Arlene Meidahl - s107106 and Danish Bangash-s104712| Digital
Communication | 21. maj 2015
Supervisor: John Aasted Sørensen
Digital Communication and Modulation
Project 3 “Satellite Link Budgets and PE”

SIDE 1
Satellite Link Budgets and PE
The objective of this project is to construct a Matlab script for link budget determination for a
satellite relay communication link, as shown in [Ziemer, 2010] Figure A.7 page 695:
The script should include the following primary parameters for a satellite-satellite link:
d Distance between the two satellites.
 Transmitted wavelength.
B Bandwidth of the transmission between the two satellites.
TP Transmitted power from satellite antenna.
TG Power gain of transmitter antenna over the isotropic radiation level.
RA Receiving aperture area.
RG Power gain of receiving antenna.
0L Atmospheric absorption.
RT Receiver noise temperature.

SIDE 2
The script should allow for the determination of the SNR at the receiver output and the probability
of bit error EP for the satellite link.
The script is used to demonstrate experimentally the relation between EP and varying d ,  , B ,
TP by appropriately selected examples. It is suggested to use the Examples A.8 and A.9 at pages
697, 698 [Ziemer, 2010] as a starting point for examples. It is suggested to augment with at least
two new examples created by yourself.
The script should allow the user to determine the received power in dBW.

SIDE 3
Problem formulation:
Considering a free-space electromagnetic-wave propagation channel as an example. For a better
understanding how it works, we set our attention on the link between the two satellites which is
the synchronous-orbit relay satellite and a low-orbit satellite or an aircraft as we can see in figure
below.
We assumed a relay satellite transmitted signal power of PT W. If we say that it is radiated
isotropically, the power density at a distance d from the satellite is given by:
𝑃𝑡 =
𝑃𝑇
4𝜋𝑑2 [
𝑊
𝑚2]
If the radiated power is directed towards the low-orbit satellite or an aircraft as an example, then
the maximum gain for the transmitting aperture antenna is:
𝐺 𝑇 =
4𝜋𝐴 𝑇
𝜆2

SIDE 4
The maximum gain for the receiving aperture antenna is:
𝐺 𝑅 =
4𝜋𝐴 𝑅
𝜆2
The power PR intercepted by the receiving antenna is given by the product of the receiving aperture
area AR and the power density at the aperture. So,
𝑃𝑅 =
𝑃𝑇 𝐺 𝑇
4𝜋𝑑2
𝐴 𝑅
The value of AR is know from the expression of ‘maximum gain for receiving aperture antenna, so
we get the following:
𝑃𝑅 =
𝑃𝑇 𝐺 𝑇 𝐺 𝑅 𝜆2
(4𝜋𝑑)2
A loss actor is included to the expression above and then we rearrange the formula, so we obtained
the following new expression:
𝑃𝑅 = (
𝜆
4𝜋𝑑
)
2
𝑃𝑇 𝐺 𝑇 𝐺 𝑅
𝐿 𝑂
Where (
𝜆
4𝜋𝑑
)
2
is referred as the free-space loss.
We then calculate the receiver power in terms of dB (with reference 1 Watt), so we get:
10𝑙𝑜𝑔10 𝑃𝑅 = 20𝑙𝑜𝑔10 (
𝜆
4𝜋𝑑
) + 10𝑙𝑜𝑔10 𝑃𝑇 + 10𝑙𝑜𝑔10 𝐺 𝑇 + 10𝑙𝑜𝑔10 𝐺 𝑅 − 10𝑙𝑜𝑔10 𝐿 𝑂

SIDE 5
Example A.8:
The following parameters for a relay-satellite-to user link are given:
- Relay satellite effective radiated power ( 30 ; 100T TG dB P W  ): 50 dBW
- Transmit frequency : 2 GHz ( 0.15 m  )
- Receiver noise temperature of user (includes noise figure of receiver and background
temperature of antenna): 700 K
- User satellite antenna gain: 0 dB
- Total system losses: 3 dB
- Relay-user separation: 41,000 km
Find the signal-to-noise power ration in a 50 kHz bandwidth at the user satellite receiver IF
amplifier output.
Solution:
We begin by finding the free-space loss in dB to be:
(
𝜆
4𝜋𝑑
)
2
= 20𝑙𝑜𝑔10 (
𝜆
4𝜋𝑑
) = 20𝑙𝑜𝑔10 (
0.15
4 ∗ 𝜋 ∗ 41 ∗ 106
) = −190.72 𝑑𝐵
Then we substitute in the formula to find the received power signal;
10𝑙𝑜𝑔10 𝑃𝑅 = 20𝑙𝑜𝑔10 (
𝜆
4𝜋𝑑
) + 10𝑙𝑜𝑔10 𝑃𝑇 + 10𝑙𝑜𝑔10 𝐺 𝑇 + 10𝑙𝑜𝑔10 𝐺 𝑅 − 10𝑙𝑜𝑔10 𝐿 𝑂
We know the value of the unknown variables in the above equation from the problem formulation,
so we put in the values in the equation:
10𝑙𝑜𝑔10 𝑃𝑅 = −190.72 + 10𝑙𝑜𝑔10(100) + 10𝑙𝑜𝑔10(30) + 10𝑙𝑜𝑔10(0) − 10𝑙𝑜𝑔10(3)
So the received power in dB is:
𝑃𝑅 = −143.72 𝑑𝐵
We know that the noise power level is:
𝑃𝑖𝑛𝑡 = 𝐺 𝑎 𝑘𝑇𝑒 𝐵
Here the Pint is the receiver output noise power caused by the internal sources. Since we are
calculating the SNR, the available gain of the receiver does not enter the calculation because both
signal and noise are multiplied by the same gain. Hence, we may set Ga to unity. And this the noise
level is:
𝑃𝑖𝑛𝑡,𝑑𝑏𝑊 = 10 𝑙𝑜𝑔10[𝐺 𝑎 𝑘𝑇𝑒 𝐵]

SIDE 6
In the equation above, all the variables are given its values excluding the constant k which is the so
called Boltsmann’s constant, which is equal to 1.38065*10-23 𝐽
𝐾
. The we substitute the given values
to the equation
𝑃𝑖𝑛𝑡,𝑑𝑏𝑊 = 10 𝑙𝑜𝑔10[1 ∗ 1.38 ∗ 10−23
∗ 700 ∗ 50 ∗ 103]
The above equation can also be stated:
𝑃𝑖𝑛𝑡,𝑑𝑏𝑊 = 10 𝑙𝑜𝑔10[1] + 10 𝑙𝑜𝑔10[1.38 ∗ 10−23] + 10 𝑙𝑜𝑔10[700] + 10 𝑙𝑜𝑔10[50 ∗ 103]
= −153.16 𝑑𝐵𝑊
So the SNR at the receiver output will be
𝑆𝑁𝑅0 = 𝑃𝑅 − 𝑃𝑖𝑛𝑡,𝑑𝑏𝑊 = −143.72 − (−153.16 ) = 9.44 𝑑𝐵
Example A.9: Continuation of example 8
From the result of the example above, we would like to interpret it in terms of the performance of
a digital communication system, so we have to convert the SNR obtained to energy-per-bit-to-noise-
spectral density ratio
𝐸 𝑏
𝑁0
.
By definition of SNR0:
𝑆𝑁𝑅0 =
𝑃𝑅
𝑘𝑇𝑒 𝐵
If we multiply the numerator and the denominator by the duration of a data bit 𝑇𝑏 , and we know
that:
PRTB = Eb and kTR = N0
So we get the following:
𝑆𝑁𝑅0 =
𝑃𝑅
𝑘𝑇𝑒 𝐵
=
𝐸 𝑏
𝑁0 𝐵𝑇𝑏
We are aware that PRTB = Eb and kTR = N0 , are the signal energy per bit and the noise power spectral
density. So, to obtain
Eb
N0
from the SNR0 we performed the following:

SIDE 7
𝐸 𝑏
𝑁0
= 𝑆𝑁𝑅0 + 10 𝑙𝑜𝑔10(𝐵𝑇𝑏)
In the previous problem A.8, SNR0 is equal to 9.44 dB. We know that the null-to-null bandwidth of
a phase-shift keyed carrier is
2
Tb
Hz. Therefore, we could say that BTb is 2 (3 dB), thus
𝐸 𝑏
𝑁0
= 9.44 + 3 = 12.44 𝑑𝐵
We now convert the above result into z-ratio since we want to calculate the bit error probability,
therefore
10 log10(z) = 12.44
z = 101.244
Now, the probability of error for a binary phase-shift keying (BPSK) digital communication system
can be express as:
PE = Q (√
2Eb
N0
) ≅ Q (√2 ∗ 101.244) ≅ 1.23 ∗ 10−9
We can conclude from the result we just found that it is a fairly small probability of error since
anything less than 10-6
would probably be considered adequate.

SIDE 8
Example 2:
Problem A.12:
Given a relay-user link with the following parameters:
- Average transmit power of relay satellite: 35 dBW = Pt
- Transmit frequency: 7.7 GHz = f
- Effective antenna aperture of relay satellite : 1 m2
= AR
- Noise temperature of user receiver (including antenna): 1000 K = TR
- Antenna gain of user: 6 dB = GR
- Total system losses: 5 dB = L0
- System bandwidth: 1 MHz = B
- Relay-user separation: 41,000 km = d
a. Find the received signal power level at the user in dBW.
b. Find the receiver noise level in dBW.
c. Compute the SNR at the receiver in decibels.
d. Find the average probability of error for the BPSK digital signaling method.
The received signal power formula is:
𝑃𝑟 = 20 𝑙𝑜𝑔 10 (
𝜆
4𝜋𝜆
)
2 𝑃 𝑇 𝐺 𝑇 𝐺 𝑅
𝐿°
where 𝑃𝑟is the received power in decibels reference to 1 W.
10 𝑙𝑜𝑔10 𝑃𝑅 = 20 𝑙𝑜𝑔 10 (
𝜆
4𝜋𝜆
)
+ 10 𝑙𝑜𝑔10(𝑃𝑇 ) + 10 𝑙𝑜𝑔10( 𝐺 𝑇 ) + 10 𝑙𝑜𝑔10(𝐺 𝑅 ) − 10 𝑙𝑜𝑔10(𝐿0)
If we look at the formula above, we can see that 𝐺 𝑇 which is the power gain of the transmitter
antenna and 𝜆 the transmission wavelength is unknown, so we need to find the value for those two
parameter.
To find the wavelength λ =
c
f
, c is the speed of light, which is equal to 3*108 m
s
and the frequency
is 7.7*109
Hz so we get the following result:
𝜆 =
3 ∗ 108
7.7 ∗ 109
= 38.96 ∗ 10−3

SIDE 9
We obtain 𝐺 𝑇 by:
𝐺 𝑇 =
4𝜋𝐴 𝑇
𝜆2
=
4𝜋 ∗ 1
(38,96 ∗ 10−3)2
= 8.228 ∗ 103
𝐺 𝑇𝑑𝐵 = 10𝑙𝑜𝑔10( 8.228 ∗ 103) = 39.15 𝑑𝐵
The free-space loss is calculated directly in dB:
20𝑙𝑜𝑔10 (
𝜆
4𝜋𝑑
) = 20𝑙𝑜𝑔10 (
38.96 ∗ 10−3
4𝜋 ∗ 41 ∗ 106) = −202.4𝑑𝐵
Now we find the power of the receiver. The given values are:
- GR = Antenna gain of user: 6 dB
- L0 = Total system losses: 5 dB
- PT = Average transmit power of relay satellite: 35 dBW
𝑃𝑅𝑑𝐵 = 20 𝑙𝑜𝑔 10 (
𝜆
4𝜋𝜆
) + 10 𝑙𝑜𝑔10(𝑃𝑇 ) + 10 𝑙𝑜𝑔10( 𝐺 𝑇 ) + 10 𝑙𝑜𝑔10(𝐺 𝑅 ) − 10 𝑙𝑜𝑔10(𝐿0)
𝑃𝑅𝐿𝑑𝐵 = −202.4 + 35 + 39.15 + 6 − 5 = 127.3 𝑑𝐵𝑊
b. Find the receiver noise level in dBW.
The noise power level is:
𝑃𝑅𝑑𝐵 = 10 𝑙𝑜𝑔10 ( 𝑇𝑅 ∗ 𝐵 ∗ 𝑘) = 10 𝑙𝑜𝑔10 ( 1 ∗ 103
∗ 1 ∗ 106
∗ 1.38 ∗ 10−23
) = −138.6𝑑𝐵𝑊
c. Compute the SNR at the receiver in decibels.
The SNR at the receiver is:
𝑆𝑁𝑅 = 𝑃𝑅 − 𝑃𝑟𝐿 = −127.3 + 138.6 = 11.4𝑑𝐵𝑊

SIDE 10
d. Find the average probability of error.
We can then calculate the probability bit error using the formula below:
𝑃𝐸 = 𝑄 (√
2𝐸 𝑏
𝑁°
) = 𝑄(√2𝑧)
We can say 𝑇𝑅 𝐵 is equal to 3 dB:
𝐸 𝑏
𝑁°
|
𝑑𝐵
= 𝑆𝑁𝑅 𝑑𝐵 + 10𝑙𝑜𝑔10(𝑇𝑅 𝐵) = 11.3 + 3 = 14.36
Then we get:
10𝑙𝑜𝑔10(𝑧) = 14.36
𝑧 = 101.436
We get the bit error probability to be:
𝑃𝐸 = 𝑄 (√
2𝐸 𝑏
𝑁°
) = 𝑄(√2𝑧) = 𝑄 (√2 ∗ 101.436) = 7.3839 ∗ 10−146

SIDE 11
Example 2: Parameters Testing
The script is used to demonstrate experimentally the relation between PE and the varying
parameters d,  , B, PT :
Every time we change a certain parameter, we assumed that all the other parameters are
constant and let it as it be.
We started changing the distance d:
d [*106] 10 20 30 40
SNR 23.6 17.59 14.06 11.57
PE 6.49*10-202 3.91*10-52 2.8*10-24 1.81*10-14
We can see on the table above that as the distance between the transmitter and the
receiver increases and the bit error probability PE, the signal-to-noise ratio(SNR)
decreases.
Then we played around the wavelength  :
 0.01 0.05 0.10 0.15
SNR -23.7 -9.75 -3.75 -0.208
PE 0.448 0.258 0.097 0.025
We can see that when we increase the wavelength, it means to say that, we decrease the
transmission frequency f, then signal-to-noise ratio increases, but the bit error probability
PE decreases.
Now, we changed the channel bandwidth B:
B 10 000 50 000 100 000 500 000
SNR 31.35 24.36 21.35 14.36
PE 0 8.98*10-240 4.65*10-121 7.33*10-26
When we increase the bandwidth B, we got a decrease in the signal-to-noise-ratio and
increase in the bit error probability.

SIDE 12
Then we changed the signal transmission power Pt:
Pt 10 20 30 40
SNR -13.65 -3.65 6.35 16.35
PE 0.34 0.094 1.62*10-5 9.70*10-40
When we increase the transmission power, we get an increase with the signal-to-noise-
ratio and decrease in bit error probability PE until it reaches 0.
So we therefore conclude, after testing and changing the values of the different
parameters:
We could simply change the 4 parameters stated above, so we could get the desired link
depending on the requirements being given. In order to get a reasonable signal-to-noise-
ratio and bit error probability, each parameters can be compensated by another
parameters.
To explain further, according from the test , we found out that when we increase distance
d between the transmitter and receiver, we get an increase in bit error probability, and in
that case we can compensate it by increasing the signal power transmission, and that
decreases the bit error probability error.

SIDE 13
Appendix:
Matlab Codes
Code 1:
% Example A.8 & A.9
clear all
close all
clc
% Parameters:
% (All parameters are in their respective SI units)
k = 1.38e-23; % Boltzmann Constant
lambda = 0.15; % Transmit Frequency
B = 50e3; % Badwidth of transmission between the satellites
T = 700; % Reciever noise temperature
Pt = 100; % Transmitted power
d = 41e6; % Distance between the two satellites
Gt = 30; % Power gain
Gr = 0; % User satellite antenna gain
L0 = 3; % Total system losses
% Example A.8
FreeSpaceLoss = 20*log10((lamda)/(4*pi*d))
Pr = FreeSpaceLoss + (10*log(Pt)) + Gt + Gr - L0
Pint_dB = 10*log10(k) + 10*log10(T) + 10*log10(B)
SNR0 = Pr - Pint_dB
% Example A.9 Page 698, Ziemer 2010
Tb = 2/B % Null-to-null bandwidth of a phase-shift-keyed
carrier
Eb_N0 = SNR0 + 10*log10(B*Tb)
z = 10^(Eb_N0/10) % The bit error probability

SIDE 14
Code 2:
% Problem A.12
clear all
clc
% The parameters:
k = 1.38e-23; % Boltzmann's Constant
d = 41*10^6; % Distance between the two satellites
B = 1*10^6; % Bandwidth of the transmission between the two satellites
Pt = 35; % Transmitted power from satellite antenna
Ar = 1 % Receiving aperture area
At = 1; % Transmitting aperture area
Gr = 6; % Power gain of receiving antenna
L0 = 5; % Atmospheric absorption
Tr = 1000 % Receiver noise temperature measured in [Kelvin]
c = 3*10^8; % Speed of light [m/s]
f = 7.7*10^9; % Transmit frequency [Hz]
lambda = c/f; % Transmitted wavelength
Gt = 4*pi*At/lambda^2; % Power gain of transmitter antenna NOT in dB
% a)
FSloss_dB = 20*log10((lambda/(4*pi*d))); % Free-space loss in dB
Pr1 = FSloss_dB + Pt + 10*log10(Gt) + Gr - L0; % Power of received signal (ref.
1W; in dB)
% b)
Pr2 = 10*log10(k*Tr*B); % Power of output signal (ref. 1W; in dB; no def.
gain);
% c)
SNRout = Pr1 - Pr2; % SNR at receiver output
% d)

SIDE 15
Tb = 2/B; % duration of data bit in terms of
bandwidth, for BPSK
EbN0 = SNRout + (10*log10(B*Tb));
z = 10^(EbN0/10);
PE = qfunc(sqrt(2*z)); % Bit error probability
Referrences:
Principles of Communications System, Modulation and Noise – Sixth Edition – Rodger E. Ziemer,
William H. Tranter