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.
Formula
Reference: Balanis, Antenna Theory, 4th ed. (2016), §4.5, §4.7 and §8.5; Stutzman & Thiele, Antenna Theory and Design, 3rd ed. (2012)
How It Works
A monopole is a single vertical conductor of height 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 , 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:
The dipole is taken from the induced-EMF method with a sinusoidal current, as Balanis gives it, with , , wire radius and the sine and cosine integrals Si and Ci:
Short whips and base loading
Below a quarter wave the input resistance falls quickly, towards , and the reactance becomes large and capacitive. A loading coil at the base cancels it:
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 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
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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