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TION NEWSLETTER Fourier analysis made Easy Part 1
Jean Baptiste Joseph, Baron de Fourier, 1768  1830
While studying heat conduction in materials, Baron Fourier (a title given to him by
Napoleon) developed his now famous Fourier series, approximately 120 years after
Newton published the first book on calculus. It took Fourier another twenty years to
develop the Fourier transform which made the theory applicable to a variety of
disciplines such as signal processing where Fourier analysis is now an essential tool.
Fourier did little to develop the concept further and most of that work was done by
Euler, LaGrange, Laplace and others. Fourier analysis is now also used in thermal
analysis, image processing, quantum mechanics and physics.
Why do we need to do Fourier analysis – In communications, we can state the
problem at hand this way; we send an informationlaced signal over a medium. The
medium and the hardware corrupt this signal. The receiver has to figure out from the
received signal which part of the corrupted received signal is the information signal and
which part the extraneous noise and distortion. The transmitted signals have well
defined spectral content, so if the receiver can do a spectral analysis of the received
signal then it can extract the information. This is what Fourier analysis allows us to do.
Fourier analysis can look at an unknown signal and do an equivalent of a chemical
analysis, identifying the various frequencies and their relative “quantities” in the signal.
Fourier noticed that you can create some really complicated looking waves by just
summing up simple sine and cosine waves. For example, the wave in Figure 1a is sum
of the just three sine waves shown in Figures 1b, 1c and 1d of assorted frequencies and
amplitudes. Easy Fourier Analysis Part 1 Complextoreal.com 1 (a)  A complicated looking wave (b)  Sine wave 1 (c) Sine wave 2 (d)  Sine wave 3 Figure 1  Sine waves
Let’s look at signal 1a in three dimensions. With time progressing to the right we
see the amplitude going up and down erratically, we are looking at the signal in Time
domain. From this angle, we see the sum of the three sine waves as shown in Fig
(1b,c,d).
When we look at the same signal from the side along the zaxis, what we see are the
three sine waves of different frequencies. We also see the amplitude but only as a line
with its maximum excursion. This view of the signal from this point of view is called
the Frequency Domain. Another name for it is the Signal Spectrum. Figure 3  Looking at signals from two different points of view The concept of spectrum came about from the realization that any arbitrary wave is
really a summation of many different frequencies. The spectrum of the composite wave
f(t) of Fig (1) is composed of just three frequencies and can be drawn as in Fig (3.1). Easy Fourier Analysis Part 1 Complextoreal.com 2 This is called a onesided magnitude spectrum. Onesided not because anything
has been left out of it, but because only positive frequencies are represented. (So what
is a negative frequency? Is there such a thing? We will discuss this in more detail in
later. For now, suffice it to say that a negative frequency is simply a frequency which is
lagging in phase.) Figure 3.1  The Frequency Domain spectrum of wave in Figure 1 Now let’s look at the signal in frequency domain. Think of it as a recipe, with xaxis showing the ingredient and the yaxis, how much of that ingredients. The xaxis
for a signal would show the different frequencies in the signal and yaxis the amplitude
of each of those frequencies.
Let’s expand on this concept. V8 juice for example has many different ingredients
such as celery juice, salt, water, spices, etc.. We can remove most of these ingredients
one by one and the remaining liquid would still taste essentially like V8. What we can
not remove and have the item still retain its primary character is called the
fundamental component. In V8, that is tomato juice.
Signals carrying information, similarly, have a fundamental frequency along with
other lesser important frequencies. A noisy signal on the other has no single
fundamental frequency. It has a flat spectrum. All frequencies are present in the signal
in the same quantities. So a spectrum does not necessarily have a fundamental
component. The spectrum of such a signal would be flat.
Let’s take the following complicated looking wave. Easy Fourier Analysis Part 1 Complextoreal.com 3 This wave is periodic with
a period = 1 sec. Figure 4  Another really complicated looking wave The first thing we notice is that the wave is periodic. Fourier analysis tells us that
any arbitrary wave such as the above that is periodic, can be represented by a sum of
other simpler waves.
Let’s try summing a bunch of sine waves to see what they look like. Figure 5a  This is a wave of frequency 1 Hz , amplitude = 1 Figure 5b  This is a wave of frequency 2 Hz , amplitude = 1 Figure 5c  This is a wave of frequency 3 Hz , amplitude = 1 Easy Fourier Analysis Part 1 Complextoreal.com 4 Figure 5d  This is a wave of frequency 4 Hz , amplitude = 1 Each of the waves here have frequencies that are integer multiples. In more
scientific words, we say that they are harmonic to each other, similar to musical notes
which are also called harmonic.
•
What is a harmonic – It is a frequency that is integer multiple of the other
frequency. Waves of frequency 2 and 4 Hz are harmonics to a wave of frequency 1 Hz
since they are both its integer multiples. Frequencies 2.4 and 3.6 Hz are harmonics to a
wave of frequency 1.2 Hz since they are both integer multiples of 1.2 Hz.
•
When the multiple factor is even, the harmonic is called an even
harmonic and when the factor is odd, it is called the odd harmonic.
•
Frequencies 66, 110, 154 Hz are odd harmonics of frequency 22 Hz,
whereas 44, 88 and 132 Hz are even harmonics.
We write the sum of N such harmonics as
N f (t ) = ∑ sin(n ω t ) (1) n =1 Each wave has a frequency that is integer multiple of the starting frequency ω,
which is equal to 2π (1) in this case since f = 1 Hz. Here is what a sum of four sine
waves of equal amplitude, each starting with a phase of 0 degrees at time 0 looks like. f (t ) = sin(1ω t ) + sin(2ω t ) + sin(3ω t ) + sin(4ω t ) Figure 6  This is the sum of all four of the above sine waves. Easy Fourier Analysis Part 1 Complextoreal.com 5 If we keep going and add a large number of sine waves of equal amplitude, the
summation approaches an impulse function as shown below for N = 25. Since we
added together 30 sine waves of amplitude 1, the maximum amplitude is 25. Figure 7  This is the sum of 25 sine waves. In the graph above, we allowed the amplitude of each harmonic to be one. Going to
the next level of abstraction, it is obvious that to represent an arbitrary wave, we need
to allow the amplitude of each component to vary. Otherwise, all we will get is the
scaled version of the signal in Fig (7). So we modify equation (1) by introducing a
coefficient an to represent the amplitude of the nth sine wave as follows:
N f (t ) = ∑ an sin(n ω t ) (2) n =1 The coefficient an allows us to vary the amplitude of each harmonic fn(t) = sin(nωt)
to create a variety of waves. Here is what one particular wave which is the sum of four
sine waves of unequal amplitude looks like. Figure 8  Sum of four sine waves of unequal amplitude But looking at the original wave, f(t) in Fig (4), we see that it starts at a nonzero
value. No matter how many sine waves we add together, we can not replicate this wave
because sine waves are always zero at time zero. But if we add some cosine waves to
the sum in equation (2) which do not start at zero, we may be able to create the wave of
Figure 2.
Easy Fourier Analysis Part 1 Complextoreal.com 6 So let’s add a bunch of cosine waves of varying amplitudes to our f(t) equation. Figure 9a  A cosine wave of frequency 1 Hz, amplitude = 1 Figure 9b  A cosine wave of frequency 2 Hz, amplitude = 1 Figure 9c  A cosine wave of frequency 3 Hz, amplitude = 1 Figure 9d  A cosine wave of frequency 4 Hz, amplitude = 1 Once again the sum of the cosine waves of equal amplitude looks like this. Easy Fourier Analysis Part 1 Complextoreal.com 7 Figure 10  Sum of four cosine waves of equal amplitude Figure 11  Sum of 30 cosine waves of equal amplitude A sum of 30 cosine waves looks like as in Fig (11). It approaches an impulse
function just as the sum of sine waves did but this one is an even function.
• Even function – The function that is symmetrical about the yaxis. Cosine wave
is an even function.
• Odd function – The function that is not symmetrical about the yaxis. Sine wave
is an odd function.
The sum of the cosines is an even function. Contrast this with Fig (7), the sum of
sines, which is an odd function. These characteristics, odd and even, are useful when
looking at real and imaginary components of signals.
Now let’s allow the amplitude of each cosine wave to vary. Here is what one
particular sum of four cosines of unequal amplitudes looks like. Figure 12  Sum of four cosine waves of unequal amplitude. Now let’s modify equation (2) to add the cosine waves. Easy Fourier Analysis Part 1 Complextoreal.com 8 N N n =1 n =1 f (t ) = ∑ an sin(n ω t ) + ∑ bn cos(n ω t ) (3) The coefficients bn allow us to vary the amplitude of each cosine wave. Putting this
equation to work, we see in the following figure the sum of four sine and four cosine
waves. Figure 13  Sum of five sine and cosine waves of unequal amplitudes We are very close to completing our equation for arbitrary periodic waves. There is
only one remaining issue. Sums of sine and cosines are always symmetrical about the
xaxis so there is no possibility of representing a wave with a dc offset. To do that we
add a constant, a0 to the equation. This constant moves the whole wave up (or down)
along the yaxis offset.
N N n =1 n =1 f (t ) = a0 + ∑ an sin(nω t ) + ∑ bn cos(nω t ) (4) The coefficient a0 provides us with the needed dc offset from zero. Now with this
equation we can fully describe any periodic wave, no matter how complicated looking
it is. All arbitrary but periodic waves are composed of just plain and ordinary sines and
cosines and can de composed in its constituent frequencies..
Equation (4) is called the Fourier Series equation. The coefficients a0, an, and
bn are called the Fourier Series Coefficients.
An equation with many faces There are several different ways to write the Fourier series. One common
representation is by linear frequency instead of the radial frequency. Replace ω by 2πf
and then write the equation as
N N n =1 n =1 f (t ) = a0 + ∑ an sin( 2π nft ) + ∑ bn cos(2π nft ) Easy Fourier Analysis Part 1 Complextoreal.com (5) 9 The Fourier series equation allows us to represent any wave, low or high frequency,
baseband or passband, large bandwidth or very small. N is the number of harmonics
used in the summation. This is a variable and we can choose it to be anything, but for
complete representation, N is set to infinity. This makes the equation completely
general and we can represent even noise signals this way. The harmonics themselves do
not have to be of integer frequencies such 1, 2, 3 etc.. The starting frequency can be any
real or imaginary number. However, the harmonics of the starting frequency ARE its
integer multiples.
f n (t ) = n f (t ) = n 1
T f(t) the smallest frequency is called the resolution frequency, determines how
finely we decompose the signal. It can be any arbitrary number, say for example 2.35.
From that point on, the next harmonic is 2 times this, next one 3 times and so on. T, is
the period of the first wave we pick, and each fn is an integer multiple of the inverse of
that period. We can also start anywhere. We can pick a small resolution frequency and
then start the analysis with the 100th harmonic for example.
Replace fn by n/T, where T is the period and replace N by ∞ to write equation (5)
in a different from.
∞ f (t ) = a0 + ∑ an sin ( 2π t n / T ) + bn cos ( 2π tn / T ) (6) n =1 We can also convert all sine waves and make them cosine waves by adding a halfperiod phase shift. The cosine representation, used often in signal processing is written
by adding a phase term to the equation.
sin(2π ft ) = cos(2π ft + π / 2)
To create the f(t) we would add two cosine waves of the same frequency, except the
one of them would have a π / 2 phase shift (that’s a sine wave, really.) Now we have
only cosines. The name of the coefficient has been changed to cn, to reduce confusion
between this term and the terms an and bn. a0 and C0 would be exactly the same as a0.
∞ f (t ) = C0 + ∑ Cn cos(2π f n t + φn )
n =1
∞ f (t ) = C0 + ∑ Cn cos( wn t + φn )
n =1
∞ f (t ) = C0 + ∑ Cn cos(
n =1 (6a) 2π n
t + φn )
T Easy Fourier Analysis Part 1 Complextoreal.com 10 In complex representation, the Fourier equation is written as f (t ) = ∞ ∑ Cne j nπ t /T
n =−∞ (7) Complex notation, first given by Euler, is most usefulalbeit scarylooking form. In
next part, we look at how it is derived and used for signal processing.
All these different representations of the Fourier Series (4), (5), (6), (6a) and (7) are
identical and mean exactly the same thing.
How to compute the Fourier Coefficients of an arbitrary wave In signal processing, we are interested in spectral components of a signal. We want
to know how many sines and cosines make up our signal and what their amplitudes are.
Alternatively, what we really want are the Fourier coefficients of our signal. Once we
know the Fourier coefficients, we know which frequencies are present in the signal and
in what quantities. This is similar to doing chemical analysis on a compound, figuring
out what elements are there and what relative quantity.
How do we compute the Fourier coefficients?
Computing a0
∞ f (t ) = a0 + ∑ (an sin ωnt + bn cos ωn t )
n =1 The constant a0 in the Fourier equation above represents the dc offset. But before
we compute it, let’s take a look at one particular property of the sine and cosine waves.
Both sine and cosine wave are symmetrical about the xaxis. When you integrate a
sine or a cosine wave over one period, you will always get zero. The areas above the xaxis cancels out the areas below it. This is always true over one period as we can see in
the figure below.
Positive and
negative area
cancel.
+Area +Area
Area Positive and
negative area
cancel. +Area Area Figure 14  The area under a sine or a cosine wave over one period is always zero.
Easy Fourier Analysis Part 1 Complextoreal.com 11 ∞ ∫o ∑ an sin wnt dt = 0 n =1 ∞
T ∫o ∑ an sin wnt + bn cos wnt dt = 0 n =1 T The same is also true of the sum of sine and cosines. Any wave made up of sum of
the sine and cosine waves also has zero area over one period. So we see that if we were
to integrate our signal over one period the area obtained will have to come from
coefficient a0 only. The harmonics can make no contribution and they fall out.
T ∫
0 6444447444448
40
4
∞
T
T f (t )dt = ∫ ao dt + ∫ ∑ an sin wn t + bn cos wnt dt
0
o n =1 (8) The second term is zero in (8), since it is just the integral of a wave made up of sine
and cosines. Now we can compute a0 by taking the integral of our complicated looking
wave over one period. The wave has nonzero area
in one period, which means it
has a DC offset. Figure 16  Signal to be analyzed, looks like it has a dc offset since there is more area
above the xaxis than below.
All area comes from the a0
coefficient. Figure 16a  The dc component Easy Fourier Analysis Part 1 Complextoreal.com 12 Area under the wave when
shifted down is zero. Figure 16b  Signal without the dc component The area under one period of this wave is equal to
T ∫
0 T f (t )dt = ∫ ao dt (9) 0 Integrating this very simple equation we get,
T ∫ f (t )dt = a T
0 (10) 0 We can now write a very easy equation for computing a0
1
a0 =
T T ∫ f (t )dt (11) 0 Since no harmonics contribute to area, we see that a0 is equal to simply the area
under our complicated wave for one period divided by T, the integral period. We can
compute this area in software and if it is zero, then there is no dc offset. This is also the
mean value of the signal. A signal with zero mean value has no dc offset.
Computing an Now we employ a slightly different trick from basic trigonometry to compute the
coefficients of the sine waves. Here is a sine wave of an arbitrary frequency nω that has
been multiplied by itself. f (t ) = sin n ω t *sin n ω t
Easy Fourier Analysis Part 1 Complextoreal.com 13 Figure 17  The area under a sine wave multiplied by itself is always nonzero. We notice that the resulting wave lies entirely above the xaxis and has a net
positive area. From integral tables we can compute the area as equal to
T ∫ an ( sin nωt ) ( sin mωt ) dt = anT / 2 for n = m (12) 0 Where T is the period of the fundamental harmonic. But now let’s multiply the sine
wave by an arbitrary harmonic of itself to see what happens to the area.
f (t ) = sin n ω t *sin m ω t Sine wave multiplied by
another of a different
harmonic Multiplying one sine wave by
any other causes the area
under the new wave to
become zero. Figure 18  The area under a sine wave multiplied by its own harmonic is always zero. The area in one period of a sine wave multiplied by its own harmonic is zero. We
conclude that when we multiply a signal by a particular harmonic, the only contribution
comes from that particular harmonic. All others harmonics contribute nothing and fall
out. Easy Fourier Analysis Part 1 Complextoreal.com 14 T ∫ a ( sin nωt ) ( sin mωt ) dt = 0 for n ≠ m n 0 (12)
T ∫ a ( sin nωt ) ( sin mωt ) dt = a T / 2
n for n = m n 0 Now let’s multiply a sine wave by a cosine wave to see what happens.
f (t ) = sin nω t *cos mω t Sine wave multiplied by a
cosine wave for any n and m Figure 19  The area under a cosine wave multiplied by a sine wave is always zero. It seems that the area under the wave which is multiplication of a sine and cosine
wave is always zero whether the harmonics are the same or not. Summarizing, by
setting ωn = nω
T ∫ a ( sin ω t ) ( sin ω t ) dt = 0
n n for n ≠ m m 0 T ∫ a ( sin ω t ) ( sin ω t ) dt = a T / 2
n n m n for n = m (13) 0 T ∫ a ( cos ω t ) ( sin ω t ) dt = 0
n n m for all n and m 0 Rules:
1. The area under one period of a sine or a cosine is zero. 2. The area under one period of a wave that is a product of two sine or cosine
waves of nonharmonic frequencies is zero. Easy Fourier Analysis Part 1 Complextoreal.com 15 3. The area under one period of a wave that is a product of two sine or cosine
waves of same harmonic frequency is nonzero and not equal to anT / 2 , where
T is the period of the resolution frequency we have chosen.
4. The area under one period of a wave that is a product of a sine wave and a
cosine wave of any frequencies (different or equal) is equal to zero. Recall that in vector representation, sine and cosines are orthogonal to each other.
So all harmonics are by definition orthogonal to each other.
A very satisfying interpretation of the above rules is that sine and cosine waves can
act as filtering signals. In essence they act as narrowband filters and take out all
frequencies except the one of interest. This forms the basic concept of a filter.
Now let’s use this information. Successively multiply the Fourier equation by a sine
wave of a particular harmonic and integrate over one period as in equation below. 64 744
40
8 T T T 64444 0
74444
8 T f (t ) sin ( nwt ) dt = ∫ a0 sin ( wt ) dt + ∫ an sin ( nω t ) sin ( nω t ) dt + ∫ bn cos ( nω t ) sin ( nω t ) dt
0
0
0 ∫
o We know that the integral of the first and the third term is zero since the first term
is the integral of a sine wave multiplied by a constant (Rule 1) and the third is a sine
wave multiplied by a cosine wave (Rule 3). This simplifies our equation considerably.
The integral of the second term is
T ∫a n sin ( nω t ) sin ( nω t ) dt = 0 anT
2 (13) From this we write the equation to obtain an, which are the coefficients of each of
the sine waves as follows
T an = 2
f (t ) sin ( n ω t ) dt
T∫
0 (14) The an is then computed by taking the signal over one period, successively
multiplying it with a sine wave of n times the starting fundamental frequency and then
integrating. This gives the coefficient for that particular harmonic.
Imagine we have a signal that consists of just one frequency, we think it is around 5
Hz (and is a sine wave from). We begin by multiplying this signal by a sine wave of
frequency .2 and each of its harmonics which are .4, .6, .8 ,…..10 and so on. Actually
Easy Fourier Analysis Part 1 Complextoreal.com 16 since we know it is in the range of 5 Hz, we can dispense with the lower harmonics say
up to 4 and start with 4.2 and go to 5.8 Hz.
Here is all the math we do.
1. Multiple the wave with a sine wave of frequencies 4.2 and integrate the result.
Most likely the result will be zero.
2. Go to next harmonic, which 4.4. This is 22nd harmonic of the resolution
frequency .2 Hz.
3. Repeat step 1 and 2 and continue until harmonic frequency is equal to 5.8 Hz.
The results will show that the integrals of all harmonics frequencies are zero, except
for the 25th harmonic, the integral of which will be equal to
a25T
= 2.5a25
2
One period integral
a25 =
2.5 = Where T = 1/f = 1/.2 = 5 sec. The coefficient can now be calculated which gives the
amplitude of the wave. (We already know its frequency, which is 5 Hz, since the
integral is nonzero for that component.).
Computing coefficient of cosines, bn Now instead of multiplying by a sine wave we multiply by a cosine wave. The
process is exactly the same as above.
T ∫
o 644 0
744
8 64444 744444
40
8
T T T f (t ) cos ( nω t ) dt = ∫ a0 cos ( nω t ) dt + ∫ an sin ( nω t ) cos ( nω t ) dt + ∫ bn cos ( nω t ) cos ( nω t ) dt
0
0
0 Now terms 1 and 2 become zero. (First term is zero from rule 1, the second term
due to rule 3.) The third terms is equal to
T ∫b n 0 cos ( nωt ) cos ( nωt ) dt = bnT
2 (14) and the equation can be written as
Easy Fourier Analysis Part 1 Complextoreal.com 17 T 2
bn = ∫ f (t ) cos ( nωt ) dt
T0 (15) So the process of finding the coefficients is multiplying our signal with
successively larger frequencies of a fundamental wave and integrating the results. This
is easy to do in software. The results obtained successively are the coefficient for each
frequency of the harmonic wave. We do the same thing for sines and cosine
coefficients.
Following this process, we compute the coefficients of the following wave Figure 20  The signal to be analyzed Without going through the math, we will give the answers in two vectors, first is the
coefficients of the sine and second the cosine waves and the dcoffset.
an = [.4 .3 .7 .3 .3 .3 .2 .3 .4]
bn = [.05 .2 .7 .5 .2 .2 .1 .05 .02]
a0 = .32
From this we can write the equation of the above wave as
f (t ) = .32 + . 4 sin(2π 1)t + .3sin(4π )t + .7 sin(6π )t
+ .3sin(8π )t + .3sin(10π )t + .3sin(12π )t + .2sin(14π )t + ....
+.05cos(2π 1)t + .2 cos(4π )t + .7 cos(6π )t + .5cos(8π )t +
+.2 cos(10π )t + .2 cos(12π )t + .1cos(14π )t + ...
The coefficients are the amplitudes of each of the harmonics. The resolution
frequency is 1 Hz and the harmonics are integer multiples of this frequency. Now we
know exactly what the components of the received wave are. If the transmitted wave
consisted only of one of these frequencies, then, we can filter this wave and get back
the transmitted signal.
Easy Fourier Analysis Part 1 Complextoreal.com 18 Summary
The Fourier series is given by
∞ ∞ n =1 n =1 f (t ) = a0 + ∑ an sin(ωnt ) + ∑ bn cos(ωnt )
where ωn = 2π nf
The coefficients of the Fourier series are given by
a0 = an = 1
T T ∫ f (t )dt
0
T 2
f (t ) sin ( nω t ) dt
T∫
0
T 2
bn = ∫ f (t ) cos ( nω t ) dt
T0
where ω is the fundamental frequency and is related to T by ωn = 2π f n = 2π n
T Coefficients become the spectrum Now that we have the coefficients, we can plot the magnitude spectrum of the
signal. Easy Fourier Analysis Part 1 Complextoreal.com 19 Magnitude 0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 Sine
Cosine Frequency Figure 21  The Fourier series coefficients for each harmonic You may now say that this spectrum is in terms of sines and cosines, and this is not
the way we see it in books. The spectrum ought to give just one number for each
frequency.
We can compute that one number by knowing that most signal are represented in
complex notation where sine and cosine waves are related in quadrature. The total
power shown on the y axis of the spectrum is the power in both the sine and cosine
waves in the real and imaginary components of the same frequency. We can compute
the magnitude by from the root sum square of the sine and cosine coefficients for each
harmonic including the dc offset of the zero frequency value.
2
Magnitude = an + bn2 Plot the modified spectrum 1.20 Magnitude 1.00 Magnitude 0.80
0.60
0.40
0.20
0.00
Frequency Easy Fourier Analysis Part 1 Complextoreal.com 20 Figure 22  A traditional looking spectrum created from the Fourier coefficients Voila! Although this is not a real signal, we see that it now looks like a traditional
spectrum. The largest component is at frequency = 3. The yaxis can easily be
converted to dB. In complex representation, the phase of the signal is defined by
φ n = tan −1 ( bn / a n ) For every frequency, we can also compute and plot the phase. Phase plays a very
important role in signal processing and particularly in complex representation and
shows useful information about the signal.
One thing you may not have noticed during this computation of the coefficients is
that they will be different depending on what you pick as the resolution frequency. We
will get different answers depending on the choice we make for this number. In essence
depending on the resolution, the signal energy leaks from one frequency to the next so
we get different answers, but the overall picture remains the same. The issues of
leakage will discussed later.
We also stated that the wave has to be periodic. But for real signals we can never
tell where the period is. Random signals do not have discernible periods. In fact, a real
signal may not be periodic at all. In this case, the theory allows us to extend the
“period” to infinity so we just pick any representative section of our signal or even the
whole signal and call it “The Period”. Mathematically this assumption works out just
fine for real signals. Figure 23  We call the signal periodic, even though we don’t know what lies at
each end. Figure 24  Our signal repeated to make it mathematically periodic, but ends
do not connect and have discontinuity Easy Fourier Analysis Part 1 Complextoreal.com 21 The part of the signal that we pick as representing the real period is only a sample
of the whole and not really the actual period. The end section of the chosen section will
most likely not match as they would for a real periodic signal. The error introduced into
our analysis due to this end mismatch is called aliasing. Windowing functions are used
to artificially shape the ends so that they are zero at the ends and so the chosen signal
portion is made artificially periodic. This introduces errors in the analysis which have
to be dealt with by other techniques.
Next the complex representation.
Copyright 1998, All rights reserved C. Langton
Revised 2002
I can be reached at
mntcastle@earthlink.net
Other tutorials at
www.complextoreal.com Easy Fourier Analysis Part 1 Complextoreal.com 22 ...
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This note was uploaded on 02/07/2011 for the course EEE EE567 taught by Professor Tutorials during the Spring '11 term at Birla Institute of Technology & Science, Pilani  Hyderabad.
 Spring '11
 Tutorials
 Signal Processing

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