Everything up to now has taken the radiation pattern as a given, something we measure or read off a datasheet. Now we compute it. That integral is the last piece of Module 1, and it is the piece that makes the rest of the course possible: steered, tapered, thinned, nulled, every array in Module 3 is this integral with a different current in it.

This is the double-slit experiment with currents in place of slits. Nothing about the antenna enters except where the elements are and what phase each one carries; the direction \(\theta\) does the rest, through the path difference. Every antenna in this course, and every array in Module 3, is this picture with more elements.

The next frame shows those two sources radiating: the field around them, with the bright and dark directions their phase difference makes. Slide the element count up and the fringes of the double slit sharpen into the lobes of a line source. Drag \(P\) around the edge to move the observer.

Every element on the wire radiates the same pattern, so that pattern factors out of the sum. What is left inside the sum is the positions and phases of the elements, and that sum is the space factor. The widget on the previous frame adds isotropic point sources, so with the box unchecked it draws \(S(\theta)\) on its own. Checking the box multiplies every element by \(\sin\theta\), and the pattern acquires the nulls on the axis that every wire along \(z\) has.

Near broadside \(\sin\theta\) is close to 1, and the space factor alone is a good approximation to the measured pattern. Near the axis the element factor dominates, and no phasing of the elements brings a lobe back there. This factorization is pattern multiplication. The element factor is set by the hardware at each position, and the space factor is what the current distribution controls. Lesson 16 replaces the integral with a sum over \(N\) discrete elements: the element factor becomes the pattern of whatever antenna sits at each position, and the space factor becomes the array factor, the part of the pattern we steer. The wire frame later in this lesson is where the \(\sin\theta\) comes out of the mathematics.

The wave from each element arrives at the observation point with an amplitude set by \(1/R\) and a phase delay set by \(kR\), and the integral adds those contributions with the bookkeeping done in phase. A definite integral over something we can write down has replaced the unknown field. \(\mathbf{A}\) is the magnetic vector potential: the field follows from it by differentiation, and the next frame shows where the integral comes from.

Two position vectors from one origin \(O\), and neither is a spherical coordinate. The prime is the whole convention: the integral sweeps \(\mathbf{r}'\) over the antenna while \(\mathbf{r}\) stays fixed at \(P\), and \(V'\) with its element \(dV'\) is the volume the current occupies.

The scalars are the lengths of these vectors. \(r = \vert\mathbf{r}\vert\) happens to be the spherical radial coordinate of \(P\), and \(\hat{\mathbf r} = \mathbf{r}/r\) is the unit vector from the origin toward \(P\), in the direction \((\theta,\phi)\). \(r' = \vert\mathbf{r}'\vert\) is simply how far a source point is from the origin; it has nothing to do with \(P\).

The exact integral is correct everywhere, near field included, but \(R = \vert\mathbf{r} - \mathbf{r}'\vert\) depends on where \(P\) is and on which source point we are looking at, and the two dependences do not separate. What we want is to write \(\mathbf{A}\) as a factor that depends only on the distance \(r\) times a factor that depends only on the direction \((\theta,\phi)\). Lesson 5 said the pattern stops changing with distance once we are far enough away; this is the same statement, seen from inside the integral.

Start with the exact distance and expand it for \(r' \ll r\):

The linear term is the path difference: the projection of the source point’s position onto the viewing direction. The quadratic term is what we drop, and the next two frames say when that is safe. \(R\) appears in two places in the integral, and the two places get different treatment.

Amplitude. In the \(1/R\) factor, the correction \(\hat{\mathbf r}\cdot\mathbf{r}'\) is a fractional change of order \(r'/r\). At any useful distance that is a fraction of a percent, and nobody can measure it. So

Phase. In \(e^{-jkR}\) the same correction is multiplied by \(k\), and what matters is whether \(k\ \hat{\mathbf r}\cdot\mathbf{r}'\) is comparable to a radian. For a source a few wavelengths across it is several radians, and it flips contributions from adding to canceling. Dropping it would destroy the pattern. So we keep the linear term and only discard the quadratic one:

This is the entire content of “far field”: the rays from every part of the antenna to the observation point are parallel, so the only thing that distinguishes one source point from another is a path difference \(\hat{\mathbf r}\cdot\mathbf{r}'\), the projection of its position onto the viewing direction.

The worst case of the quadratic term is \(r'_{\max}{}^2/2r\), and for an antenna of largest dimension \(D\) the source extends to \(r'_{\max} = D/2\), so the leftover phase error is at most

Demand that this stay under \(\pi/8\) radians and it rearranges to \(r \ge 2D^2/\lambda\). That is not a second criterion. Lesson 5 derived the same number from the geometry of a curved wavefront; here it falls out of the term we need to drop to make the integral tractable.

The integral that is left is the radiation vector \(\mathbf{N}(\theta,\phi)\), and this split is the central result of the lesson:

An antenna’s pattern, its polarization, its directivity, its sidelobes — all of it lives in \(\mathbf{N}\), and \(\mathbf{N}\) is a single integral over the currents. Note also what became of the recipe’s second and third steps: with the \(\mathbf{r}\)-dependence reduced to \(e^{-jkr}/r\), taking a curl in the far field just multiplies by \(-jk\hat{\mathbf r}\). Differentiation has collapsed into multiplication.

In the far field the wave is locally a plane wave traveling radially outward (Lesson 5), so it can have no radial field component: \(E_r \approx 0\), and only the transverse parts of \(\mathbf{A}\) survive. The magnetic field follows from the plane-wave relation, with no new integral:

Then radiation intensity, the quantity Lesson 2 built directivity and gain on, is

The normalized power pattern is \(U/U_{\max}\), and its square root is the field pattern used for the rest of this lesson and the next:

Compare this with the three-step recipe on the read-only frame above: the two curls that took \(\mathbf{A}\) to \(\mathbf{E}\) have become a projection onto the transverse directions and a constant.

The current points the same way at every source point, so the vector integral collapses to one scalar integral, and the projection onto the observation basis is one dot product, done once, afterwards: \(\hat{\boldsymbol\theta}\cdot\hat{\mathbf z} = -\sin\theta\), so \(N_\theta = -N_z\sin\theta\) and \(N_\phi = 0\). At \(\theta = 0\) the current has no transverse component to project, which is why no wire antenna radiates off its own ends.

Why Cartesian, and why afterward. The integral adds up vectors from many different source points, and we can only add vectors component by component when every component is measured against the same unit vectors. Cartesian unit vectors are the same everywhere, so integrating \(J_x\), \(J_y\) and \(J_z\) is safe. Spherical unit vectors at the source are not: \(\hat{\mathbf r}'\) and \(\hat{\boldsymbol\theta}'\) change direction as we move from one source point to the next, so an integral of \(J_{\theta'}\) adds numbers that each mean something different, and the total is not the component of anything. So integrate the Cartesian components, get \(N_x\), \(N_y\) and \(N_z\), and only then take the components along \(\hat{\boldsymbol\theta}\) and \(\hat{\boldsymbol\phi}\) of the observation direction. Those two are fixed once we choose where we look, which is why the \(\sin\theta\) can sit outside the integral.

Try it the wrong way on a dipole along \(z\) and it fails on the first line: every source point has \(\theta' = 0\), so \(J_{\theta'} = 0\) everywhere, and we would conclude that a dipole does not radiate at all.

Expanding \(\hat{\mathbf z} = \cos\theta\ \hat{\mathbf r} - \sin\theta\ \hat{\boldsymbol\theta}\) before integrating gives the same answer, because \(\theta\) is the observation angle and constant over the integral. Cartesian simply keeps the two jobs apart: the integral, which depends only on the current distribution, and the projection, which depends only on where we look. That separation is the space-factor / element-factor split later in this lesson, and all of L16.

The next frame shows the integral as a picture. The source is chopped into elements; each one contributes a phasor whose angle is \(kz'\cos\theta\), its path difference in radians. At broadside every phasor points the same way and they stack into a long straight chain. Swing off broadside and the chain curls up; when it closes on itself, we are looking at a null. It is the two-source picture with many sources, and it is the picture the rest of the lesson computes.

For a line source on the \(z\)-axis carrying current \(I(z')\), \(\hat{\mathbf r}\cdot\mathbf{r}' = z'\cos\theta\), so the whole radiation vector is the one scalar integral above, over \(-L/2 \le z' \le L/2\). The transform is evaluated over the visible region \(-k \le k_z \le +k\) and then bent onto angle by \(k_z = k\cos\theta\).

This is not a new transform with an antenna flavor. It is the Fourier transform from your signals course with the variables renamed, and every result you learned there comes across unchanged.

The same properties, row by row:

Every row on the right is the corresponding transform property applied to an antenna, and nothing on the right is derived in this course from scratch.

This is why Module 3 works the way it does. An array designer is not really solving Maxwell’s equations. They are choosing a function whose Fourier transform has the beamwidth, sidelobe level, and null placement they want, then building a current distribution that realizes it.

\(\omega\) is how fast a signal’s phase turns as time passes. \(k\) is the same thing for space: how fast a wave’s phase turns as we move through it, with \(1/\lambda\) in cycles per meter playing the part of \(f = 1/T\) in cycles per second. So the exponent in the radiation integral, \(k\ \hat{\mathbf r}\cdot\mathbf{r}'\), is \(k\) times a path difference in meters: a phase in radians, exactly as \(\omega t\) is.

Broadside demands nothing: the whole source lies on one wavefront, \(k_z = 0\), and a current that is simply in phase everywhere radiates there. Endfire demands the full \(k\): one cycle of phase every wavelength along the source. Every other direction sits in between, and no direction can demand more than \(k\), because the projection of a vector onto an axis is never longer than the vector.

So read \(N_z(k_z)\) as the answer to a question asked once per direction: does the current contain a ripple at the rate this direction demands, and how much of it? A plain in-phase current answers loudly at \(k_z = 0\) and fades from there — the broadside beam. Put a linear phase slope across the current and the loud answer moves to whatever \(k_z\) matches the slope — the beam steers.

A spectrum extends over every \(\omega\), and every \(\omega\) is a frequency we can measure. This transform extends over every \(k_z\) too, but \(k_z = k\cos\theta\) ties it to a direction, and \(\cos\theta\) is constrained to \([-1, +1]\). Only the spatial-frequency band \(\lvert k_z\rvert \le k\) is somewhere we can stand. The invisible part describes stored, non-radiating field near the aperture rather than anything we can measure at range, so the pattern is a window onto the transform, not the whole of it.

Said in wavelengths: a spatial frequency \(k_z\) is a ripple in the current with period \(2\pi/k_z\), and the band edge \(k_z = k\) is a period of exactly \(\lambda\). That idea has no counterpart in your signals course.

Keep that window in mind. When Module 3 builds an array out of separate elements, the transform repeats, and a repeat that slides inside the window is a second beam we did not ask for. Lesson 16 is about keeping it out.

This is exactly the \(1/r\) radiation term of the exact short-dipole field quoted in Lesson 5, the term that survived once \(kr \gg 1\). What took a page of exact spherical-wave algebra there comes out here in a few steps. The pattern is \(|F| = \sin\theta\): a donut, maximum broadside, null along the wire, and \(D = 1.5\) (1.76 dBi). The read-only frame below derives that 1.5.

Plotted against angle, that one expression is the pattern we compare every aperture antenna in this course against. Read off the space factor’s three headline numbers:

• Peak at \(u = 0\), i.e. \(\theta = 90^{\circ}\), broadside. All elements are equidistant from the observer, every phasor is aligned, and the sum is the full length \(L\).

• First null where \(u = \pi\), i.e. \(\cos\theta_\text{null} = \lambda/L\).

• First sidelobe at \(u \approx 4.493\), height −13.3 dB, a fixed number, independent of \(L\). Uniform illumination always gives −13.3 dB sidelobes; only tapering changes that.

The read-only frame below works the \(2\lambda\) case through.

The current has to vanish at the open ends, and the standing wave on a resonant wire of length \(L\) is well approximated by

which for \(L = \lambda/2\) is a single cosine hump, maximum at the feed and zero at both tips. Putting that into the radiation integral gives a standard (if tedious) result:

and therefore

The bracket is not \(\sin\theta\), but it is close: it is a slightly sharper donut, because the tapered current is concentrated near the middle rather than spread over the whole length. That small difference is worth real numbers: \(\theta_\text{HP} = 78.1^{\circ}\) versus 90°, and \(D = 1.64\) (2.15 dBi) versus 1.5 (1.76 dBi).

The last row’s headline number is the space factor on its own; multiply in the \(\sin\theta\) element factor and it becomes 24.8° and \(D = 4.45\) (6.5 dBi), a small correction, as promised for a broadside beam. Either way the comparison holds: a \(2\lambda\) line source is four times longer than a half-wave dipole, and it gives about 2.6 times the directivity with a beam three times narrower. For a long uniform line source \(D \to 2L/\lambda\), so directivity grows in proportion to the length in wavelengths.

The fields set the current on a conductor, and the current sets the fields. Solving that self-consistently is the hard part of antenna analysis, and the radiation integral does not touch it. What saves us is that for many antennas the current can be assumed with good accuracy:

• Thin wires: the sinusoidal standing wave used above. Excellent for \(\ell/\text{diameter} \gtrsim 100\), and it is why hand analysis of dipoles works so well.

• Apertures: assume the field in the opening (often the incident waveguide mode) and integrate that instead. This is the standard treatment of horns and reflectors in Module 2.

• Arrays: assume each element keeps its isolated pattern and only the excitation changes. That assumption is what pattern multiplication rests on in Module 3, and mutual coupling is where it starts to fray.

When the assumption fails, for thick elements, tightly coupled arrays, or antennas close to a ground plane or an airframe, we solve for the current numerically. The method of moments discretizes the structure, enforces the boundary condition on each segment, and solves a linear system for the segment currents; then it runs the very same radiation integral to get the pattern. That is what NEC does in the Lesson 8 simulation lab.

There is a measurement version of the same idea, too. Near-field scanning (Lesson 5, and Module 2’s measurement labs) samples the field on a surface close to the antenna and transforms it to the far field, which is possible for exactly one reason: the transform between the source distribution and the far-field pattern is invertible.

Beamwidth, sidelobes, and where we point the beam are all properties of that transform.

You can describe an antenna’s pattern and gain (L2), its polarization and bandwidth (L3), its terminals (L4), where its far field starts (L5), and now how to compute the pattern from the current (L6). In Module 2 each antenna is a different \(\mathbf{J}\) producing a different \(\mathbf{N}\), and the Lesson 8 simulation lab solves for the current the integral needs. Module 3 chooses the pattern we want and synthesizes the current distribution that produces it.