This document describes an experiment on Fourier theory involving the time domain and frequency domain. The objectives are to generate square and triangular waves from Fourier series, examine the difference between time and frequency domain plots, and analyze periodic pulses with different duty cycles in both domains while varying a low-pass filter's cutoff frequency. Procedures generate waves using function generators and measure them on an oscilloscope and spectrum analyzer while eliminating harmonics. The document explains Fourier analysis and how signals can be represented by sine/cosine waves of different frequencies and amplitudes in the frequency domain.

1. NATIONAL COLLEGE OF SCIENCE & TECHNOLOGY
Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite
EXPERIMENT 5
Fourier Theory – Frequency Domain and Time Domain
Bani, Arviclyn C. 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
Duty Cycle
Frequency
Bandwidth
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.

5. 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
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.

6. 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°
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
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

7. multiple of the 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.
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 f o = 1/to, there to is the
up time of the pulse train. The first zero-amplitude frequency crossing point fo) determines the
minimum bandwidth (BW) required for passing the pulse train with minimal distortion.
Therefore,
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 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

8. that the filter bandwidth can be measured. The filter is a 2-pole low-pass Butterworth active
filter using a 741 op-amp.
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 (red0 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 have the same value of frequency.

9. 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 fundamental frequency have odd values of the sine wave
harmonic frequencies.
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 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 frequency, the period and the other properties, except for the amplitude of
the signal, are still the same The amplitude increased because of the
additional dc voltage applied to the 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.

10. 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.
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.
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.

11. 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 ms
Step 12 Calculate the frequency (f) of the triangular wave from the time period (T).
f = 1 kHz
Questions: What is the relationship between the fundamental frequency and the triangular
wave frequency?
They have the same frequency.
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)?
The lowest frequency is the sine wave fundamental frequency. They are odd
function with interval of 2 kHz.
What is the relationship between the amplitude of the harmonic generators and the amplitude
of the fundamental generator?
The amplitude of each harmonic will decrease 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.

12. Question: What happened to the triangular wave curve plot? Explain.
The amplitude of the triangular wave curve increases. It is because of the additional dc
voltage added to the circuit.
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.
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.
The triangular wave curve became sine wave. It is because the harmonics sine
waves are already 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

13. 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 are almost equal.
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 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.

14. 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?
There is only odd harmonics. All even harmonics are almost approaches 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.
Was this frequency spectrum what you expected for a square wave with the duty cycle (D)
calculated in Step 19?
Yes this is what I expected
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 should be 21 square waves. Yes, I would you expect much distortion of
the input square wave at the filter, 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.
Question: Are the filter input (red) and output (blue) curve plots the same shape, disregarding
any amplitude differences?

15. No they have different shape, the filter input (red) is square wave while the
output curve 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 amplitude of the table is much lower compare with 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 5 square wave. Yes, there have much distortion in the input
square wave at the filter output. Yes, answer in Step 22 verify this onclusion.
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.

16. Question: Are the filter input (red) and the output (blue) curve plots the same shape,
disregarding any amplitude differences?
Yes, the filter input and the output curve plots 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%
Question: How did your calculated duty cycle compare with the duty cycle setting on the
function generator?
They have the same value.
Step 28 Bring down the Bode plotter enlargement to display the Bode plot of the filter.
Use the cursor to measure the cutoff frequency (f C) 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.
Step 30 Use the cursor to measure the frequency of the first zero crossing point (f o) 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

17. 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 (fc) 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, 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 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, the filter input (red) and the output (blue) curve plots 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 (f c) of the low-pass filter and
record the value.
fc = 4.239 kHz
Questions: Was the cutoff frequency (f c) 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?
As the number of harmonics increase, the amplitude of the harmonics
decreases. That is why compare with the previous frequency plot, it is much
lower.

18. Conclusion
I can therefore say that any sine wave can transform to any periodic function such as
triangular and square wave through Fourier Series. Fourier Series is obtain through the
summation of the sine and cosine function. Adding more harmonics to the sine wave the more
the waveshape becomes a square wave or triangular wave. 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. In a triangular wave, the
amplitude of each harmonic will decrease in direct proportion to the square of the odd
harmonic frequency. 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.