Link Budgets and Outage
Calculations
Dr Costas Constantinou
School of Electronic, Electrical & Computer Engineering
University of Birmingham
W: www.eee.bham.ac.uk/ConstantinouCC/
E: c.constantinou@bham.ac.uk
Decibels
• Logarithmic units of measurement suitable for describing
both very large and very small numbers conveniently
• Named by telephone engineers in honour of Alexander
Graham Bell
2
Why work with Decibels
1. Decibels can be used to express a set of values having a very
large dynamic range without losing the fine detail
2. They allow gain and signal strengths to be added and
subtracted in a link budget calculation
• The American mathematician Edward Kasner once asked his
nine-year-old nephew Milton Sirotta to invent a name for a
very large number, ten to the power of one hundred; and the
boy called it a googol. He thought this was a number to
overflow people's minds, being bigger than anything that can
ever be put into words …
3
Why work with Decibels
1 googol = 10 000 000 000 000 000 000 000 000 000
000 000 000 000 000 000 000 000 000 000 000 000
000 000 000 000 000 000 000 000 000 000 000 000
1 googol = 10100
10 log1010100 = 10 x 100
= 1000 dB
dBs are easier to write down!
4
Why work with Decibels
The figure shows a large carrier and also something else
higher up the frequency band which is hardly visible
If we plot the result in dBm (decibels relative to 1mW – see
later) we can see all of the information clearly
5
Decibels
• A power P can be expressed in decibels by
PdB

P
10 log10
Pref
where Pref is the power (unit) to which P is compared
6
Decibels
If for example
then
P
= 20 Watts
Pref = 1 Watt
P dB = 13 dBW
where the W after the dB denotes a reference value of 1 W.
If
then
Pref = 1 milliWatt
P dB = 43 dBm
where the m after the dB refers to a mW.
7
Decibels
• The decibel can also be used to refer to the power gain or
power loss of a component
Pout
Pin
GdB

Pout
10 log10
Pin
8
Decibels
Thus for an amplifier with
Pin = 0.1 W
Pout = 1 W
G dB = 10 dB
Similarly if the component is a long cable with
Pin = 1 W
Pout = 0.1 W
then
G = –10 dB
which represents a loss of 10dB.
9
Decibels
• If the input and output signals are known in voltage or current
terms, then
2
Pout
Vout
2 Zout
GdB  10log10
 10log10 2
Pin
Vin 2Z in
Vout
 20 log10
Vin
assuming that the impedances at the input and output are the
same (Zout = Zin).
10
Decibels
Decibels
1000
30
number
10
dB
number
10
dB
0.1
number
-10
dB
100
20
8
9
0.125
-9
10
10
4
6
0.25
-6
1
0
2
3
0.5
-3
-10
1
0
1
0
0.1
11
Decibels
• Previous chart is useful for converting from numbers to
dBs
• Examples
Pout/Pin
=
=
=
=
103
8 x 102
4
10-1




30 dB
29 dB
6 dB
–10 dB
• Memorising the chart will help you perform most
conversions in your head to an accuracy necessary for
estimation purposes.
12
Cascaded amplifiers
• What happens if we have two amplifiers in series?
Pint
Pin
Pout
G1
GdB

P
10 log10 out
Pin
G2
 Pint Pout 
 10 log10 

P
P
 in int 
 10 log10  G1 G2 
 10 log10  G1   10 log10  G2 
 G1dB  G2 dB
Conclusion – we add gains in dB.
13
Cascaded amplifiers
Example
Pin = 10 mW, Pint = 1 W, Pout = 100 W
Pint
Pin
So
G1
Pout
G2
G1 = 1/10x10-3 = 100 = 20 dB
G2 = 100/1
= 100 = 20 dB
And
G = 100/10x10-3 = 10,000 = 40 dB
G = G1 + G2
14
Cascaded attenuators
• What happens if we have two attenuators in series?
Pint
Pin
G1
GdB

P
10 log10 out
Pin
Pout
G2
 Pint Pout 
 10 log10 

P
P
 in int 
 10 log10  G1 G2 
 10 log10  G1   10 log10  G2 
 G1dB  G2 dB
• Conclusion – losses are negative gains in dB
• Conclusion – can add losses in dBs.
15
Cascaded attenuators
Example
Pin = 10 W, Pint = 1 W, Pout = 1 mW
So
G1 = 1/10 = 0.1 = –10 dB
G2 = 10–3/1 = 10–3 = – 30 dB
And
G = 10–3/10 = 10–4 = – 40 dB
G = G1 + G2
16
Cascaded amplifier & attenuator
• What happens if we have an amplifier followed by a
loss, such as a long cable? Pin
Pint
Pout
G1
GdB

Pout
10 log10
Pin
 Pint Pout 
 10 log10 

P
P
 in int 
G2
 10 log10  G1 G2 
 10 log10  G1   10 log10  G2 
 G1dB  G2 dB
• Conclusion – now we can proceed to do real systems
17
Cascaded amplifier & attenuator
Pin
G1
Example
Pin = 1 mW, Pint = 1 W, Pout = 1 mW
Pint
Pout
G2
So
G1 = 1/10–3 = 1000 = 30 dB
G2 = 10–3/1 = 10–3 = –30 dB
And
G = 10–3/10–3 = 1 = 0 dB
G = G1 + G2
18
Link budgets
• G = G1 + G2 is a rudimentary system link
budget
Pint
Pin
G1
Pout
G2
• Link budgets are used in all RF systems
– to get rough feel for viability
– to fine tune actual design
19
Example – submarine cable
communications
• Birmingham to Beijing
– Distance = 8171 km
– Cable attenuation = 0.3 dB/km
– Velocity of electromagnetic wave in cable = c/1.46
• Delay = 1.46 x 8191 x 103 / (3 x 108) s
• Attenuation = 0.3 x 8171 dB = 2451 dB
• Attenuation is bigger than a googol – it will never
work!
20
Simple link budget example
Pin
amp
laser
P1 diode
G1
fibre
P2
L1
P3
detector
diode
amp
P4
Pout
L2
Want a zero gain system, so they can be cascaded to cover long distance
Amp
Laser
Fibre
Diode
Amp
to get input signal power big enough to drive diode
gain = 20 dB
converts digital signal to light
conversion gain = –20 dB, (or loss = 20 dB)
100 km long gives 100 x 0.3 = 30 dB
so gain = –30 dB
converts light back to digital signal
conversion gain = –20 dB, (or loss = 20 dB)
to bring signal back to input level
gain = 50 dB
Overall gain
20
–20
–30
–20
50
0 dB
21
Example – geosynchronous satellite
link
• Birmingham to Beijing
(assuming single satellite trip, up and down)
g 
d
d
 c  3108 m / s
35,855 km
• Delay = 2 x 35,855 x 103 / 3 x 108 s
= 0.23 s
• But what is link budget?
22
Link budgets – satellite downlink
model
Transponder
Free space +
other losses
noise
antenna
Σ
Earth station Rx
23
Link budgets – downlink model
•
•
•
•
•
•
Satellite transponder output power = Pt
Antenna gain = Gt
Effective isotropic radiated power = EIRP = PtGt
Free space path loss = (λ/4πd)2 = Lp
Atmospheric loss = La
Antenna loss (feeder loss, pointing error, etc) =
Lat, Lar
• Clear air margin = Mp
• Coverage contour margin = Mc
24
25
Link budgets – downlink model
• Power at receiver
S = EIRP + Gr – Lp – La – Lat – Lar
(all terms in dBs)
• Noise at receiver
N = kTsB = k(Ta + Te)B
(dBW)
(dBW)
• Note that Ts = Noise temperature of system in Kelvin
Ta = Noise temperature of antenna in K
Te = Noise temperature of receiver in K
26
Typical link budgets
12/14 GHz link; satellite antenna = earth antenna = 1.8m, low
cost earth station
up link
down link
Pt tx power
25
20
dBW
Gt tx ant gain
46
44
dB
Lat tx ant loss
-1
-1
dB
Lp free space loss
-208
-206
dB
La atmos loss
-0.5
-0.6
dB
Gr rx ant gain
46
44
dB
Lar rx ant loss
-1
-1
dB
-93.5
-100.6
dBW
Pr rx power
Note – up/down link values different due to different frequencies
27
Typical link budgets
28
Typical link budgets
Rain loss
mm/hr
29
Typical link budgets
Rain distribution
30
Noise
• Electromagnetic noise is produced by all bodies above
absolute zero temperature (0 K)
• Examples
–
–
–
–
–
–
–
–
Earth
Sky
Atmosphere
Sun
Galaxy
Universe
Man-made noise
Interference
31
Antenna temperature
Tant

g x T L
i i i i
i
• The summation is taken over all bodies
in the field of view of the antenna
– gi = fraction of total antenna sensitivity (gain) in direction
of body i.
– xi = greyness of body i (xi = 1 for a black body)
– Ti = temperature of body i (K)
– Li = transmission factor from body i to antenna
32
Sample noise calculation for typical
satellite earth station at 20 GHz
Source
gi
xi
Li
Ti (K)
gxTL
sky
0.7
0.99
1.0
50
34.6
earth
0.3
0.3
1.0
300
27.0
sun
0.005
0.99
0.01
7000
0.4
skyearth
0.3
0.99.(1.0 – 0.3)
1.0
50
10.4
sunearth
0.3
0.99.(1.0 – 0.3)
0.01
7000
14.5
Tant
86.9
33
Receiver noise temperature
• Assuming no loss in the connection between antenna and
receiver, the total noise temperature (at input to receiver)
T  Tant  Te
 Tant  ( F  1) T0
where Te, F = effective noise temp and noise figure of receiver
T0 = reference temp for noise figure (normally 290 K)
• Noise power (at input to receiver)
N  kTB
where k = Boltzmann’s constant = 1.38 x 10-23 JK–1
B = receiver bandwidth
34
Typical link budgets
up link
down link
Pr rx power
-93.5
-100.6
dBW
T noise temp
800
1000
K
B bandwidth
36
36
MHz
N noise power
-124
-123
dBW
S/N at rx
30.5
22.4
dB
S/N required
10.0
10.0
dB
Mp clear air margin
20.5
12.4
dB
10
10
dB
S/N at rx
20.5
12.4
dB
S/N required
10.0
10.0
dB
Mp margin
10.5
2.4
dB
La atmospheric loss in
bad storm
Note – down link margin only just acceptable in storm
35
Outage calculations
• In the case of mobile radio the path loss is not known
fully; it is described by L d   L d   X
– a deterministic component L  d  and
– a stochastic (randomly varying) component X
• The overall link budget is then computed from a
desirable BER as
S
BER  f   
N
S S
 
N  N min
S
EIRP  Gr  L (d )  X  10log10  k (Tant  ( F  1)T0 ) B    
 N min
36
Area mean path loss model example
• The Hata-Okumura model, derives from extensive
measurements made by Okumura in 1968 in and
around Tokyo between 200 MHz and 2 GHz
• The measurements were approximated in a set of
simple median path loss formulae by Hata
• The model has been standardised by the ITU as
recommendation ITU-R P.529-2
Area mean path loss model example
• The model applies to three clutter and terrain
categories
– Urban area: built-up city or large town with large buildings
and houses with two or more storeys, or larger villages
with closely built houses and tall, thickly grown trees
– Suburban area: village or highway scattered with trees and
houses, some obstacles being near the mobile, but not
very congested
– Open area: open space, no tall trees or buildings in path,
plot of land cleared for 300 – 400 m ahead, e.g. farmland,
rice fields, open fields
Area mean path loss model example
urban areas:
L  dB  A  B log R  E
suburban areas:
L  dB  A  B log R  C
open areas:
L  dB  A  B log R  D
where
A  69.55  26.16 log f c  13.82 log hb
B  44.9  6.55log hb
C  2log f c 28  5.4
2
D  4.78log f c   18.33log f c  40.94
2
E  3.2log11.75hm   4.97 for large cities, f c  300MHz
2
E  8.29log1.54hm   1.1
2
for large cities, f c  300MHz
E  1.1 log f c  0.7 hm  1.56 log f c  0.8 for medium to small cities
Area mean path loss model example
• The Hata-Okumura model is only valid for:
–
–
–
–
–
Carrier frequencies: 150 MHz  fc  1500 MHz
Base station/transmitter heights: 30 m  hb  200 m
Mobile station/receiver heights: 1 m  hm  10 m
Communication range: R > 1 km
A large city is defined as having an average building height
in excess of 15 m
Local mean model
• The departure of the local mean received power from the
area mean prediction is given by a multiplicative factor which
is found empirically to be described by a log-normal
distribution
• This is the same as an additive deviation in dB from the area
mean model being described by a normal distribution
Local mean model
• Working in logarithmic units (decibels, dB), the total path loss
is given by
L  d   L  d   Xs
where Xs is a random variable obeying a lognormal
distribution with standard deviation s (again measured in dB)
1
2
pX  
exp  X 2 2s dB

s dB 2
• If x is measured in linear units (e.g. Volts)
 ln x  ln mx 
1
px  
exp

2
2s dB 
s dB x 2

where mx is the mean value of the signal given by the area
mean model
Outage calculations
• Cumulative probability density function
S
X  EIRP  Gr  L (d )  10log10  k (Tant  ( F  1)T0 ) B    
 N min
X  X max
P  BER  Threshold  
X max


1
s dB 2
2
exp  X 2 2s dB
dX
1
 X max 
 1  erfc 

2
 2 
• Xmax plays the role of the link margin that you can afford to
lose and still maintain an acceptable BER - This is called an
outage calculation
What next?
• Attempt tutorial questions on link budgets
44