This document describes an experiment on Fourier theory involving the time and frequency domains. The objectives are to: 1) Produce a square wave from sine waves of different frequencies and amplitudes using Fourier series; 2) Produce a triangular wave from cosine waves using Fourier series; 3) Examine the difference between time and frequency domain plots; 4) Examine periodic pulses with different duty cycles in both domains; and 5) Examine the effect of low-pass filtering on pulses. Circuits are provided to generate square and triangular waves from Fourier series components for analysis on an oscilloscope and spectrum analyzer.

1. NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY
Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite
EXPERIMENT 5
Fourier Theory – Frequency Domain and Time Domain
Balane, Maycen M. September 01, 2011
Signal Spectra and Signal Processing/BSECE 41A1 Score:
Engr. Grace Ramones
Instructor

2. Objectives:
1. Learn how a square wave can be produced from a series of sine waves at different frequencies
and amplitudes.
2. Learn how a triangular can be produced from a series of cosine waves at different frequencies
and amplitudes.
3. Learn about the difference between curve plots in the time domain and the frequency domain.
4. Examine periodic pulses with different duty cycles in the time domain and in the frequency
domain.
5. Examine what happens to periodic pulses with different duty cycles when passed through low-
pass filter when the filter cutoff frequency is varied.

3. Sample Computation
BW =

4. Data Sheet:
Materials:
One function generator
One oscilloscope
One spectrum analyzer
One LM 741 op-amp
Two 5 nF variable capacitors
Resistors: 5.86 kΩ, 10 kΩ, and 30 kΩ
Theory:
Communications systems are normally studies using sinusoidal voltage waveforms to simplify the
analysis. In the real world, electrical information signal 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 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 frequencies and amplitudes
determined by the particular signal waveshape. Any, nonsinusoidal periodic waveform can be broken
down into sine or cosine wave equal to the frequency of the periodic waveform, called the fundamental
frequency, and a series of sine or cosine waves that are integer multiples of the fundamental frequency,
called the harmonics. This series of sine or cosine wave 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 of time. 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 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.
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, eliminate of the
harmonics does not significantly alter the original waveform. The more information contained in a
signal voltage waveform (after 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

5. 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
the harmonics of a pulse waveshape, in reality, relatively few of the harmonics are need to preserve the
waveshape.
The duty cycle of a series of periodic pulses is equal to the ratio of the pulse up time (t O) to the time
period of one cycle (T) expressed as a percentage. Therefore,
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,
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
V6
15 R1 1 J1 Ext T rig
+
_ 0
10.0kΩ A B
10 V _ _
+ +
Key = A
V1
R2 J2
9 2
10 Vpk 10.0kΩ 6
1kHz Key = B
0° V2
10 R3 3 J3 R7
100Ω
3.33 Vpk 10.0kΩ
3kHz 0
V3 Key = C
0°
12 R4 4 J4
2 Vpk 10.0kΩ
5kHz Key = D
0° V4
14 R5 5 J5
1.43 Vpk 10.0kΩ
7kHz
0° Key = E
V5 J6
R6 8
0 13
1.11 Vpk 10.0kΩ
9kHz Key = F
0°
The circuit in Figure 5-2 will generate a triangular 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

6. fundamental with no even harmonics. The fundamental frequency will be equal to the frequency of the
triangular wave, the amplitude of each harmonic will decrease in direct proportion to the square of the
odd harmonic frequency. Therefore,
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
V6
12 R1 1 J1 Ext T rig
+
_ 0
10.0kΩ A B
10 V _ _
+ +
Key = A
V1
R2 J2
13 2
10 Vpk 10.0kΩ
1kHz Key = B
90° V2
8 R3 3 J3 R7
1.11 Vpk 100Ω
10.0kΩ
3kHz 0
90° V3 Key = C
R4 J4 6
9 4
0.4 Vpk 10.0kΩ
5kHz
90° V4 Key = D
0 11 R5 5 J5
0.2 Vpk 10.0kΩ
7kHz
90° Key = E
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 of the 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 peaks if the individual frequency components is
called envelope of the frequency spectrum. The first zero-amplitude frequency crossing point is labelled
fo = 1/to, there to is the up time of the pulse train. The first zero-amplitude frequency crossing point fo)
determines the minimum bandwidth (BW0 required for passing the pulse train with minimal distortion.
Therefore,

7. A
f=1/to 2/to f
Notice than 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 (1/T) of the pulse train. Therefore a higher frequency pulse train requires a
wider bandwidth (BW) because f = 1/T
The circuit in Figure 5-3 will demonstrate the difference between the time domain and the frequency
domain. It will also determine how filtering out some of the harmonics effects the output waveshape
compared to the original3 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 (70, 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 Butterworth active filter using a 741 op-amp.
Figure 5-3 Time Domain and Frequency Domain
XFG1
XSC1
C1 XSA1
Ext T rig
+
2.5nF 50% _
Key=A A
_
B
_
IN T
+ +
R1 R2 741
30kΩ 30kΩ
42
OPAMP_3T_VIRTUAL
0
6
0
31
R3
C2 R4
5.56kΩ
10kΩ
XBP1
2.5nF 50%
Key=A
R5 IN OUT
10kΩ

8. Procedure:
Step 1 Open circuit file FIG 5-1. Make sure that the following oscilloscope settings are
selected: Time base (Scale = 200 µs/Div, Xpos = 0, Y/t), Ch A (Scale = 5V/Div, Ypos
= 0, DC), Ch B (Scale = 50 mV/Div, 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.
Step 2 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 on the plot
and the fundamental sine wave (red) curve plot in the space provided.
Step 3 Use the cursors to measure the time periods for one cycle (T) of the square wave (blue)
and the fundamental sine wave (red) and show the value of T on the curve plot.
T1 = 1.00 ms T2 = 1.00 ms
Step 4 Calculate the frequency (f) of the square wave and the fundamental sine wave from the
time period.
f = 1 kHz
Questions: What is the relationship between the fundamental sine wave and the square wave frequency
(f)?
They are both 1 kHz. They have the same value.
What is the relationship between the sine wave harmonic frequencies (frequencies of sine wave
generators f3, f5, f7, and f9 in figure 5-1) and the sine wave fundamental frequency (f1)?
The sine wave harmonic frequency is different with the sine wave fundamental. The
harmonics frequency has lot of ripples.
What is the relationship between the amplitude of the harmonic sine wave generators and the amplitude
of the fundamental sine wave generator?
The amplitude of the odd harmonics will decrease in direct proportion to odd harmonic
frequency.
Step 5 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

9. pressing the A key). Run the simulation again. Change the oscilloscope settings as
needed. Draw the new square wave (blue) curve plot on the space provided.
Question: What happened to the square wave curve plot? Explain why.
The amplitude increased. This is due to the dc voltage applied to the previous circuit.
Step 6 Press the F and E keys to open the 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.
Step 7 Press the D key to open the switch D to eliminate the fifth harmonics sine wave. Run
the simulation again. Draw the new curve plot (blue) in the space provided. Note any
change on the graph.

10. Step 8 Press the C key to open switch C and eliminate the third harmonic sine wave. Run the
simulation again.
Question: What happened to the square wave curve plot? Explain.
It became sinusoidal because the harmonic frequency generators had been eliminated.
Step 9 Open circuit file FIG 5-2. Make sure that the following oscilloscope settings are
selected: Time base (Scale = 200 µs/Div, Xpos = 0, Y/t), Ch A (Scale = 5V/Div, Ypos
= 0, DC), Ch B (Scale = 100 mV/Div, 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 sine waves called a Fourier series.
Step 10 Run the simulation. Notice that you have generated a triangular wave curve plot on the
oscilloscope screen (blue curve) from the 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.
Step 11 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 of T on the curve plot.
T = 1.00 ms
Step 12 Calculate the frequency (f) of the triangular wave from the time period (T).
f = 1 kHz

11. Questions: What is the relationship between the fundamental frequency and the triangular wave
frequency?
They are the same.
What is the relationship between the harmonic frequencies (frequencies of generators f 3, f5, and f7 in
figure 5-2) and the fundamental frequency (f1)?
Each harmonic frequency is an odd multiple of the fundamental.
What is the relationship between the amplitude of the harmonic generators and the amplitude of the
fundamental generator?
The amplitude of the harmonic generators decreases in direct proportion to the square
of the odd harmonic frequency
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 on the
space provided.
Question: What happened to the triangular wave curve plot? Explain.
The waveshape shifted up. It is because dc voltage is added to a periodic time varying
voltage; the waveshape will be shifted up by the amount of the dc voltage.
Step 14 Press the E and D keys to open switches E and D to eliminate the seventh and fifth
harmonic sine waves. Run the simulation again. Draw the new curve plot (blue) in the
space provided. Note any change on the graph.

12. Step 15 Press the C key to open the switch C to eliminate the third harmonics sine wave. Run
the simulation again.
Question: What happened to the triangular wave curve plot? Explain.
It became sine wave, because the harmonic sine waves had already been eliminated.
Step 16 Open circuit FIG 5-3. Make sure that 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
µs/Div, Xpos = 0, Y/T), Ch A (Scale = 5 V/Div, Ypos = 0, DC), Ch B (Scale = 5
V/Div, 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
plot 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 the output (blue) plots the same shape disregarding any
amplitude differences?
Yes
Step 18 Use the cursor to measure the time period (T) and the time (f o) of the input curve plot
(red) and record the values.
T= 1 ms to = 500.477µs
Step 19 Calculate the pulse duty cycle (D) from the to and T
D = 50.07%.
Question: How did your calculated duty cycle compare with the duty cycle setting on the function
generator?
They have difference of 0.07%.
Step 20 Bring down the Bode plotter enlargement to display the Bode plot of the 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

13. completion. Use the cursor to measure the cutoff frequency (fC) of the low-pass filter
and record the value.
fC = 21.197
Step 21 Bring down the analyzer enlargement. Make sure that the following spectrum analyzer
settings are selected: Freq (Start = 0 kHz, Center = 5 kHz, End = 10 kHz), Ampl (Lin,
Range = 1 V/Div), 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.
Table 5-1
Frequency (kHz) Amplitude
f1 1 5.048 V
f2 2 11.717 µV
f3 3 1.683 V
f4 4 15.533 µV
f5 5 1.008 V
f6 6 20.326 µV
f7 7 713.390 mV
f8 8 25.452 µV
f9 9 552.582 mV
Questions: What conclusion can you draw about the difference between the even and odd harmonics for
a square wave with the duty cycle (D) calculated in Step 19?
The wave consists of odd harmonics while the even harmonics are almost zero.
What conclusions can you draw about the amplitude of each odd harmonic compared to the
fundamental for a square wave with the duty cycle (D) calculated in Step 19?
The amplitude of odd harmonics decreases in direct proportion with the odd harmonic
frequency. Also, the plot in the frequency domain consist of a fundamental and all odd
harmonics, with the even harmonics missing
Was this frequency spectrum what you expected for a square wave with the duty cycle (D) calculated in
Step 19?
Yes.
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? Did your answer in Step 17
verify this conclusion?
There are 21 square waves. Yes, because the more number of harmonics square wave
the more distortion in the input square wave.
Step 22 Adjust both filter capacitors (C) to 50% (2.5 nF) each. (If the 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.

14. Question: Are the filter input (red) and output (blue) curve plots the same shape, disregarding any
amplitude differences?
No, the input is square wave while the output is a sinusoidal wave.
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 = 2.12 kHz
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 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) Amplitude
f1 1 4.4928 V
f2 2 4.44397µV
f3 3 792.585 mV
f4 4 323.075 µV
f5 5 178.663mV
f6 6 224.681 µV
f7 7 65.766 mV
f8 8 172.430 µV
f9 9 30.959 mV
Questions: How did the amplitude of each harmonic in Table 5-2 compare with the values in Table 5-1?
The result is lower than the previous table.
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?
There should be less than 5 square wave harmonics to be passed by this filter. Yes,
there have much distortion in the input square wave at the filter output.
Step 25 Change the both capacitor (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 of time).
The red curve plot is the filter input and the blue curve plot is the filter output.
Question: Are the filter input (red) and the output (blue) curve plots the same shape, disregarding any
amplitude differences?
Yes, they have the same shape.
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= 1 ms to = 198.199 µs
Step 27 Calculate the pulse duty cycle (D) from the to and T.
D = 19.82%

15. Question: How did your calculated duty cycle compare with the duty cycle setting on the function
generator?
The values of both D are the same.
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 = 21.197 kHz
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 duty cycle less than
50%?
Yes, it is what I expected for 50% duty cycle.
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 the to measured in Step 26, calculate the expected first zero
crossing point (fo) of the spectrum envelope.
fo = 5.045 kHz
Question: How did your calculated value of fo compare the measured value on the curve plot?
They have a difference of 117 Hz
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.
BW = 4.719 kHz
Question: Based on this answer and the cutoff frequency (f c) of the low-pass filter measure 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?
No, because BW is inversely proportion to the distortion formed. Then, the higher the
bandwidth, the lesser the distortion formed.
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

16. 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 the output (blue) curve plots the same shape, disregarding any
amplitude differences?
No, they do not have the same shape.
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 = 4.239 kHz
Questions: Was the cutoff frequency (fc) less than or greater than the minimum bandwidth (BW)
required to pass the input waveshape with minimal distortion as determined in Step 32?
fc is greater than BW required.
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?
No, if the bandwidth is reduced, there will occur much distortion of the input pulse
waveshape at the filter output .
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?
It is inversely proportional. As the number of the harmonics increase, the amplitude decrease.

17. Conclusion:
Based on the experiment, more harmonics added to the sine wave will result or generate a more
complex waveshape. Square wave is used in the bode plot so harmonics are easily observed. Triangular
wave is used in the spectrum analyzer to see the difference of the frequencies from each switch. Square
wave consists of only fundamental frequency and the odd harmonics. In direct proportion to the odd
harmonic frequency, the amplitude of each odd harmonic will decrease. For triangular wave, the
amplitude of each harmonic will decrease in direct proportion to the square of the odd harmonic
frequency. Moreover, decreasing the number of harmonic, the more ripple will appear to the generated
square wave curve plot.