An Introduction To Fourier Analysis
Introduction
The Fourier Series
Calculating The Coefficients
An Example - Repetitive Pulse
Problems
You are at: Basic Concepts - System Models - Fourier Series
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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,

• In a system, if we have the time function, we can solve an input-output differential equation to get the output, and
• We can plot functions of time and get information about them, and
• etc., etc.
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.
• In a system, if we have the time function, we can solve an input-output differential equation to get the output, but if we use frequency-based methods we may only need to do some algebra to get the output.  Most of us would rather do algebra than solve differential equations.
• Information about frequency content of a signal has often proved to give more insight into how to process a signal to remove noise.  Often it is easier to characterize the frequency content of a noise signal than it is to give a time description of the noise.
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.

• Start with a single term in the series and plot the response.  A single term should give you a sine wave signal with an amplitude of 1.0.
• Slowly increment the number of terms so that you include the third harmonic (two terms), the fifth harmonic (three terms), etc.
• Does the peak value increase or decrease as you increase the number of terms? • Determine if you can get the series to a point where the approximation is always within 5% of the ideal square wave.
• When the series looks like it has converged, determine the value of the square wave amplitude.  Compare that to the amplitude of the sine wave you started with in the first step.

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.

• Start with a single term in the series and plot the response.  A single term should give you a sine wave signal with an amplitude of 1.0.
• Slowly increment the number of terms so that you include the third harmonic (two terms), the fifth harmonic (three terms), etc.
• Does the function approach a square wave. • Is there anything you notice about the approximation, especially near the discontinuity?

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.

• What kind of functions can be represented using these types of series?
• Actually, most periodic signals can be represented with a series composed of sines and cosines.  Even discontinuities (like in the square wave function or the sawtooth function in the simulations) will not present an insurmountable problem, although you might expect (from the simulation results) that there are some phenomena we need to account for right at the discontinuities.
• How do you figure out what the series is for any given function?
• That's an interesting question, and we will discuss that soon.  There are some mathematical results we will need, but you should be prepared for that.
• Are there any practical implications to all of this?
• Since functions can be thought of as being composed of sines at cosines at different frequencies, and since various linear systems process sinusoidal signals in a way that is frequency dependent, these two facts mean that the response of a system with a periodic input can be predicted using frequency response methods.
• Many signals are now analyzed using frequency component concepts.  Special computational techniques (particularly the FFT, or Fast Fourier Transform) have been developed to calculate frequency components quickly for various signals.  Signals that have been analyzed include sound signals in earthquakes, bridge vibrations under dynamic load (as well as stress vibrations in many different structures from tall buildings to aircraft vibrations) and communication signals (including the signals themselves as well as the noise that interferes with the signals).
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:

• f = 1/T,
• wo = 2pf.
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.

• The triangle function above is even.  In other words, if T(t) is the triangle signal above.
• T(t) = T(-t).
• Since the function is even, there can be no odd functions in the Fourier expansion.
• In other words, there are not sine terms since sin(x) = -sin(-x), i.e. the sine function is odd.
• The triangle function above has odd symmetry around the quarter period point.  In other words, if you look at the wave as though it were centered at t = T/4, you find odd symmetry.  Even harmonic cosine functions have even symmetry around this point.  (That would be the second harmonic cosine, the fourth harmonic cosine, etc.)  In this figure, you can see the triangle wave, a second harmonic cosine wave, and the product of the two.  The product of the even harmonic cosine signal and this triangle signal do not have any net area - in a period, or even in half a period. • The conclusion that we reach is that there will be no even harmonic cosine terms.  Since there are no sine terms whatsoever, that means that the entire Fourier Series is composed of odd harmonic cosines only.  Thus, we need only compute those terms.
• Doing the computation, we note that we need only do the computation for one half of a period, and double the result. • The result is:  (NOTE:  You can check the integral as a problem for the interested student.)
• bk = 8A/(p2k2) as long as k is odd
• bk = 0 for even k.
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

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