Wireless Communications and Networks

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Transcript Wireless Communications and Networks

EC 723 Satellite Communication Systems Mohamed Khedr http://webmail.aast.edu/~khedr

Syllabus

 Tentatively Week 1 Week 2 Week 3 Week 4 Week 5 Week 6 Week 7 Week 8 Week 9 Week 10 Week 11 Week 12 Week 13 Week 14 Week 15 Overview Orbits and constellations: GEO, MEO and LEO Satellite space segment, Propagation and satellite links , channel modelling

Satellite Communications Techniques

Satellite error correction Techniques Multiple Access I Multiple access II Satellite in networks I INTELSAT systems , VSAT networks, GPS GEO, MEO and LEO mobile communications INMARSAT systems, Iridium , Globalstar, Odyssey Presentations Presentations Presentations Presentations Presentations

Radio Propagation: Atmospheric Losses

 Different types of atmospheric losses can perturb radio wave transmission in satellite systems: 

Atmospheric absorption;

 

Atmospheric attenuation; Traveling ionospheric disturbances.

Radio Propagation:

Atmospheric Absorption

   Energy absorption by atmospheric gases, which varies with the frequency of the radio waves.

Two absorption peaks are observed (for 90º elevation angle):   22.3 GHz from resonance absorption in water vapour (H 2 O) 60 GHz from resonance absorption in oxygen (O 2 ) For other elevation angles:  [AA] = [AA] 90 cosec  Source:

Satellite Communications

, Dennis Roddy, McGraw-Hill

Radio Propagation:

Atmospheric Attenuation

  Rain is the main cause of atmospheric attenuation (hail, ice and snow have little effect on attenuation because of their low water content).

Total attenuation from rain can be determined by:  A =  L [dB]   where  [dB/km] is called the specific attenuation, and can be calculated from specific attenuation coefficients in tabular form that can be found in a number of publications; where L [km] is the effective path length of the signal through the rain; note that this differs from the geometric path length due to fluctuations in the rain density.

Signal Polarisation:

Cross-Polarisation Discrimination

  Depolarisation can cause interference where orthogonal polarisation is used to provide isolation between signals, as in the case of frequency reuse.

The most widely used measure to quantify the effects of polarisation interference is called Cross-Polarisation Discrimination (XPD): 

XPD = 20 log (E 11 /E 12 )

  To counter depolarising effects circular polarising is sometimes used.

Alternatively, if linear polarisation is to be used, polarisation tracking equipment may be installed at the antenna.

Source:

Satellite Communications

, Dennis Roddy, McGraw-Hill

Illustration of the various propagation loss mechanisms on a typical earth-space path

The ionosphere can cause the electric vector of signals passing through it to rotate away from their original polarization direction, hence causing signal depolarization.

The absorptive effects of the atmospheric constituents cause an increase in sky noise to be observed by the receiver the sun (a very “hot” microwave and millimeter wave source of incoherent energy), an increased noise contribution results which may cause the C/N to drop below the demodulator threshold.

Refractive effects (tropospheric scintillation) cause signal loss.

The ionosphere has its principal impact on signals at frequencies well below 10 GHz while the other effects noted in the figure above become increasingly strong as the frequency of the signal goes above 10 GHz

Atmospheric attenuation

Attenuation of the signal in % 50 Example: satellite systems at 4-6 GHz e 40 30 20 10 rain absorption fog absorption atmospheric absorption 5 ° 10° 20 ° 30 ° elevation of the satellite 40 ° 50 °

Signal Transmission

Link-Power Budget Formula

   Link-power budget calculations take into account all the gains and losses from the transmitter, through the medium to the receiver in a telecommunication system. Also taken into the account are the attenuation of the transmitted signal due to propagation and the loss or gain due to the antenna.

The decibel equation for the received power is: 

[P R ] = [EIRP] + [G R ] - [LOSSES]

Where:   [P R ] = received power in dBW [EIRP] = equivalent isotropic radiated power in dBW   [G R ] = receiver antenna gain in dB [LOSSES] = total link loss in dB dBW = 10 log 10 (P/(1 W)), where P is an arbitrary power in watts, is a unit for the measurement of the strength of a signal relative to one watt.

Link Budget parameters

         Transmitter power at the antenna Antenna gain compared to isotropic radiator EIRP Free space path loss System noise temperature Figure of merit for receiving system Carrier to thermal noise ratio Carrier to noise density ratio Carrier to noise ratio

Signal Transmission

Equivalent Isotropic Radiated Power

    An isotropic radiator is one that radiates equally in all directions.

The power amplifier in the transmitter is shown as generating P T watts.

A feeder connects this to the antenna, and the net power reaching the antenna will be P T minus the losses in the feeder cable, i.e. P S .

The power will be further reduced by losses in the antenna such that the power radiated will be P RAD (< P T ).

(a) Transmitting antenna

Source:

Satellite Communications

, Dennis Roddy, McGraw-Hill

Antenna Gain

  We need directive antennas to get power to go in wanted direction.

Define Gain of antenna as increase in power in a given direction compared to isotropic antenna.

G

(  ) 

P

( 

P

0 / ) 4  • P(  ) is variation of power with angle.

• G(  ) is gain at the direction  .

• P 0 is total power transmitted.

• sphere =

4

 solid radians

EIRP - 1

     An isotropic radiator is an antenna which radiates in all directions equally Antenna gain is relative to this standard Antennas are fundamentally passive  No additional power is generated  Gain is realized by focusing power Effective Isotropic Radiated Power (EIRP) is the amount of power the transmitter would have to produce if it was radiating to all directions equally Note that EIRP may vary as a function of direction because of changes in the antenna gain vs. angle

EIRP - 2

    The output power of a transmitter HPA is: P out watts Some power is lost before the antenna: P t =P P t out / L t watts reaches the antenna = Power into antenna The antenna has a gain of:

P out

HPA

L t

G t relative to an isotropic radiator This gives an effective isotropic radiated power of: EIRP = P t G t watts relative to a 1 watt isotropic radiator

EIRP P t

Received Power

• We can rewrite the power flux density now considering the transmit antenna gain: 

F

EIRP

4 

R

2 

P t G t

4 

R

2 W/m 2

The power available to a receive antenna of area

A r

m

2

we get:

P r

F

x

A r

P t G t

4 

R A r

2

Effective Aperture

 Real antennas have effective flux collecting areas which are LESS than the physical aperture area.

 Define Effective Aperture Area Ae:

A

e 

A phy

x  Where A phy is actual (physical) aperture area.

 = aperture efficiency Very good: 75% Typical: 55% • Antennas have (maximum) gain G related to the effective aperture area as follows:

Gain

 4   2

A e

Back to Received Power…

 The power available to a receive antenna of effective area A r = A e m 2 is:

P r

F

x

A r

P t G t

4 

R A e

2 Where A r = receive antenna effective aperture area = A e

G r

 4   2

A e

Inverting…

A e

G r

4   2

Back to Received Power…

P r

P t G t G r

 4 

R

2 Friis Transmission Formula • The inverse of the term at the right referred to as “Path Loss”, also known as “Free Space Loss” (Lp):

L p

 4 

R

 2 Therefore…

P r

P t G t G r L p

More complete formulation

   Demonstrated formula assumes idealized case.

Free Space Loss (L p ) represents spherical spreading only.

Other effects need to be accounted for in the transmission equation:       L a L ta L ra L pol = Losses due to attenuation in atmosphere = Losses associated with transmitting antenna = Losses associates with receiving antenna = Losses due to polarization mismatch L other = (any other known loss - as much detail as available) Lr = additional Losses at receiver (after receiving antenna)

P r

P t G t G r L p L a L ta L ra L pol L other L r

Signal Transmission

Link-Power Budget Formula Variables

     Link-Power Budget Formula for the received power [P R ]: 

[P R ] = [EIRP] + [G R

] - [LOSSES] The equivalent isotropic radiated power [EIRP] is:  

[G R

] is the receiver antenna gain in dB.

[LOSSES] = [FSL] + [RFL] + [AML] + [AA] + [PL], where:  [EIRP] = [P S ] + [G] dBW, where: [P S ] is the transmit power in dBW and [G] is the transmitting antenna gain in dB.

(in watts)     [FSL] = free-space spreading loss in dB = P T /P R [RFL] = receiver feeder loss in dB [AML] = antenna misalignment loss in dB [AA] = atmospheric absorption loss in dB [PL] = polarisation mismatch loss in dB The major source of loss in any ground-satellite link is the free-space spreading loss.

Link Power Budget

Tx

EIRP

Transmission:

HPA Power Transmission Losses (cables & connectors) Antenna Gain Antenna Pointing Loss Free Space Loss Atmospheric Loss (gaseous, clouds, rain) Rx Antenna Pointing Loss

Reception:

Antenna gain Reception Losses (cables & connectors) Noise Temperature Contribution Rx

P r

Translating to dBs

 The transmission formula can be written in dB as:

P r

EIRP

L ta

L p

L a

L pol

L ra

L other

G r

L r

 This form of the equation is easily handled as a spreadsheet (additions and subtractions!!)  The calculation of received signal based on transmitted power and all losses and gains involved until the receiver is called “Link Power Budget”, or “Link Budget”.

 The received power Pr is commonly referred to as “Carrier Power”,

C

.

Link Power Budget

Tx

EIRP

Transmission: +

HPA Power - Transmission Losses (cables & connectors) + Antenna Gain - Antenna Pointing Loss - Free Space Loss - Atmospheric Loss (gaseous, clouds, rain) - Rx Antenna Pointing Loss Now all factors are accounted for as additions and subtractions

Reception:

+ Antenna gain - Reception Losses (cables & connectors) + Noise Temperature Contribution Rx

P r

Easy Steps to a Good Link Power Budget

    First, draw a sketch of the link path   Doesn’t have to be artistic quality Helps you find the stuff you might forget Next, think carefully about the system of interest   Include all significant effects in the link power budget Note and justify which common effects are insignificant here Roll-up large sections of the link power budget    Ie.: TXd power, TX ant. gain, Path loss, RX ant. gain, RX losses Show all components for these calculations in the detailed budget Use the rolled-up results in build a link overview Comment the link budget  Always, always,

always

use units on parameters (dBi, W, Hz ...)  Describe any unusual elements (eg. loss caused by H 2 0 on radome)

Simple Link Power Budget

Parameter

Frequency

Transmitter

Transmitter Power Modulation Loss Transmission Line Loss

Transmitted Power Transmit Antenna Value

11.75 40.00 3.00 0.75

Totals Units

GHz dBm dB dB

36.25 dBm

Diameter Aperture Efficiency

Transmit Antenna Gain Slant Path

Satellite Altitude Elevation Angle Slant Range Free-space Path Loss Gaseous Loss Rain Loss (allocated)

Path Loss

0.5 0.55 35,786 14.5 41,602 206.22 0.65 m none

33.18 dBi

km degrees km dB dB 3.50 dB

210.37 dB Parameter Receive Antenna

Random Loss Diameter Aperture Efficiency

Value

0.50 1.5 0.6

Totals Units

dB m none Gain Polarization Loss

Effective RX Ant. Gain Received Power Summary

Transmitted Power Transmit Anntenna Gain EIRP Path Loss Effective RX Antenna Gain Received Power 43.10 0.20 36.25 33.18 dBi dB

42.40 dB -98.54 dBm

dBm dBi 69.43 dBmi 210.37 dB 42.4 dBi -98.54 dBm

Why calculate Link Budgets?

   System performance tied to operation thresholds.

Operation thresholds C work properly.

min tell the minimum power that should be received at the demodulator in order for communications to Operation thresholds depend on:  Modulation scheme being used.

     Desired communication quality.

Coding gain.

Additional overheads.

Channel Bandwidth.

Thermal Noise power.

We will see more on these items in the next classes.

Closing the Link

 We need to calculate the Link Budget in order to verify if we are “closing the link”.

P r >= C min  Link Closed P r < C min  Link not closed  Usually, we obtain the “Link Margin”, which tells how tight we are in closing the link: Margin = P r – C min  Equivalently: Margin > 0 Margin < 0   Link Closed Link not closed

Carrier to Noise Ratios

  C/N:   carrier/noise power in RX BW (dB) Allows simple calculation of margin if: Receiver bandwidth is known  Required C/N is known for desired signal type C/N o :  carrier/noise p.s.d. (dbHz) Allows simple calculation of allowable RX bandwidth if required C/N is known for desired signal type  Critical for calculations involving carrier recovery loop performance calculations

System Figure of Merit

 G/T s :   RX antenna gain/system temperature Also called the System Figure of Merit, G/T s Easily describes the sensitivity of a receive system  Must be used with caution:   Some (most) vendors measure G/T s G/T s under ideal conditions only degrades for most systems when rain loss increases  This is caused by the increase in the sky noise component  This is in addition to the loss of received power flux density

System Noise Power - 1

     Performance of system is determined by C/N ratio.

Most systems require C/N > 10 dB.

(Remember, in dBs: C - N > 10 dB) Hence usually: C > N + 10 dB We need to know the noise temperature of our receiver so that we can calculate N, the noise power (N = P n ).

T n (noise temperature) is in Kelvins (symbol K):

T

  

T

0    273

T

   

T

0    32  5 9  273

 

System Noise Power - 2

System noise is caused by thermal noise sources   External to RX system   Transmitted noise on link Scene noise observed by antenna Internal to RX system The power available from thermal noise is:

N

kT s B

(dBW) where T B s k = Boltzmann’s constant = 1.38x10

-23 J/K(-228.6 dBW/HzK), is the effective system noise temperature, and is the effective system bandwidth

Noise Spectral Density

N = K.T.B per hertz):

N/B = N

0

is the noise spectral density (density of noise power

N N kT s B

0 

B

B

kT s

(dBW/Hz)  

N

0

= noise spectral density is constant up to 300GHz.

All bodies with Tp >0K radiate microwave energy.

System Noise Temperature

1) 2)   System noise power is proportional to system noise temperature Noise from different sources is uncorrelated (AWGN) Additive White Gaussian Noise (AWGN) Therefore, we can   So: Add up noise powers from different contributions Work with noise temperature directly

T s

T

T antenna

T LNA

T lineloss

T RX

But, we must:   Calculate the effective noise temperature of each contribution Reference these noise temperatures to the same location

Typical Receiver

(Source: Pratt & Bostian Chapter 4, p115)

Noise Model

(Source: Pratt & Bostian Chapter 4, p115) Noise is added and then multiplied by the gain of the device (which is now assumed to be noiseless since the noise was already added prior to the device)

Equivalent Noise Model of Receiver

(Source: Pratt & Bostian Chapter 4, p115) Equivalent model: Equivalent noise Ts is added and then multiplied by the equivalent gain of the device, G RF G m G IF (noiseless).

  

Calculating System Noise Temperature - 1

Receiver noise comes from several sources.

We need a method which reduces several sources to a single equivalent noise source at the receiver input.

Using model in Fig. 4.5.a gives:

P n

G

 

IF kT IF G IF G IF B

(IF)

G m kT m G m G RF B

(Mixer)

kB

T RF

T in

 (Front End)

Calculating System Noise Temperature - 2

 Divide by G IF G m G RF kB:

P n

G IF G m G RF kB

 

T RF

T in

T m G RF

T IF G m G RF

   If we replace the model in Fig. 4.5.a by that in Fig. 4.5b

P n

G IF G m G RF kT s B

Calculating System Noise Temperature - 3

Equate Eqns :

T

S    

T RF

T in

T m G RF

T IF G m G RF

    

Since C is invariably small, N must be minimized.

How can we make N as small as possible?

Reducing Noise Power

   Make B as small as possible – just enough bandwidth to accept all of the signal power (C ).

Make T S  as small as possible Lowest T RF   Lowest T in High G RF (How?) If we have a good low noise amplifier (LNA), i.e., low T RF , high G RF , then rest of receiver does not matter that much.

T

S    

T RF

T in

T G m RF

G T m IF G RF

   

T RF

T in

Reducing Noise Power

Discussion on T in

  Earth Stations: Antennas looking at space which appears cold and produces little thermal noise power (about 50K).

Satellites: antennas beaming towards earth (about 300 K):   Making the LNA noise temperature much less gives diminishing returns. Improvements aim reduction of size and weight.

Antenna Noise Temperature

   Contributes for T in Natural Sources (sky noise):      Cosmic noise (star and inter-stellar matter), decreases with frequency, (negligible above 1GHz). Certain parts of the sky have punctual “hot sources” (hot sky).

Sun (T  12000 f -0.75

K): point earth-station antennas away from it.

Moon (black body radiator): 200 to 300K if pointed directly to it.

Earth (satellite) Propagation medium (e.g. rain, oxygen, water vapor): noise reduced as elevation angle increases.

Man-made sources:   Vehicles, industrial machinery Other terrestrial and satellite systems operating at the same frequency of interest.

Antenna Noise Temperature

Useful approximation for Earth Station antenna temperature on clear sky (no rain):

Earth Station Antenna - Noise Temperature

50 45 40 35 30 25 20 15 0 10 20 30 40 50 60

Elevation Angle (degrees)

70 80 90 100

So many trade-offs !!!

G

 

D

 2  

C N

P r KT s B P r

P t G t G r L p L a L ta L ra L pol L other L r L p

 4 

R

 2

L a

F T

S    

T RF

T in

T m G RF

T IF G m G RF

  

Power Budget Example - 1

4.1.1 Satellite at 40,000 km (range) Transmits 2W Antenna gain Gt = 17 dB (global beam) Calculate: a. Flux density on earth’s surface b. Power received by antenna with effective aperture of 10m 2 c. Gain of receiving antenna.

d. Received C/N assuming Ts =152 K, and Bw =500 MHz 

EIRP

4 

R

2 

P t

4 

G t R

2  2 x 50 4  x(4x10 7 ) 2 (Solving in dB…)  4.97

x 10 15 W/m 2   143 dBW/m 2

EIRP

 (

Pt

Gt

)  3  17  20 dBW R 4  2  2 x log 10 ( 4 x 10 7  11

dB

) dB[meter] 2

F

 20  11  152   143 dBW/m2

Power Budget Example - 1

b. Received Power P r  F x A  (4.97x10

P r  4.97

x 10 14

W

  -15 ) x 10 133

dBW

(Solving in dB…)

P r

F

A

 (  143 )  10

P r

  133 dBW c. Gain given Ae = 10 m 2 and Frequency = 11GHz ( eqn. 4.7)

G r

 4   2

A e

 4π x 10 0 .

0273  52 .

3

dB

Power Budget Example - 1

b. System Noise Temperature N  P n 

KTB

 1 .

38 x 1 0  23 x 152 x 500 x 10 6

or

K dB

T dB

B dB

  228 .

6  21 .

82  86 .

99   119 .

7 9

dBW

C

C

/ 

N

P r  4.97

x 10 14

W

C

N

  133    133

dBW

(  119 .

79 )

C

/

N

  13 .

2

dB

Power Budget Example - 2 Generic DBS-TV:

Received Power

Transponder output power , 160 W Antenna beam on-axis gain Path loss at 12 GHz, 38,500 km path Receiving antenna gain, on axis Edge of beam Miscellaneous losses Received power, C 22.0 dBW 34.3 dB -205.7 dB 33.5 dB -3.0 dB -0.8 dB -119.7 dBW

Power Budget Example - 2

Noise power

Boltzmann’s constant, k -228.6 dBW/K/Hz System noise temperature, clear air, 143 K 21.6 dBK Receiver noise bandwidth, 20MHz dBHz Noise power, N 73.0 -134.0 dBW

C/N in clear air

Link margin over 8.6 dB threshold Link availability throughout US 14.3 dB 5.7 dB Better than 99.7 %

Thank you