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RF

Wavelength & Frequency Calculator

Convert between frequency, wavelength, and wavenumber in free space or any dielectric. Calculate half-wave and quarter-wave lengths for antenna design. Free, instant results.

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Formula

λ=cfεr\lambda = \frac{c}{f\sqrt{\varepsilon_r}}

Reference: Balanis, "Antenna Theory" 3rd ed.

λ— Wavelength in medium (m)
c— Speed of light (299.792458 mm/ns) (m/s)
f— Frequency (Hz)
εᵣ— Relative permittivity of medium

How It Works

This calculator converts wavelength to frequency for RF engineers, optical designers, and physicists working across the electromagnetic spectrum from radio waves to gamma rays. The fundamental relationship f = c / lambda uses c = 299,792,458 m/s exactly (SI Brochure 9th edition, 2019 - a defined constant with zero uncertainty). This spans 15 orders of magnitude: radio waves (lambda = 1 km, f = 300 kHz) to visible light (lambda = 500 nm, f = 600 THz) to X-rays (lambda = 0.1 nm, f = 3 × 10^18 Hz). In dielectric media, the effective wavelength shortens: lambda_eff = lambda_0 / sqrt(epsilon_r). FR-4 PCB (epsilon_r = 4.3) reduces wavelength to 48% of free-space, critical for microstrip filter and antenna design per IPC-2141.

Worked Example

Problem

A 5.8 GHz ISM-band microstrip filter is designed on FR-4 substrate (epsilon_r = 4.2). Calculate the free-space wavelength, effective wavelength, and quarter-wave stub length.

Solution
  1. Free-space wavelength: lambda = c/f = 299,792,458 / (5.8 × 10^9) = 51.69 mm
  2. Velocity factor: VF = 1/sqrt(4.2) = 0.488
  3. Effective wavelength: lambda_eff = 51.69 × 0.488 = 25.22 mm
  4. Quarter-wave stub: lambda_eff/4 = 25.22/4 = 6.31 mm
  5. Per IPC-2141: traces > lambda_eff/10 = 2.52 mm need impedance control
  6. Physical stub length with fringing: ~6.0 mm (5% shorter due to edge effects)

Practical Tips

  • ✓Quick formula: f_GHz = 300/lambda_mm for free space (0.07% error). Reverse: lambda_mm = 300/f_GHz. At optical wavelengths: f_THz = 300/lambda_um
  • ✓Per Rogers Corp application notes: measure substrate epsilon_r at your operating frequency - FR-4 varies from 4.7 at 100 MHz to 4.2 at 10 GHz due to dielectric dispersion
  • ✓For precision RF: use vector network analyzer to measure actual electrical length rather than calculating from nominal epsilon_r; substrate thickness tolerance of +/-10% causes +/-5% wavelength error

Common Mistakes

  • ✗Using approximation c = 3 × 10^8 instead of exact 299,792,458 m/s - this 0.069% error causes 35 um positioning error per 50 mm at mmWave, exceeding typical PCB tolerances of +/-25 um
  • ✗Neglecting dielectric constant in PCB calculations - assuming free-space wavelength on FR-4 makes a quarter-wave stub 2.05x too long, causing resonance at 2.83 GHz instead of 5.8 GHz
  • ✗Mixing up effective permittivity with bulk permittivity - microstrip epsilon_eff depends on geometry; a 50-ohm trace on FR-4 has epsilon_eff = 3.3, not 4.3

Frequently Asked Questions

Wavelength is inversely proportional to frequency: lambda = c/f. Doubling frequency halves wavelength. Per SI: at 1 GHz lambda = 299.8 mm, at 2 GHz lambda = 149.9 mm, at 10 GHz lambda = 30.0 mm. This inverse relationship is why high frequencies enable smaller antennas but suffer higher path loss (Friis: loss proportional to f^2).
Wave velocity in a medium is v = c/sqrt(epsilon_r × mu_r). For non-magnetic materials (mu_r = 1), velocity factor VF = 1/sqrt(epsilon_r). Since lambda = v/f, wavelength scales by VF. Per IPC-2141: FR-4 epsilon_r = 4.3 gives VF = 0.48, so a 2.4 GHz signal has lambda_eff = 60 mm instead of 125 mm free-space.
Yes, with the correct dielectric constant. Common values per IEEE/IPC: vacuum/air epsilon_r = 1.0, FR-4 = 4.3, Rogers RO4350B = 3.66, PTFE = 2.1, silicon = 11.7, GaAs = 12.9. For waveguides, use cutoff-adjusted wavelength: lambda_g = lambda_0/sqrt(1 - (f_c/f)^2).
Wavelength governs all RF physical dimensions: antenna elements (dipole = lambda/2 per Balanis), transmission line stubs (quarter-wave = lambda/4), filter cavities (half-wave resonator), and PCB layout rules (IPC-2141: impedance control for traces > lambda/10). At 28 GHz 5G, lambda = 10.7 mm, so even 1 mm traces are electrically significant.
The value 299,792,458 m/s is exact by SI definition (2019 redefinition). The meter is now defined as the distance light travels in 1/299,792,458 second, making c a defined constant with zero uncertainty. This provides 9-digit precision for all wavelength calculations, far exceeding typical manufacturing tolerances.
lambda = c/f = 299,792,458 / 2.4e9 = 124.9 mm in free space. On FR-4 PCB (epsilon_r = 4.3): lambda_eff = 60.2 mm. Per IPC-2141: traces > 6 mm at 2.4 GHz need impedance control. Half-wave dipole = 62.4 mm per element; quarter-wave patch on FR-4 = 15.0 mm. WiFi channel spacing of 5 MHz corresponds to 0.26 mm wavelength difference.
In free space: lambda_mm = 300/f_GHz (0.07% approx). In medium: lambda_eff = lambda_0/sqrt(epsilon_r). Velocity factor VF = 1/sqrt(epsilon_r): FR-4 VF = 0.48, Rogers RO4003C VF = 0.53, foam coax VF = 0.83, solid PE coax VF = 0.66 (per Belden specs). For microstrip, use effective permittivity which is geometry-dependent, typically 60-80% of bulk epsilon_r.

Methodology & References

References

  • Microwave Engineering, 4th ed. — David M. Pozar (2011), Chapter 1.3 — Wavelength and wave velocity
  • ITU Radio Regulations — International Telecommunication Union (2020), Article 2 — Frequency and wavelength definitions link

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