Wavelength Calculator

Wavelength Calculator Tutorial: Step-by-Step Analysis of Wavelength Problems

1. How the Wavelength Calculator Works

Calculator inputs and outputs

Wavelength equation

A time-harmonic plane wave in a lossless, isotropic region travels at v = 1/√(με) and repeats every wavelength λ = v/f.^[1][2][5] Writing μ = μ_rμ₀ and ε = ε_rε₀ gives the equation the calculator uses:

The half- and quarter-wavelength outputs are Equation 1 divided by 2 and 4:

Velocity and velocity factor

The velocity output is the phase velocity of the plane wave in the region. Dividing it by c gives the velocity factor (VF) quoted on coaxial cable data sheets:^[1][4]

Wave impedance

The characteristic impedance output is the intrinsic (wave) impedance of the region: the ratio of the electric to magnetic field of a plane wave.^[1][2] In vacuum it is η₀ ≈ 376.730 Ω.^[7]

Two features of these equations drive every problem below. Wavelength and velocity depend on the product ε_rμ_r, while impedance depends on the ratio μ_r/ε_r. In free space, Equation 1 reduces to a quick rule of thumb:

2. Tutorial 1: Find the Free-Space Wavelength at 2.45 GHz

Problem: A 2.45 GHz ISM-band system (Wi-Fi, Bluetooth or a microwave oven) radiates into air. What are the wavelength, half-wavelength, quarter-wavelength, wave impedance and phase constant?

1. Convert the frequency to hertz. The calculator's frequency field is in Hz. Enter f = 2.45E9 (2,450,000,000 Hz). Entering 2.45 instead would describe a 2.45 Hz wave about 122,000 km long.
2. Enter the material properties. For air or vacuum, enter ε_r = 1 and μ_r = 1. Dry air has ε_r ≈ 1.0006, which changes λ by only 0.03%.
3. Read the outputs. The calculator returns the values below.

   Calculator output: 2.45 GHz, ε_r = 1, μ_r = 1

   Output	Calculator value	In practical units
   Wavelength λ	0.1223643 m	122.36 mm
   Half-wavelength λ/2	0.0611821 m	61.18 mm
   Quarter-wavelength λ/4	0.0305911 m	30.59 mm
   Wave impedance η	376.730 Ω	≈ 377 Ω
   Velocity v	2.99792×108 m/s	VF = 1.000

4. Verify the wavelength by hand. Substitute into Equation 1:
   Equation 6 — Wavelength check at 2.45 GHz in air $$\lambda =\frac{2.99792\times {10}^8}{(2.45\times {10}^9)\sqrt{\left(1\right)\left(1\right)}}=0.12236\text{ m}$$ λ = (2.99792×10^8) / [(2.45×10^9)·√(1·1)] = 0.12236 m
   The shortcut in Equation 5 agrees: 299.79 / 2.45 = 122.36 mm.
5. Compute the phase constant if you need phase shift per unit length. The wavenumber (phase constant) converts any distance into phase:^[1][5]
   Equation 7 — Wavenumber (phase constant) [1, Sec. 1.4], [5, Ch. 8] $$k=\frac{2\pi }{\lambda }=\omega \sqrt{\mu \epsilon }=\frac{2\pi }{0.12236}=51.35\text{ rad/m}$$ k = 2π/λ = ω·√(μ·ε) = 2π / 0.12236 = 51.35 rad/m
   A 10 mm path difference between two antennas is therefore 51.35 × 0.010 = 0.514 rad, or 29.4°.
6. Work the problem in reverse when the length is fixed. To find the frequency at which a given length equals one wavelength, solve Equation 1 for f:
   Equation 8 — Frequency for a given wavelength (Eq. 1 rearranged) $$f=\frac{c}{\lambda \sqrt{{\epsilon }_r{\mu }_r}}$$ f = c / (λ·√(ε_r·μ_r))
   Enter the trial frequency in the calculator to confirm the result.

Result: At 2.45 GHz in air, λ = 122.36 mm, λ/2 = 61.18 mm and λ/4 = 30.59 mm. These set the scale of 2.45 GHz antennas, the spacing of standing-wave hot spots in a microwave oven cavity (λ/2), and the minimum far-field distance estimates used in antenna measurements.

3. Tutorial 2: Size a Quarter-Wave Monopole and Half-Wave Dipole

Problem: Build a quarter-wave ground-plane monopole and a half-wave dipole for the center of the 144–148 MHz (2 m) amateur band. What are the theoretical and practical element lengths?

1. Pick the design frequency. Use the band center, f = 146 MHz. Enter 146E6 in the calculator.
2. Enter free-space properties. The wire radiates into air, so enter ε_r = 1 and μ_r = 1. Wire insulation is handled by the shortening factor in step 4, not by ε_r.
3. Read the theoretical lengths. The calculator returns λ = 2.053373 m, λ/2 = 1.026687 m and λ/4 = 0.513343 m. Repeating for the band edges gives the tuning range:

   Calculator output across the 2 m band (air)

   Frequency	λ (m)	λ/2 (mm)	λ/4 (mm)
   144 MHz	2.081892	1,040.9	520.5
   146 MHz	2.053373	1,026.7	513.3
   148 MHz	2.025625	1,012.8	506.4

4. Apply the end-effect shortening factor. A real wire has finite diameter and fringing fields at its ends, so it resonates when it is slightly shorter than λ/2 (dipole) or λ/4 (monopole).^[3][9] Multiply by a shortening factor K:
   Equation 9 — Practical resonant element length [3, Ch. 4], [9] $${\ell }_{\text{monopole}}\approx K\frac{\lambda }{4},{\ell }_{\text{dipole}}\approx K\frac{\lambda }{2},K\approx 0.95–0.98$$ ℓ_monopole ≈ K·λ/4; ℓ_dipole ≈ K·λ/2; K ≈ 0.95–0.98 (thicker conductors need smaller K)
   For the 146 MHz monopole, the practical length is 0.95 × 513.3 = 487.7 mm to 0.98 × 513.3 = 503.1 mm. For the dipole, start near 0.95 × 1,026.7 = 975 mm tip to tip.
5. Cut long, then trim. Cut the element at the long end of the range, measure the resonant frequency with an antenna analyzer, and trim. If an element of length ℓ resonates too low, the quarter-wave relation in Equation 8 tells you how far to move: a 500.0 mm monopole is a quarter-wave at c/(4 × 0.500) = 149.9 MHz before end effects. Each 1 mm of trim raises resonance by roughly 0.2% (about 0.3 MHz here).
6. Check the feed impedance. A thin resonant half-wave dipole has an input impedance near 73 Ω, and a quarter-wave monopole on a large ground plane is about half that, near 36.5 Ω.^[3] Sloping the ground radials downward raises the monopole impedance toward 50 Ω.

Note: The 376.73 Ω impedance output is the wave impedance of free space. It is not the input impedance of the antenna. Use the antenna input impedances in step 6 when matching to a feed line.

4. Tutorial 3: Cut a Coaxial Quarter-Wave Stub

Problem: A shorted quarter-wave stub of solid-polyethylene coaxial cable (ε_r ≈ 2.25) must notch a 100 MHz FM interferer. How long is the stub, what is its velocity factor, and what delay does it add?

1. Find the dielectric constant of the cable. A coaxial TEM line carries its fields entirely in the insulating dielectric, so the plane-wave wavelength in that dielectric is the wavelength on the line.^[1][4] Solid polyethylene has ε_r ≈ 2.25 and μ_r = 1. For foam or PTFE cable, use the data sheet velocity factor to back out ε_r = 1/VF².
2. Enter the inputs. Enter f = 100E6, ε_r = 2.25, μ_r = 1.
3. Read the outputs.

   Calculator output: 100 MHz, solid polyethylene vs. air

   Output	Polyethylene (εr = 2.25)	Air (εr = 1)
   Wavelength λ	1.998616 m	2.997925 m
   Quarter-wavelength λ/4	0.499654 m	0.749481 m
   Wave impedance η	251.154 Ω	376.730 Ω
   Velocity v	1.99862×108 m/s	2.99792×108 m/s

4. Compute the velocity factor. From Equation 3, VF = 1.99862×10⁸ / 2.99792×10⁸ = 1/√2.25 = 0.667. The stub is 66.7% of its free-space length: 499.7 mm instead of 749.5 mm.
5. Confirm the electrical length and delay. Electrical length and propagation delay follow from λ and v:^[1]
   Equation 10 — Electrical length and propagation delay [1, Ch. 2] $$\theta =\beta \ell =360°\frac{\ell }{\lambda },\tau =\frac{\ell }{v}$$ θ = β·ℓ = 360°·ℓ/λ; τ = ℓ/v
   For ℓ = 0.499654 m: θ = 360° × 0.499654 / 1.998616 = 90.0°, and τ = 0.499654 / 1.99862×10⁸ = 2.50 ns (one quarter of the 10 ns period).
6. Apply the stub. A shorted λ/4 stub looks like an open circuit at 100 MHz, so connected in series it blocks the interferer; an open-circuited λ/4 stub looks like a short circuit, so connected in shunt it shorts the interferer to ground.^[1] Include the connector length in ℓ and trim while watching the notch on a network analyzer.

Note: The 251.15 Ω output is the wave impedance of polyethylene, not the 50 Ω or 75 Ω characteristic impedance of the cable. Line impedance depends on the conductor diameters; use the Coax TEM Mode Calculator for that.

5. Tutorial 4: Find the Wavelength on an FR-4 Circuit Board

Problem: A 1 GHz quarter-wave transformer must be laid out on 1.6 mm FR-4 (ε_r ≈ 4.4). How long is λ/4 for a stripline buried in the board, and for a 50 Ω microstrip on its surface?

1. Stripline: use the substrate ε_r directly. A stripline is fully embedded in the dielectric, so it supports a TEM wave in a uniform region.^[1] Enter f = 1E9, ε_r = 4.4, μ_r = 1.
2. Read the stripline outputs. The calculator returns λ = 0.142920 m, λ/2 = 0.071460 m and λ/4 = 0.035730 m (35.73 mm). VF = 1/√4.4 = 0.477.
3. Microstrip: compute the effective permittivity first. On microstrip, part of the field is in the air above the trace, so the wave sees an effective permittivity ε_e between 1 and ε_r.^[1][8] For a 50 Ω line on 1.6 mm FR-4, the trace width is W ≈ 3.06 mm (W/d ≈ 1.91), and the quasi-static formula gives:
   Equation 11 — Microstrip effective permittivity [1, Sec. 3.8], [8] $${\epsilon }_e=\frac{{\epsilon }_r+1}{2}+\frac{{\epsilon }_r-1}{2}\frac{1}{\sqrt{1+12d/W}}=2.70+1.70\times 0.370=3.33$$ ε_e = (ε_r + 1)/2 + [(ε_r − 1)/2]·1/√(1 + 12d/W) = 2.70 + 1.70 × 0.370 = 3.33
4. Enter ε_e in place of ε_r. Enter f = 1E9, ε_r = 3.33, μ_r = 1. The calculator returns λ = 0.164285 m and λ/4 = 0.041071 m (41.07 mm), with VF = 0.548.

   Quarter-wavelength at 1 GHz on 1.6 mm FR-4

   Structure	Value entered for εr	λ (mm)	λ/4 (mm)	VF
   Air (reference)	1.00	299.79	74.95	1.000
   50 Ω microstrip	3.33 (εe)	164.29	41.07	0.548
   Stripline	4.40	142.92	35.73	0.477

5. Allow for material tolerance. FR-4 permittivity varies from about 4.0 to 4.7 with resin content and frequency. Re-run the calculator at both limits: a ±0.3 change in ε_r moves the stripline λ/4 by about ±3.4%. For tight tolerances, use a controlled-permittivity laminate and the value the manufacturer specifies at your frequency.

Result: The same 1 GHz quarter-wave section is 35.73 mm as stripline and 41.07 mm as microstrip. Using the substrate ε_r for microstrip would make the line 13% too short.

6. Tutorial 5: Compare Dielectric and Magneto-Dielectric Loading

Problem: A 300 MHz antenna must shrink by a factor of four. Compare a pure dielectric (ε_r = 16, μ_r = 1) with a magneto-dielectric (ε_r = 4, μ_r = 4). Which loading reflects less energy at the material surface?

1. Find the reference wavelength. Enter f = 300E6 with ε_r = μ_r = 1. The calculator returns λ₀ = 0.999308 m.
2. Run both materials. Enter each material in turn. Both have ε_rμ_r = 16, so both shrink the wavelength by √16 = 4.

   Calculator output: 300 MHz, two materials with ε_rμ_r = 16

   Material	λ (m)	λ/4 (mm)	η (Ω)	v (m/s)
   Air (reference)	0.999308	249.83	376.730	2.99792×108
   Dielectric: εr = 16, μr = 1	0.249827	62.46	94.183	7.49481×107
   Magneto-dielectric: εr = 4, μr = 4	0.249827	62.46	376.730	7.49481×107

3. Compare the wave impedances. Equation 4 shows why the outputs differ: the dielectric lowers η by √16 to 94.18 Ω, while ε_r = μ_r leaves η equal to free space.
4. Compute the reflection at the air–material boundary. For a plane wave at normal incidence, the reflection coefficient depends only on the two wave impedances:^[2][4]
   Equation 12 — Normal-incidence reflection coefficient [2, Ch. 5], [4, Ch. 6] $$\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1},P_{\text{refl}}/P_{\text{inc}}={\left|\Gamma \right|}^2$$ Γ = (η_2 − η_1)/(η_2 + η_1); P_refl/P_inc = |Γ|²
   Dielectric: Γ = (94.18 − 376.73)/(94.18 + 376.73) = −0.600, so 36% of the incident power reflects. Magneto-dielectric: Γ = 0, so none reflects.

Result: Both materials give the same 4:1 size reduction, but only the matched magneto-dielectric keeps the wave impedance at 376.73 Ω. That is why magneto-dielectric substrates can shrink antennas with less loss of bandwidth than high-ε_r dielectrics, which store more energy and raise Q.^[10] In practice, magnetic loss in ferrites limits this approach to frequencies where tan δ_m stays low.

7. Tutorial 6: Find the Wavelength in Water for Microwave Heating

Problem: At 2.45 GHz, water has ε′ ≈ 78 and ε″ ≈ 10 near room temperature.^[11] What is the wavelength inside water, how much power reflects at an air–water surface, and how good is the calculator's lossless result?

1. Enter the real part of the permittivity. The calculator accepts real values only. Enter f = 2.45E9, ε_r = 78, μ_r = 1.
2. Read the outputs. The calculator returns λ = 0.013855 m (13.86 mm), λ/2 = 6.93 mm, λ/4 = 3.46 mm, η = 42.656 Ω and v = 3.39448×10⁷ m/s (VF = 0.113). The wavelength is 8.8 times shorter than the 122.36 mm free-space value from Tutorial 1.
3. Estimate surface reflection. With Equation 12, Γ = (42.66 − 376.73)/(42.66 + 376.73) = −0.797, so about 63% of a normally incident plane wave reflects from a flat water surface. In a closed oven cavity the reflected energy is not lost; it bounces back to the load.
4. Correct the wavelength for loss. For a lossy material with loss tangent tan δ = ε″/ε′, the phase constant increases by a factor that depends only on tan δ:^[2][5]
   Equation 13 — Wavelength in a lossy dielectric [2, Ch. 4], [5, Ch. 8] $${\lambda }_{\text{lossy}}=\frac{\lambda }{\sqrt{\frac{1}{2}(\sqrt{1+{\tan \delta }^2}+1)}},\tan \delta =\frac{{\epsilon }^{″}}{{\epsilon }^{\prime }}$$ λ_lossy = λ / √[(√(1 + tan²δ) + 1)/2], tan δ = ε″/ε′ (λ from the calculator with ε_r = ε′)
   For water, tan δ = 10/78 = 0.128, the correction factor is 1.0020, and λ_lossy = 13.855 / 1.0020 = 13.83 mm. The calculator's lossless result is within 0.2%.
5. Estimate how deep the heating reaches. Loss also makes the wave decay. A standard estimate of the power penetration depth (power falls to 1/e) for ε″ ≪ ε′ is:^[12]
   Equation 14 — Power penetration depth (low-loss approximation) [12] $$D_p\approx \frac{{\lambda }_0\sqrt{{\epsilon }^{\prime }}}{2\pi {\epsilon }^{″}}=\frac{\left(0.12236\right)\sqrt{78}}{2\pi \left(10\right)}=17.2\text{ mm}$$ D_p ≈ λ_0·√ε′ / (2π·ε″) = (0.12236)(√78) / (2π·10) = 17.2 mm
   Use λ₀ from the calculator with ε_r = μ_r = 1. For heating rates and electric field strength, continue with the Dielectric Heating Calculator.

Result: Inside water at 2.45 GHz, λ ≈ 13.8 mm and power penetrates about 17 mm. Food pieces several centimeters thick therefore heat from the outside in, and features near λ/2 (about 7 mm) in size can form internal hot spots. Permittivity of water falls with temperature, so repeat the calculation with temperature-specific data for process design.^[11]

8. Common Input Mistakes

• Entering MHz or GHz in the Hz field. The frequency input is in hertz. Type 146E6 for 146 MHz and 2.45E9 for 2.45 GHz.
• Misreading the output units. Wavelength outputs are in meters. Multiply by 1,000 for millimeters.
• Treating η as a line or antenna impedance. The impedance output is the wave impedance of the material. Coaxial, microstrip and antenna impedances depend on geometry.
• Using substrate ε_r for microstrip or coplanar waveguide. These lines need the effective permittivity from Equation 11 or a field solver.
• Using it for hollow waveguide. The guided wavelength in a hollow waveguide is longer than the plane-wave wavelength. Use the Rectangular Waveguide Calculator or Circular Waveguide Calculator.
• Cutting antennas to exactly λ/4 or λ/2. Real conductors resonate 2–5% short. Apply Equation 9, cut long and trim.
• Ignoring frequency dependence of ε_r. Use the permittivity at your operating frequency and temperature, not a low-frequency or nominal value.

9. Wavelength Equation Summary

Need help with an antenna, matching network, dielectric-loaded component or microwave heating design? Contact SiberSci, LLC. Related tools: Coax TEM Mode Calculator, Skin Depth Calculator and Dielectric Heating Calculator.

10. References

1. D. M. Pozar, Microwave Engineering, 4th ed. Hoboken, NJ: Wiley, 2012. Sec. 1.4, "The Wave Equation and Basic Plane Wave Solutions"; Ch. 2, "Transmission Line Theory"; Sec. 3.8, "Microstrip Lines."
2. C. A. Balanis, Advanced Engineering Electromagnetics, 2nd ed. Hoboken, NJ: Wiley, 2012. Ch. 4, "Wave Propagation and Polarization," and Ch. 5, "Reflection and Transmission."
3. C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. Hoboken, NJ: Wiley, 2016. Ch. 4, "Linear Wire Antennas" (half-wavelength dipole and quarter-wavelength monopole).
4. S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, 3rd ed. New York: Wiley, 1994. Ch. 6, "Propagation and Reflection of Plane Waves."
5. D. K. Cheng, Field and Wave Electromagnetics, 2nd ed. Reading, MA: Addison-Wesley, 1989. Ch. 8, "Plane Electromagnetic Waves."
6. National Institute of Standards and Technology, "CODATA Value: speed of light in vacuum," c = 299,792,458 m/s (exact). https://physics.nist.gov/cgi-bin/cuu/Value?c
7. National Institute of Standards and Technology, "CODATA Value: characteristic impedance of vacuum," Z₀ ≈ 376.730 Ω. https://physics.nist.gov/cgi-bin/cuu/Value?z0
8. E. O. Hammerstad, "Equations for microstrip circuit design," in Proc. 5th European Microwave Conference, Hamburg, Germany, 1975, pp. 268–272.
9. J. D. Kraus and R. J. Marhefka, Antennas for All Applications, 3rd ed. New York: McGraw-Hill, 2002. Chapters on dipole and monopole antennas (resonant length and end effects).
10. R. C. Hansen and M. Burke, "Antennas with magneto-dielectrics," Microwave and Optical Technology Letters, vol. 26, no. 2, pp. 75–78, 2000.
11. U. Kaatze, "Complex permittivity of water as a function of frequency and temperature," Journal of Chemical & Engineering Data, vol. 34, no. 4, pp. 371–374, 1989.
12. A. C. Metaxas and R. J. Meredith, Industrial Microwave Heating. London: Peter Peregrinus/IEE, 1983. (power penetration depth).

Tutorial by Dr. Scott Best, SiberSci, LLC. Values computed with c = 299,792,458 m/s; calculator outputs rounded to six or seven significant figures.