# Any demonstration of ITU-R SA.2183 (Method for calculating link performance in the space research service)?

I'am looking for demonstration of ITU-R SA.2183 (2010-10 edition), in MATLAB or Mathematica.

I have written some Mathematica code, any help would be appreciated.

(*ITU-R SA.2183, Calculations for atmospheric attenuation (no rain or \
clouds)*)

WaterVapourDensity = 7.5;(*in g/m^3*)
StationHeightAboveSeaLevel = 0.81;(*in km*)
f = 1;(*frequency,in GHz*)
p = 1.023*10^8;(*pressure (hPa) at the station location*)
Subscript[r, "p"] = p/1013;
Subscript[t, "1"] =
4.64/(1 + 0.066 Subscript[r, "p"]^(-2.3))*
Exp[-((f - 59.7)/(2.87 + 12.4*Exp[-7.9*Subscript[r, "p"]]))^2];
Subscript[t, "2"] =
0.14 Exp[2.12 Subscript[r, "p"]]/((f - 118.75)^2 +
0.031 Exp[2.2 Subscript[r, "p"]]);
Subscript[t, "3"] =
0.0114/(1 + 0.14*Subscript[r, "p"]^(-2.6))*
f*(-0.0247 + 0.0001 f + 1.61*10^(-6)*f^2)/(1 - 0.0169 f +
4.1*10^(-5)*f^2 + 3.2*10^(-7)*f^3);
Subscript[h, "0"] =
6.1/(1 + 0.17*Subscript[r, "p"]^(-1.1))*(1 + Subscript[t, "1"] +
Subscript[t, "2"] +
Subscript[t,
"3"]);(*The equivalent oxygen height from sea level up to an \
t = 23;(*temperature, in ℃*)
Subscript[r, "t"] = 288/(273 + t);
PhiFunction[rp_, rt_, a_, b_, c_, d_] :=
rp^a*rt^b*Exp[c (1 - Subscript[r, "p"]) + d (1 - Subscript[r, "t"])];
Subscript[\[Xi], "1"] =
PhiFunction[Subscript[r, "p"], Subscript[r, "t"], 0.0717, -1.8132,
0.0156, -1.6515];
Subscript[\[Xi], "2"] =
PhiFunction[Subscript[r, "p"], Subscript[r, "t"], 0.5146, -4.6368,
0.1921, -5.7416];
Subscript[\[Xi], "3"] =
PhiFunction[Subscript[r, "p"], Subscript[r, "t"], 0.3414, -6.5851,
0.2130, -8.5854];
Subscript[\[Gamma],
"0"] = (7.2*
Subscript[r, "t"]^2.8/(f^2 +
0.34*Subscript[r, "p"]^2*Subscript[r, "t"]^2) +
0.62*Subscript[\[Xi],
"3"]/((54 - f)^(1.16 Subscript[\[Xi], "1"]) +
0.83*Subscript[\[Xi], "2"]))*f^2*
Subscript[r,
"p"]^2*10^(-3);(*in dB/km, Specific attentuation at ground level \
due to dry air, from sea level up to an altitude of 10km, and for \
frequencies<= 54 GHz*)
Subscript[\[Sigma], "w"] =
1.013/(1 + Exp[-8.6 (Subscript[r, "p"] - 0.57)]);
Subscript[h, "w"] =
1.66 (1 +
1.39*Subscript[\[Sigma],
"w"]/((f - 22.235)^2 + 2.56 Subscript[\[Sigma], "w"]) +
3.37*Subscript[\[Sigma],
"w"]/((f - 183.31)^2 + 4.69 Subscript[\[Sigma], "w"]) +
1.58*Subscript[\[Sigma],
"w"]/((f - 325.1)^2 +
2.89 Subscript[\[Sigma],
"w"]));(*in km, Equivalent water vapour height at the \
earth station elevation. For frequencies < 350GHz*)

\[Rho] = WaterVapourDensity; (*the water density (g/m^3) at the \
station location, I think the 2 is equal*)

Subscript[\[Eta], "1"] =
0.955*Subscript[r, "p"]*Subscript[r, "t"]^0.68 + 0.006 \[Rho];
Subscript[\[Eta], "2"] =
0.735*Subscript[r, "p"]*Subscript[r, "t"]^0.5 +
0.03553*Subscript[r, "t"]^4*\[Rho];
gFunction[f_, fi_] := 1 + ((f - fi)/(f + fi))^2;

Subscript[\[Gamma],
"w"] = (3.98*Subscript[\[Eta], "1"]*
Exp[2.23*(1 - Subscript[r, "t"])]/((f - 22.235)^2 +
9.42*Subscript[
\[Eta], "1"]^2))*
gFunction[f, 22] + (11.96*Subscript[\[Eta], "1"]*
Exp[0.7*(1 - Subscript[r, "t"])]/((f - 183.31)^2 +
11.14*Subscript[
\[Eta], "1"]^2)) + (0.081*Subscript[\[Eta], "1"]*
Exp[6.44*(1 - Subscript[r, "t"])]/((f - 321.226)^2 +
6.29*Subscript[
\[Eta], "1"]^2)) + (3.66*Subscript[\[Eta], "1"]*
Exp[1.6*(1 - Subscript[r, "t"])]/((f - 325.153)^2 +
9.22*Subscript[
\[Eta], "1"]^2)) + (25.37*Subscript[\[Eta], "1"]*
Exp[1.09*(1 - Subscript[r, "t"])]/((f - 380)^2)) + (17.4*
Subscript[\[Eta], "1"]*
Exp[1.46*(1 - Subscript[r, "t"])]/((f - 448)^2)) + (844.6*
Subscript[\[Eta], "1"]*
Exp[0.17*(1 - Subscript[r, "t"])]/((f - 557)^2))*
gFunction[f,
557] + (290*Subscript[\[Eta], "1"]*
Exp[0.41*(1 - Subscript[r, "t"])]/((f - 752)^2))*
gFunction[f,
752] + (8.3328*10^4*Subscript[\[Eta], "2"]*
Exp[0.99*(1 - Subscript[r, "t"])]/((f - 1780)^2))*
gFunction[f,
1780];(*in dB/km, Specific  attenuation at ground level due to \
water vapour, from sea level up to an altitude of 10km, and \
frequencies< 360 GHz*)


Subscript[h, "0"]
Subscript[\[Gamma], "0"]
Subscript[h, "w"]
Subscript[\[Gamma], "w"]

• Hi! So, as you might know, we have a close reason that says "questions asking for code written to the specification of the asker are off-topic". So, your question can't be "could you give me code for this?". If that question is ruled out, what is your actual signal processing question? Oct 17, 2023 at 14:39