Radio Link budget

CHAPTER 12

System noise andcommunications link budg ets

12.3 System noise calculations (REVISION EEE 3)

Gain, G, often expressed in decibels, the ratio of the output to

input voltages (or powers): GdB = 20 log10Vo

Vi

.

There is a logarithmic unit of power which defines absolutepower in dB above a specified reference level.

If the reference power is 1 mW the units are dBm and if thereference is 1 W the units are dBW.

2 System noise and communications link budgets

Table 12.1 Signal power measures.

dBW Power level (W) dBm30 dBW 1,000 60 dBm20 dBW 100 50 dBm10 dBW 10 40 dBm0 dBW 1 30 dBm

−10 dBW 1/10 20 dBm−20 dBW 1/100 10 dBm−30 dBW 1/1,000 0 dBm−40 dBW 1/10,000 −10 dBm

If T = 290 K and B = 1 Hz, then the available noise power kTBis 1. 38 × 10−23 × 290 × 1 = 4 × 10−21 W/Hz.

Expressing this (in a 1 Hz bandwidth) in dBm gives a noisepower spectral density or thermal noise floor of −174 dBm/Hz.

Absolute power, in dBm, can be scaled by the dB gain ofamplifiers or loss of attenuators to give output power, again indBm.

System noise calculations (REVISION EEE 3) 3

12.3.1 Noise temperature

A convenient way to specify noise power is via thermal noisetemperature, Te,:

Te =N

kB(K) (12.29)

Note the system noise temperature is not the actual temperature.

It is the temperature which an ideal resistor would need to be atin order to account for the extra noise observed at the device’soutput over and above the actual input noise.

The total available output noise is thus:

N = (kTs B + kTe B)G (12.30)

where the first term is the input noise contribution and thesecond term is the subsystem contribution, see below.

Figure 12.12 Equivalent noise temperature (Te) of an amplifier.


EXAMPLE 12.2Calculate the output noise of the amplifier with gain 20 dB,noise bandwidth 1 MHz, equivalent noise temperature 580 Kand source noise temperature 290 K.

The available NPSD from the matched source at temperature T0

is given by:

Gn,s( f ) = kT0 (W/Hz)

(T0 denotes 290 K.) The available NPSD from the equivalentresistor at temperature Te is:

Gn,e( f ) = kTe (W/Hz)

The total equivalent noise power at the input is:

Nin = k(T0 + Te)B (W)

The total noise power at the amplifier output is thus:

Nout = Gk(T0 + Te)B (W)

= 1020/10 × 1. 38 × 10−23 × (290 + 580) × 1 × 106

= 1. 2 × 10−12 W = − 119. 2 dBW or − 89. 2 dBm

Figure 12.12 Equivalent noise temperature (Te) of an amplifier.


12.3.3 Noise factor and noise figure

The noise factor, f , is the ratio of SNR at the system input toSNR at the output when the input noise corresponds to atemperature of 290 K, i.e.:

f =(S/N )i

(S/N )o

Ni = k 290 B

(12.43)

(S/N )i =signal power at input, Si

k 290 B(12.44(a))

(S/N )o =signal power at output, So

k(290 + Te)B G(12.44(b))

f is thus a figure of merit for performance comparison.

f =

Si

(k 290 B)GSi

[Gk(290 + Te)B]

=290 + Te

290(12.45(a))

f = 1 +Te

290(12.45(b))


This makes it possible to calculate system noise temperatureev en when Ts ≠ 290 K by converting the noise factor into anequivalent noise temperature:

Te = ( f − 1)290 (K) (12.46)

making no assumption about source temperature.

The noise effects due to several subsystems can be cascadedbefore converting to noise temperatures.

The equivalent noise temperature of cascaded subsystems is:

Te = Te1 +Te2

G1+

Te3

G1G2(K) (12.47)

or

f = f1 +f2 − 1

G1+

f3 − 1

G1G2(12.49)

This is called the Friis noise formula. The final system noisetemperature is then calculated from equations (12.46) andTsys in = Ts + Te.

Traditionally noise factor is quoted in decibels, i.e.:

F = 10 log10 f (dB) (12.50)

and in this form is called the noise figure. You must rememberto convert F to a ratio before using any of the abovecalculations.


The noise factor of lossy (passive) devices is related to their‘gain’, Gl ,:

f = 1/Gl (12.51)

when f and Gl are expressed as ratios, or by dB values:

F = − Gl (dB) (12.52)

A transmission line with 10% power loss has a noise factor of:

f = 1/Gl = 1/0. 9 = 1. 11 (12.53)

A mixer with a conversion loss of 6 dB has a noise figure of:

F = − Gl = 6 dB (12.54)


Table 12.2 Comparison of noise performance measures.

Te f F Comments0 K 1.00 0 dB Perfect (i.e. noiseless) device

10 K 1.03 0.2 dB Excellent LNA100 K 1.34 1.3 dB Good LNA290 K 2.00 3.0 dB Typical LNA500 K 2.72 4.4 dB Typical amplifier

1000 K 4.45 6.5 dB Poor quality amplifier10,000 K 35.50 15.5 dB Noise source temperature

All practical amplifiers introduce some noise.


EXAMPLE 12.3Figure 12.15 shows a simple superheterodyne receiver with afront end low noise amplifier (LNA), a mixer and two IF stages.The characteristics of the individual subsystems are:

Device Gain (or conversion loss) Te

LNA 12 dB 50 KMixer −6 dB

IF Amp 1 20 dB 1000 KIF Amp 2 30 dB 1000 K

Calculate the noise power at the second IF amplifier output in a5.0 MHz bandwidth.

Figure 12.15 Superheterodyne receiver.


Using equation Tsys in = Ts + Te:

Tsyst in = Ts + Te1 +Te2

G1+

Te3

G1G2+

Te4

G1G2G3

G1 = 12 dB = 15.8, G2 = −6 dB = 0.25 (= Gl),G3 = 20 dB = 100, Ts = 100 K, Te1 = 50 K,Te2 = T ph(1 − Gl)/Gl = 290(1 − 0. 25)/0. 25 = 870 K,Te3 = 1000 K, Te4 = 1000 K,

Therefore:

Tsyst in = 100 + 50 +870

15. 8+

1000

15. 8 × 0. 25+

1000

15. 8 × 0. 25 × 100= 461 K

In the 5.0 MHz bandwidth the output noise power is:

Nout = kTsyst in BG1G2G3G4

= 1. 38 × 10−23 × 461 × 5 × 106 × 15. 8 × 0. 25 × 100 × 1000 = − 49. 0 dBm

Reversing the LNA and mixer gives:

Tsyst in = 100 + 870 +50

0. 25+

1000

0. 25 × 15. 8+

1000

0. 25 × 15. 8 × 100= 1426 K

Degrading the result by a factor of 3, indicating the importanceof incorporating the LNA at the input.


EXAMPLE 12.4The output noise of the EXAMPLE 12.3 system is recalculatedusing noise factors instead of noise temperatures.

The noise factor of the entire system, equation (12.49), is:

f = f1 +( f2 − 1)

G1+

( f3 − 1)

G1G2+

( f4 − 1)

G1G2G3

= 100.7/10 +106/10 − 1

15. 8+

106.5/10 − 1

15. 8 × 0. 25+

106.5/10 − 1

15. 8 × 0. 25 × 100= 1. 17 + 0. 19 + 0. 88 + 0. 01 = 2. 25

(or F = 3. 5 dB)

The equivalent system noise temperature at the LNA input is:

Te = ( f − 1)290

= (2. 25 − 1)290 = 362 K

and the total output noise power in 5.0 MHz is:

Nout = k(Ts + Te)BGsyst

= 1. 38 × 10−23 × (100 + 362) × 5 × 106 × 1056/10

= 1. 27 × 10−8 W = − 49. 0 dBm


12.4 Radio communication link budg ets

Link budgets refer to the calculation of received signal-to-noiseratio given a specification of transmitted power, transmissionmedium attenuation and/or gain, and all sources of noise.

The calculation is based on applying Friis formula. For radiosystems the signal energy is usually lost due radiation otherthan towards the receiver.

First we must review important antenna concepts.

Radio communication link budgets 13

12.4.1 Antenna gain, effective area and efficiency

The simplified relation between antenna gain and its effectivearea is:

G =4πae

λ 2(12.62)

where ae is the effective antenna receiving area.

We focus on the gain and effective area in the direction ofmaximum radiation.

The feed struts, etc reduce the physical area, Aph, by anefficiency factor, η ap and give the effective aperture, Ae, i.e.:

Ae = η ap Aph (m2) (12.63(a))

These aperture losses are in addition to the ohmic losses whichwhen applied to the effective aperture gives the antenna’seffective area, i.e.:

ae = ηΩ Ae (m2) (12.63(b))

Effective area is thus related to physical area by:

ae = η apηΩ Aph (m2) (12.64)


12.4.2 Free space and plane earth signal budg ets

For a free space radio communication link the received powerdensity radiated from an isotropic radiator is:

Wlossless isotrope =PT

4πR2(W m−2) (12.65)

For a transmitter with gain GT the receiver power density is:

W =PT

4πR2GT (W m−2) (12.66)

The carrier power at the receive antenna terminals is:

PR = C = W ae (W) (12.67)

substituting equation (12.66) for W , giv es:

C =PT

4πR2GT ae (W) (12.68)

Using equation (12.62) to replace the ae term by GR:

C =PT

4πR2GT GR

λ 2

4π(W) = PT GT GR

λ4πR

2

= PR (W)

where GR is the gain of the receiving antenna.

Figure 12.20 Fr ee space propagation.


This equation (12.69) can be rewritten as:

C = PT GT

λ4πR

2

GR (W) (12.70)

To giv e the basic radio free space transmission loss formula.

PT GT is the effective isotropic radiated power (EIRP) and the[λ /(4πR)]2 is the free space path loss (FSPL).

FSPL is a function of wav elength because it converts thereceiving antenna area to gain, as well as incorperating thespreading loss.

Equation (12.70) expressed in decibel is:

C = EIRP − FSPL + GR (dBW) (12.71(a))

where:

EIRP = 10 log10 PT + 10 log10 GT (dBW)

and:

FSPL = 20 log10

4πR

λ

(dB) (12.71(c))


SIMPLIFIED EXAMPLE 12.5Calculate the power density at a distance of 20 km from amicrowave antenna with a directivity of 42.0 dB, an ohmicefficiency of 95% for a 4 GHz 25 dBm transmitter.

G = ηΩD

= 10 log10 ηΩ + DdB (dB)

= 10 log10 0. 95 + 42. 0 = − 0. 2 + 42. 0 = 41. 8 dB

The received power density is EIRP − rad. loss so:

W =PT

4πR2GT

= PT + GT − 10 log10 (4πR2) dBm/m2

= 25 + 41. 8 − 10 log10 (4π × 20, 0002)

= − 30. 2 dBm/m2 = − 60. 2 dBW/m2

The radiation intensity is given as:

I =Prad

4πD =

ηΩPT4π

D =0. 95 × 10

25

104π

× 10

42

10 mW/rad2

= 3. 789 × 105 mW/rad2

ERMS = √ W Zo = √ 9. 52 × 10−7 × 377 V/m = 18. 9 mV/m

For an identical receiving antenna 20 km from transmitter, thereceived carrier power, C, is calculated via:

λ =c

f=

3 × 108

4 × 109= 0. 075 m

The antenna effective area, equation (12.62), is:

ae = GRλ 2

4π= 10

41.8

10

(0. 075)2

4π

= 6. 775 m2

C = W ae = 9. 52 × 10−7 × 6. 775 = 6. 45 × 10−6 W = − 21. 9 dBm


12.4.3 Antenna temperature and radio noise budg ets

The receiving antenna temperature, Tant , is:

Tant =available NPSD at antenna terminals

Boltzmann′s constant=

GN ( f )

k(K)

where GN ( f ) is white and the NPSD is one sided.

Below 30 MHz antenna noise is dominated by the radiationfrom lightning discharges.

Above 1 GHz galactic noise is relatively weak. Atmosphericand ground noise is approximately flat with frequency up toabout 10 GHz.

As elevation increases from 0° to 90° the thickness ofatmosphere through which the beam passes decreases as doesthe influence of the ground, leading to a decrease in receivedthermal noise.

In the 1 - 10 GHz frequency range a zenith-pointed antenna inclear sky may have a noise temperature close to the cosmicbackground temperature of 3 K.

Above 10 GHz resonance effects (of water vapour molecules at22 GHz and oxygen molecules at 60 GHz) increase atmosphericattenuation and thermal noise.


Figure 12.25 Antenna sky noise temperature as a function of frequency and antenna elevation

angle (source: Kraus, 1966, reproduced with his permission).


SIMPLIFIED EXAMPLE 12.7For a 10 GHz terrestrial LOS link The received antenna signalpower is −40. 0 dBm, overall receiver noise figure is 5.0 dB andthe bandwidth is 20 MHz.Estimate the clear sky antenna terminal CNR for an antennaohmic efficiency of 95% at 280 K.

From Fig. 12.25 the aperture temperature, T A, at 10 GHz for ahorizontal link is ˜100 K. Including the physical temperatureT ph gives:

Tant = T AηΩ + T ph(1 − ηΩ)

= 100 × 0. 95 + 280 (1 − 0. 95) = 95 + 14 = 109 K

N = kTB = 1. 38 × 10−23 × 109 × 20 × 106 = 3. 01 × 10−14 W

= − 135. 2 dBW

The clear sky antenna CNR for a received −70 dBW is:

C

N= − 70. 0 − (− 135. 2) = 65. 2 dB

The receiver noise temperature, equation (12.46) is:

Te = ( f − 1)290 = (105

10 − 1)290 = 627 K

The system noise temperature (at the antenna output) is:

Tsyst in = Tant + Te = 109 + 627 = 736 K

The effective system noise power (at the antenna output):

N = kTsyst in B = 1. 38 × 10−23 × 736 × 20 × 106 = − 126. 9 dBW

The effective carrier to noise ratio is thus:

C

N eff

= C − N (dB) = − 70. 0 − (− 126. 9) = 56. 9 dB


12.4.5 Multipath fading and diversity reception

Multipath fading occurs to varying extents in many differentradio applications. Figure shows how multipath propagationmay occur on a point-to-point line-of-sight link with reflectionsfrom ground and atmosphere.

The fading process is governed by atmospheric changes. If thefrequency response of the channel is essentially constant overthe operating bandwidth fading is called flat since all frequencycomponents of a signal are subjected to the same fade.

Figure 12.27 Multipath in line of sight terrestial link due to: (a) direct path plus (b) ground

reflection and/or (c) reflection from (or refraction through) a tropospheric layer.


When several or more propagation paths exist the fading ofsignal amplitude obeys Rayleigh statistics then the frequencyresponse of the channel may change rapidly on a frequencyscale comparable to signal bandwidth.

Here the fading is said to be frequency selective and thereceived signal exhibits amplitude and phase distortions.

Figure 12.28 Amplitude response of a frequency selective channel for 3-ray multipath

propagation with ray amplitudes and delays of: 1.0, 0 ns; 0.9, 0.56 ns; 0.1, 4.7 ns.


Diversity reduces the necessary dB fade margin allocation bycompensating for the fades. Three types of diversity system,are space, frequency, and angle. In all cases the essentialassumption is that it is unlikely that both main and diversitychannel will suffer severe fades at the same instant.

Selecting the channel with largest CNR, or combining channelswill result in improved overall CNR.

Figure 12.29 Three types of diversity arrangements to combat multipath fading.


CHAPTER 14

Fixed point microwavecommunications

24 Fixed point microwave communications

14.2 Terrestrial microwave links

Many wideband point-to-point radio communications linksemploy microwave carriers in the 1 to 20 GHz frequency range.Figure shows the trunk routes of the UK microwave network.

Antennas are located on high ground to avoid large buildings orhills and repeaters are used every 40 to 50 km to compensatefor path loss.

Figure 14.1 The UK microwave communications wideband distribution network.

Terrestrial microwave links 25

The following points can be made about these links:

1. Microwave energy does not follow the curvature of the earthin the way that MW and LW transmissions do. Microwavetransmissions are thus restricted to line of sight (LOS).

2. Microwave transmissions are particularly well suited topoint-to-point communications using narrow beam, highgain, antennas.

3. At about 1 GHz circuit design techniques change from usinglumped to distributed elements. Above 20 GHz it becomesexpensive to generate big amounts of microwave power.

The bands in current use are near 2 GHz, 4 GHz, 6 GHz, 11GHz and 18 GHz.

(On a 6 GHz link with a hop distance of 40 km the free spacepath loss is 140 dB. With transmit and receive antenna gains of40 dB, however, the transmission loss reduces to 60 dB.)


14.2.1 Analogue systems

In the UK microwave links were widely installed during the1960s for analogue FDM telephony. In these systems eachallocated frequency band is subdivided into 30 MHz radiochannels.

Figure shows how the 500 MHz band is divided into 16separate channels on 29.65 MHz centre spacings. Adjacentradio channels use orthogonal antenna polarisations, horizontal(H) and vertical (V), to reduce crosstalk.

The 500 MHz bandwidth microwave link can accommodatetraffic simultaneously in all sixteen 30 MHz channels.

Figure 14.2 Splitting of a microwave frequency allocation into radio channels.


In Figure 14.4 f1 is used for the first hop but f ′1, Figure 14.2, is

used on the second hop, etc.

Figure 14.4 Fr equency allocations on adjacent repeaters.

Figure 14.2 Splitting of a microwave frequency allocation into radio channels.


In the superheterodyne receiver a microwave channel filter,centred on the appropriate radio channel, extracts the 30 MHzchannel from the transmissions. The signal is mixed down toan intermediate frequency (IF) for additional filtering andamplification.

Figure 14.3 Extraction (dropping) of a single radio channel in a microwave repeater.


14.2.2 Digital systems

Digital microwave (PSTN) links operate at 140 Mbit/s on acarrier frequency of 11 GHz using QPSK modulation.

The 64-QAM system offers practical spectral efficiency of 4 to5 bit/s/Hz implying that the 30 MHz channel can support a 140Mbit/s of multiplexed telephone traffic signal.


Figure is a schematic of a digital regenerative repeater,assuming DPSK modulation, for a single 30 MHz channel.

Here the circulator and channel filter access the part of themicrowave spectrum where the signal is located.

Digital systems have a major advantage over analogue systemsas they operate at a much lower CNR.

Figure 14.7 Digital DPSK reg enerative repeater for a single 30 MHz radio channel.


14.3 Fixed point satellite communications

Satellites are one of the three most important developments intelecommunications. Geostationary satellites have an orbitalradius of 42,164 km and earth surface altitude of 35,786 km.

Other classes of satellite orbit include highly inclined highlyelliptical (HIHE) orbits, polar orbits and low earth orbits(LEOs).

Figure 14.25 Selection of especially useful satellite orbits: (a) geostationary (GEO); (b) highly

inclined highly elliptical (HIHEO); (c) polar orbit; and (d) low earth (LEO).


For fixed point communications the geostationary orbit is themost important, for the following reasons:

1. Its high altitude means that a single satellite is visible froma large fraction of the earth’s surface.

2. No earth station tracking of the satellite is needed.3. No handover from one satellite to another is necessary

since the satellite never moves or sets.4. Three satellites give almost global coverage.5. As there is no relative motion no Doppler shifts occur in

the received carrier.

Figure 14.26 Coverage areas as a function of elevation angle for a satellite with global beam

antenna (from CCIR Handbook, 1988, reproduced with the permission of ITU).

Fixed point satellite communications 33

The following advantages apply to geostationary satellites:

1. The communications channel can be either broadcast orpoint-to-point.

2. New links can be made by pointing an antenna at thesatellite.

3. The cost of transmission is independent of distance.4. Wide bandwidths are available, limited only by the

transponder electronics and noise performance.

Figure 14.27 Global coverage (excepting polar regions) from 3 geostationary satellites.

(Approximately to scale, innermost circle represents 81° parallel.)


Despite their very significant advantages, geostationarysatellites do suffer some disadvantages:

1. Polar regions are not covered.2. High altitude means large (200 dB) FSPL.3. High altitude results in long propagation delays

(approximately 1/8 s).


14.3.3 Satellite link budg ets

On the uplink the earth pointing satellite antenna has a noisetemperature of 290 K.

The noise temperature of the earth station antenna, which looksat the sky, is usually in the range 5 to 100 K for frequenciesbetween 1 and 10 GHz, Figure 12.25.

Above 10 GHz resonance effects of water vapour and oxygenmolecules, at 22 GHz and 60 GHz respectively degrade thereceiver performance.

Figure 14.30 Antenna aperture temperature, T A, in clear air (pressure one atmosphere, surface

temperature 20°C, surface water vapour concentration 10 g/m3). (Source: ITU-R

Handbook of Radiometeorology, 1996, reproduced with the permission of the ITU.)


For wideband multiplex telephony, the front end of the earthstation receiver is cooled so that its noise temperature, Te, isvery much smaller than 290 K. This achieves a noise floor thatis very much lower than −174 dBm/Hz.

The gain, G, of the cooled low noise amplifier boosts thereceived signal so that the following receiver amplifiers canoperate at room temperature.

Operating frequencies for satellite communication systems arelimited at the low end to >1 GHz by galactic noise and at thehigh end to <15 GHz by atmospheric noise and rain attenuation.


The satellite transponder is the critical component as itstransmitter is power limited by the onboard satellite powersupply giving the downlink the worst power budget.

EIRP is also dependent on satellite antenna design and inparticular the earth ‘footprint’.

Spot beams covering only a small part of the earth have highgain giving EIRP values of 30 to 40 dBW.

The satellite transponder must also be small and light. Thetransponder’s high power amplifier (HPA) must thus operate ata high power to maintain adequate downlink CNR, i.e. in a non-linear saturated region.

Figure 14.31 Contours of EIRP with respect to EIRP on antenna boresight.


EXAMPLE 14.5An 11.7 GHz geostationary satellite downlink has 25 W outputpower, a 20 dB gain antenna with 2 dB feeder losses. The earthstation, 38,000 km from the satellite, has a 15 m diameterreceive antenna, with 55% efficiency, a low noise (cooled)amplifier with a receiver noise temperature of 100 K.

If an Eb/N0 of 20 dB is required for adequate BER performancewhat is the maximum BPSK bit rate?

Transmitter power = 14 dBWEIRP = 14 + 20 − 2 = 32 dBW = 62 dBmFree space loss = 20 log10 (4π38 × 106)/0. 0256 = 205.4 dB

Receiver antenna GR =4πλ 2

πd2

40. 55 = 62.7 dB

Received power level = 62 − 205. 4 + 62. 7 dBm = −80. 7 dBm

Receiver noise at 100 K noise temperature = −178. 6 dBm/HzIf BTo = 1. 0 then Eb/N0 = C/N .

Av ailable margin for Eb/N0 and modulation bandwidth= 178. 6 − 80. 7 = 97. 9 dB Hz.

If Eb/N0 = 20 dB then the margin for modulation = 77.9 dB Hz= 61.6 MHz.With BPSK at 1 bit/s/Hz then the modulation rate can be 61.6Mbit/s.


Alternatively using G/T calculations:If GR = 62.7 dB and Tsyst = 100 K then G/T = 42.7 dB/KThen radiated power at receiver antenna = + 62 − 205. 4 dBm= −143. 4 dBmPower at receiver input = − 143. 4 + 42. 7 dBm/K = −100. 7dBm/KBoltzmann’s constant = − 198. 6 dBm/Hz/KDifference = 97.9 dBHzAllowing 20 dB for acceptable Eb/N0 leaves 77.9 dBHz whichwill support a 61.6 Mbit/s BPSK symbol rate.

Radio Link budget

Documents

Transcript of Radio Link budget