Link Budget and Satcoms Training

Introduction

The aim of this presentation is to provide an overview of the considerations necessary when calculating a satellite link budget.

INVSAT as a VSAT provider must implement solutions ensuring link budget integrity, whilst maintaining simple and inexpensive solutions where possible.

The link budget, no matter what part of the world the system is operating in, is driven by hardware performance, as well as atmospheric effects. The four main contributions from hardware to the overall performance are:

Satellite performance

Antenna performance

MODEM performance

HPA performance

The satellite performance is out of our hands, so it up to us to identify satellites with a suitable footprint for a required link and choose the appropriate provider based on cost and performance. Once a satellite has been identified a link budget calculation can be performed.

Introduction

There are often constraints on the size of antennas that can be deployed at a given site. For VSAT applications cost, mobility, type approval, location and appearance often feature in antenna choice.

In order for a satellite link to be maintained the MODEMS must ‘keep lock’. MODEM manufacturers specify signal strength, required to ‘keep lock’, in terms of power received as a carrier or error to noise ratio.

VSAT HPAs are sized in order to achieve a given performance (Eb/No). For VSAT applications solid state parametric amplifiers, SSPAs, are used to provide efficient , reliable, low cost amplification.

When performing a link budget calculation it is often necessary to trade performance, with antenna size or HPA size. The aim being to provide a reliable, compliant low cost solution.

UNITS

Link budget calculations express power in dBW

1W = 0dBW

2W = 3 dBW

10W = 10dBW

100W = 20dBW

1000W = 30dBW

x W = 10*log(x/1) dBW

x dBW = 10x/10 W

Noise calculations use both noise temperatures and noise figure

Noise Temperature = 290(10NF/10 - 1)

Antennas

Invsat 2400SE - C Band

Invsat 1200SE - Ku Band

Antenna Gain

The gain of an antenna is the ratio of power radiated (or received) by an antenna per unit solid angle in a given direction to the power radiated (or received) by an isotropic antenna (radiates evenly an all directions) fed with the same power

The gain is maximum in the direction of peak radiation (antenna boresight) and has a value expressed as:

Gmax = (4πD / λ2)Aeff

Aeff is the equivalent electromagnetic surface area of the antenna. For circular aperture antennas of diameter, D, A= πD2/4

Aeff= ηA, where η is efficiency of the antenna.

Gmax = η(πD / λ)2 = η(π D f / c)2

c is the speed of light

Question:

The 1200SE has a gain of 43.4dB at 14GHz, what is it’s efficiency?

Polarisation

Vertical

Horizontal Horizontal

Vertical

a ac

b bc

bx

ax

At TransmittingAntenna

At ReceivingAntenna

Cross Polar IsolationXPI = ac / bx or bc / ax

or in dBXPI(dB) = 20log(ac / bx ) or 20log(bc / ax )

Cross Polar DiscriminationXPD = 20log ( ac / ax )

A typical value for XPD is 30dBc

Equivalent Isotropic Radiated Power

The power radiated per solid angle by an isotropic antenna fed from a an RF source of power PT is given by, PT / 4π

In a direction where the transmit gain is GT , any antenna radiates a power per unit solid angle equal to, PT GT / 4π

PTPT

GT = 1

ISOTROPIC ANTENNA

GT

Area ASolid Angle = A / R2

REAL ANTENNA

R

Power Flux Density

A satellite antenna of effective area A, at a distance R from the earth station subtends a solid angle A / R2, at the earth station antenna. The satellite antenna receives a power equal to

PR = (PT GT / 4π)(A / R2) = φ A

φ = PT GT / 4π R2 is called the power flux density (W / m2)

Typical values of φ are -90dBW/m2 to -80dBW/m2

PTPT

GT = 1

ISOTROPIC ANTENNA

GT

Area ASolid Angle = A / R2

REAL ANTENNA

R

Received Signal Power

For a receive antenna of effective area, AReff, from the transmit antenna, the received power is:

PR= φAReff = (PT GT / 4πR)AReff

The equivalent area of the antenna is a function of the gain, GR, and can be expressed as:

AReff = GR/(4π/λ2)

Therefore received power:

PR= (PT GT / 4π R2) (λ2 /4π) GR

= (PT GT )(λ2 /4πR)2 GR

= (PT GT )(1/LFS) GR

LFS is free space loss and represents the ratio of the received and transmitted powers in a link between two isotropic antennas and is of the order of 200dB.

Free Space Loss

Geostationary satellites are placed at a range, R0, 35786km above the equator.

LFS= (4πR/λ)2 = (4πR0/λ)2 (R/R0)= LFS(R0) (R/R0) 2

The range from an earth station to a satellite can be calculated from:

(R/R0)2 = (1+0.42(1 - cosLcosl))

l is the earth station latitude

L is the earth station to satellite relative longitude

(R/R0)2 is between 1 and 1.356 (0 to 1.3dB) depending on the earth station location.

Example Uplink

The 1200SE has a maximum antenna gain 43.4dB at 14GHz

It is fed by a 25W Kstar transceiver operating at 1dB output back off (OBO).

The RF loss between the Kstar output and the antenna interface is 1.5dB.

Calculate the uplink EIRP.

EIRP = PTGTmax/L

=((10*log(25))-1)+ 43.4 -1.5

= ? dBW

Calculate the power flux density , assuming the uplink is from Aberdeen to Telstar 12.

φ = (PTGTmax/L) / (4πR2)

= EIRP / (10*log(4*3.14*39100))

= ? dB(W/m2)

Example Downlink

BP operate a VSAT link to Aberdeen from the Jack Bates platform located West of Shetland. The range from the platform to the satellite is 39477km.

Calculate the power flux density at the earth station. Assume the link is via Telstar 12. The EIRP over the Shetland area is 49dBW.

φ = (PTGTmax/L) / (4πR2)

= 49 - (10*log(4*3.14*(39477000)2)

= -114 dB(W/m2)

If the downlink is to a 1200SE at 11.4GHz, calculate the free space loss, the gain of the 1200SE at the receive frequency and the power received by the 1200SE at Westhill.

LFS= 10*Log((4*3.14*39100000*11.4*10^9)/(3*10^8)) = 205.4dB

Grmax=10*log(0.67((3.14*1.2*11.4*10^9)/(3*10^8)^2) = 41.4

PR= EIRP - LFS + Grmax = 49 - 205.4 + 41.4= -115dBW = = 3.16pW

Sources of RF Loss

As well as free space loss and thermal losses in the output run from a VSAT HPA, there are additional losses arising from varied sources including:

atmospheric attenuation, rain, scattering, etc.

thermal losses in the equipment

mis-alignment of Tx and/or Rx antenna boresight

polarisation mismatch losses

Sources of RF Loss

Attenuation Of RF signal in the Atmosphere

Attenuation by rain and free space loss gives a total path loss that can be expressed as L = LFSLA

Transmit Losses, LFTX, between the HPA and the antenna

To provide an antenna with a power, PT, it is necessary to provide a power, PTX, at the HPA output so that PTX = PTLFTX

EIRP is, therefore, expressed as, EIRP = PTGT = (PTXGT)/ LFTX

Receive path losses, LFRX ,between the antenna and the receiver (LNB)

The signal power at the input to the receiver is PRX = PR LFTX / LFRX

Losses due to antenna misalignment

If the Tx and Rx antenna boresights are not aligned perfectly, there is a resultant loss in antenna gain. The losses are a function of the misalignment angles. Their value in dB is given by:

LT = 12(αT / θ3dB)2 and LR = 12(αR / θ3dB)2

Sources of RF Loss

Losses due to polarisation mismatch, LPOL

These occur when the receiving antenna is not oriented with the polarisation of the received wave.

At C Band we commonly use circular polarisation. The transmitted wave is only circularly polarised on the boresight axis of the antenna. Off axis the polarisation becomes elliptical. Propagation through the atmosphere can also change circular into elliptical polarisation.

At Ku Band we commonly use linear polarisation, as the wave propagates through the atmosphere the the of polarisation can rotate.

If the receive antenna is not properly aligned to the received waves plane of polarisation, and the angle of offset is γ, then LPOL = 20logcos γ

Typically, when installing a VSAT antenna feeds are adjusted to align the receive antenna boresight to the polarisation of the received wave.

γ

Summary of RF Loss

Considering all sources of loss, the signal power at the receiver becomes:

PRX = (PTX Gtmax / LTLFTX)(1/LFSLA)(Grmax/ LRLFRXLPOL)

The first bracket characterises the transmitting equipment EIRP

– This expression accounts for losses between the HPA and the antenna and the reduction in gain due to misalignment of the transmitting antenna.

The second bracket (1/L) characterises the transmitting medium

– The path loss takes account of attenuation of free space and the attenuation of the atmosphere due to rain etc.

The third bracket characterises the gain of the receiving equipment

– This expression accounts for losses between the receive antenna and the receiver (LNB), the loss of antenna gain due to misalignment and the losses due to polarisation mismatch.

Noise at the Receiver Input

Noise is unwanted signal without information content which is added to the useful signal. Noise reduces the ability of a receiver to reproduce the information content of the wanted signal correctly.

Noise originates from:

natural sources of radiation located within the antenna reception area

electronic components in the equipment

Only noise which lies in the signal bandwidth degrades receiver performance. Noise is commonly modelled assuming the power spectral density, N0 (W/Hz), is constant across the bandwidth. This is white noise. The equivalent noise power N (W) measured in a bandwidth BN (Hz) has a value: N = N0 BN (W)

B

N0(f)

N0

(W/Hz)

Frequency (Hz)

Spectral White Noise Density

Noise at the Receiver Input

The noise temperature of a noise source with noise power N is

T = N / kB = N0 / k

k is Boltzman’s constant = 1.379*10-23

T represents the thermodynamic temperature of a resistance which delivers the same available noise power as the source under consideration e.g. an LNB

If we consider the receive path of a VSAT terminal, we have several elements in series, each of these element has a gain, G j (j-1, 2, ….N) and an associated noise Tej. The overall noise of the receive path receives a contribution from each of the elements in the path.

The noise temperature of the device is

Te = Te1 + Te2/G1 + Te3/G1 G2 + …..+ TiN/G1 G2….GN-1

The noise figure of the device is

F = F1 + (F2 -1)G1 + (F3 -1)G1 G2 + …..+ (FN -1)G1 G2 ….GN-1

Application to noise temperature of Receiving Equipment

The receiving equipment above shows an antenna connected to an LNB.

There connection is lossy and is at a thermodynamic temperature, TF (which will be close to 290K). The connection has a loss value, LFRX, which corresponds to the equipment gain, GFRX= 1/ LFRX and is less than 1. The noise temperature of the system is determined at:

– the antenna output, before connection losses, temperature T1

– the LNB input, after connection losses, temperature T2

TA , Antenna

LFRX Feeder

T1

T2

TR , LNB

TF

Application to noise temperature of Receiving Equipment

At the antenna output, T1, the noise temperature, is the sum of the antenna noise temperature, and the receive subsystem consisting of the connection and the LNB in series. The noise temperature of the receive subsystem is (LFRX - 1) TF+ TR / GFRX. Adding the contribution of the antenna, which should be considered as a noise source, this becomes

T1 = TA + (LFRX - 1)TF + TR/GFRX

Now consider the LNB input, this noise must be attenuated by a factor LFRX. Replacing GFRX with 1/ LFRX , the noise temperature, T2, is:

T2 = T1 / LFRX = TA / LFRX + TF(1 - 1/LFRX ) + TR

TA , Antenna

LFRX Feeder

T1

T2

TR , LNB

TF

Consider a system with and without connection loss, LFRX.

Antenna noise temperature: TA = 100K

Thermodynamic temperature of the connection: TF = 290K

LNB noise temperature: TR = 90K

If LFRX, is zero

T2 = T1 / LFRX = TA / LFRX + TF(1 - 1/LFRX ) + TR

= 100 + 90 = 190K

If LFRX, is 1dB

= 100/101/10 + 290(1 - 1/101/10) + 90 = 79.43 + 59.6 + 90 = 229K

The connection losses reduce the antenna noise contribution but cause an overall increase in system noise temperature. Every 0.1dB of loss in the connection causes a 6.6K increase in system noise temperature

Note that the antenna noise temperature is a function of the direction in which it is pointing, its environment and its radiation pattern.

Noise Temperature Example

The Carrier to Noise Ratio, (Hz), of a link is calculated from

C/N0 = [(PTX Gtmax / LTLFTX)(1/LFSLA)(Grmax/ LRLFRXLPOL)]

/ [TA / LFRX + TF(1 - 1/LFRX ) + TR ](1/k)

C/N0 can also be interpreted as:

C/N0 = (transmitter EIRP)(1/path loss)(receiver gain/noise temperature)(1/k)

C/N0 can also be expressed in terms of power flux density

C/N0 = φ(λ2/4π)(receiver gain/noise temperature)(1/k)

Where, φ = (transmitter EIRP)(4πR2)

Carrier to Noise Ratio

The Carrier to Noise Ratio, (Hz), of a link is calculated from

C/N0 = [(PTX Gtmax / LTLFTX)(1/LFSLA)(Grmax/ LRLFRXLPOL)]

/ [TA / LFRX + TF(1 - 1/LFRX ) + TR ](1/k)

C/N0 can also be interpreted as:

C/N0 = (transmitter EIRP)(1/path loss)(receiver gain/noise temperature)(1/k)

C/N0 can also be expressed in terms of power flux density

C/N0 = φ(λ2/4π)(receiver gain/noise temperature)(1/k)

Where, φ = (transmitter EIRP)(4πR2)

The (receiver gain/noise temperature) expression in C/N0 the receive equipment G/T and is a figure of merit for the quality of performance of the receiving equipment.

Carrier to Noise Ratio

The noise temperature of a receiver (LNB) depends on the contributions from the components within it. Thus, there is a contribution from the front end LNA, a contribution from the downconverter stage and a contribution from any output IF amplifiers. This can be expressed as:

TR = TLNA + TMX/GLNA + TIF/GLNA GMX

Calculate the LNB system noise temperature for an LNB comprising:

TLNA = 110K, GLNA = 50dB

TMX = 850K, GMX = -10dB or LMX = 10dB

TIF = 400K, GIF = 30dB

TR = 110 + 850/105 + 400/ 105 10-1

=110K

It is the high gain of the LNA which limits the overall noise temperature of the LNB to that of the LNA.

Receiver Noise Temperature Example

Calculate the uplink C/N0 for the following system:

VSAT parameters:

– Frequency, 14.4GHz– Transmit power, 25W– Loss between HPA and

antenna, 1.5dB– Antenna diameter, 1.2m– Antenna efficiency, 0.67

– Maximum pointing error, 0.2°

– VSAT-Satellite distance, 39100km

– Atmospheric wave attenuation, 0.3dB

Example C/N0 Uplink

Satellite Parameters:

– Antenna efficiency, 0.65

– Receiving beam aperture, 2.5°

– Receiver noise figure, 2.5dB

– Input losses, 1dB– Polarisation mismatch

loss, 0dB

– Thermodynamic temperature of connection, 290K

– Antenna noise temperature, 290K

From the VSAT, EIRP = (PTX Gtmax / LTLFTX)

PTX = 25W = 10*log(25) = 14dBW

GTmax = η(πD / λ)2 = η(π D f / c)2 = 0.67[(π*1.2*14.4*109)/(3*108)]2 = 29139

GTmax = 10*log(29139) = 43.4dB

LT = 12(αT / θ3dB)2 =12 (αTDfu / 70c)2 = 0.3

EIRP = 43.4 + 14 - 0.3 - 1.5 = 55. 6dBW

Attenuation in the uplink is LU = LFS LA

LFS = (4πRfu / c)2 = 5.55*1020 = 10*log(5.55*1020 ) = 207.4dB

LU = 207.4 + 0.3 = 207.7dB

For the satellite,

G/T = (Grmax/ LRLFRXLPOL)] / [TA / LFRX + TF(1 - 1/LFRX ) + TR ]

GTmax = η(πD / λ)2 = η(π70/θ3dB)2 = 0.65(3.14*70/2.5)2 = 37dBi

Example C/N0 Uplink

G/T = 37 - 2.5 -1 - 10log(290/101/10 + 290(1 - 1/ 101/10) + 290) = 5.9dBK-1

For the uplink, (C/N0)U = (EIRP)VSAT(1/LU)(G/ T)SL(1/k)

(C/N0)U = 55.5dBW - 207.7dB + 5.9dBK-1 + 10*log(1.38*10-23) = 82.3dBHz

In order to calculate / achieve a given link availability it is necessary to include rain margin in the link budget. In a temperate climate (Northern Europe), attenuation due to rain, ARAIN, is typically 10dB. This figure, at a frequency of 14GHz, would typically be exceeded for 0.01% of the time for an average year.

This gives LU = 207.7 + 10 = 217.7dB

(C/N0)U = 55.5dBW - 217.7dB + 5.9dBK-1 + 10*log(1.38*10-23) = 72.3dBHz

The ratio (C/N0)U for the uplink would be greater than the 72.3dBHz for 99.99% of the time in an average year.

Example C/N0 Uplink

Calculate the downlink C/N0 for the following system:

Satellite parameters:

– Frequency, 11.6GHz– Transmit power, 125W– Satellite Output Losses,

2 dB– Satellite EIRP, 50dBW

VSAT Parameters:

– VSAT-Satellite distance, 39400km

– Atmospheric attenuation, 0.3dB

– Antenna diameter, 1.2m

Example C/N0 Downlink

VSAT Parameters (Contd.):

– Antenna efficiency, 0.67

– Receiving beam aperture, 2.5°

– Receiver noise figure, 1.4dB

– Input losses, 0.5 dB– Polarisation mismatch

loss, 0dB– Thermodynamic

temperature of connection, 290K

– Antenna noise temperature, 290K

From the Satellite, EIRP = (PTX Gtmax / LTLFTX) = 50dBW

Attenuation in the uplink is LD = LFS LA

LFS = (4πRfu / c)2 = 3.66*1020 = 10*log(5.55*1020 ) = 205.6dB

LU = 205.6 + 0.3 = 205.9dB

For the VSAT,

G/T = (GRmax/ LRLFRXLPOL)] / [TA / LFRX + TF(1 - 1/LFRX ) + TR ]

GRmax = η(πD / λ)2 = η(π D f / c)2 = 0.67[(π*1.2*11.6*109)/(3*108)]2 = 14236

GRmax = 10*log(14236) = 41.5dB

LT = 12(αT / θ3dB)2 =12 (αTDfu / 70c)2 = 0.2

LFRX = 0.5dB

Antenna noise temperature, TA = TSKY / ARAIN + Tm(1-1/ ARAIN) + TGROUND

– = 20/1010/10 + 275(1-1/1010/10) + 45 = 295K G/T = 41.5 - 0.2 - 0.6 -10*log(295/100.5/10 + 290(1-1/ 100.5/10)+110 =

14.8dBK-1

Example C/N0 Downlink

For the downlink, (C/N0)D = (EIRP)SL(1/LU)(G/ T)ES(1/k)

(C/N0)D = 50 - 216.2 + 14.8 +228.6 = 77.1 dBHz

The ratio C/ND for the downlink will be greater than 77.1dBHz for 99.99% of the time during an average year.

Example C/N0 Downlink

For the complete VSAT to VSAT link

(C/N0)-1T = (C/N0)-1

U + (C/N0)-1D + (C/N) )-1

SAT

This link has been calculated for a single carrier case.

For multicarriers, the HPA, of the VSAT uplink and the Satellite are backed off from saturation. This is done so that Intermodulation products resulting from interaction between the carriers are reduced. When the HPA is backed off sufficiently, it is said to be operating in the linear region.

Example C/N0 Total

Digital Transmission - Modulation

AnalogSource

DigitalSource

SourceEncoder

Encryptionand / orScrambling

Time DivisionMultiplexer

ChannelEncoding

DigitalModulation VSAT

Rb, Bit Rate

Rc, Bit Rate (bits/s)

R, Symbol Rate (baud)

Digital Transmission - Channel Encoding

N = 2n

information data symbolsN = 2n

Encoded data symbols

rredundancy bits

ChannelEncoder

Input rateRb

Output rateRc

Code Rate, ρ = n / (n + r)

Channel encoding inserts redundancy (r redundant bits for n information bits) for purposes of error control and error correction.

Rb = information bit rate

Rc = channel (encoded) bit rate

Rc = Rb / ρ

Rc is larger than Rb

Digital Transmission - Demodulation

Symbol error probability is a function E/N0

E = energy per bit (J)

N0 = noise power spectral density (W/Hz)

Eb/N0 = (C/N0) / Rb

VSAT DEMOD &DETECTION

SYMBOL DETECTION

BIT DETECTION

Symbol errorProbability

Bit errorProbability

R