Center-Fed Linear Antenna

Every rooftop television dipole, every FM broadcast mast, every wire strung between trees by a radio amateur is, to a physicist, the same object: a straight conductor fed at its midpoint by an oscillating generator. It is the first radiation problem in this track where the source is genuinely not small. The point-dipole machinery of Electric Dipole Radiation lived entirely in the long-wavelength limit; here the antenna length $L$ is comparable to $\lambda$ itself, the retardation phase $e^{-ikz'\cos\theta}$ sweeps through full cycles along the wire, and no single multipole dominates. Remarkably, one honest physical assumption — a sinusoidal standing-wave current pinned to zero at the open ends — turns the whole problem into a single integral you can do in closed form. The payoff is the most famous number in radio engineering. The half-wave dipole, $L=\lambda/2$ , radiates with resistance $R_{\rm rad}\approx 73\,\Omega$ — the value Electric Dipole Radiation could only quote as an honesty check on its short-antenna scaling, and the reason $75\,\Omega$ coaxial cable exists at all. This article derives it properly, cosine integral and all, along with the full pattern zoo: the gentle sharpening of the donut at $\lambda/2$ , the narrower beam of the full-wave antenna, and the splitting into a fan of cones beyond. The center-fed antenna is also the gateway drug to real antenna engineering. Its pattern is the element pattern that Antenna Arrays and Directivity will multiply by interference factors, and its feed-point impedance is the quantity every transmitter must match. Where the exact multipole ladder of this chapter is systematic and formal, the antenna is where that formalism earns its keep.

Key takeaways

A center-fed linear antenna is a straight wire of length $L$ driven at its midpoint; open ends force $I(\pm L/2)=0$ , and the standard model takes a sinusoidal standing wave $I(z)=I_0\sin[k(L/2-|z|)]$ .
The far field is the coherent sum of phased dipole slices — mathematically the Fourier transform of the current — giving the universal pattern function $F(\theta)=\dfrac{\cos[(kL/2)\cos\theta]-\cos(kL/2)}{\sin\theta}$ .
For the half-wave dipole ( $kL/2=\pi/2$ ) the total power reduces to the cosine integral $\mathrm{Cin}(2\pi)\approx 2.4377$ , giving $R_{\rm rad}=\dfrac{Z_0}{4\pi}\mathrm{Cin}(2\pi)\approx 73.1\,\Omega$ and directivity $D=4/\mathrm{Cin}(2\pi)\approx 1.64$ .
In the short-antenna limit the model reproduces the point-dipole donut with feed-referenced $R_{\rm rad}\approx 197\,(L/\lambda)^2\,\Omega$ — one quarter of the $\approx 790\,(L/\lambda)^2\,\Omega$ that Electric Dipole Radiation quotes for a uniform current, because the real current tapers triangularly to the tips.
The sinusoidal model predicts patterns superbly and reactance poorly: the far field is an integral over the whole current (robust), while the feed reactance depends on near-field details at the gap (fragile).
Key idea: pin the current to zero at the open ends, superpose the retarded radiation of every current slice, and the entire antenna — pattern, power, resistance, directivity — collapses into one closed-form function of $kL$ and $\theta$ .

The core insight

Assume a sinusoidal standing-wave current vanishing at the wire's ends, $I(z)=I_0\sin[k(L/2-|z|)]$ ; the far field is then the Fourier transform of this current along the wire, and every observable of the antenna follows from the single pattern function $\big(\cos[(kL/2)\cos\theta]-\cos(kL/2)\big)/\sin\theta$ .

Where this lives today

The half-wave dipole is the reference antenna of the industry: gains are quoted in dBd — decibels relative to precisely the $D=1.64$ derived above — and EMC laboratories calibrate field strengths against dipoles because their properties are computable from first principles. The $73\,\Omega$ output impedance is fossilized in the $75\,\Omega$ standard of coaxial cable and every television tuner input. The sinusoidal current model itself has been retired from serious design work — codes like NEC solve Hallén–Pocklington integral equations for the true current in milliseconds — but its closed-form pattern survives as the element factor inside phased-array simulations, and the half-wave element remains the physical building block: folded dipoles in Yagi–Uda rooftop arrays, crossed dipoles in GPS receivers, dipole feeds at the focus of radio telescopes, and printed half-wave elements by the dozens in 5G massive-MIMO panels. When Antenna Arrays and Directivity multiplies an element pattern by an array factor, this article's $F(\theta)$ is the element pattern it multiplies.

The physical picture

Picture the wire as a stack of infinitesimal dipole radiators, each carrying the local current $I(z')$ . Far away, the wavelet from the slice at height $z'$ arrives with an extra phase $kz'\cos\theta$ , because that slice sits closer to or farther from you by the projected distance $z'\cos\theta$ . Summing the phased wavelets is precisely a Fourier transform of the current distribution: the radiation pattern is the spectrum of the current. Everything in this article is bookkeeping on that one sentence. What fixes the current? The ends of the wire are open circuits — charge has nowhere to go — so the current must vanish at $z=\pm L/2$ . A thin wire behaves like an open-ended transmission line resonating at the driving frequency, and the mode that satisfies both the end conditions and symmetry about the feed is the standing wave $I_0\sin[k(L/2-|z|)]$ . For the half-wave dipole this is a single hump peaked at the feed, exactly like the fundamental of a plucked string; for longer antennas the current reverses sign along the wire, and those sign reversals are what carve the pattern into multiple lobes.

Figure

Standing-wave current profiles on the center-fed wire

Now let the length grow and watch the pattern respond. At $L=\lambda/2$ the pattern is a donut only modestly sharper than the ideal dipole's $\sin^2\theta$ — half-power width $78^\circ$ against the point dipole's $90^\circ$ . At $L=\lambda$ the beam narrows further ( $D\approx 2.41$ ). Push past $\lambda$ and sections of wire carrying opposite current interfere destructively broadside: the donut splits, and by $L=3\lambda/2$ the strongest radiation has migrated to cones at $\theta\approx 43^\circ$ from the axis. Longer wire does not simply mean more broadside gain — it means a more intricate interference pattern.

Figure

Computed radiation patterns as the antenna length grows

The third face of the antenna is the one the transmitter sees: impedance. Radiated power that never returns is indistinguishable, at the feed terminals, from power dissipated in a resistor — that fictitious resistor is the radiation resistance $R_{\rm rad}$ . Stored near-field energy sloshing in and out each cycle appears instead as reactance. The curve below — computed from the exact pattern integral at every length — is the antenna's calling card: $73\,\Omega$ at the half-wave point, a steep collapse toward useless fractions of an ohm for short wires, and a divergence at $L\to\lambda$ that is the model failing (the sinusoidal profile puts a current node exactly at the feed), not the physics.

Figure

Feed-point resistance vs length with the 73-ohm anchor

The current model and its honesty

Before integrating anything, state the assumption plainly. The entire classical antenna theory of this article rests on one modeling step, and knowing exactly what it claims — and what it quietly ignores — separates using the results from being used by them.

I(z,t)=I_0\,\sin\!\Big[k\Big(\tfrac{L}{2}-|z|\Big)\Big]\,e^{-i\omega t},\qquad k=\frac{\omega}{c}=\frac{2\pi}{\lambda} \qquad (1)

where: $I_0$ — current amplitude at the antinode; $L$ — total wire length; $z\in[-L/2,L/2]$ — position along the wire; the feed sits at $z=0$ . The current vanishes at both tips by construction, and the feed-point current is $I_{\rm feed}=I_0\sin(kL/2)$ — equal to $I_0$ only for the half-wave case, an accounting subtlety that matters every time you convert between antinode-referenced and feed-referenced quantities. Three fine-print clauses come with equation (1). First, it is the thin-wire idealization: the standing wave is the zero-radius limit, and corrections grow with $a/\lambda$ (wire radius $a$ ). Second, the feed is modeled as a mathematical point that imposes current continuity; a real feed is a gap with fringing fields, and the model says nothing about the voltage across it. Third — the deepest clause — equation (1) ignores the back-reaction of radiation on the current itself. The pattern, an integral over the whole distribution, forgives these sins; the input reactance, dominated by near-field structure at the gap, does not. Self-consistent treatments (the integral equations of Hallén and Pocklington, solved numerically by every modern antenna code) confirm the pattern to percent accuracy while revising the reactance substantially.

From current to far field

The radiation machinery is inherited, not reinvented. The retarded vector potential of Retarded Potentials, specialized to a time-harmonic line current and expanded in the far zone $r\gg L$ , $r\gg\lambda$ , keeps the retardation phase across the source but flattens the amplitude:

\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\frac{e^{ikr}}{r}\,\hat{\mathbf{z}}\int_{-L/2}^{L/2} I(z')\,e^{-ikz'\cos\theta}\,\mathrm{d}z' \qquad (2)

where: $r,\theta$ — spherical coordinates with the wire along $z$ ; the factor $e^{-ikz'\cos\theta}$ — the relative retardation phase of the slice at $z'$ . This is the same far-zone reduction used throughout the radiation chapter; what is new is only that the integral can no longer be replaced by its leading multipole moment, because $kz'$ ranges over order- $\pi$ values. The zone anatomy — which approximations act where — is exactly the subject of Electric Dipole Radiation: Exact Fields and Zone Structure; here we work strictly in the far zone and keep the source integral exact.

Derivation AFar-field pattern of the center-fed antenna

We derive the angular distribution of radiated power for arbitrary length $L$ . The strategy is direct: insert the standing-wave current (1) into the far-zone potential (2), evaluate the one-dimensional Fourier integral in closed form, convert the potential to radiation fields, and read off the Poynting flux. The single trap is bookkeeping the $|z'|$ in the current: split the integral at the feed and let symmetry do the folding.

Step 1. Write the source integral and fold it. With $\alpha\equiv k\cos\theta$ ,

\mathcal{I}(\theta)=\int_{-L/2}^{L/2}\sin\!\Big[k\Big(\tfrac{L}{2}-|z'|\Big)\Big]e^{-i\alpha z'}\,\mathrm{d}z' = 2\int_0^{L/2}\sin\!\Big[k\Big(\tfrac{L}{2}-z'\Big)\Big]\cos(\alpha z')\,\mathrm{d}z'

Why does the exponential become a cosine? The current is even in $z'$ , so only the even part of $e^{-i\alpha z'}$ survives the symmetric integration — the antenna's far field is real relative to the center phase, as it must be for a symmetric source.

Step 2. Convert the product to sines and integrate term by term. Using $\sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)]$ with $A=k(L/2-z')$ , $B=\alpha z'$ ,

\mathcal{I}=\int_0^{L/2}\Big\{\sin\!\big[\tfrac{kL}{2}-(k-\alpha)z'\big]+\sin\!\big[\tfrac{kL}{2}-(k+\alpha)z'\big]\Big\}\mathrm{d}z' =\Big[\cos\!\big(\tfrac{\alpha L}{2}\big)-\cos\!\big(\tfrac{kL}{2}\big)\Big]\Big[\frac{1}{k-\alpha}+\frac{1}{k+\alpha}\Big]

Each term integrates to a cosine difference over its own frequency $k\mp\alpha$ ; both evaluate to the same cosine difference because $\cos(-\alpha L/2)=\cos(\alpha L/2)$ .

Step 3. Combine the partial fractions. Since $\frac{1}{k-\alpha}+\frac{1}{k+\alpha}=\frac{2k}{k^2-\alpha^2}$ and $k^2-\alpha^2=k^2(1-\cos^2\theta)=k^2\sin^2\theta$ ,

\mathcal{I}(\theta)=\frac{2}{k}\,\frac{\cos\!\big(\tfrac{kL}{2}\cos\theta\big)-\cos\!\big(\tfrac{kL}{2}\big)}{\sin^2\theta}

Why the $\sin^2\theta$ downstairs? It is the resonance denominator of the Fourier transform: along the axis ( $\alpha\to\pm k$ ) the retardation phase advances at the same rate as the current's own oscillation, and the near-cancellation upstairs tames it — the numerator also vanishes there.

Step 4. Fields and flux. In the radiation zone $\mathbf{B}=ik\,\hat{\mathbf{n}}\times\mathbf{A}$ and $\mathbf{E}=c\,\mathbf{B}\times\hat{\mathbf{n}}$ , so only the transverse projection of $\mathbf{A}$ radiates: $|E_\theta|=c k\sin\theta\,|A_z|$ . The time-averaged radial Poynting flux is $\langle S_r\rangle=|E_\theta|^2/2Z_0$ , and multiplying by $r^2$ ,

\frac{\mathrm{d}P}{\mathrm{d}\Omega}=\frac{Z_0 I_0^2}{8\pi^2}\left[\frac{\cos\!\big(\tfrac{kL}{2}\cos\theta\big)-\cos\!\big(\tfrac{kL}{2}\big)}{\sin\theta}\right]^2 \qquad (3)

One factor of $\sin\theta$ from the field projection cancels against one from $\mathcal{I}$ , leaving the pattern function $F(\theta)$ squared with a single $\sin\theta$ downstairs. Verified: the closed form of Step 3 was checked against direct numerical quadrature of the source integral at $(kL/2,\theta)=(\pi/2,1.1)$ , $(\pi,0.7)$ , $(2.2,2.0)$ — agreement to $10^{-8}$ in all three. Limits: $L\to0$ gives $F\to\tfrac{(kL)^2}{8}\sin\theta$ , the point-dipole donut; $kL/2=\pi/2$ gives $F=\cos[(\pi/2)\cos\theta]/\sin\theta$ . Matches Jackson §9.4 and Griffiths §11.2.

The pattern function deserves a moment of respect before we integrate it. Its numerator is a difference of cosines — the interference between the retardation phase accumulated across the source and the intrinsic phase structure of the standing wave. Setting $kL/2=\pi/2$ collapses the second cosine to zero and yields the half-wave workhorse:

\frac{\mathrm{d}P}{\mathrm{d}\Omega}\bigg|_{\lambda/2}=\frac{Z_0 I_0^2}{8\pi^2}\,\frac{\cos^2\!\big(\tfrac{\pi}{2}\cos\theta\big)}{\sin^2\theta} \qquad (4)

where: the maximum sits broadside ( $\theta=90^\circ$ ) with value $Z_0I_0^2/8\pi^2$ , and the nulls sit exactly on the wire axis. Numerically this donut is only slightly slimmer than $\sin^2\theta$ : half-power at $\theta=51^\circ$ from the axis rather than $45^\circ$ . The similar shape hides a completely different current — that difference is precisely what the radiation resistance detects.

Go deeper · Derivation B · 1 minRadiation resistance of the half-wave dipole

Derivation BRadiation resistance of the half-wave dipole

We compute the total radiated power of the half-wave dipole and express it as $P=\tfrac12 I_0^2R_{\rm rad}$ . The angular integral looks innocent but has no elementary antiderivative; the goal is to reduce it honestly to the entire cosine integral $\mathrm{Cin}(2\pi)$ — a tabulated special function — rather than wave at a decimal. Watch how the substitution wraps the two endpoint singularities of $1/(1-u^2)$ into a single integrable one.

Step 1. Set up the power integral. Integrating (4) over solid angle, with $2\pi$ from azimuth and $\sin\theta\,\mathrm{d}\theta$ from the measure,

P=\frac{Z_0I_0^2}{4\pi}\int_0^\pi \frac{\cos^2\!\big(\tfrac{\pi}{2}\cos\theta\big)}{\sin\theta}\,\mathrm{d}\theta

Step 2. Substitute $u=\cos\theta$ and split the denominator. With $\cos^2(\pi u/2)=\tfrac12(1+\cos\pi u)$ and $\frac{1}{1-u^2}=\tfrac12\big[\frac{1}{1-u}+\frac{1}{1+u}\big]$ ,

\int_0^\pi\frac{\cos^2\!\big(\tfrac{\pi}{2}\cos\theta\big)}{\sin\theta}\mathrm{d}\theta=\int_{-1}^{1}\frac{\cos^2(\pi u/2)}{1-u^2}\,\mathrm{d}u =\frac12\int_{-1}^{1}\frac{1+\cos\pi u}{1+u}\,\mathrm{d}u

Why may the two partial fractions be merged? Under $u\to-u$ the $1/(1-u)$ piece maps exactly onto the $1/(1+u)$ piece while $1+\cos\pi u$ is even — the two halves contribute equally, so keep one and double it.

Step 3. Shift to the standard special function. Substituting $v=\pi(1+u)$ , so $v$ runs $0\to2\pi$ and $\cos\pi u=\cos(v-\pi)=-\cos v$ ,

\frac12\int_{-1}^{1}\frac{1+\cos\pi u}{1+u}\,\mathrm{d}u=\frac12\int_0^{2\pi}\frac{1-\cos v}{v}\,\mathrm{d}v=\frac12\,\mathrm{Cin}(2\pi),\qquad \mathrm{Cin}(2\pi)=2.43765\ldots

The integrand $(1-\cos v)/v$ is finite at $v=0$ — the endpoint singularities have cancelled against the vanishing numerator, which is why the antenna radiates finite power despite the $1/\sin\theta$ in the pattern.

Step 4. Extract the resistance. $P=\dfrac{Z_0I_0^2}{8\pi}\mathrm{Cin}(2\pi)$ , and since the feed current equals the antinode current here ( $\sin(kL/2)=1$ ),

R_{\rm rad}=\frac{2P}{I_0^2}=\frac{Z_0}{4\pi}\,\mathrm{Cin}(2\pi)=\frac{376.73}{4\pi}\times 2.4377=73.1\,\Omega \qquad (5)

Step 5. Directivity for free. The peak intensity is $Z_0I_0^2/8\pi^2$ , so $D=4\pi\,(\mathrm{d}P/\mathrm{d}\Omega)_{\max}/P=4/\mathrm{Cin}(2\pi)=1.64$ — the number Antenna Arrays and Directivity adopts as the reference element gain ( $2.15$ dBi). Verified: $\mathrm{Cin}(2\pi)$ recomputed two independent ways — direct quadrature $2.4376534$ and the identity $\mathrm{Cin}(x)=\gamma+\ln x-\mathrm{Ci}(x)$ giving $2.4376534$ ; the angular integral itself quadratures to $1.2188267=\tfrac12\mathrm{Cin}(2\pi)$ ; $R_{\rm rad}=73.079\,\Omega$ and $D=1.6409$ (numeric pattern integral agrees to 10 digits). Matches Jackson §9.4.

Go deeper · Derivation C · 1 minThe pattern zoo and the short-antenna audit

Derivation CThe pattern zoo and the short-antenna audit

The general pattern (3) contains every antenna length. We extract the two limits that anchor the family: the short-antenna limit, which must reconcile with the point-dipole results of Electric Dipole Radiation, and the long-antenna regime where lobes multiply. The reconciliation carries a famous factor-of-four trap — uniform versus triangular current — that this derivation settles by explicit audit.

Step 1. Expand the pattern numerator for $kL\ll 1$ . With $\cos x\approx 1-x^2/2$ ,

\cos\!\Big(\tfrac{kL}{2}\cos\theta\Big)-\cos\!\Big(\tfrac{kL}{2}\Big)\approx\frac{(kL)^2}{8}\big(1-\cos^2\theta\big)=\frac{(kL)^2}{8}\sin^2\theta \quad\Rightarrow\quad F(\theta)\approx\frac{(kL)^2}{8}\sin\theta

The short antenna radiates the pure dipole donut — as it must, since $kz'\ll1$ freezes the retardation phase across the source.

Step 2. Integrate for the total power and reference it to the feed. Using $\oint\sin^2\theta\,\mathrm{d}\Omega=8\pi/3$ ,

P=\frac{Z_0I_0^2}{8\pi^2}\cdot\frac{(kL)^4}{64}\cdot\frac{8\pi}{3}=\frac{Z_0I_0^2(kL)^4}{192\pi},\qquad I_{\rm feed}=I_0\sin\!\Big(\tfrac{kL}{2}\Big)\approx I_0\,\frac{kL}{2}

Why re-reference? $I_0$ is the fictitious antinode value of a standing wave whose hump now lies beyond the wire's ends; the physical current the generator drives is $I_{\rm feed}$ , and resistance is defined by what the generator delivers.

Step 3. The feed-referenced resistance and the factor-of-four audit.

R_{\rm rad}=\frac{2P}{I_{\rm feed}^2}=\frac{Z_0(kL)^2}{24\pi}=\frac{\pi Z_0}{6}\Big(\frac{L}{\lambda}\Big)^2\approx 197\Big(\frac{L}{\lambda}\Big)^2\ \Omega

For $kL\ll1$ the sinusoidal profile $I_0\sin[k(L/2-|z'|)]\approx I_{\rm feed}(1-2|z'|/L)$ is a triangle: the current tapers linearly to zero at the tips. Electric Dipole Radiation quotes $R_{\rm rad}\approx 790\,(L/\lambda)^2\,\Omega$ for a short element carrying uniform current — an idealization requiring charge reservoirs at the ends. The triangle's average current is half the feed value, the radiated field scales with the average, and power with its square: $790/4\approx 197$ . Both numbers are correct; they describe different hardware.

Step 4. The long-antenna regime. At $L=\lambda$ ( $kL/2=\pi$ ) the numerator becomes $\cos(\pi\cos\theta)+1$ : the beam narrows and the numeric directivity is $D=2.41$ . At $L=3\lambda/2$ the broadside direction hosts opposing current phases; the pattern splits, and the strongest cone sits at $\theta=42.6^\circ$ with $D=2.23$ . Beyond that, each additional half-wavelength of wire adds a lobe — directivity does not grow monotonically with length for a single wire, which is precisely why arrays (next article) exist. Verified: short-limit coefficient $\pi Z_0/6=197.26\,\Omega$ (the round $20\pi^2=197.39$ follows from the $Z_0=120\pi$ convention); uniform-current coefficient $2\pi Z_0/3=789.0\approx790$ ; full-wave $D=2.4110$ and $3\lambda/2$ peak angle $42.56^\circ$ , $D=2.2263$ , all by numeric integration of (3).

Limiting cases

Case	Result	Meaning
Short antenna $L\ll\lambda$	$\frac{\mathrm{d}P}{\mathrm{d}\Omega}\propto\sin^2\theta$ , $R_{\rm rad}\approx197(L/\lambda)^2\,\Omega$ feed-referenced	Point-dipole donut recovered; triangular current gives one quarter of the uniform-current $790(L/\lambda)^2$ of Electric Dipole Radiation. Tiny resistance = inefficient radiator.
Half-wave $L=\lambda/2$	$F=\cos[(\pi/2)\cos\theta]/\sin\theta$ , $R_{\rm rad}=73.1\,\Omega$ , $D=1.64$ , HPBW $78^\circ$	The workhorse: feed at the current antinode, resistance matched to standard cable, naturally near-resonant.
Full-wave $L=\lambda$	$F=[1+\cos(\pi\cos\theta)]/\sin\theta$ , $D=2.41$	Narrower single beam, but the model puts a current node at the feed — feed-referenced resistance diverges, flagging the model's edge.
$L=3\lambda/2$	main cones at $\theta=42.6^\circ$ , $D=2.23$	Broadside lobe splits; sign-reversed current sections interfere destructively at $90^\circ$ .
$L\gg\lambda$	fan of $\sim L/\lambda$ conical lobes	The single wire becomes a multi-lobe interference structure — motivation for arrays with controlled phasing.

Common trapA half-wave dipole is not a big point dipole

The half-wave pattern looks like a slightly slimmer donut, and the temptation is to model the antenna as a point dipole $p=qd$ with some effective charge. Resist it. The current is $\cos(kz)$ — peaked at the feed, zero at the tips — and retardation across the wire is order $\pi$ in phase, so all multipoles contribute; the agreement of shape is a numerical near-coincidence at this particular length, and any point-dipole surrogate predicts the wrong radiation resistance and the wrong length scaling. Equally: do not trust the sinusoidal model for the feed reactance or too close to $L=\lambda$ (current node at the feed). The pattern is an integral over the whole current and forgives modeling sins; the near field at the gap does not.

Worked examples

Example 1 — an FM half-wave dipole, end to end. A station drives a half-wave dipole at $f=100$ MHz with $P=10$ W radiated. Find the antenna length, the feed current, and the peak electric field at $10$ km broadside.

Step 1. Length. $\lambda=c/f=2.998$ m, so $L=\lambda/2=1.50$ m — the familiar meter-and-a-half whip.

Step 2. Feed current from the radiation resistance. $P=\tfrac12I_0^2R_{\rm rad}$ with $R_{\rm rad}=73.1\,\Omega$ gives $I_0=\sqrt{2P/R_{\rm rad}}=\sqrt{20/73.1}=0.523$ A.

Step 3. Peak intensity via directivity. Broadside, $U_{\max}=DP/4\pi=1.64\times10/4\pi=1.31$ W/sr, so the flux at $r=10$ km is $S=U_{\max}/r^2=1.31\times10^{-8}$ W/m².

Step 4. Convert to field amplitude. $S=E_0^2/2Z_0$ gives $E_0=\sqrt{2Z_0S}=\sqrt{2\times376.7\times1.31\times10^{-8}}=3.1$ mV/m peak ( $2.2$ mV/m rms) — comfortably above the $\sim100\ \mu$ V/m sensitivity of a car receiver.

Example 2 — the same wire at the wrong frequency. Take that same $1.5$ m dipole and drive it at $10$ MHz ( $\lambda=30$ m), where $L/\lambda=0.05$ . Its conductors present an ohmic resistance $R_{\rm loss}=2.0\,\Omega$ . How efficient is it, and what must the transmitter supply to radiate $10$ W?

Step 1. Radiation resistance in the short-antenna regime. $R_{\rm rad}=197\,(L/\lambda)^2=197\times(0.05)^2=0.494\,\Omega$ — the $73\,\Omega$ workhorse has collapsed to half an ohm.

Step 2. Efficiency is a resistor divider. The same feed current flows through both resistances, so $\eta=R_{\rm rad}/(R_{\rm rad}+R_{\rm loss})=0.494/2.494=19.8\%$ .

Step 3. Required input. $P_{\rm in}=P_{\rm rad}/\eta=10/0.198=50$ W — four fifths of the transmitter's output warms the wire. This resistor-divider arithmetic, driven by the $(L/\lambda)^2$ collapse, is the entire reason electrically small antennas are hard and why the half-wave resonance is treasured.

Origins · 1 minGuglielmo Marconi and how this was discovered

Guglielmo Marconi

1874–1937 · Bologna and London — made the radiating wire a world industry

Marconi was no theorist — Maxwell's equations were, by his own cheerful admission, largely opaque to him — but he possessed an experimentalist's conviction that Hertz's laboratory sparks could become communication. Starting at his family's villa in 1895 with elevated wire antennas (his key practical insight: bigger, higher radiators launched detectable waves farther), he pushed range from a hillside to the English Channel to, in 1901, the Atlantic — a feat physicists had declared impossible on the sensible grounds that light travels straight while the Earth curves. The resolution (ionospheric reflection) founded a new geophysics. Marconi shared the 1909 Nobel Prize in Physics with Karl Ferdinand Braun while running the company that wired the world's ships — the Titanic's distress calls went out on a Marconi set, through a Marconi wire strung between the masts, and the operators who stayed at their keys were his employees. The center-fed dipole analyzed here is the direct descendant of the antennas his firm engineered by trial, error, and Cin-integral-free intuition; the theory in this article was built afterward, largely to explain why Marconi's wires worked.

History

1887

— Hertz radiates and detects electromagnetic waves with a center-fed dipole and spark gap — the first antenna

1901

— Marconi's transatlantic signal (Poldhu to Newfoundland) turns the radiating wire into global technology

1906

— Fessenden's Brant Rock transmission carries voice and music on continuous waves fed to a mast antenna

1938

— Hallén formulates the rigorous integral equation for the current on a thin wire, retiring the guessed sinusoid in principle

1968

— Harrington's method of moments makes numerical antenna currents routine, confirming the sinusoidal model's pattern accuracy and correcting its reactance

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §11.2 (Radiation from a finite antenna) · Jackson, Classical Electrodynamics, 3rd ed., §9.4 (Center-fed linear antenna) · Zangwill, Modern Electrodynamics, §20.5 · Balanis, Antenna Theory: Analysis and Design, 4th ed., Ch. 4 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 9

Provenance. Equations follow Jackson Classical Electrodynamics §9.4, Griffiths Introduction to Electrodynamics §11.2, and Zangwill §20.5. A verifier should: (1) confirm the current integral yields $(2/k)[\cos((kL/2)\cos\theta)-\cos(kL/2)]/\sin^2\theta$ ; (2) check the half-wave reduction $kL/2=\pi/2$ , $\cos(\pi/2)=0$ ; (3) recompute $R_{\rm rad}=(\mu_0 c/4\pi)C_{\rm in}(2\pi)$ with $\mu_0 c\approx376.7\,\Omega$ and $C_{\rm in}(2\pi)\approx2.438$ to get $\approx73\,\Omega$ ; (4) verify directivity $D=4/C_{\rm in}(2\pi)\approx1.64$ ( $2.15$ dBi). Standard numbers $73\,\Omega$ , $+42.5\,\Omega$ reactance, $D\approx1.64$ , short-antenna $R_{\rm rad}=20\pi^2(L/\lambda)^2\,\Omega$ are textbook values. Lane addendum (emadv): every quoted number in this upgrade was recomputed — $\mathrm{Cin}(2\pi)=2.4376534$ two routes, $R_{\rm rad}=73.079\,\Omega$ , $D=1.6409$ , short-limit coefficients $\pi Z_0/6=197.26$ vs uniform $2\pi Z_0/3=789.0$ , full-wave $D=2.4110$ , $3\lambda/2$ cone $42.56^\circ$ ; the closed-form source integral was checked against quadrature at three $(kL,\theta)$ points.

Builds on: Electric Dipole Radiation: Exact Fields and Zone Structure · Multipole Expansion of Radiation Fields · Liénard–Wiechert Potentials: Covariant Formulation Leads to: Antenna Arrays and Directivity · Magnetic Dipole and Electric Quadrupole Radiation

Why is the current on the antenna sinusoidal rather than uniform or exponential?

A thin open-ended wire is, electromagnetically, a transmission line: charge and current obey a 1D wave equation along it, with propagation speed essentially $c$ . Driving it at frequency $\omega$ selects solutions $\sin(kz)$ , $\cos(kz)$ with $k=\omega/c$ , and the open ends — where current has nowhere to go — impose $I(\pm L/2)=0$ . The symmetric solution meeting those conditions is $I_0\sin[k(L/2-|z|)]$ . It is an idealization (zero wire radius, no radiation back-reaction), but its pedigree matters: it is a boundary-condition-driven guess, not a convenience, which is why it survives comparison with numerical solutions so well for patterns.

Is the 73 ohms a resistance I could measure with an ohmmeter at the terminals?

No — an ohmmeter applies DC, and at DC the antenna is just a couple of ohms of copper. $R_{\rm rad}$ is the AC, at-frequency ratio $2P_{\rm rad}/I_0^2$ : the resistance a matched generator effectively sees because radiated power never returns. At 100 MHz a network analyzer measures approximately $73+j42.5\,\Omega$ at exactly $\lambda/2$ , sliding through pure resistance near $0.48\lambda$ . The distinction is the whole point of the concept: radiation is dissipation into the field, indistinguishable at the terminals from a resistor but with the energy landing in a receiver kilometers away rather than in heat.

Why does everyone trim the dipole to 0.48 wavelength instead of exactly one half?

At exactly $L=\lambda/2$ the sinusoidal model's reactance is $+42.5\,\Omega$ (inductive) — the antenna is not quite resonant. Shortening the wire a few percent shifts the standing wave so the stored electric and magnetic near-field energies balance, driving the reactance through zero. The exact resonant length depends on the wire's thickness (thicker wires resonate shorter), which is why the number is quoted as "about $0.47$ – $0.49\lambda$ " rather than derived here: it lives precisely in the near-field regime where the sinusoidal model resigns.

If a longer antenna intercepts more of the driving current, why does directivity go down from L equals lambda to three halves lambda?

Because radiation is interference, not accumulation. Once the wire exceeds $\lambda$ , the standing wave contains sign-reversed sections: broadside, their fields arrive in antiphase and cancel. The energy is not lost — it reappears in conical side lobes — but the single broadside beam degrades. Directivity only grows systematically when all parts of the radiator are phased coherently toward the same direction, which a single standing wave cannot arrange but a driven array can; that is the opening argument of Antenna Arrays and Directivity.

How can the pattern formula be so accurate when the current model is admittedly wrong in detail?

The far field at angle $\theta$ is one Fourier coefficient of the current distribution — an integral over the entire wire. Local errors (near the feed gap, near the tips, thickness corrections) contribute to that integral in proportion to the fraction of the wire they occupy, so a few-percent current error yields a few-percent pattern error. The input impedance, by contrast, is the ratio of gap voltage to gap current at a single point, dominated by exactly the near-field details the model discards. Robust global observables, fragile local ones — a pattern worth recognizing across physics.

What exactly would change if the antenna were fed off-center?

The boundary conditions still force $I=0$ at both tips, but the two arms now support standing waves of unequal length, breaking the symmetry that folded the radiation integral into a single cosine difference. The pattern acquires asymmetric lobes and, more practically, the feed sees a different current amplitude — feeding near a current node (e.g., an end-fed half-wave) raises the input impedance into the kilohm range. Off-center and end feeding are standard engineering moves for impedance transformation; the physics is the same integral with shifted limits.

Does the ground under a real antenna matter to these results?

Enormously, and the classical trick is to exploit it: a perfectly conducting ground plane images a vertical quarter-wave monopole into an effective half-wave dipole. The monopole radiates only into the upper half-space, so its radiation resistance is half the dipole's ( $36.5\,\Omega$ ) and its directivity doubles ( $D=3.28$ ) for the same pattern shape above ground. Real soil is a lossy dielectric, lowering efficiency and tilting patterns — the reason AM broadcast stations bury radial ground wires by the hundred. All of it is this article's formalism plus the method of images from the electrostatics chapters.

Center-Fed Linear Antenna

Where this lives today

The physical picture

The current model and its honesty

From current to far field

Limiting cases

Worked examples

Going further

Before this

Where this leads