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Worked Examples
These examples are intentionally compact. Each one shows the equation → substitution → number → interpretation pattern.
1. Wavelength at 5.8 GHz
\[\lambda=\frac{c}{f}=\frac{2.998\times10^8}{5.8\times10^9}\approx5.17\ \text{cm}.\]A quarter-wave is about $1.29$ cm. This immediately sets the physical scale of antennas, PCB features and chamber objects that are electrically significant.
2. Free-space path loss at 5.8 GHz over 10 m
\[FSPL=20\log_{10}\left(\frac{4\pi R}{\lambda}\right).\]With $R=10$ m and $\lambda=0.0517$ m,
\[FSPL\approx67.7\ \text{dB}.\]If $P_t=10$ dBm and both antennas have $10$ dBi gain, ideal Friis gives about
\[P_r\approx10+10+10-67.7=-37.7\ \text{dBm}.\]3. Thermal noise in 1 MHz bandwidth
At room temperature,
\[P_n\approx-174+10\log_{10}(10^6)=-114\ \text{dBm}.\]With a 5 dB receiver noise figure,
\[P_{n,receiver}\approx-109\ \text{dBm}.\]This is why bandwidth is a first-class sensitivity parameter.
4. Copper skin depth at 100 MHz
Using $\sigma\approx5.8\times10^7$ S/m and $\mu_r\approx1$,
\[\delta=\frac{1}{\sqrt{\pi f\mu_0\sigma}}\approx6.6\ \mu\text{m}.\]At 100 MHz, most current resides within only a few tens of micrometres of a copper surface.
5. WR-90 dominant-mode cutoff
For $a=22.86$ mm,
\[f_{c,10}=\frac{c}{2a}\approx6.56\ \text{GHz}.\]This helps explain why WR-90 is used in X-band rather than far below 6.5 GHz.
6. Radar range resolution from 1 GHz bandwidth
\[\Delta R\approx\frac{c}{2B} =\frac{2.998\times10^8}{2\times10^9}\approx0.15\ \text{m}.\]Carrier frequency affects antenna size and Doppler scale; waveform bandwidth sets this ideal range resolution.
7. 77 GHz automotive-radar Doppler shift at 30 m/s
At 77 GHz,
\[\lambda\approx3.89\ \text{mm}.\]For a monostatic radar,
\[f_D=\frac{2v}{\lambda}\approx\frac{60}{3.89\times10^{-3}}\approx15.4\ \text{kHz}.\]8. Far-field distance of a 30 cm aperture at 10 GHz
At 10 GHz, $\lambda\approx3$ cm. With $D=0.30$ m,
\[R_{FF}\gtrsim\frac{2D^2}{\lambda} \approx\frac{2(0.3)^2}{0.03}\approx6\ \text{m}.\]The correct far-field distance depends on electrical aperture size, not just “a few wavelengths.”
9. Ideal Halbach-cylinder field
Suppose $B_r=1.2$ T, $R_i=20$ mm and $R_o=40$ mm. Then
\[B\approx B_r\ln(R_o/R_i)=1.2\ln2\approx0.83\ \text{T}.\]A practical segmented, finite-length assembly will differ and must be simulated/measured for homogeneity.
10. Ground-state atomic Larmor frequency
For an effective gyromagnetic ratio of $3.5$ GHz/T and a $50\ \mu$T field,
\[f_L=(3.5\times10^9)(50\times10^{-6})\approx175\ \text{kHz}.\]Frequency measurement becomes a magnetic-field measurement once the relevant atomic $\gamma$ is known.
11. Electric-dipole Rabi frequency
Take a projected transition dipole $d=1000\,ea_0$ and $E=0.1$ V/m.
\[\Omega=\frac{dE}{\hbar}.\]Since $ea_0\approx8.48\times10^{-30}$ C·m,
\[\Omega/2\pi\approx1.28\ \text{MHz}.\]The actual Rabi rate depends on the full angular matrix element and polarization projection.
12. Shockley–Ramo current between parallel plates
For plate spacing $d=6$ mm, a single electron moving at $v_x=10^5$ m/s gives the ideal magnitude
\[|i|=\frac{ev_x}{d} \approx\frac{1.602\times10^{-19}\times10^5}{6\times10^{-3}} \approx2.7\ \text{pA}.\]A measured signal from many charges is the sum of their instantaneous $q\mathbf v\cdot\mathbf E_w$ contributions, filtered by the readout electronics.
13. MRI proton Larmor frequency at 3 T
For protons, $\gamma/2\pi\approx42.58$ MHz/T, so
\[f_L\approx42.58\times3\approx127.7\ \text{MHz}.\]This is why a 3-T MRI scanner requires RF hardware in the VHF range.
14. Link budget with mismatch and cable loss
Suppose the ideal Friis result is $-37.7$ dBm, but transmitter and receiver cable losses are 1.5 dB each and polarization mismatch contributes 3 dB:
\[P_r\approx-37.7-1.5-1.5-3=-43.7\ \text{dBm}.\]The “engineering reality” corrections can be comparable to, or larger than, the effect you are trying to measure.
Use the Interactive Calculators to vary the numbers and the Measurements page to see how each quantity is obtained experimentally.