Atmospheric Effects on Millimeter and Sub-millimeter (THz) Satellite Communication Paths

Abstract

Satellite communications require more bandwidth due to the necessity of increasing the capacity of communication channels and bandwidth to end-users. As a result, looking for new bands is required in the electromagnetic spectrum including millimeter and sub-millimeter wavelengths. Recent technological developments made the extremely high frequencies (EHF) above 30 GHz as a candidate for wireless applications such as the fifth generation (5G) of mobile communications, high resolution radars, and remote sensing. The EHF communication systems are becoming more and more commercial, cheap, and compact. However, the fact that the atmospheric medium is not completely transparent to millimeter waves requires considerations of the frequency selective absorption and dispersion effects emerging in this band. These phenomena affect also remote sensing in the millimeter and sub-millimeter waves (the terahertz frequencies). The atmospheric effects on the propagation of millimeter and sub-millimeter wave transmission from land to satellite are discussed. It is shown that not only atmospheric absorption plays a significant role on the received signal strength but also the refraction index of the atmospheric medium. The inhomogeneous refractivity causes the beam to “bend” along the propagation path, and it may even “miss” its destination. This phenomenon should be considered in the design of a link operating in extremely high frequencies involving highly directive antennas.

Appendix. Derivation of the ray path (Eq. (10)) in spherical coordinates

Looking at Fig. 14, an expression for the angle θ(r) can be written in terms of the angle of incidence γ:

$$ \frac{d\theta}{d r}=\frac{d\theta}{d s}/\frac{d r}{d s}=\frac{\sin \gamma /r}{\cos \gamma }=\frac{\sin \gamma }{r\sqrt{1-{\sin}^2\gamma }} $$

(13)

Fig. 14

Illustration of the ray path

Here, ds is an infinitesimal element at a position r along the path of the ray. Noting that γ₀ = γ(r = R) = 90^∘ − ψ₀ (see Fig. 5), the Snell’s law (Eq. (9)) in spherical coordinates can be rewritten:

$$ \sin \gamma =\frac{R\cdot {n}_0\cdot \cos \left({\psi}_0\right)}{r\cdot {n}_r(r)} $$

(14)

Substitution of Eq. (14) in Eq. (13), we get:

$$ \frac{d\theta}{d r}=\frac{\frac{R\cdot {n}_0\cdot \cos \left({\psi}_0\right)}{r\cdot {n}_r(r)}}{r\sqrt{1-{\left[\frac{R\cdot {n}_0\cdot \cos \left({\psi}_0\right)}{r\cdot {n}_r(r)}\right]}^2}}=\frac{R\cdot {n}_0\cdot \cos \left({\psi}_0\right)}{r\sqrt{{\left[r\cdot {n}_r(r)\right]}^2-{\left[R\cdot {n}_0\cdot \cos \left({\psi}_0\right)\right]}^2}} $$

(15)

Integrating the last expression results in:

$$ \theta (r)=\underset{R}{\overset{r}{\int }}\frac{R\cdot {n}_0\cdot \cos \left({\psi}_0\right)}{r^{,}\sqrt{{\left[{r}^{,}\cdot {n}_r\left({r}^{,}\right)\right]}^2-{\left[R\cdot {n}_0\cdot \cos \left({\psi}_0\right)\right]}^2}}d{r}^{,} $$

(16)

Which is Eq. (10) for θ(r).