How a dipole radiates

Connect a transmitter to two straight wires, one on each side of a small gap. Leave the far ends open. A receiver across the room can pick up the signal. How does energy get from those wires to the receiver?

A dipole is one of the simplest antennas to build. Yet changing its length can turn a useful antenna into a difficult load, or move its strongest radiation away from the receiver. To understand why, start with what happens in the wires.

Two wires, an alternating current

The two wires are the dipole’s arms. The transmitter connects across the gap between them, at the feed, and applies an alternating voltage. It moves charge from one arm to the other, then back again.

The open tips do not stop this back-and-forth current. Charge can accumulate on an arm without leaving its tip. When current flows into the upper arm, that arm becomes more positive and the lower arm more negative. Here the arrows follow conventional current: the direction positive charge would move. After the current reverses, the accumulated charge starts to decrease. It takes time to remove it, so the upper arm remains positive for a while.

Watch that reversal below. The charge signs change after the current arrows reverse, and the change in the electric field takes time to reach the probe.

From wire current to a radio wave

Set the slider just past ¼ cycle. Is the upper arm still positive?

Highlighted field line

Slow-motion visualization. Large arrows show current; small arrowheads show field direction. The wire and nearby field are enlarged. The probe is 0.75 wavelengths from the feed.

One cycle at the feed Relative to each peak

Cycle position

Current reverses at ¼ cycle. The upper arm stays positive until ½ cycle.

The electrons move back and forth within the metal; they do not fly across the room. The transmitter makes them speed up, slow down and reverse direction. This acceleration produces a radiating field that travels outward at the speed of light, carrying energy from the transmitter. At the receiving antenna, the wave’s electric field moves charge and produces a voltage.

Why start with a short dipole?

“Short” means short compared with the wavelength, the distance a radio wave travels in one cycle. If the whole dipole is much shorter than that distance, current rises and falls at almost the same time along both arms. We can understand its radiation before dealing with different parts of a longer wire carrying current in opposite directions.

The current is not equally strong everywhere. It is largest at the feed and falls to zero at the open tips, where charge cannot continue along the wire. For a short, thin dipole, this current distribution is nearly triangular. A total length of about one tenth of a wavelength or less is a common working range for the approximation; there is no sudden physical boundary at one tenth.

We will start with that simple case, then lengthen the wire. First, where does its radiation go?

Where does the signal go?

A vertical short dipole radiates most strongly to its sides and has a radiation null along its axis, above and below the tips. Its three-dimensional pattern looks like a ring around the wire. The ring shows relative strength by direction, not the shape of a wave moving through space.

Imagine walking around the antenna at the height of its feed, always the same distance away. In open space, the signal stays the same. This is what omnidirectional means here: equal radiation around the wire’s axis. It does not mean equal radiation in every direction.

Now move the receiver over a tip while keeping its distance fixed. You leave the strongest direction, called broadside, and approach the null. The radiated electric field is transverse: it points across the direction of travel. Only the component of wire current perpendicular to your line of sight contributes to this far field. Seen end-on, that component is zero.

Try both paths below. The receiving antenna stays aligned with the local electric field so you can isolate the effect of position.

Walk around the short dipole

Move the receiver while keeping its wire aligned with the local electric field. The transmitter current stays fixed.

Angle from the upper wire axis 90°

Electric field100.0%

Power density100.0%

The copper dot marks the receiver. Broadside at 3 m, both readings are 100%. The wire is drawn larger so you can see it.

Reading the pattern

Measure the angle θ from the upward wire axis: 0° is above the tip and 90° is broadside. For the short dipole, field amplitude varies as sin θ and power density as sin² θ. Power density is power passing through a unit area, measured in W/m².

At 45° and 135°, power density is half its broadside value. Those directions are 90° apart, giving a half-power beamwidth of 90°. Broadside power density is 1.5 times what an antenna radiating the same total power equally in all directions would produce. That ratio is the short dipole’s peak directivity: 1.5, or 1.76 dBi. Directivity describes concentration by direction, not transmitter power or antenna efficiency.

Position and orientation are different

Stay broadside and rotate the receiving wire. The field is still strong, but the received signal falls because the wire responds to the electric field along its own length. A vertical dipole produces a vertically oriented field broadside. The field reverses along that same line, so its polarization is linear.

Tilt a small receiving dipole by an angle ψ away from the field and its voltage falls by cos ψ. With the same load, power falls by cos² ψ. A 60° tilt halves the voltage and leaves one quarter of the power; a 90° tilt gives an ideal null. This polarization loss is separate from the transmitting antenna’s directional null.

Why distance weakens the signal

Double the distance without changing direction. The radiated power spreads over four times the area, so power density falls to one quarter. Electric and magnetic field amplitudes each fall to one half. These are far-field rules: far enough away, the pattern no longer changes shape with distance, and the electric and magnetic fields rise and fall together.

The short-dipole equations collect those relationships. E_θ is the electric field in the plane containing the wire and observer; H_φ is the perpendicular magnetic field, directed around the wire axis. Their far-field ratio in free space is η₀ ≈ 377 Ω.

E_\theta\simeq j\eta_0\frac{k I_{\mathrm{feed}}\ell_{\mathrm{eff}}}{4\pi r}e^{-jkr}\sin\theta

H_\phi\simeq\frac{E_\theta}{\eta_0}

Here r is distance, I_feed is peak feed current and k = 2π/λ. The effective length, ℓ_eff, is about L/2 for this triangular current distribution: the average current amplitude along the wire is half the feed amplitude. Thus a longer short dipole produces a stronger field at the same feed current. The sin θ and 1/r factors describe direction and distance. The factors j and exp(−jkr) specify phase; they do not change the amplitude.

Phase notation and a field calculation

The equations use complex amplitudes for sinusoidal fields. The imaginary unit j represents a quarter-cycle phase shift; exp(−jkr) represents propagation delay. Multiply by exp(jωt) and take the real part to recover the instantaneous field, with ω = 2πf.

For a numerical example, choose 2.45 GHz. The free-space wavelength is c/f ≈ 122.4 mm, where c is the speed of light. A wire 0.1λ long has an effective length of about 6.12 mm. With 10 mA peak feed current, the peak electric field three meters broadside is 377 × 51.35 × 0.010 × 0.00612 / (4π × 3) ≈ 31.4 mV/m. The magnetic field is E/377 ≈ 83.3 µA/m. Both amplitudes halve at six meters.

Close to the wire, some energy comes back

Near the antenna, energy also moves into and out of the surrounding fields during each cycle. The electric and magnetic fields can reverse at different times, so their instantaneous power flow can briefly point toward the antenna even while the average is outward.

The following comparison uses a separate, very small ideal current element: 1 mm at 300 MHz. It isolates the effect of distance. Watch the power trace at 0.05λ cross below zero, then compare it with the trace at 1λ.

Energy flow near and far from the wire

Slow-motion visualization · Two probes broadside to the same source

Eη₀HAverage outward power density

H direction: ⊙ out of the page · ⊗ into the page.

Near 0.05λ from the source

Cycle average: 189 µW/m² outward

Farther 1λ from the source

Cycle average: 0.472 µW/m² outward

One source clock · travel delay included

Watch when E and H cross zero. Near the source they reverse at different times. Farther away the crossings nearly line up. The average flow is outward at both probes.

Compare timing across the two rows. Each row has its own scale. To compare average power density, use the values in µW/m² below the plots.

What the plots measure

The source is a 1 mm ideal current element at 300 MHz with 0.1 A peak current. Both probes are far from the element compared with its length. The average outward power density is 400 times as large at the near probe.

Multiplying H by η₀ puts it in the same units as E. Each row uses the peak E at that probe as the scale for both fields. The power trace uses its own positive peak. The dotted line marks the average over one cycle, and the horizontal arrow shows the direction of flow.

The plots show the fields and power flow at two points. Even at 1λ, the exact solution shows a brief interval of inward flow. The change from near field to far field is gradual.

For a field measurement near the wire, you need both E and H and their relative timing. The outward component of their cross product, E × H, gives instantaneous outward power density. Average it over a cycle. In the far field, their known relationship lets you use the simpler E_RMS²/377 Ω instead. A large electric-field reading close to an antenna is therefore not, by itself, evidence of strong radiation.

What changes when the wire gets longer?

So far, the current has had one simple shape. As the wire becomes a larger fraction of a wavelength, propagation along the arms and reflection from the open ends become important. They produce a standing-wave current distribution: its shape stays in place while the current oscillates. The tips remain current zeros, but longer wires can have additional zeros between them.

To put dimensions on the comparison, keep the frequency at 2.45 GHz. One wavelength in free space is 122.4 mm. A half-wave dipole is therefore 61.2 mm from tip to tip, with two quarter-wave arms. All lengths below refer to the whole dipole, not one arm.

Compare 0.25λ with 0.5λ in the explorer. Current is strongest at the feed in both. Both radiate most strongly broadside, but the half-wave beam is narrower. Next try 1λ and 1.5λ: the current peaks move, and the pattern changes much more.

Wire current and radiation pattern

Move the slider to change the wire length. The copper dot marks the feed, and the copper curve shows the current along the wire.

Total dipole length 0.50 λ · 61.2 mm

Wire current

Current at feed +1.00

Distances along the wire start at the feed.

Power cut

Thin-line 3D power surface

Selected power cutFixed 0.5λ power cut

Peak directivity2.15 dBi

Peak direction from wire90.0°

The current scale stays fixed as the wire changes length. Parts of the curve on opposite sides of the wire show current flowing in opposite directions at the same instant. Each power pattern uses its own peak as 1. The wire in the 3D view is drawn on a separate scale.

The model sets the current along the wire. In a real dipole, the feed current depends on the source and the input impedance.

Why the beam narrows—and then splits

Every region of the wire contributes to the field at a distant receiver. The result depends on both the current in that region and the time its field takes to reach the receiver.

On a full-wave dipole, current has a peak midway along each arm. The two peaks are half a wavelength apart and rise and fall together. Broadside, their fields travel essentially equal distances and reinforce each other. Move away from broadside and one field has farther to travel. Its extra delay makes the fields add less strongly.

With the strong-current regions farther apart, a smaller change in angle produces a significant difference in arrival phase. That is why the full-wave beam is narrower: its half-power points are about 24° either side of broadside, compared with 39° for a half-wave dipole.

At 1.5λ, current near the center flows opposite to current farther out. Broadside, their fields partly cancel. In other directions, different travel delays reduce that cancellation, so the fields can add more strongly. The strongest radiation moves to about 43° and 137° from the wire axis. Compare the two 1.5λ presets below and follow the bold total field.

How fields from the wire combine

Three sections of wire · one receiver direction · slow motion

Total electrical length 0.50λAngle from the wire axis 90.0°

Lower (L)Center (C)Upper (U)Total field

Each length uses its own peak current as the scale.

Diagram B. At the cursor, add the three field readings with their signs to get the bold trace. The traces from the two outer sections overlap here. Count both. Even when current is zero at the feed, current elsewhere on the wire can still radiate.

The sketch shows paths from three points on the wire. Each field trace adds up the field from its whole section. The plot subtracts the travel time from the wire center so you can compare the delays from different parts of the wire.

How the model adds the fields

For each section, the model adds the field from every point using the current there, the angle to the receiver and the travel time. The three fields and their sum share one scale at each length. That scale is proportional to the wire length in wavelengths. A negative current means it flows in the opposite direction along the same straight wire. The parallel paths end at a shared plane to show the difference in travel distance to a far-away receiver.

Longer does not always mean higher peak directivity. In this model it rises from 2.15 dBi at 0.5λ to 5.16 dBi at 1.25λ, then falls to 3.48 dBi at 1.5λ as radiation is redistributed among lobes. The ideal straight wire remains omnidirectional around its axis and linearly polarized wherever its field is nonzero.

Each displayed power pattern is scaled to its own peak. It tells you where the antenna radiates most strongly, not how many watts it radiates. That second question needs the feed impedance and losses.

The model behind the curves

For a thin straight wire, the following sinusoidal current is a useful approximation. L is total length, z is distance along the wire from the feed, and I_s sets the current scale. With k = 2π/λ, the expression gives zero current at both tips. For a very short wire it approaches the triangular shape we started with.

I(z)=I_s\sin\!\left[k\left(\frac{L}{2}-|z|\right)\right]

-\frac{L}{2}\leq z\leq\frac{L}{2}

Adding the fields from the whole wire, including their travel delays, gives the finite-dipole field:

E_\theta\simeq\frac{j\eta_0 I_s e^{-jkr}}{2\pi r}F(\theta)

F(\theta)=\frac{\cos[(kL/2)\cos\theta]-\cos(kL/2)}{\sin\theta}

F(θ) describes variation with direction; |F|² gives the power-pattern shape. E_θ, r and η₀ have the same meanings as before, and H_φ = E_θ/η₀ in the far field. I_s equals feed current only at particular lengths, including 0.5λ. The model prescribes current; it does not establish how much a real transmitter can drive into the antenna.

Check the half-wave field at broadside

For L = λ/2 and θ = 90°, F = [cos 0 − cos(π/2)]/1 = 1. With 10 mA peak feed current and r = 3 m, the peak electric field is 377 × 0.010 / (2π × 3) ≈ 0.200 V/m. This is larger than the short-wire result at the same feed current. It is not a comparison at equal transmitter power.

What does the transmitter have to drive?

The radiation pattern assumes a current distribution. The transmitter has to produce that current. How much voltage is needed, and does the current rise at the same time as the voltage? The antenna’s input impedance, Z = R + jX, answers both questions at the feed.

The resistance R accounts for average power leaving the circuit as radiation or heat. Radiation resistance represents the radiated part; it is not a physical resistor heating the wire. Loss resistance represents heat. Reactance X describes the effect of energy stored in the fields and returned during the cycle.

Why a short wire looks capacitive

Remember the charge building up on the two arms. More separated charge means more voltage across the gap. Current adds charge fastest before the charge—and therefore the voltage—reaches its peak. This is capacitor-like behavior, giving negative reactance. A very short dipole can require a high feed voltage for a small current.

Lengthen the wire toward half a wavelength and its reactance moves toward zero. Where X = 0, voltage and current are in phase: the antenna is resonant. Fields extend beyond the wire tips, making the antenna electrically a little longer than its physical length suggests. The first resonance is consequently slightly below a geometric half wavelength. Wire diameter, feed geometry and nearby materials affect the exact value. End effects.

Resonance is not the same as a match

A 50 Ω feed line reflects no power only when it sees 50 + j0 Ω. Removing the reactance is only part of that requirement. A resonant antenna with a different resistance still reflects some power.

The voltage standing-wave ratio (VSWR) describes the voltage maxima and minima formed by forward and reflected waves on the line. A value of 1:1 means no reflection; larger values mean more reflection. Compare First resonance with Exactly 0.5λ below. This reference calculation uses a lossless wire with a fixed radius of 0.0001λ to isolate the effect of length.

Resonance and a 50 Ω match

Thin-wire calculation · wire radius 0.0001λ · 50 Ω line

Total electrical length 0.5000λ

Full Length Range

Near First Resonance · Enlarged

1 · Resonance · 0.4846λ2 · Half-wave · 0.5000λ

Outlined points follow your selection. A 50 Ω match needs R = 50 Ω and X = 0 together.

Input impedance73.1 + j42.5 Ω

VSWR on 50 Ω2.182:1

Incident power accepted86.2%

Positive reactance means the dipole looks inductive at the feed. The model keeps the radius fixed as a fraction of a wavelength. In a frequency sweep on a real wire, the wavelength changes while the radius stays the same.

At first resonance, the model gives 66.7 + j0 Ω. At exactly half a wavelength it gives 73.1 + j42.5 Ω: already inductive. These are different conditions, and neither is a perfect 50 Ω match. The half-wave wire accepts about 86.2% of incident power.

The lowest VSWR occurs very slightly below resonance in this calculation. The difference is small—1.332:1 versus 1.335:1—but the distinction matters: find resonance using zero reactance, and find the best match using minimum reflection.

Calculate reflection, accepted power and VSWR

For Z = R + jX connected to a 50 Ω line, Γ is reflected voltage divided by forward voltage. Its squared magnitude is the reflected power fraction, so the remaining fraction enters the antenna:

|\Gamma|^2=\frac{(R-50\,\Omega)^2+X^2}{(R+50\,\Omega)^2+X^2},\qquad P_{\mathrm{acc}}=P_{\mathrm{inc}}(1-|\Gamma|^2)

\mathrm{VSWR}=\frac{1+|\Gamma|}{1-|\Gamma|}

For 73.1 + j42.5 Ω, |Γ| ≈ 0.371, giving VSWR ≈ 2.18:1 and accepted power ≈ 86.2% of incident power. These quantities describe the same connection; none tells you how much accepted power becomes heat.

Why a full-wave dipole is hard to feed at the center

At 1λ, the current plot has strong peaks along the arms but a minimum at the feed. A small feed current can therefore coexist with strong current elsewhere. Since input impedance is feed voltage divided by feed current, this center connection has high impedance and is a poor direct match to 50 Ω.

The ideal sinusoidal model gives exactly zero feed current at 1λ and 2λ. Real antennas have high, finite impedance peaks near these lengths; wire thickness and the feed gap prevent the ideal infinity. A narrower beam alone is therefore not a reason to choose a full-wave dipole. You also need a practical way to feed it. Full-wave measurements.

From transmitter power to a working antenna

An antenna can be efficient with the power it accepts and still radiate very little of the power arriving from the transmitter. Our short dipole makes that distinction especially clear.

At 2.45 GHz, take a copper dipole 12.2 mm tip to tip (0.1λ), made from 0.25 mm diameter wire. Connect it directly to a lossless 50 Ω line with 1 W incident at the feed. The short-wire model predicts that almost all of that watt reflects. Only 0.334 mW enters the antenna; 0.323 mW radiates and 0.011 mW heats the wire.

Its radiation efficiency is nevertheless about 97%: radiated power divided by accepted power. Efficiency excludes the power that never entered the antenna. This is why matching and efficiency must be checked separately.

What makes a short antenna difficult?

Radiation resistance falls roughly with the square of electrical length. Halve a short dipole’s length at the same frequency and feed current, and it radiates about one quarter as much power. The copper loss resistance falls only in proportion to length at fixed diameter. As the wire gets shorter, a greater share of accepted power therefore becomes heat.

For this wire, radiation resistance is about 1.97 Ω, copper loss resistance 0.067 Ω, and reactance −1104 Ω. The copper loss is small. The large negative reactance is the immediate problem: the feed impedance, 2.04 − j1104 Ω, is very far from 50 Ω.

Calculate radiation resistance, copper loss and reactance

Use total length L = 0.1λ = 12.236 mm, radius a = 0.125 mm and copper conductivity σ = 5.8 × 10⁷ S/m. For the triangular current distribution, radiation resistance referred to the feed is

R_{\mathrm{rad}}\simeq\frac{\eta_0\pi}{6}\left(\frac{L}{\lambda}\right)^2\approx1.97\,\Omega.

At RF, current is concentrated near the copper surface. With free-space permeability μ₀ ≈ 4π × 10⁻⁷ H/m, skin depth δ and surface resistance R_s are

\delta=\frac{1}{\sqrt{\pi f\mu_0\sigma}}\approx1.34\,\mu\mathrm{m},\qquad R_s=\frac{1}{\sigma\delta}\approx0.0129\,\Omega.

Integrating the heat loss along the triangular current profile gives

R_{\mathrm{wire}}\simeq\frac{R_sL}{6\pi a}\approx0.067\,\Omega,\qquad \eta_{\mathrm{rad}}=\frac{R_{\mathrm{rad}}}{R_{\mathrm{rad}}+R_{\mathrm{wire}}}\approx96.7\%.

For a thin, straight, short wire with a small feed gap in free space, the approximate capacitive reactance is

X\simeq-\frac{120\,\Omega}{\pi L/\lambda}\left[\ln\left(\frac{L}{2a}\right)-1\right]\approx-1104\,\Omega.

Here ln is the natural logarithm. Reactance formula and assumptions. Use Z = R_rad + R_wire + jX in the reflection calculation: the accepted fraction is about 0.000334. Multiplying 1 W by that fraction and by 0.967 gives 0.323 mW radiated. These are estimates for the stated geometry, not measured hardware performance.

Matching helps—but the components can waste power

A series coil can supply positive reactance to cancel the antenna’s negative reactance. For this teaching model, a 71.7 nH coil does so at 2.45 GHz. Give it a quality factor Q = 100: Q is reactance divided by series loss resistance at this frequency, so the coil adds about 11 Ω of loss. That is much larger than the antenna’s radiation resistance.

Adding the coil lets more power enter, raising radiation to about 99 mW. An ideal transformer can then match the remaining resistance to 50 Ω. The complete assembly accepts the whole watt and radiates about 151 mW. These are circuit-model results; a real coil’s parasitic capacitance and self-resonance matter at 2.45 GHz. Follow the three connections below; most of the newly accepted power heats the coil.

Power at three feed connections

Each bar starts with 1 W arriving at the connection shown. The 50 Ω line has no loss.

Bare wireAt the wire terminals

Radiated: 0.323 mW
Wire heat: 0.011 mW
Coil heat: 0 mW
Reflected: 999.7 mW

Wire and series coilAt the coil and wire input

Radiated: 99.14 mW
Wire heat: 3.371 mW
Coil heat: 555 mW
Reflected: 342.5 mW

Wire, coil and transformerAt the transformer input

Radiated: 150.8 mW
Wire heat: 5.127 mW
Coil heat: 844.1 mW
Reflected: 0 mW

Compare the last two bars. The transformer helps the antenna and coil accept more power. In both cases, 15.1% of the accepted power radiates.

Size the tuning coil and include its loss

The coil must supply X_coil = −X_antenna. Its inductance and loss are

L_{\mathrm{coil}}=\frac{X_{\mathrm{coil}}}{2\pi f},\qquad R_{\mathrm{coil}}=\frac{X_{\mathrm{coil}}}{Q}.

Using 1104 Ω at 2.45 GHz gives 71.7 nH and, with Q = 100, about 11 Ω loss. Include that loss in the efficiency: R_rad/(R_rad + R_wire + R_coil) ≈ 15.1%. Even a lossless tuning coil would leave about 2.04 Ω of resistance, so canceling reactance alone would still not give a 50 Ω match.

The coil and transformer are circuit models. At 2.45 GHz, a real 71.7 nH component may be limited by its self-resonance and parasitic capacitance. Check component data or measured impedance; the example is not a component recommendation.

A match at one frequency is not a bandwidth

Keep the network fixed and change frequency. Below the tuned frequency, the coil no longer cancels all the capacitive reactance; above it, the combination becomes inductive. Reflection grows on either side. A short antenna stores substantial field energy relative to the power it radiates, which makes a low-loss match sensitive to frequency. This sweep holds inductance, Q and the ideal transformer ratio fixed.

Define the usable match bandwidth here as the range with VSWR ≤ 2:1. At that limit, one ninth of incident power reflects. Sweep the frequency to find the two edges.

VSWR at the transformer input

Change the frequency without retuning the coil or transformer. The model uses Q = 100 across the sweep.

VSWR

1:12:13:14:15:1

2.4252.4502.475

Frequency (GHz)

Frequency 2.4500 GHzVSWR 1.00:1

The two dots mark VSWR = 2. The shaded band runs from 2.4398 to 2.4603 GHz, a width of 20.5 MHz or 0.84% of 2.45 GHz.

This model gives about 20.5 MHz, or 0.84% of 2.45 GHz. That is a match bandwidth, not a guarantee of useful radiation throughout it. Loss can broaden a good-looking match by absorbing power. Check radiation efficiency and directional performance as well as VSWR.

If there is room, start near half a wavelength

For a first 2.45 GHz prototype, a straight dipole near half a wavelength is easier to feed than our very short wire. Half a wavelength is 61.2 mm total. Allowing for end effects gives a slightly shorter resonant length. For wire diameter d, this estimate gives a useful first cut:

L_{\mathrm{res}}\simeq\frac{\lambda}{2}\left[1-\frac{0.2257}{\ln(\lambda/d)-1}\right]

With 0.5 mm diameter wire, the estimate is 58.1 mm tip to tip, about 29 mm per arm with a small feed gap. Start slightly long and trim both arms equally. Shortening the wire raises its first resonant frequency. The estimate assumes a straight dipole in free space; supports, an enclosure and nearby metal or ground can change both resonance and pattern. First-resonance estimate.

Measure the antenna, not just a good-looking match

Measure with the final feed and mounting in place. A vector network analyzer (VNA) measures reflection; its calibration determines where that reflection is reported. Calibrate at the cable connector that will attach to the antenna. Anything beyond that connector—including an adapter or balun—remains part of the measurement.

Cable loss can be deceptive. Suppose the antenna has 10 dB return loss, meaning 10% of incident power reflects. Add a matched cable with 3 dB one-way loss, but leave the calibration at the VNA port. The return loss now reads about 16 dB: the signal loses 3 dB going out and another 3 dB coming back. The antenna has not improved. Less power reaches it, and less reflection returns to the instrument.

The feed cable can become part of the antenna

Inside the coax, signal current flows on the center conductor and the inner surface of the shield in opposite directions. At the dipole, the center conductor connects to one arm and the shield to the other. That connection can also drive common-mode current on the shield’s outer surface. The cable then radiates too, changing the pattern you intended to build.

Follow the two paths below. A current balun or choke at the feed can reduce the outside current.

Current inside and outside the coax

Interactive connection schematic · arrows show one illustrative instant

Current can flow on the outside of the shield while the coax carries the signal inside. A choke at the feed can reduce that outside current. Measure it at the frequency you plan to use.

If moving the cable changes the VNA trace, investigate the feed arrangement. To test a choke, use an RF current probe around the whole coax and compare readings with and without it. Keep the frequency, drive level, probe position and cable routing fixed. The internal currents cancel in that measurement; the probe detects the outside current. A change in reflection alone does not tell you whether the choke worked.

Finally, check radiation toward the intended receiver. Realized gain combines the pattern, radiation efficiency and input mismatch: it describes directional radiation relative to the power incident at the antenna’s input. Length helps establish the current distribution; the pattern tells you where radiation goes; impedance determines what the feed line delivers; losses determine what becomes heat. A useful antenna has to satisfy all of them in its actual installation.