Lesson 5 told us where the far field starts, and only there does the pattern stop changing with range. Lesson 6 gave us the radiation integral: assume a current, transform it into the radiation vector \(\mathbf{N}\), and read the pattern from it. The radiation intensity goes as the square of the transverse part of \(\mathbf{N}\), the field pattern \(\vert F\vert\) is the square root of the intensity normalized to its peak, and directivity compares that peak to the average over the sphere. Lesson 6 also handed us the half-wave dipole pattern from an assumed sinusoidal current. Lesson 4 gave us matching, VSWR, and baluns. Today we need all of these on the same antenna.

Lesson 6 closed on this factorization, and we use it once today, in Step 3 of the pattern derivation. The radiation integral over the sinusoidal current is the space factor of that distribution. Projecting \(N_z\) onto the observer’s transverse direction multiplies by \(\sin\theta\), which is the element factor of every short piece of the wire. Module 3 keeps the same product with a sum in place of the integral. The element factor becomes the pattern of the antenna at each position, and the space factor becomes the array factor, the part of the pattern you steer.

An isotropic radiator radiates equally in every direction. Wrap a sphere around it and every square meter of that sphere receives the same power density. Its radiation intensity is simply the radiated power spread over the whole sphere, so its directivity is exactly \(D = 1\), which is 0 dBi — the “i” is literally there to say relative to isotropic.

An isotropic antenna is a physical impossibility. In order to create an antenna, we have to separate charge, which inherently produces curved electric field lines, which means at some physical location the field has to vanish, and a place where the field vanishes is a null.

The gain of this theoretical antenna is most useful as a unit of measurement. Recall that dB is a power ratio, so directivity, gain, and effective aperture are all defined as ratios against isotropic. This is why a horn is “16 dBi” rather than “16 dB”. Two related conventions come out of the same reference. Because a half-wave dipole is 2.15 dBi, the two decibel scales differ by a constant.

EIRP is the standard measure of antenna power performance. Feeding 5 W into a half-wave dipole produces the same peak power density as feeding \(5 \times 1.64 = 8.2\ \text{W}\) into an isotropic radiator, so the EIRP is 8.2 W, or 39.1 dBm. One number now describes the transmitter and the antenna together, which is exactly what a link budget needs and exactly what the FCC specifies in their licensing regulations.

Lesson 6 handled the infinitesimal dipole — a current element so short that the current is essentially constant along it. Its field pattern is \(\vert F(\theta)\vert = \sin\theta\), a donut with its maximum broadside at \(\theta = 90^\circ\) and nulls off the wire ends. Its beamwidth is 90° and its directivity is \(D = 1.5\), or 1.76 dBi. Its radiation resistance, however, is

about \(2\ \Omega\) for a wire a tenth of a wavelength long. \(2\ \Omega\) against a \(50\ \Omega\) line is a hopeless match, and that, not the pattern, is why nobody feeds a short dipole directly.

Moving away from the infinitesimal wire, the current is no longer a constant as the wire becomes longer. So before we can use Lesson 6’s machinery, we need to understand the effects of a variable current feeding our antenna.

To construct our more realistic dipole, start with a two-wire transmission line, open-circuited at the far end. You already know the mathematical form of its current; on an open-circuited line the current must be zero at the open end. Furthermore, the standing wave varies sinusoidally as you move from the open circuit load back towards the source. At a distance \(s\) back from the open end, the current is:

Now take the open end of that line and fold the two conductors apart until they lie in a straight line, one arm up and one arm down. Congratulations, you have built a center-fed dipole, and it inherits the standing wave that came with it.

Each arm still ends in an open tip. A point at height \(z\) on the upper arm sits a distance \(s = L/2 - z\) back from its tip, so

The lower arm is the mirror image, so replacing \(z\) by \(\vert z \vert\) covers both. Confirm the boundary conditions:

• At the tips, \(z = \pm L/2\), the sine argument is zero, so \(I = 0\). Current vanishes at both open ends — charge has nowhere further to go.

• At the feed, \(I(0) = I_m \sin(kL/2)\). For \(L = \lambda/2\) that is \(I_m \sin(\pi/2) = I_m\): the current maximum lands exactly at the feed. For \(L = \lambda\) it is \(I_m \sin(\pi) = 0\): a current null at the feed. Remember that contrast — it determines the impedance later in this lesson.

Measurement and numerical solvers both confirm that the standing wave is very close to the truth for a thin resonant wire. For thick or non-resonant wires it is visibly wrong, and we will be exploring the tradeoffs. The impedance, which depends on the current right at the feed, is where the error shows first.

We start with the current, and the radiation integrals of Lesson 6 provide the rest. This is the one antenna in the course where we will run that computation end to end. For a thin wire lying on \(z\), Lesson 6’s radiation vector collapses to one scalar integral,

and the three steps from there are worked in full on the derivation frame below. The radiation intensity goes as \(U \propto \vert N_\theta\vert^2\), so the field pattern is

The result holds within a constant, which is what the proportionality sign is for. The calculus is elementary; the hard part was choosing the current.

The product-to-sum identity turns the integrand into two plain sines, both of which integrate directly, and the projection onto the observation direction removes one power of \(\sin\theta\). In the language of the recap frame, the integral is the space factor of the sinusoidal current and the \(\sin\theta\) from the projection is the element factor. The derivation frame below works both steps in full, with \(kL/2\) written out.

Set \(L = \lambda/2\), so that \(kL/2 = \pi/2\) and the second cosine vanishes. The broadside peak is then \(1 - 0 = 1\), so this one is already normalized and can be written with an equals sign. This is the half-wave dipole pattern. The \(\sin\theta\) in the denominator looks like trouble at \(\theta = 0\), but the numerator vanishes there too and the ratio goes quietly to zero. The nulls are still straight off the ends of the wire.

Solving \(\vert F(\theta)\vert^2 = 1/2\) numerically gives half-power points at \(\theta = 51.0^\circ\) and 129.0°, so

Integrating \(\vert F\vert^2\) over the sphere gives the directivity,

Copper is a good conductor, so radiation efficiency for a wire dipole is above about 98% and \(G = \eta_{\text{rad}} D\) is within a tenth of a dB of \(D\). For this antenna we can quote gain and directivity interchangeably — but it is still helpful to be in the habit of specifying which one you mean, because for the lossy antennas in Module 4 they diverge significantly.

Now look at what lengthening the wire from a small fraction of a wavelength to \(\lambda/2\) actually gained us: the beamwidth went from 90° to 78° and the directivity went from 1.76 to 2.15 dBi. Lengthening the wire brought 0.39 dB of directivity.

The reason to use a half-wave dipole is its input impedance rather than its 2.15 dBi. At \(\lambda/2\) the current maximum sits at the feed, so the feed sees a useful resistance. A short dipole’s \(2\ \Omega\) is hopeless; \(73\ \Omega\) is close enough to a \(75\ \Omega\) line to nearly match it without a network. The resistive part is real radiation. The \(+42.5\ \Omega\) is inductive: stored energy in the reactive near field of Lesson 5, the term that never made it into the far field. Nothing is wrong with the antenna at exactly \(\lambda/2\); it simply is not resonant there, and trimming fixes that.

The next frames show where both numbers come from: the resistance from the pattern we just derived, and the reactance from one standard near-field result that we name but do not re-derive.

The pattern is only half of the analysis. What the transmitter actually sees is the input impedance, and for a thin half-wave dipole we can calculate both parts of it — the resistance comes from the pattern we just derived and the reactance from one standard result we will name but not re-derive.

Radiation resistance is defined by asking what resistor, carrying the same current, would dissipate the power the antenna radiates. If we can compute \(P_\text{rad}\), we can find \(R_r\). Start from the half-wave pattern, and this time keep the constants. Lesson 6 gave the radiation intensity in terms of the radiation vector,

and for the dipole \(N_\phi = 0\) while Step 3 gave \(N_\theta = -N_z\sin\theta\). With the Step 2 result for \(N_z\) at \(L = \lambda/2\),

Square it and multiply by the Lesson 6 constant. The \(k^2\) cancels, which is why the radiation resistance of a half-wave dipole depends on the shape \(\lambda/2\) and not on the frequency:

The \(\vert I_m\vert^2\) out front is what makes the last step work: dividing \(P_\text{rad}\) by \(\tfrac{1}{2}\vert I_m\vert^2\) leaves a resistance with no current in it.

The radiated power is that intensity integrated over the whole sphere. Nothing depends on \(\phi\), so the azimuth integral simply contributes \(2\pi\):

Here the derivation stops being algebra. That integral has no elementary antiderivative. You cannot write the answer in terms of sines, logs, and powers — this is the point where a number enters instead of a formula.

That is not a failure, and it is not unusual. It is the same situation as \(\text{erf}\) in probability: the integral is important enough that somebody tabulated it, gave it a name, and moved on. Antenna work leans on three such special functions, and we will use all of them again:

\(Si\) is the sine integral, \(Ci\) the cosine integral, and \(C_{in}\) the modified cosine integral; the last two are one function apart, \(C_{in}(x) = \gamma + \ln x - Ci(x)\) with \(\gamma = 0.5772\), Euler’s constant. Look them up, or let a calculator evaluate them; do not try to integrate them. The solution function we need for this integral is:

Equate the two expressions for \(P_\text{rad}\); the \(\vert I_m \vert^2\) cancels and the antenna’s current level drops out, as it must. Since \(\eta_0/4\pi = 29.98\ \Omega\), which is \(30\ \Omega\) to better than one tenth of a percent, the result is easy to remember as 30 times 2.4376.

The power calculation used \(I_m\), which is the largest current anywhere on the wire. A resistance referred to that point equals the resistance measured at the feed terminals only when the current maximum is physically located at the feed. For a half-wave dipole it is, so the two numbers agree. For a full-wave dipole they do not: a full-wave dipole has a current minimum at the feed, and its feed resistance is enormous — hundreds to thousands of ohms — even though the current-maximum resistance is a moderate \(199\ \Omega\).

A far-field power integral can only ever produce the real part — it accounts for power that leaves. The reactance describes cyclical energy in the near field which never crosses the far-field sphere at all, so no amount of pattern integration will produce it.

Finding the reactance requires a different technique. We will use the induced-EMF method, in which we integrate the field the antenna reflects back against its own current, along the wire. That is a near-field calculation, and the derivation frame below sets it up and states its general result. That result contains \(Si\), \(Ci\), and the wire radius \(a\) — meaning that the reactance of a dipole normally depends on how thick the wire is.

At exactly \(L = \lambda/2\) that dependence disappears. Every term containing the wire radius is multiplied by \(\sin(kL)\), and at \(kL = \pi\),

The wire radius factors out, and what survives is a single special function.

Put the two halves side by side and the whole impedance is one line, built from two tabulated numbers:

So read the two parts separately. The \(73\ \Omega\) is power leaving and never coming back — the whole point of the antenna. The \(+j42.5\ \Omega\) is the reactive near field of Lesson 5, which does no useful work, and wrecking your match.

Every step above assumed the sinusoidal current from the standing-wave frame. How much that matters depends on what you are computing. The pattern shape barely cares: the radiation integral smooths over small errors in the current, so the 78° beamwidth and 2.15 dBi hold up well. The impedance cares a great deal, because it depends on the current right at the feed and on the near fields close to the wire. This is the main reason a simulator will not return exactly \(73 + j42.5\ \Omega\). Measuring how large that disagreement is, and deciding whether it matters, is the work of Lesson 8.

At exactly \(\lambda/2\) we are \(42.5\ \Omega\) away from resonance. The figure is computed, not sketched: the induced-EMF impedance evaluated at hundreds of lengths for three wire radii. The reactance passes through zero below \(\lambda/2\) for every radius, and the crossing moves left as the element gets thicker. The next frame says why.

The reason is that the wire is electrically longer than it is physically.

• End effect. Capacitance between the tips of the dipole — and to the insulators, mast, and everything else nearby — lets charge accumulate past where the metal stops. The standing wave behaves as though the wire continued a little further than it does.

• Wire thickness. A thicker element has more end capacitance and a lower characteristic impedance, so it shortens further. This is why a thin wire resonates near \(0.480\lambda\) while a thick tubular element can drop to \(0.46\lambda\).

Both effects slow the wave traveling on the wire relative to free space, and a slower wave needs less physical length to fit the same electrical half wavelength. For ordinary wire the answer lands in the range \(0.47\lambda\) to \(0.48\lambda\), and the resistance drops with the length, to roughly \(70\ \Omega\). The induced-EMF derivation frame above tabulates both against wire radius from the general result.

The 5% is an average. It depends on wire gauge, insulation, and what is nearby, which is exactly why you trim rather than compute to four digits.

With a resonant dipole in hand, the match is a one-line calculation. Using Lesson 4’s

the table compares the untrimmed and resonant dipole on each line. On a \(50\ \Omega\) line, removing the \(+j42.5\ \Omega\) of reactance takes the VSWR from 2.18 to 1.40. Leaving the reactance in place and switching to a \(75\ \Omega\) cable only reaches 1.76. Trimming to resonance is worth more than changing the cable.

One practical matter remains before you connect anything. Wire a dipole and a coax together directly and current flows on the outside of the shield, the feedline joins the radiating structure, and your pattern and VSWR both start depending on where you are standing. Lesson 4 gave you the fix: put a balun at the feed point.

Keep stretching the wire past \(\lambda/2\) and the standing wave develops phase reversals: sections of the wire carry current in the opposite direction. Reversed current radiates out of phase with the rest, the contributions interfere, and the single donut breaks into lobes. The radiation integral is linear, so this is exact in the language of the recap frame: a \(1.5\lambda\) wire is three half-wave sections carrying \(+, -, +\), the half-wave pattern is the element factor, and the three sections at \(\lambda/2\) spacing with alternating signs are a space factor, which Lesson 16 would call a three-element array with 180° progressive phase. The element factor’s null on the axis is why the lobes lean toward the wire without reaching it. Below \(\lambda/2\) nothing changes; above \(1.25\lambda\) the main lobes move off broadside, so a single broadside beamwidth is meaningless at \(1.5\lambda\).

The resistance column is referred to the current maximum, and it sweeps through \(50\ \Omega\) twice before the wire is a wavelength long, which is why non-resonant lengths are hard to match.

Design a resonant half-wave dipole for 146 MHz. The two routes agree to within a centimeter, which is well inside the accuracy of the rule, and 49 cm is a length you can eyeball. The worked example below carries the full calculation.

Every number here is unremarkable, which is what you want. A dipole calculation that returns 12 dBi or \(8\ \Omega\) is outside the physically reasonable range and indicates an arithmetic error. If the analyzer disagrees by more than about 10%, suspect the balun before the theory.

• A dipole’s pattern is set by how many wavelengths of current fit on the wire.

• Its impedance is set by where the current maximum sits relative to the feed.

• Half a wavelength is the optimal length because it puts the current maximum at the feed. It costs only 0.39 dB of directivity to implement.

• The resistance comes from a far-field power integral; the reactance does not and cannot.

• A resonant dipole is about \(0.475\lambda\), near \(70\ \Omega\), 2.15 dBi, 78° wide.

• Every number here rests on the assumed sinusoidal current.

Lesson 8 puts this exact antenna into nec_lab, the course simulator, which acts as a front end to NEC-2, the Numerical Electromagnetics Code — a method-of-moments engine written in the 1970s and still the standard tool for wire antennas. What it does is compute the radiation pattern numerically, which is the one thing this lesson could not do. It divides the wire into short segments, treats each as a small radiator whose pattern we already know, and adds those patterns up; the weights in that sum come from enforcing the boundary condition that the total tangential electric field must vanish on a perfect conductor, which also hands back the current on each segment along the way. So the simulator is not doing different physics from this lesson. It is running the same superposition without anyone having to prescribe the current first, and that is exactly why comparing the two is informative.

In Lesson 8 you will model a wire, set its length, sweep the frequency, and read back impedance, VSWR, gain, and pattern — then compare each one against the numbers you produced by hand today. The numbers you just predicted are the ones you will check against simulation. A simulator that agrees with a hand calculation on a dipole can be trusted a little further on a structure you cannot solve by hand; a disagreement usually points to a setup error in the model, and the only way to notice it is to bring predictions with you. The 915 MHz dipole you soldered onto an SMA connector closes the same loop with hardware instead of software: in Lesson 10 you will put it on a vector network analyzer and see how far a real balun-less wire lands from the length, resonance, and impedance you wrote down today.

After that, Lesson 12 takes the same wire apart. Cut a dipole in half and stand it on a ground plane and you have a monopole — half the impedance, double the directivity, and only the upper half-space to radiate into. Bend it into a circle and you have a loop, whose behavior depends entirely on whether the circumference is small or comparable to a wavelength. Further out, in Module 3, the dipole stops being an antenna and becomes an element: line up many of them, control the phase of each, and pattern multiplication turns 2.15 dBi into whatever the array is long enough to deliver.