Mumbai University > EXTC > Sem 8 > Satellite Communication and Networks

Marks: 10M

1.Brad ford UK station (54°N,2°W)

2.Blacksburg USA (37°N, 80°W)

Given:

Longitude of satellite $(ϕ_s )$=$35^0 W$

brad ford UK station (54°N,2°W)

Latitude of earth station $(θ_e)$=$(54)^0N$

Longitude of earth station $(ϕ_e)$=$2^0W$

To find: Look Angles=?

Slant Range=?

Solution:

Hence

$θ_e$=$54^0$

$ϕ_e$=$-2^0$

$ϕ_s$=$-35^0$

$ϕ_es$=$ϕ_e-ϕ_s$

=-2+35

$∴ϕ_es=33^0 $ i.e earth station is to the east of subsatellite point

To calculate Azimuth angle A_z or ζ

$A=\Big|tan^-1 \Big(\frac{tan \phi_{es}}{sin (\theta_e)}\Big)\Big|$

$A=\Big|tan^-1 \Big(\frac{tan (33^0)}{sin(54^0)}\Big|$

$A=\Big|tan^{-1⁡}(0.8027) \Big|$

$∴A=38.754^0$

Since the satellite is in the Northern hemisphere and it is west of the earth station (i.e ϕ_es>0,θ_e>0)

∴ Azimuth angle (A_z or ζ)=$180+A^0$

i.e. $A_z=180^0+38.754^0$

$A_z=218.754^0$

To calculate elevation angle (η)

$η=tan^-1\Big( \frac {cos⁡(ψ)-σ}{sin⁡ ψ }\Big)$

$ψ=cos^-1 ⁡(cos⁡(θ_e ) cos⁡(ϕ_{es} ) )$

$=cos^-1(cos⁡(54^0 ) cos⁡(33^0 ) )$

$=cos^-1⁡(0.4929)$

$∴ψ=60.4647^0$

$η=tan^-1\Big( \frac {cos⁡ (60.4647)-0.151}{sin⁡ (60.4647)}\Big)$

$η=tan^{-1⁡}(0.39303)$

$η=21.456^0$

Slant Range (d):

$d=35786[1+0.4199{1-cos⁡(ψ) }]^{0.5}Km$

$=35786[1+0.4199{1-cos⁡(60.4647) }]^{0.5}Km$

d=35786×1.101=39411.845Km

∴Slant Range (d)=39411.845Km

Blacksburg USA (37°N, 80°W)

Longitude of satellite $(ϕ_s )=35^0 W$

Latitude of earth station $(θ_e) =37^0N$

Longitude of earth station $(ϕ_e)=80^0 W$

To find:

Look Angles=?

Slant Range=?

Solution:

Hence $θ_e=37^0$

$ϕ_e=-80^0$

$ϕ_s=-35^0$

$ϕ_es=ϕ_e-ϕ_s$

=-80+35

$∴ϕ_{es}=-45^0$ i.e earth station is to the west of subsatellite point

To calculate Azimuth angle A_z or ζ

$A=\Big|tan^-1⁡ \Big(\frac{ tan ϕ_{es}}{sin⁡(θ_e )} \Big) \Big|$

$A=\Big|tan^{-1}\Big(\frac{tan (-45^0)}{sin(37^0)}\Big)\Big|$

$A=|tan^{-1}⁡(-1.6616) |$

$\therefore A= 58.959^0$

Since the satellite is in the Northern hemisphere and it is east of the earth station (i.e ϕ_es<0 & θ_e>0)

∴ Azimuth angle (A_z or ζ)=$180-A^0$

i.e. $A_z=180^0+58.959^0=121.041$

$A_z=121.041^0$

To calculate elevation angle (η)

$η=tan^{-1}⁡\Big(\frac {cos⁡(ψ)-σ}{sin⁡(ψ)} \Big)$

$ψ=cos^{-1}⁡(cos⁡(θ_e ) cos⁡(ϕ_{es} ) )$

$ =cos^{-1}⁡(cos⁡(37^0 ) cos⁡(-45^0 ) )$

$=cos^{-1}(0.56472)$

$∴ψ=55.617^0$

$η=tan^{-1}\Big(\frac {cos⁡(55.617)-0.151}{sin⁡(55.617)}\Big )$

$η=tan^{-1}⁡(0.5013)$

$η=26.625^0$

Slant Range (d):

$d=35786[1+0.4199 (1-cos⁡(ψ) )]^{0.5}Km$

$=35786[1+0.4199(1-cos⁡(55.617) )]^{0.5}Km$

d=35786×1.0875=38919.198Km

∴Slant Range (d)=38919.198K