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.
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.
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
Given a signal as a time function,
Be able to compute the frequency components of the signal.
