An Introduction To Fourier Analysis
Introduction
The Fourier Series
Calculating The Coefficients
An Example - Repetitive Pulse
Problems
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What are you trying to do in this lesson?

        Here are some goals for this lesson

   Given a signal as a time function,
   Be able to compute the frequency components of the signal.
   Be able to predict how the signal will interact with linear systems and circuits using frequency response methods.
        The first goal is really to be able to express a periodic signal in frequency response language.  The second goal is to be able to take a frequency representation of a signal and use that representation to predict how the signal will interact with systems.

Why Use Frequency Representations When We Can Represent Any Signal With Time Functions?

        Signals are functions of time. A frequency response representation is a way of representing the same signal as a frequency function. Why bother - especially when we can represent the signal as a function of time and manipulate it any way we want there? For example,

Frequency response methods give a different kind of insight into a system. Those insights can have unexpected results.

        Frequency methods focus on how signals of different frequencies are represented in a signal. We think in terms of the spectrum of the signal.  Here is a rainbow.  In a rainbow, white sunlight - composed of many different colors or parts of the spectrum - is spread into its spectrum.  Here the atmosphere is a filter that treats the different parts of the light spectrum - the different light frequencies - in different ways.  For a rainbow, the different parts of the light spectrum - the different colors - are bent differently as they enter the atmosphere.  In many electrical circuits and systems, the different parts of the signal spectrum  are treated differently.  Different treatment of different parts of the electromagnetic spectrum means that you can separate out different radio, television and cell phone signals.  That gives you one very strong reason to learn about frequency methods.  In a linear system, frequency methods may be easier to apply, and may give insight you would not get otherwise.         So, give it a shot and try learning about frequency response methods. They can save you time and money in the long run.


Goals:  What are you trying to do in this lesson?
   Given a signal as a time function,
   Be able to compute the frequency components of the signal.
Be able to predict how the signal will interact with linearsystems and circuits using frequency response methods.

The Fourier Series

       Some time ago, Fourier, doing heat transfer work, demonstrated that any periodic signal can be viewed as a linear composition of sine waves. Lets look at a periodic wave. Here is an example plot of a signal that repeats every second.

        Clearly this signal is not a sinusoid - and it looks as though it has no relationship to sinusoidal signals. However, over a century ago, Fourier showed that a periodic signal can always be represented as a sum of sinusoids (sines and cosines, or sines with angles). That representation is now called a Fourier Series in his honor.

        Fourier not only showed that it was possible to represent a periodic signal with sinusoids, he showed how to do it.  Assuming this signal repeats every T seconds, then we can describe it as a sum of sinusoids.  Here is the form of the sum.  Fourier gave an explicit way to get the coefficients in a Fourier Series and we need to look at that in a while.  First we are going to look at how a signal can be built from a sum of sinusoids.

Here's that signal again.  Is this signal a sum of sinusoids? We will examine that question here now, starting with a single sine signal.

Here is a single sine signal.
The expression for this signal is just:

Sig(t) = 1 * sin(2pt/T) and T = 1 second.

Now, we are going to add one other sine to our original sine signal. The sine we add will be at three times the frequency of the original and it will be one third as large.

Sig(t) = 1 * sin(2pt/T) + (1/3) * sin(6pt/T)

This looks a little different.  Continue by adding one more sine signal - at five times the original frequency and one-fifth of the original size.

Sig(t) = 1 * sin(2pt/T) + (1/3) * sin(6pt/T) + (1/5) * sin(10pt/T)

This is getting interesting. We are just adding in terms at odd multiples of the original frequency. Here's what the signal looks like with the terms up to the 11th multiple.
This looks like a fairly lousy square wave. Let's add a lot more terms and see what happens.
Here is the signal with terms up to the 49th multiple.

        At this point is seems that this process is giving us a signal that is getting closer and closer to a square wave signal.  However, this looks like a fairly lousy square wave.   Let's add a lot more terms and see what happens.

        Here is the signal with odd terms up to the 79th multiple.  Now we're getting a pretty clear indication of a square wave with an amplitude a little under 0.8.  In fact, the way we are building this signal we are using Fourier's results.  We know the formula for the series that converges to a square wave.

        In fact, the way we are building this signal we are using Fourier's results. We know the formula for the series that converges to a square wave.  Here's the formula. For a perfectly accurate representation, let N go to infinity.

Now, we're going to give you a chance to do this kind of experiment yourself.  Shown below is an interactive demo that will let you control the number of terms in the summation above.  In the demo you can also control the frequency.


Experiments

E1   In the demo above, do the following.


Let's examine another case.  Here is another simulator.  However, here the function that is implemented is given by the sum below.
Here is the simulator

Experiments

E2   In the demo above, do the following.



Calculating The Fourier Series Coefficients

       At this point there are a few questions that we need to address.  Here are some questions that need to be answered.

Now, let's try to answer some of these questions, starting with the computation of the frequency components.

        In general, a periodic signal can be represented as a sum of both sines and cosines.  Also, since sines and cosines have no average term, periodic signals that have a non-zero average can have a constant component.  Altogether, the series becomes the one shown below.  This series can be used to represent many periodic functions.  The function, f(t), is assumed to be periodic.

The coefficients, an and bn, are what you need to know to generate the signal.

        To compute the coefficients we take advantage of some properties of sinusoidal signals.  The starting point is to integrate a product of f(t) with one of the sinusoidal components as shown below.

Now, if we assume that the function, f(t), can be represented by the series above, we can replace f(t) with the series in the integral.

Here, we note the following:

Then, when we do the integration, we can use some properties of the sinusoidal functions.  In all cases here, the integral is take over exactly one period of the periodic signal, f(t).


So, when we do the integration of the function, f(t), multiplied by any sine or cosine function, they almost all work out to be zero.  The only one that doesn't work out to be zero is the one where n and m are equal.

Realizing all of this, we can conclude:

or:

Which gives us a way to compute any of the b's in the Fourier Series.

        At this point we have half of our problem solved because we can compute the b's, but we still need to compute the a's.  However, we can do the same thing for the a's that we did for the b's (and we will let you do that yourself) and we get the following expressions for the coefficients.


and these expressions are good for n>0 and m> 0.  The only coefficient not covered is ao which is given by:

So, now we have a way to find all of the coefficients in a Fourier Series expansion.  Let us apply what we know to an example.


Example/Experiment

E3      We will compute the Fourier Series of a general pulse that repeats.  The pulse sequence is shown below.  The pulse signal varies between zero volts and one volt.

Now, to evaluate the coefficients, we do the integrations indicated above.  We have the following.

or:
an = 2Asin(nwoTp)/(nTwo)

an = Asin(nwoTp)/(np)

Similarly,

or:
bn = 2A[-cos(nwoTp) + 1]/(nTwo)

bn = A[-cos(nwoTp) + 1]/(np)

and,

ao = (Tp/T)

Now, we can compute some of the coefficients for a particular case.  We will examine the situation where the pulse is high for one-fourth of the period, i.e. when Tp = T/4.  In that situation we have:

nwoTp = (n2p/T)Tp = np/2

Note that the a's (the cosine coefficients) will all be zero for even n's, while the b's (the sine coefficients) will be zero for every fourth n.  That being said, the coefficients we have computed are given in the table below.  For this table we have assumed a period of 4 seconds.  We'll show that later in a real-time simulator.
 

n
an
bn
0
.25
-
1
.31831
.31831
2
0
.31831
3
-.10610
.10610
4
0
0
5
.06366
.06366
6
0
.10610
7
-.04547
.04547
8
0
0
9
.03537
 .03537
10
0
.06366

Now, we can check whether these coefficients actually produce a pulse.  Here is a real-time simulator that will let you check that.  It has been pre-loaded with the coefficients we calculated above to produce a pulse.  However, since we are only using harmonics up to the 10th harmonic, it will not be an exact representation.

Run the simulator to check whether we are close.  Then do the following.

Questions/Problems

        Using the simulator, answer the following questions

Q1   Does the waveform with 10 harmonics look like - with more harmonics - it will converge to the pulse we started with?


P1   Determine the average value (i.e. DC component) of the signal.

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

Your grade is:


Example/Experiment

E4      Next, we will compute the Fourier Series of a triangle wave, as pictured below.

Now, to evaluate the coefficients, we need to do the integrations indicated above.  However, we know a few things about this function.

Now, if you click here you can get to an example problem where you can check this calculation.  Go to that example problem and answer the questions there.


        At this point you have the basic knowledge you need to compute Fourier Series representations for periodic signals.  Moreover, when you encounter the Fast Fourier Transform (FFT) you should be able to understand the concepts you will encounter there.  Fourier Series concepts are useful in their own right, but they are also the building blocks you need to be able to understand Fourier and laPlace transforms, and, in turn, those concepts are the fundamental concepts you need when you encounter linear systems and control systems.

Problems
Tools Useful Link(s)
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