Name
----------~-----------------
Date
------------------
Fourier Theory-Frequency Domain and
Time Domain
OBJECTIVES
1. Learn how a square wave can be produced from a series of sine waves at different
2.
3.
4.
5.
frequencies and amplitudes.
Learn how a triangular wave can be produced from a series of cosine waves at different
frequencies and amplitudes.
Learn about the difference between curve plots in the time domain and the frequency
domain.
Examine periodic pulses with different duty cycles in the time domain and in the frequency
domain.
Examine what happens to periodic pulses with different duty cycles when passed through a
low-pass filter when the filter cutoff frequency is varied.
MATERIALS
One function generator
One oscilloscope
One spectrum analyzer
One LM 741 op-amp
Two 5 nF variable capacitors
Resistors: 5.86 kn, 10 kn, and 30 kQ
THEORY
Communications systems are normally studied using sinusoidal voltage waveforms to simplify
the analysis. In the real world, electrical information signals are normally nonsinusoidal voltage
waveforms, such as audio signals, video signals, or computer data. Fourier theory provides a
powerful means of analyzing communications systems by representing a nonsinusoidal signal
as a series of sinusoidal voltages added together. Fourier theory states that a complex voltage
waveform is essentially a composite of harmonically related sine or cosine waves at different
53
frequencies and amplitudes determined by the particular signal voltage waveshape. Any
nonsinusoidal periodic waveform can be broken down into a sine or cosine wave equal to the
frequency ofthe periodic waveform, called the fundamental frequency, and a series of sine or
cosine waves that are integer mUltiples of the fundamental frequency, called harmonics. This
series of sine or cosine waves is called a Fourier series.
Most of the signals analyzed in a communications system are expressed in the time domain,
meaning that the voltage, current, or power is plotted as a function oftime. The voltage,
current, or power is represented on the vertical axis and time is represented on the horizontal
axis. Fourier theory provides a new way of expressing signals in the frequency domain,
meaning that the voltage, current, or power is plotted as a function of frequency. Complex
signals containing many sine or cosine wave components are expressed as sine or cosine
wave amplitudes at different frequencies, with amplitude represented on the vertical axis and
frequency represented on the horizontal axis. The length of each of a series of vertical
straight lines represents the sine or cosine wave amplitudes, and the location of each line
along the horizontal axis represents the sine or cosine wave frequencies. This is called a
frequency spectrum. In many cases the frequency domain is more useful than the time
domain because it reveals the bandwidth requirements of the communications system in
order to pass the signal with minimal distortion. Test instruments for displaying signals in
both the time domain and the frequency domain are available. The oscilloscope is used to
display signals in the time domain and the spectrum analyzer is used to display the
frequency spectrum of signals in the frequency domain.
,.
i
In the frequency domain, normally the harmonics decrease in amplitude as their frequency gets
higher until the amplitude becomes negligible. The more harmonics added to make up the
composite waveshape, the more the composite waveshape will look like the original
waveshape. Because it is impossible to design a communications system that will pass an
infinite number of frequencies (infinite bandwidth), a perfect reproduction of an original signal
is impossible. In most cases, eliminating some ofthe harmonics does not significantly alter the
original waveform. The more information contained in a signal voltage waveform (faster
changing voltages), the larger the number of high-frequency harmonics required to
reproduce the original waveform. Therefore, the more complex the signal waveform (the
faster the voltage changes), the wider the bandwidth required to pass it with minimal
distortion. A formal relationship between bandwidth and the amount of information
communicated is called Hartley's law, which states that the amount of information
communicated is proportional to the bandwidth of the communications system and the
transmission time.
Because much of the information communicated today is digital, the accurate transmission of
binary pulses through a communications system is important. Fourier analysis of binary pulses
is especially useful in communications because it provides a way to determine the bandwidth
required for the accurate transmission of digital data. Although, theoretically, the
communications system must pass all of the harmonics of a pulse waveshape, in reality,
relatively few of the harmonics are needed to preserve the waveshape.
54
Tf
..
The duty cycle of a series of periodic pulses is equal to the ratio ofthe pulse up time (to) to the
time period of one cycle (T) expressed as a percentage. Therefore,
t
D=...E...xlOO%
T
In the special case where a series of periodic pulses has a 50% duty cycle, called a square
wave, the plot in the frequency domain will consist of a fundamental and all odd harmonics,
with the even harmonics missing. The fundamental frequency will be equal to the frequency of
the square wave. The amplitude of each odd harmonic will decrease in direct proportion to the
odd harmonic frequency. Therefore,
v(t) = Asin27ifi + A sin(3)27ifi + A sin (5)27ifi + A sin(7)27tfi- - __ -- __ _
3 5 7
The circuit in Figure 5-1 will generate a square wave voltage by adding a series of sine wave
voltages as specified above. As the number of harmonics is decreased, the square wave that is
produced will have more ripples. An infinite number of harmonics would be required to
produce a perfectly flat square wave.
Figure 5-1
Square Wave Fourier Series
XSC1
10V
oc
III
Key=A
10kOhm
'VV"v
0---0-
10kOhm
Key=B
03.33V 3kHz OOeg
f3
10kOhm
Key=C
01000hm
2V 5kHz ODeg
f5
1OkOhm
Key = 0
01.43V 7kHz OOeg
f7
10kOhm
Key=E
r----------~~v_------~
0-
1.11V 9kHz ODeg
f9
10kOhm
Key=F
0-
55
The circuit in Figure 5-2 will generate a triangular wave voltage by adding a series of cosine
wave voltages. In order to generate a triangular wave, each harmonic frequency must be an odd
multiple of the fundamental frequency with no even harmonics. The fundamental frequency
will be equal to the frequency ofthe triangular wave. The amplitude of each odd harmonic will
decrease in direct proportion to the square of the odd harmonic frequency. Therefore,
A
A
v(t) = Acos27ift + )Tcos(3)27ift + "52COS(5)27tjt + - ---- ----
Whenever a dc voltage is added to a periodic time varying voltage, the waveshape will be
shifted up by the amount of the dc voltage.
Figure 5-2
Triangular Wave Fourier Series
XSC1
15V
DC
r-----Illt-I-----~
10kOhm
Key=B
01.11V 3kHz 90Deg
f3
10kOhm
Key=C
01000hm
O.4V 5kHz 90Deg
f5
10kOhm
Key=D
0O.2V 7kHz 90Deg
f7
1OkOhm
Key=E
0-
For a series of periodic pulses with other than a 50% duty cycle, the plot in the frequency
domain will consist of a fundamental and even and odd harmonics. The fundamental frequency
will be equal to the frequency ofthe periodic pulse train. The amplitude (A) of each harmonic
will depend on the value of the duty cycle. A general frequency domain plot of a periodic pulse
train with a duty cycle other than 50% is shown in the figure on page 57. The outline of the
peaks of the individual frequency components is called the envelope of the frequency
spectrum. The first zero-amplitude frequency crossing point is labeled fo = lito, where to is the
up time of the pulse train. This first zero-amplitude frequency crossing point (fo) detennines
the minimum bandwidth (BW) required for passing the pulse train with minimal distortion.
56
Therefore,
BW
1
=;;o =t
o
A
I~ I
fo=l/to
2/to
f
Frequency Spectrum of a Pulse Train
Notice that the lower the value of to, the wider the bandwidth required to pass the pulse train
with minimal distortion. Also note that the separation of the lines in the frequency spectrum is
equal to the inverse of the time period (lIT) of the pulse train. Therefore a higher frequency
pulse train requires a wider bandwidth (BW) because f= lIT.
The circuit in Figure 5-3 will demonstrate the difference between the time domain and the
frequency domain. It will also demonstrate how filtering out some of the harmonics affects
the output waveshape compared to the original input waveshape. The frequency generator
(XFG1) will generate a periodic pulse waveform applied to the input of the filter (5). At the
output of the filter (7), the oscilloscope will display the periodic pulse waveform in the time
domain, and the spectrum analyzer will display the frequency spectrum of the periodic pulse
waveform in the frequency domain. The Bode plotter will display the Bode plot of the filter
so that the filter bandwidth can be measured. The filter is a 2-pole low-pass Butterwoth
active filter using a 741 op-amp.
57
Figure 5-3
Time Domain and Frequency Domain
XFG1
5nF
Key = a
. C 5%
3
5
30kOhm
30kOhm
R
R
6
2
1-=-~
· ;~~=b
7
741
5.86kOhm
R1
C 5%
10kOhm
R2
PROCEDURE
Step 1
Open circuit file FIGS-I. Make sure that the following oscilloscope settings are
selected: Time base (Scale = 200 uslDiv, Xpos = 0, Yff), Ch A (Scale = 5 VlDiv,
Ypos = 0, DC), Ch B (Scale = 50 mVlDiv, Ypos = 0, DC), Trigger (pos edge,
Level = 0, Auto). You will generate a square wave curve plot on the oscilloscope
screen from a series of sine waves called a Fourier series.
Note: It is difficult to perform these steps in a hardwired lab environment because a number of
frequency generators are needed to generate a series of sinusoidal functions at different
frequencies.
Step 2
58
Run the simulation. Notice that you have generated a square wave curve plot on the
oscilloscope screen (blue curve) from a series of sine waves. Notice that you have
also plotted the fundamental sine wave (red). Draw the square wave (blue) curve
plot and the fundamental sine wave (red) curve plot in the space provided on
page 59.
I
Step 3
Use the cursors to measure the time period for one cycle (T) of the square wave
(blue) and the fundamental sine wave (red) and show the value ofT on the curve
plot.
Step 4
Calculate the frequency (f) of the square wave and the fundamental sine wave from
the time period (T).
Questions: What is the relationship between the fundamental sine wave frequency and the
square wave fr,equency (f)?
What is therelationship between the sine wave harmonic frequencies (frequencies of sine wave
generators f3, fs, f7, and f9in Figure 5-1) and the sine wave fundamental frequency (fl)?
59
What is the relationship between the amplitude of each ofthe hannonic sine wave generators
and the amplitude of the fundamental sine wave generator?
StepS
Press the A key to close switch A to add a dc voltage level to the square wave curve
plot. (If the switch does not close, click the mouse arrow in the circuit window
before pressing the A key). Run the simulation again. Change the oscilloscope
settings as needed. Draw the new square wave (blue) curve plot in the space
provided.
.
Question: What happened to the square wave curve plot? Explain why.
60
Step 6
Press the F and E keys to open switches F and E to eliminate the ninth and seventh
harmonic sine waves. Run the simulation again. Draw the new curve plot (blue) in
the space provided. Note any change on the graph.
1I
----+-
.
I
I
i
11
II
I
i
1--1-
.
I
I
I
I
I
-1---1---- ----+------f--+---+------f--+---+---!---t-I'
I
-l-------r--+-+----+---+-+---+---+-+---+--
-J--r---J----+-I-
--+---!-----+--+---+--+----+----+-___+__
Step 7
Press the D key to open switch D to eliminate the fifth harmonic sine wave. Run
the simulation again. Draw the new curve plot (blue) in the space provided. Note
any change on the graph.
--~-r
I
I
I
--- -----t----+----+-- +---+ --
Step 8
Press the C key to open switch C and eliminate the third harmonic sine wave. Run
the simulation again.
61
Question: What happened to the square wave curve plot? Explain.
Step 9
Open circuit file FIG5-2. Make sure that the following oscilloscope settings are
selected: Time base (Scale = 200 uslDiv, Xpos = 0, Y/T), Ch A (Scale = 5 VlDiv,
Ypos = 0, DC), Ch B (Scale = 100 mVlDiv, Ypos = 0, DC), Trigger (pos edge,
Level = 0, Auto ). You will generate a triangular wave curve plot on the
oscilloscope screen from a series of cosine waves called a Fourier series.
Note: It is difficult to perform these steps in a hardwired lab environment because a number of
frequency generators are needed to generate a series of sinusoidal functions at different
frequencies.
Step 10
Run the simulation. Notice that you have generated a triangular wave curve plot on
the oscilloscope screen (blue curve) from a series of cosine waves. Notice that you
have also plotted the fundamental cosine wave (red). Draw the triangular wave
(blue) curve plot and the fundamental cosine wave (red) curve plot in the
space provided.
I
Step 11
62
Use the cursors to measure the time period for one cycle (T) of the triangular wave
(blue) and the fundamental (red), and show the value ofT on the curve plot.
Step 12
Calculate the frequency (f) of the triangular wave from the time period (T).
Questions: What is the relationship between the fundamental frequency and the triangular
wave frequency?
What is the relationship between the harmonic frequencies (frequency of generators f3, f5, and
f7 in Figure 5-2) and the fundamental frequency (fJ)?
What is the relationship between the amplitude of each of the harmonic generators and the
amplitude of the fundamental generator?
Step 13
Press the A key to close switch A to add a dc voltage level to the triangular wave
curve plot. Run the simulation again. Draw the new triangular wave (blue) curve
plot in the space provided.
63
Question: What happened to the triangular wave cmve plot? Explain.
Step 14
Press the E and D keys to open switches E and D to eliminate the seventh and fifth
hannonic sine waves. Run the simulation again. Draw the new cmve plot (blue) in
the space provided. Note any change on the graph.
i
I
I
I
Ii
I
I
I
Ii
J
I
I
I
I
,
Step 15
I!
J
Press the C key to open switch C to eliminate the third hannonic sine wave. Run
the simulation again.
Question: What happened to the triangular wave cmve plot? Explain.
Step 16
64
Open circuit file FIG5-3. Make sure that the following function generator settings
are selected: Square wave, Freq = 1 kHz, Duty cycle = 50%, Ampl = 2.5 V, Offset
= 2.5 V. Make sure that the following oscilloscope settings are selected: Time base
(Scale = 500 uslDiv, Xpos = 0, YIT), Ch A (Scale = 5 VlDiv, Ypos = 0, DC), Ch B
(Scale = 5 VlDiv, Ypos = 0, DC), Trigger (pos edge, Level = 0, Auto). You will
plot a square wave in the time domain at the input and output of a two-pole
low-pass Butterworth filter.
Step 17
Bring down the oscilloscope enlargement and run the simulation to one full screen
display, then pause the simulation. Notice that you are displaying square wave
curve plots in the time domain (voltage as a function of time). The red curve plot is
the filter input (5) and the blue curve plot is the filter output (7).
Question: Are the filter input (red) and output (blue) curve plots the same shape,
disregarding any amplitude differences?
Step 18
Use the cursors to measure the time period (T) and the up time (to) of the input
curve plot (red) and record the values.
T= - - - - -
Step 19
to= - - - - -
Calculate the pulse duty cycle (D) from to and T.
Question: How did your calculated duty cycle compare with the duty cycle setting on the
function generator?
Step 20
Bring down the Bode plotter enlargement to display the Bode plot ofthe filter.
Make sure that the following Bode plotter settings are selected: Magnitude,
Vertical (Log, F = 10 dB, I = -40 dB), Horizontal (Log, F = 200 kHz, I = 100
Hz). Run the simulation to completion. Use the cursor to measure the cutoff
frequency (fe) of the low-pass filter and record the value.
fe= _ _ _ __
Note: The following steps carmot be performed in a hardwired lab environment unless a
spectrum analyzer is available.
Step 21
Bring down the spectrum analyzer enlargement. Make sure that the following
spectrum analyzer settings are selected: Freq (Start = 0 kHz,Center = 5 kHz, End
= 10kHz), Ampl (Lin, Range = 1 VlDiv), Res = 50 Hz. Run the simulation until
the Resolution Frequencies match, then pause the simulation. Notice that you
have displayed the filter output square wave frequency spectrum in the frequency
domain. Use the cursor to measure the amplitude of the fundamental and each
harmonic to the ninth and record your answers in Table 5-1.
65
Table 5-1
Frequency (kHz)
f1
1
f2
2
f3
3
f4
4
f5
5
f6
6
fl
7
f8
8
f9
9
Amplitude
Questions: What conclusions can you draw about the difference between the even and odd
hannonics for a square wave with the duty cycle (D) calculated in Step 19?
What conclusion can you draw about the amplitude of each odd hannonic compared to the
fundamental for a square wave with the duty cycle (D) calculated in Step 19?
Was this frequency spectrum what you expected for a square wave with the duty cycle (D)
calculated in Step 19?
Based on the filter cutoff frequency (fc) measured in Step 20, how many of the square wave
harmonics would you expect to be passed by this filter? Based on this answer, would you
expect much distortion of the input square wave at the filter output? Did your answer in Step
17 verify this conclusion?
66
Step 22
Adjust both filter capacitors (C) to 50% (2.5 nF) each. (Ifthe capacitors won't
change, click the mouse arrow in the circuit window). Bring down the oscilloscope
enlargement and run the simulation to one full screen display, then pause the
simulation. The red curve plot is the filter input and the blue curve plot is the filter
output.
Question: Are the filter input (red) and output (blue) curve plots the same shape,
disregarding any amplitude differences?
Step 23
Bring down the Bode plotter enlargement to display the Bode plot of the filter.
Use the cursor to measure the cutoff frequency (fc) of the low-pass filter and record
the value.
fc= _ _ _ __
Step 24
Bring down the spectrum analyzer enlargement to display the filter output
frequency spectrum in the frequency domain. Run the simulation until the
Resolution Frequencies match, then pause the simulation. Use the cursor to
measure the amplitude of the fundamental and each harmonic to the ninth and
record your answers in Table 5-2.
.
Table 5-2
Frequency (kHz)
fl
1
f2
2
f3
f4
3
4
f5
5
f6
6
f7
7
f8
8
19
9
Amplitude
Questions: How did the amplitude of each harmonic in Table 5-2 compare with the values in
Table 5-1?
67
Based on the filter cutoff frequency (fc), how many of the square wave harmonics should be
passed by this filter? Based on this answer, would you expect much distortion of the input
square wave at the filter output? Did your answer in Step 22 verify this conclusion?
Step 25
Change both capacitors (C) back to 5% (0.25 nF). Change the duty cycle to 20% on
the function generator. Bring down the oscilloscope enlargement and run the
simulation to one full screen display, then pause the simulation. Notice that you
have displayed a pulse curve plot on the oscilloscope in the time domain (voltage as
a function oftime). The red curve plot is the filter input and the blue curve plot is
the filter output.
Question: Are the filter input (red) and output (blue) curve plots the same shape,
disregarding any amplitude differences?
Step 26
Use the cursors to measure the time period (T) and the up time (to) of the input
curve plot (red) and record the values.
T= - - - - Step 27
to= - - - - -
Calculate the input pulse duty cycle (D) from to and T.
Question: How did your calculated duty cycle compare with the duty cycle setting on the
function generator?
Step 28
Bring down the Bode plotter enlargement to display the Bode plot of the filter.
Use the cursor to measure the cutoff frequency (fc) of the low-pass filter and record
the value.
fc = - - - - -
68
Step 29
Bring down the spectrum analyzer enlargement to display the filter output
frequency spectrum in the frequency domain. Run the simulation until the
Resolution Frequencies match, then pause the simulation. Draw the frequency
plot in the space provided. Also draw the envelope .of the frequency spectrum.
Question: Is this the frequency spectrum you expected for a square wave with a duty cycle less
than 50%?
Step 30
Use the cursor to measure the frequency of the first zero crossing point (fo) of the
spectrum envelope and record your answer on the graph.
Step 31
Based on the value of to measured in Step 26, calculate the expected first zero
crossing point (fo) of the spectrum envelope.
Question: How did your calculated value of fa compare the measured value on the curve plot?
Step 32
Based on the value of fo, calculate the minimum bandwidth (BW) required for the
filter to pass the input pulse waveshape with minimal distortion.
Question: Based on this answer and the cutoff frequency (fc) of the low-pass filter measured
in Step 28, would you expect much distortion of the input square wave at the filter output?
Did your answer in Step 25 verify this conclusion?
Step 33
Adjust the filter capacitors (C) to 50% (2.5 nF) each. Bring down the oscilloscope
enlargement and run the simulation to one full screen display, then pause the
simulation. The red curve plot is the filter input and the blue curve plot is the filter
output.
Question: Are the filter input (red) and output (blue) curve plots the same shape,
disregarding any amplitude differences?
Step 34
Bring down the Bode plotter enlargement to display the Bode plot of the filter.
Use the cursor to measure the cutoff frequency (fc) of the low-pass filter and record
the value.
fc= _ _ _ __
Questions: Was the cutoff frequency (fc) less than or greater than the minimum bandwidth
(BW) required to pass the input pulse waveshape with minimal distortion as determined in
Step 32?
Based on this answer, would you expect much distortion of the input pulse waveshape at the
filter output? Did your answer in Step 33 verify this conclusion?
Step 35
Bring down the spectrum analyzer enlargement to display the filter output
frequency spectrum in the frequency domain. Run the simulation until the
Resolution Frequencies match, then pause the simulation.
Question: What is the difference between this frequency plot and the frequency plot in
Step 29?
70