Free-space path loss comparison – 0.5 GHz versus 5 GHz Assuming the same link distance and antennas, how does free-space transmission loss at 5 GHz compare with that at 0.5 GHz?

Introduction:Free-space path loss (FSPL) quantifies spreading loss as a radio wave propagates without obstacles. It increases with frequency for a fixed distance and antenna aperture. Comparing 0.5 GHz and 5 GHz (a 10× frequency increase) illustrates the logarithmic dependence of FSPL on frequency.

Given Data / Assumptions:

• Same transmitter power, range, and antenna gains (in dBi).
• Line-of-sight free-space propagation.
• Frequencies f1 = 0.5 GHz and f2 = 5 GHz.

Concept / Approach:

The standard formula in decibels is FSPL(dB) = 20 * log10(4πd / λ) = 32.44 + 20 * log10(d_km) + 20 * log10(f_MHz). At the same distance, the difference depends only on frequency ratio. When frequency increases by a factor of 10, FSPL increases by 20 * log10(10) = 20 dB.

Step-by-Step Solution:

Verification / Alternative check:

Using wavelength form: FSPL ∝ 1/λ^2, and λ shrinks by 10× at 5 GHz, yielding 100× power density loss, which is 20 dB.

Why Other Options Are Wrong:

Less loss contradicts the formula. 700 dB is unrealistic for a 10× change. No change ignores the frequency term in FSPL.

Common Pitfalls:

Confusing antenna aperture scaling; if physical antenna size is constant, gain may rise with frequency, partially offsetting FSPL, but FSPL itself still increases by 20 dB.

Final Answer:

20 dB more at 5 GHz.