Radio Link budget

CHAPTER 12
System noise and
communications link budgets
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 log10


Vo
Vi


.
There is a logarithmic unit of power which deﬁnes absolute
power in dB above a speciﬁed reference level.
If the reference power is 1 mW the units are dBm and if the
reference is 1 W the units are dBW.

2 System noise and communications link budgets
Table 12.1 Signal power measures.
dBW Power level (W) dBm
30 dBW 1,000 60 dBm
20 dBW 100 50 dBm
10 dBW 10 40 dBm
0 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 kTB
is 1. 38 × 10−23
× 290 × 1 = 4 × 10−21
W/Hz.
Expressing this (in a 1 Hz bandwidth) in dBm gives a noise
power spectral density or thermal noise ﬂoor of −174 dBm/Hz.
Absolute power, in dBm, can be scaled by the dB gain of
ampliﬁers or loss of attenuators to give output power, again in
dBm.

System noise calculations (REVISION EEE 3) 3
12.3.1 Noise temperature
A convenient way to specify noise power is via thermal noise
temperature, 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 at
in order to account for the extra noise observed at the device’s
output 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 the
second term is the subsystem contribution, see below.
Figure 12.12 Equivalent noise temperature (Te) of an ampliﬁer.

EXAMPLE 12.2
Calculate the output noise of the amplifier with gain 20 dB,
noise bandwidth 1 MHz, equivalent noise temperature 580 K
and 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 equivalent
resistor 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 to
SNR at the output when the input noise corresponds to a
temperature 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 temperature
even when Ts ≠ 290 K by converting the noise factor into an
equivalent noise temperature:
Te = ( f − 1)290 (K) (12.46)
making no assumption about source temperature.
The noise effects due to several subsystems can be cascaded
before 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 noise
temperature is then calculated from equations (12.46) and
Tsys 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 remember
to convert F to a ratio before using any of the above
calculations.

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 Comments
0 K 1.00 0 dB Perfect (i.e. noiseless) device
10 K 1.03 0.2 dB Excellent LNA
100 K 1.34 1.3 dB Good LNA
290 K 2.00 3.0 dB Typical LNA
500 K 2.72 4.4 dB Typical amplifier
1000 K 4.45 6.5 dB Poor quality amplifier
10,000 K 35.50 15.5 dB Noise source temperature
All practical amplifiers introduce some noise.

EXAMPLE 12.3
Figure 12.15 shows a simple superheterodyne receiver with a
front 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 K
Mixer −6 dB
IF Amp 1 20 dB 1000 K
IF Amp 2 30 dB 1000 K
Calculate the noise power at the second IF ampliﬁer output in a
5.0 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 = Tph(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 importance
of incorporating the LNA at the input.

EXAMPLE 12.4
The output noise of the EXAMPLE 12.3 system is recalculated
using 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 budgets
Link budgets refer to the calculation of received signal-to-noise
ratio given a speciﬁcation of transmitted power, transmission
medium attenuation and/or gain, and all sources of noise.
The calculation is based on applying Friis formula. For radio
systems the signal energy is usually lost due radiation other
than 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 effective
area is:
G =
4π ae
λ2
(12.62)
where ae is the effective antenna receiving area.
We focus on the gain and effective area in the direction of
maximum radiation.
The feed struts, etc reduce the physical area, Aph, by an
efficiency factor, ηap and give the effective aperture, Ae, i.e.:
Ae = ηap Aph (m2
) (12.63(a))
These aperture losses are in addition to the ohmic losses which
when applied to the effective aperture gives the antenna’s
effective area, i.e.:
ae = ηΩ Ae (m2
) (12.63(b))
Effective area is thus related to physical area by:
ae = ηapηΩ Aph (m2
) (12.64)

12.4.2 Free space and plane earth signal budgets
For a free space radio communication link the received power
density radiated from an isotropic radiator is:
Wlossless isotrope =
PT
4π R2
(W m−2
) (12.65)
For a transmitter with gain GT the receiver power density is:
W =
PT
4π R2
GT (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, gives:
C =
PT
4π R2
GT ae (W) (12.68)
Using equation (12.62) to replace the ae term by GR:
C =
PT
4π R2
GT GR
λ2
4π
(W) = PT GT GR


λ
4π R


2
= PR (W)
where GR is the gain of the receiving antenna.
Figure 12.20 Free space propagation.

This equation (12.69) can be rewritten as:
C = PT GT


λ
4π R


2
GR (W) (12.70)
To give the basic radio free space transmission loss formula.
PT GT is the effective isotropic radiated power (EIRP) and the
[λ/(4π R)]2
is the free space path loss (FSPL).
FSPL is a function of wavelength because it converts the
receiving antenna area to gain, as well as incorperating the
spreading 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


4π R
λ


(dB) (12.71(c))

SIMPLIFIED EXAMPLE 12.5
Calculate the power density at a distance of 20 km from a
microwave antenna with a directivity of 42.0 dB, an ohmic
efﬁciency of 95% for a 4 GHz 25 dBm transmitter.
G = ηΩ D
= 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
4π R2
GT
= PT + GT − 10 log10 (4π R2
) 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
4π
D =
η
ΩPT
4π
D =
0. 95 × 10
25
10
4π
× 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, the
received 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 = GR
λ2
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 budgets
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 radiation
from lightning discharges.
Above 1 GHz galactic noise is relatively weak. Atmospheric
and ground noise is approximately ﬂat with frequency up to
about 10 GHz.
As elevation increases from 0° to 90° the thickness of
atmosphere through which the beam passes decreases as does
the inﬂuence of the ground, leading to a decrease in received
thermal noise.
In the 1 - 10 GHz frequency range a zenith-pointed antenna in
clear sky may have a noise temperature close to the cosmic
background temperature of 3 K.
Above 10 GHz resonance effects (of water vapour molecules at
22 GHz and oxygen molecules at 60 GHz) increase atmospheric
attenuation 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 12.7
For a 10 GHz terrestrial LOS link The received antenna signal
power is −40. 0 dBm, overall receiver noise ﬁgure is 5.0 dB and
the bandwidth is 20 MHz.
Estimate the clear sky antenna terminal CNR for an antenna
ohmic efﬁciency of 95% at 280 K.
From Fig. 12.25 the aperture temperature, TA, at 10 GHz for a
horizontal link is ˜100 K. Including the physical temperature
Tph gives:
Tant = TAηΩ + Tph(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 = (10
5
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 different
radio applications. Figure shows how multipath propagation
may occur on a point-to-point line-of-sight link with reflections
from ground and atmosphere.
The fading process is governed by atmospheric changes. If the
frequency response of the channel is essentially constant over
the operating bandwidth fading is called flat since all frequency
components 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 of
signal amplitude obeys Rayleigh statistics then the frequency
response of the channel may change rapidly on a frequency
scale comparable to signal bandwidth.
Here the fading is said to be frequency selective and the
received 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 by
compensating for the fades. Three types of diversity system,
are space, frequency, and angle. In all cases the essential
assumption is that it is unlikely that both main and diversity
channel will suffer severe fades at the same instant.
Selecting the channel with largest CNR, or combining channels
will result in improved overall CNR.
Figure 12.29 Three types of diversity arrangements to combat multipath fading.

CHAPTER 14
Fixed point microwave
communications

24 Fixed point microwave communications
14.2 Terrestrial microwave links
Many wideband point-to-point radio communications links
employ 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 or
hills and repeaters are used every 40 to 50 km to compensate
for 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 earth
in the way that MW and LW transmissions do. Microwave
transmissions are thus restricted to line of sight (LOS).
2. Microwave transmissions are particularly well suited to
point-to-point communications using narrow beam, high
gain, antennas.
3. At about 1 GHz circuit design techniques change from using
lumped to distributed elements. Above 20 GHz it becomes
expensive to generate big amounts of microwave power.
The bands in current use are near 2 GHz, 4 GHz, 6 GHz, 11
GHz and 18 GHz.
(On a 6 GHz link with a hop distance of 40 km the free space
path loss is 140 dB. With transmit and receive antenna gains of
40 dB, however, the transmission loss reduces to 60 dB.)

14.2.1 Analogue systems
In the UK microwave links were widely installed during the
1960s for analogue FDM telephony. In these systems each
allocated frequency band is subdivided into 30 MHz radio
channels.
Figure shows how the 500 MHz band is divided into 16
separate channels on 29.65 MHz centre spacings. Adjacent
radio channels use orthogonal antenna polarisations, horizontal
(H) and vertical (V), to reduce crosstalk.
The 500 MHz bandwidth microwave link can accommodate
trafﬁc 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 Frequency 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 MHz
channel from the transmissions. The signal is mixed down to
an intermediate frequency (IF) for additional filtering and
amplification.
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 a
carrier frequency of 11 GHz using QPSK modulation.
The 64-QAM system offers practical spectral efﬁciency of 4 to
5 bit/s/Hz implying that the 30 MHz channel can support a 140
Mbit/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 the
microwave spectrum where the signal is located.
Digital systems have a major advantage over analogue systems
as they operate at a much lower CNR.
Figure 14.7 Digital DPSK regenerative repeater for a single 30 MHz radio channel.

14.3 Fixed point satellite communications
Satellites are one of the three most important developments in
telecommunications. Geostationary satellites have an orbital
radius of 42,164 km and earth surface altitude of 35,786 km.
Other classes of satellite orbit include highly inclined highly
elliptical (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 the
most important, for the following reasons:
1. Its high altitude means that a single satellite is visible from
a 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 or
point-to-point.
2. New links can be made by pointing an antenna at the
satellite.
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, geostationary
satellites 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 budgets
On the uplink the earth pointing satellite antenna has a noise
temperature of 290 K.
The noise temperature of the earth station antenna, which looks
at the sky, is usually in the range 5 to 100 K for frequencies
between 1 and 10 GHz, Figure 12.25.
Above 10 GHz resonance effects of water vapour and oxygen
molecules, at 22 GHz and 60 GHz respectively degrade the
receiver performance.
Figure 14.30 Antenna aperture temperature, TA, 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 earth
station receiver is cooled so that its noise temperature, Te, is
very much smaller than 290 K. This achieves a noise floor that
is very much lower than −174 dBm/Hz.
The gain, G, of the cooled low noise amplifier boosts the
received signal so that the following receiver amplifiers can
operate at room temperature.
Operating frequencies for satellite communication systems are
limited at the low end to >1 GHz by galactic noise and at the
high end to <15 GHz by atmospheric noise and rain attenuation.

The satellite transponder is the critical component as its
transmitter is power limited by the onboard satellite power
supply giving the downlink the worst power budget.
EIRP is also dependent on satellite antenna design and in
particular the earth ‘footprint’.
Spot beams covering only a small part of the earth have high
gain giving EIRP values of 30 to 40 dBW.
The satellite transponder must also be small and light. The
transponder’s high power ampliﬁer (HPA) must thus operate at
a 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 14.5
An 11.7 GHz geostationary satellite downlink has 25 W output
power, a 20 dB gain antenna with 2 dB feeder losses. The earth
station, 38,000 km from the satellite, has a 15 m diameter
receive 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 performance
what is the maximum BPSK bit rate?
Transmitter power = 14 dBW
EIRP = 14 + 20 − 2 = 32 dBW = 62 dBm
Free space loss = 20 log10 (4π 38 × 106
)/0. 0256 = 205.4 dB
Receiver antenna GR =
4π
λ2
π d2
4
0. 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/Hz
If BTo = 1. 0 then Eb/N0 = C/N.
Available 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 61.6
Mbit/s.

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

Radio Link budget

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