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SatelliteLinkDesi n
R.N.Mutagi
ElectronicsandCommunicationEngineeringDepartmentn us ns u eo ec no ogyan ng neer ng
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po n source n space
transmitting powerPtwatts
in all the directions, alon 4
radians of a sphere is called
an isotropic radiator
.
The flux density in unit area
(1 m2) at a distance R
me ers rom s source s
2
2W/m
PF t=
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A practical antenna radiates P0watts of total power thepower radiated in unit solid angle in the direction is P().
If this power is transmitted by an ideal isotropic radiator
the power transmitted along unit solid angle would be
P /4W/m2.
The gain of the practical antenna is therefore
directioninantennapracticalbydtransmittePower
directionsameinradiatorisotropicbydtransmittePower
4/
)(
0P
G =
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e power ra a e y an an enna a ong e ore s g
is the maximum
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Antenna gain depends of the ability of the antenna to concentrate
the energy in particular direction
hwavelen tisandareaa erturetheiswhere4
AA
G
=
An effective aperture areaAe = AA, is defined to account for theA
22
4
4 eA
AAG ==
For a circular parabolic dishA=D2/4, where D is diameter
2
2A2
===
G
AA
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sa e e rece v ng s a on as a s an enna w an
aperture area of 7.5 m2. It receives a signal at 4.2GHz. If
the a erture efficienc is assumed to be 70% find the
gain of the antenna. If the service is switched to Ku band
and the new frequency of reception is 11.5 GHz find the.
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=2
G AAntenna gain is given by
We have A=7.5 m2, =0.7 and
=c/f=3x108/4.2x109=0.071m
dBdBG 12.41)12941log(10129410714.0
.7.0
2===
=
For 11.5 GHz the wavelength is =3x108/11.5x109=0.026m
dBidBG 89.49)97594log(1097594
026.0
.7.0
2===
=
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reateanequat ontoca cu atet eantennaga n n w enfrequency
in
GHz
and
diameter
in
meter
are
given
2
=
G A
/
2
=c
DG A
16054.9180lo20lo2094.9lo10
)103log(20)10log(20log20log20log10 89
++++=
+++=
D
fDG AdB
4.20log20log20log10++=
fDA
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or a ransm er ransm ng power t a s w an
antenna gain Gt the EIRP is defined as
ttGPEIRP=
2W/mGP
F tt=
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4 eA r
effectivearea
si nalwavelen th
2
r =
Foragivenantennathegainishigherfor
higher
frequency
TxPin
EIRP
R
FRx
Pr
Prad
t r
( ) 22 GGPEIRP rtrad
12
22 4444 =
==
RRRrtrader
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- ,
expansion of the spherical wave front as the wave
.
22
44 == fRR c
p
losspathspacefreewhere =Lp
(Hz)frequency
(m)distance
==
f
R
m/s)10(3space-freeinlightofvelocityc8
=
=
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orageostat onarysate teca cu atet e reespace oss n
dBfor
the
signal
received
on
ground
at
frequencies
4GHz
and11GHz
Create
an
equation
for
measuring
free
space
loss
in
dB
when
22
=
=c
Lp
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2 += LGEIRPP
4 = RGGPPrtradr
(dBW)powerreceived=Where Pr
4
2
== R
tt
p
Considering atmospheric attenuation, La, Transmit Antennalosses Lta, receiving antenna losses Lra we can write
dBWrataaprr
LLLLGEIRPP +=
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o ec s a p ys ca empera ure n >o genera e
electrical noise at receiver frequencies. The noise poweris iven b
Pn = N0B = kTnB
n
N0 is noise power spectral density in W/Hz
k is Boltzmanns constant =1.38 x 10-23 J/K
=-228.6 dBW/K/Hz
Tn is system noise temperature in0K(0C+273)
Fall2008 16
B is the system noise bandwidth in Hz
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Ideal IFIdealIdeal RF
Pn
amplifiermixeramp er
Gain
GRF
Gain
G
Gain
GTin
TRF
TM TIF
Gain E uivalent
GRF.GM.GIF ns system
R.N.Mutagi ELN5597 17
)( inRFMIFRFMMIFIFIF TTkBGGGBkTGGBkTGPn +++=
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TTkBkTkTPn +++=
inRFMIF
IFMIF
n
TTG
T
GG
TkBGGG
+++= )(
SIFMIF kBTGGG=
TS isthesingleequivalentnoisetemperatureproducingsame
noisepower
+++= )( inRFMIF
S TTTT
TRFRFM
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an rece ver assu systemsw t o ow ng
specifications
dBGKT MRF
RFin
350
==
==
KTIF
IFM
1000
=
a cu atet e
system
no se
temperature
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+++= )( inRFRF
M
RFM
IFS TT
GT
GGTT dBGKT
dBGKT
MRF
RFin
5.0350
2002350
===
===
KT
dBGKT
IF
IFM
1000
100030500
=
===
5001000 S
1005.210
2002005.0
++=
K5.112=
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andis
defined
as
( )in
NS Si/Ni
Anamplifier
will
add
its
own
noise
and
the
output
S/N
will
be
( )out
NSo o
lowerthaninputS/N
( )AAiiiii NNANNSNS +/
( ) iiAiioo ANANNANASNS +/
,
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e re a on e ween no se gure an no se empera ure
is give by
Where T0
is the reference noise temperature = 290K
0 =d
Example, for an amplifier with noise figure of 3 dB the noise
empera ure s
( ) KTd 29012290 ==
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e carr er o no se ra o a e rece ver npu s
===
S
rtt
S
rtt
T
G
RkB
GP
BkT
RGGP
powerNoise
powerSignal
N
C2
4
4
The terms in first bracket depend on the satellite
parame ers an ose n e secon epen on ereceiving earth station
r Scalled the Figure of Merit of the earth station
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n ear s a on as an an enna o m ame er an
has an efficiency of 65%. The system noise temperatureof 80K. The receive fre uenc is 4GHz. Find the G/T
ratio of the earth station.
DG Ar 256609
104/103
1565.0
2
98
2
=
=
=
TS 80
==
=
Sr
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equation
dBW+= LLLLGEIRPP
sationearthofeirp
=
raaaprr
EIRP
frequencyuplinkatlosspath=
=
p
r
L
lossantennasatellite
lossantennastationearth
==
ra
ta
L
L
lossescatmospheri=aL
( ) ( )
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( ) ( ){ {
UL
M
UL
p
Sat
sr
ESSat
ULLLTGEIRPNC ++= 6.228// 0
43421321
43421
( ) ( ) MpSrUL LLTGEIRPNC ++= 3214341
3214341
6.228// 0ULkULSat
ESSatelliteat
AlltermsareexpressedindB
Thelasttermincludesalltheantennalosses,
atmosphericattenuation
loss
and
also
any
other
marginrequired(likerainfadingmargin)
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e sa e e n cons s s o up n an own n
In a frequency translation type transponder the downlinkis affected b the u link
The total C/N received is given by
111
+
=
DLUL NNN
If (C/N)UL>> (C/N)DL then the link is downlink limited, else
it is uplink limited
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In addition to thermal noise interference si nals within
the band also impair the carrier
The total uplink noise power isth
+= k kuINN
k
11111
+
=
+
=
CCCCC
Similarly for downlink
UUkUUL
11111 CCCCC
Overall satellite link is
=
=
DDk kDDL ININN
1
+
+
+
=
DUDUTotal I
C
I
C
N
C
N
C
N
C
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sate tetransm ts s gna n u an Transmit
frequency
12
GHz
Transmit oweris 160W
gainofsatelliteantennais35dBi ontheboresight.
Receiveantennagainis33dBi .
Thereceivesystemnoisetemperatureis145K Signalbandwidthis20MHz.
Calculate theC/Natthereceiverandthelinkmarginifthe
demodulatorrequiresaminimumC/Nof10dB.
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rataaprr ==
====
4422
=
=fRR
L
lo20lo204
lo20 ++
= RdBL
c
)000,000,36log(20)1012log(206.147103
9 ++=
13.20513.311206.211806.147 =++++=
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= - . - + =- .
N=k(dB)+Ts(dB)+B(dB)= -228.6+10lo 145+10lo 20x106
= -228.6 + 21.61 + 70+3
= -133.99
C/N dB = -116.13 (-133.99) = 17.86 dB
arg n = va a e res o= 17.86 10 = 7.86 dB