Fourier Analysis - Redes de Computadores - UPV

Laboratory
2
Fourier Analysis
Study of the distortion produced by a physical transmission
medium on the transmission of a digital signal.
Objetive
Fourier analysis is a fundamental concept that allows the understanding the distortion
effects suffered by a digital signal injected in a physical transmission medium (i.e.
copper wires). With this lab session, you will learn how a digital signal may be
decomposed in a sum of basis functions (sines and cosines) by using Fourier series.
Also, you will be able to recognize the impact of transmission medium characteristics
over the signal received at the other end.
Overview
When a signal is delivered through a physical transmission medium, a sort of different
phenomena may introduce perturbation on the signal (i.e. attenuation, noise, etc.). Some
of these effects may be observed in the time domain, however other effects are better
analysed in the frequency domain. So, we are going to introduce Fourier analysis as a
math tool for study and analyze the transmitted digital signal..
Fourier Fundamentals.
In this section we are going to introduce some theoretical concepts about Fourier
decomposition. For more information, refer to course documentation.
Fourier analysis is based in the fact that every periodic signal s(t) with period T
(s(t)=s(t+T)) and frequency f=1/T, may be decomposed in an infinite sum of sines and
cosines following this expression:
Where:
∑ ∞
n=
1
s( t)
= a + a cos( 2π
nft)
+ b sen( 2πnft)
(1)
a
a
b
0
n
n
1
=
T
2
=
T
2
=
T
0
∫
T
0
∫
∫
T
0
T
0
n
s(
t)
dt
n
(2)
s(
t)
cos( 2π
nft)
dt (3)
s(
t)
sen( 2π
nft)
dt (4)

In expression (1), f is the fundamental frequency of signal s(t). At the other hand, a0 is a
constant, we call DC component of signal s(t), that is calculated with expression (2).
The rest of terms in (1) correspond to the infinite components of signal s(t), we call
them harmonics or frequencial components. So, the n-th harmonic of signal s(t) is given
by the following expression
sn n
n
( t)
= a cos( 2π
nft)
+ b sen( 2πnft)
(5)
Related to (5), if nf=1, then cos( 2π
t) + sen( 2πt)
corresponds with the sum of one sine
and cosine of period 1. So, nf components define the sine and cosine frequency. As a
consequence, the frequency of n-th harmonic will be f n = nf
Finally, the amplitude of a sine (or cosine) of n harmonic is given by coefficients an and
bn. These coefficients weight the contribution of the corresponding harmonic in the
original signal s(t), and are calculated by mean of expressions (3) and (4). The power or
amplitude of harmonic n may be calculated as:
P = a + b
(6)
n
2
n
2
n
The graphical representation of all harmonics of original signal in the frequency domain
is known as spectral domain of signal s(t). Latter, we will show some examples.
Impact of physical medium in signal transmission
When using a wire for transmitting a signal, the resulting signal at the other end is
received as a attenuated version of the original signal. However, in the frequency
domain, not all harmonics are uniform attenuated. The resulting behaviour is similar to
a low-pass filter, where low frequency harmonics cross the filter without attenuation,
but as harmonic frequency increase they are more attenuated and even filtered.
So, we will consider that a physical transmission medium transport all signal harmonics
with frequency inferior to fc (cut frequency), and the rest of signal harmonics are totally
filtered. This assumption is not realistic because there is a progressive attenuation of
harmonics up to the cut frequency. Anyway, under the above assumption we can say
that a physical transmission medium has a bandwidth BW corresponding to the
maximum frequency it is able to transport (fc).
A digital signal is made up of an infinite number of harmonics. This implies that we
would need an infinite bandwidth channel for transmitting the signal without distortion.
Unfortunately, there is no physical medium with infinite bandwidth, so in practice the
received signal will be a distorted version from the original one (not all harmonics will
be received).
It is obvious that as more harmonics be received the reconstructed signal will have
better quality. However, not all harmonics contribute in the same manner to the
reconstruction of the signal. IN particular, the first harmonics carry the most part of the
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total signal power. So, we have to determine the minimum number of harmonics
required by the received to recognize and reconstruct the signal.
How to compute the Fourier series of an ASCII character.
Suppose that we are going to continuously transmit one character (one byte). If T is the
repetition period, we can formally define the transmitted signal s(t) as:
⎧
⎡ T ⎡
⎪ bit( 0)
when t ∈ ⎢0
+ Tn,
+ Tn⎢
⎪
⎣ 8 ⎣
⎪
⎡T
2T
⎡
s(
t)
=
bit( 1)
when t ∈
⎨
⎢ + Tn,
+ Tn⎢
⎣8
8 ⎣
⎪
M
⎪
⎪
⎡7T
⎡
bit( 7)
whent
∈
⎪
⎢ + Tn,
T + Tn⎢
⎩
⎣ 8
⎣
∀n∈
?
(7)
Where bit(i) are constants that define the value of bit i (of transmitted character).
If the signal is not periodic, we can not use Fourier Series, instead we are obliged to use
the Fourier transform which is out of the course scope.
Now, we are ready to calculate the terms a0, an and bn de (1) for the signal defined in
(7). So, we are going to use the equations (2), (3) and (4).
Let start with the DC component a0,
a
0
1
=
T
∫
T
0
s(
t)
dt
As function s(t) is defined as (7), we can perform the integral calculation by intervals:
a
0
T
2T
1
⎛
⎜ 8
8
= bit ( 0)
dt + bit ( 1)
dt + K +
⎜∫
∫T
T 0
∫
⎝
8
3
T
7T
8
⎞
bit ( 7)
dt⎟
⎟
⎠
Remember that bit(i) are constants, so we can consider the following
a
0
T
2T
1
⎛
⎜ 8
8
= bit ( 0)
dt + bit ( 1)
dt + K + bit ( 7)
⎜ ∫ ∫T
T
0
∫
⎝
8
Solving the simple integral
and
a
0
1 ⎛
= ⎜bit
( 0)
T ⎜
⎝
T
2T
[ t]
8 + bit ( 1)
[ t]
8 + K + bit ( 7)
[ t]
0
1 ⎛ T T
T
a0 = ⎜ bit ( 0)
+ bit ( 1)
+ K
+ bit ( 7)
T ⎝ 8 8
8
T
8
⎞
⎟
⎠
T
7T
8
T
7T
8
⎞
⎟
⎟
⎠
⎞
dt⎟
⎟
⎠

Finally, after removing T, we can observe that the DC component is independent from
signal period (T), in other words, it is independent from transmission speed. We can
define a0 as
7
∑
i=
0
1
a 0 = bit ( i)
(8)
8
The DC component a0 of signal s(t) is the arithmetic mean of all its bits.
The an and bn calculation can be done in a similar way, using equations (3) and (4). The
result, you have to perform the development by yourself, is
a
b
n
n
1
=
nπ
1
=
nπ
7
∑
i=
0
7
∑
i=
0
⎡ ⎛ ⎛ nπ
⎞ ⎛ nπ
⎞⎞⎤
⎢bit(
i)
⎜ sen⎜
( i + 1)
⎟ − sen⎜
i ⎟⎟⎥
⎣ ⎝ ⎝ 4 ⎠ ⎝ 4 ⎠⎠⎦
⎡ ⎛ ⎛ nπ
⎞ ⎛ nπ
⎞⎞⎤
⎢bit(
i)
⎜cos⎜
i⎟
− cos⎜
( i + 1)
⎟⎟⎥
⎣ ⎝ ⎝ 4 ⎠ ⎝ 4 ⎠⎠⎦
sen(
kt)
cos( kt)
Clue: Notice that cos( kt)
dt = ∫
and sen(
kt)
dt = −
k ∫ k
Procedure
Firstly, we are going to develop in C source code the computation of coefficients a0, an
and bn by means of expressions (8), (9) and (10). Then, we will analyze the graphical
representation of the transmitted character .
Required resources
We supply a uncompleted C source file “Fourier.cpp”. Also we require a C compiler.
We may use whatever ANSI C compiler (MS Visual C/C++, Borland C/C++, gnu C++,
etc.). In particular, we suggest the use of GNU C++ compiler, gcc, bundled with
Unix/Linux distributions.
The software corresponds with a console program that calculates the Fourier series
analysis. If we prefer using other compilers as MS Visual C++ or Borland C/C++, it is
necessary to create a new Console Proyect and add the source file Fourier.cpp to it.
In order to represent the reconstructed signal (in time and frequency domains), we need
a drawing application. We propose the use of GNUPlot (version 3.6 or later). GNUPlot
is a multiplatform software available in near all Linux distributions and other platforms
like Windows (gp373w32.zip).
In Herodes server, \\herodes.redes.upv.es\practicas\redesFI\Fourier, we supply all
the required resources for this lab. Also, you can find a binary version of Fourier
program for both Linux (Fourier) and Windows (Fourier.exe) systems.
4
(9)
(10)

Execution example
Before start programming, we will show an example of the output generated by the
available solution. So, copy the binary file to a local temporal directory. Then, open a
system console window and run the program without specifying command line
arguments. The default parameters are fixed to a predefined character, transmission
speed and channel bandwidth.
The generated output is composed by two files: An ASCII file “armonico.txt” that
contains the frequency (in Hertzs) and power of the harmonics that cross the channel,
and a GNUPlot script file with the commands required for drawing the original signal
(input signal), the signal received at the other channel end (output signal), and the
frequency domain of the latter signal.
The resulting GNUPlot script file should be the following:
# REDES DE COMPUTADORES - FACULTAD DE INFORMATICA DE VALENCIA
# PRACTICA SOBRE ANALISIS DE SEÑALES MEDIANTE SERIES DE FOURIER
reset
set size 1.0, 1.0
set label "- Caracter: 97 'a' (61h)\n\n- Vel. txon: 1200 bps\n --> Duracion
de un bit:\n 0.8333 ms \n --> Periodo de un byte:\n 6.6667 ms \n -
-> Frec. fundamental:\n 150.00 Hz \n\n- Ancho de banda del medio \n(ideal):
1000.00 Hz\n --> Numero de armonicos\n pasantes: 6\n\n"at screen 0.7,0.8
left
set multiplot
set nokey
set size 0.7, 0.5
set origin 0.0, 0.5
set yrange [-0.5:1.5]
set ylabel "Amplitud"
set xlabel "Tiempo (seg)"
f=150.000000
# Funcion que define el byte original que se transmite
byte(x)=(x

Alter that, s(x) is defined as the sum of six harmonics s1(x)… s6(x) that cross the
channel. In order to calculate the a0, an and bn coefficients, the program implements
several functions that we have to develop a bit latter.
Finally, The firs plot command draws functions byte(x) and s(x) in overlapped manner,
while the second plot command draws the frequency components of s(x) by using the
information included in “armonico.txt”.
For more information about GNUPlot commands check the internal help system (help
command).
Now, we are ready to produce the curves, so we have to:
1. Check if the execution attribute of Fourier binary is activated. If not, use the
chmod command: chmod +x Fourier
2. Introduce ./Fourier > comandos.txt in order to redirect the Standard output to
an ASCII file. This file will be the GNUPlot script file.
3. Next, we call GNUPlot tool passing it the script file generated just before
(introduce gnuplot comandos.txt)
GNUPlot will show the following window:
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In the right, the required information to perform the Fourier analysis is shown (The
transmitted character, transmission speed and channel bandwidth (BW). Also additional
info is presented as the time of one bit, the signal period, the fundamental frequency and
the number of passing harmonics.
In the top of the window the input signal (in blue) an the output signal (received signal)
are shown. The input and output signals are defined by byte(x) and s(x) functions,
respectively.
In the bottom, the frequency domain of the received signal is shown. Every harmonic is
represented by a vertical red line. The position of each harmonic in the X-axis
represents its frequency and its amplitude corresponds with the power of this harmonic.
We can observe that channel bandwidth is marked by a vertical blue arrow, just on the
cut frequency (fc).
Programming section: How to compute the a0, an and bn coefficients.
Firstly, you have to edit the Fourier.cpp file and take a first look. The first functions
have only defined their header, being their body empty:
double a(int n, int caracter)
{
// IMPLEMENTAR
}
double b(int n, int caracter)
{
// IMPLEMENTAR
}
double a0(int caracter)
{
// IMPLEMENTAR
}
The two first functions are the ones that calculate the an and bn coefficients. For that
purpose, use the equations (9) and (10), where the input argument, n, specifies the
harmonic order. Using the following expression bit(caracter,i) you can get the value of
bit(i). The third function is about the a0 calculation which can be done through the use
of expression (8). In order to return the results in all these functions you have to use the
return command.
Alter implementing the above functions, we will debug the program and verify its
correctness. To call compiler in Linux:
gcc Fourier.cpp –o MiFourier –lm
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With –o option we specify the name of the resulting binary file. The –lm option is
necessary to link with math library (sin and cos trigonometric functions)
Impact of physical medium on the transmitted signal.
Here we are going to use our program to analyze how the transmitted signal is distorted
when crossing an specific transmission channel. For that purpose, we are going to
produce four different scenarios corresponding with the transmission of one signal with
the same transmission speed under different channel bandwidths. In particular, we will
use channels with bandwidths of 400, 1000, 2000 and 9000 Hz.
Run the program with the following parameters:
./MiFourier –c a –v 1200 –b 400 –a 10000 > bw400.txt
Option –c specifies the character we are going to transmit (also we can specify the
ASCII code: –c 97). Option –v fixes the transmission speed (in bits per second). With
option --b we can define the desired channel bandwidth. Finally, option –a is aesthetic
option used to determine width of the frequency domain plot (in Hz).
As the character and transmission speed are just the same than the default ones, we do
not need to specify them. Run the program to get results for the rest of proposed
bandwidths.
./MiFourier –b 1000 –a 10000 > bw1000.txt
./MiFourier –b 2000 –a 10000 > bw2000.txt
./MiFourier –b 9000 –a 10000 > bw9000.txt
Compare the four graphics, using the GNUPlot program. You can observe that as
bandwidth increases, more harmonics cross the channel and as a consequence the
reconstructed signal is better (closer to the original one). Do you think that a bandwidth
of 400 Hz is enough for transmitting this character with a speed of 1.200 bps. ? Should
it be necessary a channel bandwidth of 9 KHz or it would be enough with 2 KHz?
Impact of transmission speed on the transmitted signal
En general, todos tenemos la idea de que al transmitir por un medio físico una señal
digital, la velocidad máxima de transmisión está acotada. Hay muchos factores que
introducen esta limitación, pero uno de los principales es que el ancho de banda del
medio es limitado. Como ya hemos visto, esto provoca que a partir de una determinada
frecuencia, el resto de armónicos se atenúan.
We can notice that transmission speed is very related with signal period T. So that, as
transmission speed increases the signal period becomes shorter. The fundamental
frequency that determines the distance between harmonics also increases, so a lower
number of harmonics will cross the channel and, as a consequence, the reconstructed
signal will be poorer .
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To verify this effect we will use our program with the following transmission speeds
1200, 2400, 4800 and 9600 bps, and a fixed channel bandwidth of 6 KHz.
./MiFourier –b 6000 –a 10000 –v 1200 > v1200.txt
./MiFourier –b 6000 –a 10000 –v 2400 > v2400.txt
./MiFourier –b 6000 –a 10000 –v 4800 > v4800.txt
./MiFourier –b 6000 –a 10000 –v 9600 > v9600.txt
Analyze the results shown in the four resulting graphics. The difference between them
stay on the distance between two consecutive harmonics. So, when the transmission
speed increases the number of harmonics that arrive at the other end decreases. Is it
acceptable a transmission speed of 9600 bps for this physical channel ?
In general, is not obvious to determine the maximum transmission speed for a particular
physical channel. In theory, we can say that the maximum transmission speed will be
the one that allows crossing the channel the enough harmonics to properly reconstruct
the signal at the other end. This would be true in this example when the first six
harmonics cross the channel.
Next, use the program to transmit a character using a channel with a 6 KHz. bandwidth
and a transmission speed of 7600 bps. Then, do the same with a channel of 45 KHz.
bandwidth and a transmission speed of 56000 bps. Is there any difference between both
reconstructed signals? What do you think is more important channel bandwidth,
transmission speed o the relation between both ? Justify the answers.
Problem
We have a physical channel that behaves as an ideal low-pass filter with a 10 KHz.
bandwidth (0..10 KHz.). We continuously transmit a signal composed of an ASCII
character (whatever char) of 8 bits with a transmission speed of 1000 bps. How many
harmonics cross the channel? Idem if transmission speed is 2000 bps ?. Solve the
problem analytically, and then run program to verify the results.
Analytic solution: The signal period covers the eight character bits. If transmission
speed is 1000 bps, then one bit lasts 1 ms, and T = 8 ms. The fundamental frequency
(the inverse of period) will be f = 1/0,008 = 125 Hz. If the channel bandwidth is 10000
Hz, the number of passing harmonics will be (10000 / 125) = 80. If transmission speed
doubles, the fundamental frequency also doubles (250 Hz) and the number of harmonics
will be just the half (40).
Additional questions
One interesting improvement to the program will be the following: Allow a physical
channel with a band-pass behaviour like the local-loop phone lines (with a passing band
between 300 and 3300 Hz.)
Another proposed addition is modelling the channel taking in a more realistic approach.
So harmonics will be attenuated from one frequency fa up to the cut frequency fc. To
perform this change, find the attenuation tables (as a function of frequency) for different
wire categories, and apply the attenuation over a specified number of harmonics (i.e. the
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first 100 harmonics). If you do not find these tables, you can build them from the
attenuation graphics shown in lecture slides (signal theory topic).
10