Fourier Series |
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.
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?
-
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:
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.
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?
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