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Antenna

Monopole / Whip Antenna Calculator

Input impedance, radiation resistance and directivity of a monopole or whip over a ground plane at any height, from the induced-EMF model by image theory, plus the quarter-wave length and the base-loading inductance a short whip needs.

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Formula

Zinmono(h)=12Zindip(2h)=Rr+jXm2sin⁡2(kh),Dmono=2Ddip=4FmaxQ,L=−Xin2πfZ_{in}^{mono}(h) = \tfrac{1}{2} Z_{in}^{dip}(2h) = \frac{R_r + jX_m}{2\sin^2(kh)},\quad D^{mono} = 2D^{dip} = \frac{4F_{max}}{Q},\quad L = \frac{-X_{in}}{2\pi f}

Reference: Balanis, Antenna Theory, 4th ed. (2016), §4.5, §4.7 and §8.5; Stutzman & Thiele, Antenna Theory and Design, 3rd ed. (2012)

h— Element height above the ground plane (m)
k— Free-space wavenumber 2π/λ (rad/m)
R_r, X_m— Image dipole's resistance and reactance at the current maximum (induced EMF, Si/Ci) (Ω)
Q— Radiation integral of the image dipole, in Si and Ci
F_max— Peak of the dipole pattern factor
L— Base-loading inductance when X_in is capacitive (H)

How It Works

A monopole is a single vertical conductor of height hh fed against a ground plane: a car whip, a handheld's rubber duck over its own case, a quarter-wave ground-plane antenna on a mast. Over a perfectly conducting, infinite plane the ground acts as a mirror. The current on the element and its image form a centre-fed dipole of length 2h2h, so the monopole radiates into the upper half-space exactly as that dipole does.

That image gives the whole calculation. The monopole has half the dipole's input impedance and twice its directivity, because the same field fills half the space:

Zinmono(h)=12 Zindip(2h),Dmono=2 DdipZ_{in}^{\mathrm{mono}}(h) = \tfrac{1}{2}\,Z_{in}^{\mathrm{dip}}(2h), \qquad D^{\mathrm{mono}} = 2\,D^{\mathrm{dip}}

The dipole is taken from the induced-EMF method with a sinusoidal current, as Balanis gives it, with k=2π/λk = 2\pi/\lambda, l=2hl = 2h, wire radius aa and the sine and cosine integrals Si and Ci:

Rr=η2π[γ+ln⁡kl−Ci(kl)+12sin⁡kl (Si(2kl)−2 Si(kl))+12cos⁡kl (γ+ln⁡kl2+Ci(2kl)−2 Ci(kl))]R_r = \frac{\eta}{2\pi}\Big[\gamma + \ln kl - \mathrm{Ci}(kl) + \tfrac{1}{2}\sin kl\,\big(\mathrm{Si}(2kl) - 2\,\mathrm{Si}(kl)\big) + \tfrac{1}{2}\cos kl\,\big(\gamma + \ln\tfrac{kl}{2} + \mathrm{Ci}(2kl) - 2\,\mathrm{Ci}(kl)\big)\Big]
Xm=η4π[2 Si(kl)+cos⁡kl (2 Si(kl)−Si(2kl))−sin⁡kl (2 Ci(kl)−Ci(2kl)−Ci(2ka2/l))]X_m = \frac{\eta}{4\pi}\Big[2\,\mathrm{Si}(kl) + \cos kl\,\big(2\,\mathrm{Si}(kl) - \mathrm{Si}(2kl)\big) - \sin kl\,\big(2\,\mathrm{Ci}(kl) - \mathrm{Ci}(2kl) - \mathrm{Ci}(2ka^2/l)\big)\Big]
Zindip=Rr+jXmsin⁡2(kl/2)Z_{in}^{\mathrm{dip}} = \frac{R_r + jX_m}{\sin^{2}(kl/2)}
RrR_r and XmX_m are referred to the current maximum. Dividing by sin⁡2(kh)\sin^2(kh) refers them to the base, where the feed is. For a quarter-wave monopole the two are the same, and the result is the familiar 36.5+j21.3 Ω36.5 + j21.3\ \Omega and 5.16 dBi: half the half-wave dipole's 73.1+j42.5 Ω73.1 + j42.5\ \Omega, twice its directivity of 1.64. The positive reactance is why a resonant quarter-wave whip is cut a few percent shorter than λ/4\lambda/4; the thicker the wire, the more it must be shortened.

Short whips and base loading

Below a quarter wave the input resistance falls quickly, towards Rin≈40π2(h/λ)2R_{in} \approx 40\pi^2 (h/\lambda)^2, and the reactance becomes large and capacitive. A loading coil at the base cancels it:

L=−Xin2πfL = \frac{-X_{in}}{2\pi f}

The calculator reports that inductance whenever the reactance is capacitive, and zero when the element is already inductive, which calls for shortening instead.

Validity

The model assumes a thin wire: the conductor radius should not exceed 0.0159 λ, the thin-wire limit 2πa/λ≤0.12\pi a/\lambda \le 0.1 that NEC-2 also uses. Near a height of half a wavelength, or any multiple of it, the base current vanishes and the modelled impedance grows without bound. Within 0.08 λ of such a height the calculator warns that the impedance is not a design value, and within 0.01 λ it does not report one at all. The ground is perfect and infinite. Radials, finite ground planes, car roofs and lossy earth are not modelled; for those, open the antenna simulator from this page. The velocity factor only shortens the quarter-wave cutting length; the impedance is computed for the element as entered.

Worked Example

Problem: A 27 MHz CB whip has to be shortened to 1.2 m of 10 mm tube on a car roof. What does the base see, and what loading coil resonates it?

Step 1 - Electrical height: λ = c/f = 299792458 / 27e6 = 11.103 m h/λ = 1.2 / 11.103 = 0.1081

Step 2 - Input impedance, from the image dipole of 2.4 m (induced EMF): Z_in = 4.909 − j327.73 Ω

Step 3 - Cross-check against the short-whip limit: 40π² × 0.1081² = 4.61 Ω At 0.108 λ the element is no longer very short, so the full model is 6.5% higher.

Step 4 - Base-loading inductance: L = 327.73 / (2π × 27e6) = 1.932 µH

Step 5 - Directivity over perfect ground: D = 4.84 dBi, against 5.16 dBi for a quarter wave

Step 6 - Compare a full-size quarter wave of the same tube, 2.65 m: Z_in = 31.96 − j3.58 Ω, which needs only 0.0211 µH The quarter-wave cutting length at a velocity factor of 0.95 is 2637.1 mm.

The shortened whip trades a 32 Ω, nearly resonant feed for 4.9 Ω behind a 1.93 µH coil. The coil's own loss resistance and the car body's ground losses add to the 4.9 Ω, so efficiency, not the match, is what shortening costs.

Practical Tips

  • ✓Use the electrical height h/λ as the first check: below about 0.1 λ the resistance is only a few ohms and efficiency, not matching, limits the design.
  • ✓Put the loading coil at the base for the simplest build, or higher up the whip for more radiation resistance; this calculator models base loading only.
  • ✓Four quarter-wave radials sloping downward at about 45° bring a ground-plane antenna's resistance up from 36 Ω towards 50 Ω.
  • ✓Open the antenna simulator from this page to replace the perfect ground with real soil or radials, and to sweep the element across the band.

Common Mistakes

  • ✗Assuming the car roof or a few radials is the infinite perfect ground of the model. A finite or lossy ground raises the input resistance, adds loss and tilts the pattern upward; simulate the real ground before trusting the figures.
  • ✗Cutting a quarter-wave whip to exactly λ/4. A quarter-wave element is inductive, about +21 Ω for a thin wire, so it resonates a few percent shorter, and more so as the wire gets thicker.
  • ✗Ignoring the loading coil's loss. A short whip's radiation resistance is a few ohms, so a coil with a loss resistance of an ohm or two already wastes a large fraction of the power.
  • ✗Designing a half-wave vertical from this model. Near h = λ/2 the base current vanishes and the induced-EMF impedance is unbounded; real half-wave verticals are fed through a matching network whose values must be measured or simulated.

Frequently Asked Questions

Over a perfect ground plane the element and its mirror image form a dipole twice as long. The monopole drives only the upper half of that dipole with the same current, so it needs half the voltage, which halves the impedance. It also puts all its power into half the space, which doubles the directivity.
About 36.5 + j21.3 ohms over a perfect ground plane. The positive reactance means the element is slightly long for resonance; shortening it by a few percent removes the reactance and leaves a little over 30 ohms in this model.
Find the input reactance at the base; for a whip shorter than a quarter wave it is negative, or capacitive. The coil that resonates it has an inductance equal to minus the reactance divided by two pi times the frequency. The calculator reports that value directly.
No. It assumes a perfectly conducting, infinite ground plane. Radials, finite ground planes, vehicle bodies and lossy earth all change the impedance and pattern. The antenna simulator linked from this page can model real ground.
At a height of half a wavelength the current at the base falls to zero in the sinusoidal-current model, so the predicted impedance becomes infinite. Real antennas show a high but finite impedance there that depends on the wire thickness and the feed gap, which this model does not include, so the calculator warns near that height and withholds the value within 0.01 wavelength of it.

Methodology & References

References

  • Antenna Theory: Analysis and Design, 4th ed. — Constantine A. Balanis (2016), §4.5 finite dipole, §4.7 ground effects and image theory, §8.5 induced-EMF self-impedance
  • Antenna Theory and Design, 3rd ed. — Warren L. Stutzman & Gary A. Thiele (2012), ch. 3 — monopoles and image theory
  • NIST Handbook of Mathematical Functions — F. W. J. Olver et al. (eds.), NIST and Cambridge University Press (2010), §6.2 and §6.6 — sine and cosine integrals (online as DLMF ch. 6) link

Si and Ci are checked against tabulated values to 1e-9. The quarter-wave impedance is half the half-wave dipole's, and a short whip tends to 40π²(h/λ)². For radials, a finite ground or real soil, open the antenna simulator.

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