorporating Equations [12], [14], [15] into Equation [5], we have the solution:

final result - radiated electric fields [16]

Since the selected observation point was arbitrary, we found the radiated fields for all directions from this surface electric current distribution! Also, having this description of the electric field in vector form also tells us the polarization of the radiated fields.

To cement the example, let's look at the normalized (forgetting the terms in front of the sinc functions) magnitude of E:

radiation pattern [17]

This function is also known as the radiation pattern, if we view the surface current as an antenna.

We can plot the fields of Equation [17] for versus theta, for a=b=. Using wavevector , we can plot the magnitude of the electric field:

plot of radiated electric fields versus theta

Figure 5. Radiated Electric Fields as a function of theta, for .

Horn Antennas

We can also use this technique to calculate the radiation pattern for horn antenna. Horn antennas are basically apertures with an electric field across them, which gives rise to the radiation. Now, we can model an electric field as a "magnetic surface current" Ms:

magnetic surface current [18]

In [18], is a unit vector perpendicular to the surface. The far field results can then be calculated using [18] in place of Js, and replacing the electric field with the magnetic field. That is, we can calculate E directly from Js, and H directly from Ms. In addition, since we are looking at radiated fields, we can calculate E from H, or H from E as:

E and H are related for radiated fields [19]

In Equation [19], is a constant (about 377 Ohms), known as the characteristic impedance of free space, and is the direction of propagation for the plane wave.

Radar Cross Section - RCS

Radar works by transmitting an electromagnetic wave from an antenna. The wave then bounces off some object, and the returned energy is measured. The amount of energy returned is a function of the radar cross section of the object.

If we can estimate the fields radiated by an antenna, we can estimate the surface electric current induced on an object (such as an airplane). From these fields, we can use Equation [5] (which we now know how to solve) to determine the reflected power as a function of angle. In this way, the radar cross section of objects can be determined, particularly as they fly across the sky. This application is common in military defense programs.

If you managed to stay with me through this whole page, I am very impressed. I hope you enjoyed the ride as much as I did.

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